View Full Version : intro to differential forms
don't want my words on PF
a couple of weeks ago there was a thread over there about Stoke s theorems (http://www.sciforums.com/showthread.php?s=&threadid=17219). one of the prerequisites for understanding how to calculate with stoke s theorems is the knowledge of differential forms. maxwell s equations can then be formulated in a coordinate independent way, using differential forms, and elucidating the dependence on geometry/ differential forms are also heavily used in describing yang-mills gauge theory.
i m hoping to keep the prerequisites for this thread at a minimum. you definitely need to know some calculus to follow this thread. you have to know what a derivative is, and you should probably know a little bit of multivariable calculus, what a partial derivative is, basically.
also, some familiarity with vector spaces is useful, although everything we need to know about vector spaces is included somewhere. if you are familiar with vectors in \mathbb{R}^3, and the vector cross product, that will probably suffice. oh and i assume that you know what it means for vectors to be linearly independent, and what a basis of a vector space is. those concepts could certainly be explained here, if anyone wishes.
i do make quite a few references to the concept of a manifold throughout. i don t expect you to know the technical definition of a manifold. we won t need it here, and it is not really taught in any undergraduate math or physics curricula, as far as i know. so let me just describe generally what i mean here when i say manifold.
a manifold is basically a generalization of \mathbb{R}^n. for example, \mathbb{R}^n is itself a manifold, albeit a flat one, but we want to extend our idea of a space to include curved spaces, so lemme just give a few examples: a parabola is a curved 1 dimensional manifold, that extends to infinity. a circle is a 1 dimensional manifold that folds back on itself. a sphere is a 2 dimensional manifold, or actually, any surface you would want to think of is a manifold. a manifold is just a space that is not necessarily flat. that is about all we need to know about them.
i encourage anyone to ask questions about any parts of this thread that are unclear, if you re interested to learn this stuff. or correct me if you find any mistakes, if you already know this stuff.
i am going to assume that you are a little familiar with euclidean vectors. a euclidean vector is an arrow between two points. it has direction and magnitude1. mathematically, we can specify a vector in euclidean space with a pair of points in the space, and let the vector be the arrow directed from one point to the other. or you can assume that the first point is always the origin, and specify the vector with just a single point. by doing this, you are essentially moving the vector from it s basepoint, to the origin. this is possible because euclidean space is both a manifold and a vector space.
this won t be true when we move to noneuclidean manifolds. for example, there is no sensible way to make points on a sphere into a vector space. there is no sensible way to define addition on these points.
1Well, the vectors don t have magnitude or direction until we endow the space with a metric. almost everything we are going to talk about here is independent of metric, and we will not need to specify a metric on this space. when using metric dependent quantities, this is the differential geometry, and when dealing with the more general metric independent quantities, this is differential topology. if you don t know what any of this means, ignore it.
now, hopefully we are all pretty comfortable with what a normal euclidean vector is. it s basically just an arrow between two points. it has a magnitude and a direction. right? euclidean space also comes endowed with a way to calculate the length of a vector. the pythagorean theorem. hopefully we re all familiar with this concept. if anyone wants to hear a little more about euclidean vectors, just holler, and we ll let you know.
Brad_Ad23
Jun12-03, 02:00 AM
This should be interesting! Keep it up!
Tom Mattson
Jun12-03, 05:49 PM
Originally posted by lethe
oh and i assume that you know what it means for vectors to be linearly independent,
Vectors vi (i=1,2,3,...) in Rn are independent iff
a1v1+a2v2+a3v3+...=0
implies that
a1+a2+a3+...=0
and what a basis of a vector space is.
A set of vectors (v1,v2,v3,...) is a basis for a vector space V iff
1. v1,v2,v3,... span* V.
2. v1,v2,v3,... are independent.
*span(v1,v2,v3,...)={a1v1+a2v2+a3v3+...|for ai in R}
edit: fixed subscript bracket
Tom Mattson
Jun12-03, 06:08 PM
Originally posted by lethe
now, hopefully we are all pretty comfortable with what a normal euclidean vector is. it s basically just an arrow between two points. it has a magnitude and a direction. right?
I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.
When something is said to be a "vector", one has to specify a set of transformations with respect to which that object is a vector. In the case of Euclidean 3-space, that set of transformations is rotations and parity.
Definition: A vector in Euclidean 3-space (E3) is a mathematical object that transforms under rotations R and parity Π as follows.
x-->x'=Rx
x-->x'=Πx=-x
where R is an orthogonal matrix (RTR=1). Orthogonality is important because the norm of the vector must be preserved under the rotation.
Explicitly, we must have:
v'.v'=v.v
in terms of row and column vectors (vT and v, respectively):
v'Tv'=vTRTRv
For the equality of the inner products to hold, we can see that we must have RTR=1.
IMO, when vectors are defined in terms of transformations, the extension to other vector spaces and to higher rank tensors in the same vector space is most natural.
Lethe, if you don't mind, could you wait to post the next section for another day? I would like to pick a few exercises out of my linear algebra to reinforce this stuff.
edit: fixed superscript bracket, various typos
Originally posted by Tom
I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.
yes, i do want to de-emphasize the notion of a vector as an arrow with direction. in the next post to come, i will write the definition of an abstract vector space, and ask that the reader abandon any preconceptions about vectors as arrows, and think of a vector as mathematical object obeying certain alebraic rules.
When something is said to be a "vector", one has to specify a set of transformations with respect to which that object is a vector. In the case of Euclidean 3-space, that set of transformations is rotations and parity.
Definition: A vector in Euclidean 3-space (E3) is a mathematical object that transforms under rotations R and parity Π as follows.
x-->x'=Rx
x-->x'=Πx=-x
where R is an orthogonal matrix (RTR=1). Orthogonality is important because the norm of the vector must be preserved under the rotation.
Explicitly, we must have:
v'.v'=v.v
in terms of row and column vectors (vT and v, respectively):
v'Tv'=vTRTRv
For the equality of the inner products to hold, we can see that we must have RTR=1.
IMO, when vectors are defined in terms of transformations, the extension to other vector spaces and to higher rank tensors in the same vector space is most natural.
well, i m not sure that i want to emphasize a definition that relies on transformations, we want to delay any introduction of coordinates and metrics/inner products as much as possible. orthogonality relies on the metric, so i don t want to talk about it. transformation rules of vectors rely on the introduction of coordinates on the manifold, so i don t want to talk about that either, at least to start. i want to define, e.g. tangent vectors to a manifold without any reference to local coordinates, and then derive the transformation rule for coordinate transformations, including rotations.
the definition of a (contravariant) vector as an object which transforms one way, and a covariant vector as one that transforms another way is what i was taught when i first learned GR, and i don t like it and am not going to use it. those are derivable, once you choose local coordinates, and should not be considered fundamental to the definition of the vector.
also, there is confusion about the terms contravariant and covariant, and to add to the confusion, i am going to define them oppositely of most textbooks. and then to alleviate some of the confusion, i will agree to never use those terms anymore.
Lethe, if you don't mind, could you wait to post the next section for another day? I would like to pick a few exercises out of my linear algebra to reinforce this stuff.
of course.
I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.
Interesting; I was going to suggest exactly the opposite; I was going to suggest that students think of vectors as bound to a point in space and having a direction and magnitude, to try and quell the notion of a vector as a displacement, and to emphasize that we cannot slide the arrows around like we can in Euclidean space.
IMHO, the magnitude and direction interpretation helps with the understanding of a tangent space. At least it helped me understand a tangent space when I was trying to figure out what it was.
Tom Mattson
Jun13-03, 08:48 PM
Here are two worked examples and one exercise to reinforce the prerequisites that were touched on earlier. This is not meant to be comprehensive, it is only meant to show you how the definitions I posted earlier ("Prerequisite Review Sheet") are used.
Linear Independence
Example:
Determine whether the set {v1,v2,v3} is linearly independent or linearly dependent, where
vT1=[1 2 3]
vT2=[2 -1 4]
vT3=[0 5 2]
Solution:
We must determine whether the vector equation:
a1v1+a2v2+a3v3=0
has a nontrivial solution (where 0 is the zero vector).
Note that the above is equivalent to Va=0, where V is the 3x3 matrix [v1,v2,v3]. The augmented matrix [V|0] for the system is:
[1 2 0|0]
[V|0]=[2 -1 5|0]
[3 4 2|0]
This reduces to:
[1 2 0|0]
[V|0]=[0 -5 5|0]
[0 0 0|0]
Backsolving, we get:
x1=-2x3
x2=x3
x3 is arbitrary.
Let x3=1, which gives x1=-2 and x2=1, and we have:
-2v1+v2+v3=0
Thus, the set {v1,v2,v3} is linearly dependent.
Spanning Sets
Example:
In R3 (regular Euclidean 3-space) let S={u1,u2,u3}, where
uT1=[1 -1 0]
uT2=[-2 3 1]
uT3=[1 2 4]
Determine whether S is a spanning set for R3.
Solution:
We must determine whether an arbitrary vector v in R3 can be constructed as a linear combination of the ui. In other words, we must determine whether the equation
a1u1+a2u2+a3u3=v
always has a solution. Note that the above is equivalent to the system Ax=v, where A os the 3x3 matrix [u1,u2,u3]. The augmented matrix [A|v] is:
[1 -2 1| a]
[A|v]=[-1 3 2| b]
[0 1 4| c]
Solving this system yields:
[1 0 0|10a+9b-7c]
[A|v]=[0 1 0|4a+4b-3c]
[0 0 1|-a-b+c]
Thus, our original vector equation indeed always has a solution, and so S is a spanning set for R3.
Basis
Exercise
Use the definition of a basis to determine whether S in the above Example is a basis of R3.
Lethe, that's all I wanted to say. The floor is yours.
edit: fixed a greivous error
Originally posted by Tom
Lethe, that's all I wanted to say. The floor is yours.
should we let someone solve your exercise before we move on?
Tom Mattson
Jun13-03, 09:02 PM
All that remains to be done on that is to check for linear independence, and I gave an example of that already. I don't think it's necessary to have the solution posted, but this is your show, so you can decide.
The notion of vector arithmetic and linear spaces turns out to be very useful in many different areas of mathematics and physics. so let s write down those mathematical properties of vector spaces that make them useful, and forget any notion of vectors as arrows with direction and magnitude. let v and w be vectors, and here i mean it in the abstract sense.
in other words they are just elements of a set which i am going to call vectors. they are not necessarily arrows. let a and b be real numbers2. these are the properties that the vectors must satisfy to be called a vector space.
Abelian Group Properties
[list=1]
v+w = w+v. i.e. vector addition is commutative.
there is a vector in my set 0 such that v+0=v.
in other words, there is a zero vector
for any vector v, -v is also a vector.
[/list=1]
Distributivity and Associativity
[list=1]
(a+b)v=av+bv
a(v+w)=av+aw
a(bv)=(ab)v
[/list=1]
this all seems a little abstract, but i assure you, the abstraction will pay off, when we can use all the theorems we know for vectors for all kinds of things that look nothing like our euclidean arrows with magnitude and direction.
any set of objects which satisfy these axioms, together with the set of numbers, is called a vector space over those numbers.
2 To be abstract and general, i do not need to require that a and b are real numbers. they can be members of any field. in fact, to be completely general, most, but not all, of the useful properties of a vector space also hold if i use a ring, instead of a field. this is called a module, instead of a vector space. if you don t know what any of this means, then ignore it.
Yikes, lethe, is this thread intended for those of us who haven't taken linear algebra yet, or is that another prerequisite that you neglected to mention? I would like to try to follow it, but it looks like it will require a fair amount of "outside" reading.
Please try to clarify the concept of a manifold. You said that "a manifold is just a space that is not necessarily flat." I guess that doesn't really help me until I fully understand the concept of a flat space. I mean, it's clear enough that a plane is a 2 dimensional flat space (I hope), but what is a flat 3-space, a flat 4-space, etc.?
The examples you gave of curved lines and curved surfaces don't really convey the essence of "manifoldness" (whatever that is). Presumably a space is not necessarily flat. So what distinguishes a manifold from a space?
Next problem:
what course would cover rotations, parity and orthogonality? I haven't come across any of these terms before (at least not in this context). Should I be trying to read about them now, or is that unnecessary?
Also, your summary of the properties that a set of vectors must have to be called a vector space is clear enough, but can you define or explain the concept of a vector space in words?
Brad_Ad23
Jun17-03, 02:47 AM
Aha, finally a rather clear explination of what a module is. Thanks. This is proving to be rather refreashing to go over some principles, but also educational to learn new stuff!
Originally posted by gnome
Yikes, lethe, is this thread intended for those of us who haven't taken linear algebra yet, or is that another prerequisite that you neglected to mention? I would like to try to follow it, but it looks like it will require a fair amount of "outside" reading.
well, this is supposed to be a suicidal crash course in linear algebra, and cover all the prerequisites. i don t know if that is too ambitious a hope, but at least you re trying. that is promising.
Please try to clarify the concept of a manifold. You said that "a manifold is just a space that is not necessarily flat." I guess that doesn't really help me until I fully understand the concept of a flat space. I mean, it's clear enough that a plane is a 2 dimensional flat space (I hope), but what is a flat 3-space, a flat 4-space, etc.?
OK, yes, a plane is flat 2-dimensional space. an example of a non-flat 2-dimensional space would be the surface of a sphere, or torus (doughnut).
also, a flat 1-dimensional space is just a straight line, whereas a non-flat 1-dimensional space could basically be any curve, like a circle.
flat 3-dimensional space would just be R3. it would just be a big straight volume. draw the x-, y-, and z- axes, if they go straight in each direction off to infinity, then the space is flat. so it doesn t mean flat in the sense that it s flat like a pancake. it still has volume. i just mean flat as in not curvy. flat means that the pythagorean theorem still holds (the pythagorean theorem is not true on the surface of a sphere, in case you didn t know).
but what do i mean by curvy (non-flat)? what is an example of a curvy 3-dimensional space? well this is a bit tricky to explain. recall that my examples of non-flat 2-dimensional spaces were surfaces drawn in R3. it is the case that you always need to draw your space in some Rn, where n is more than the dimension of your space. the reason is, that the space neds extra room to bend. like when you bend the line into a circle, you need the second dimension to bend into, even though your space (the circle) is only 1-dimensional. when you bend a plane into a sphere, you need the third dimension to bend around, even though the space (the sphere) is only 2-dimensional. so non-flat 3-dimensional space can only fit in some Rn if n is 4 or more.
problem is, i don t know how to draw that kind of thing. i can t draw R4. i can t even imagine R4. it s very hard to imagine what is meant by a non-flat 3-dimensional space, so the best advice i can give you is just use your 2-dimensional analogies, and try to just generalize that in your mind. just think of a 3-dimensional space, with 3 coordinate axes, that bend, and all 3 meet back on themselves. this is the 3-sphere, the 3-dimensional version of the sphere.
if you can swallow that story for 3 dimensional spaces, it is no harder to go to higher dimensions.
The examples you gave of curved lines and curved surfaces don't really convey the essence of "manifoldness" (whatever that is). Presumably a space is not necessarily flat. So what distinguishes a manifold from a space?
not much. for example, the flat space is itself a manifold. so the line, and the plane, those are both manifolds. but a manifold is not necessarily flat. so the circle and the sphere are also manifolds, but they are not flat spaces. what makes them a manifold is that if you are standing very very close to a sphere, and you forgot your glasses, and you don t look around you, you re just looking at one point on the sphere, right on top of your nose, then it will look like it is flat space, and you might not realize that it is actually a sphere.
kind of like planet earth. she is actually a sphere, but from where i m standing, she looks pretty flat.
Next problem:
what course would cover rotations, parity and orthogonality? I haven't come across any of these terms before (at least not in this context). Should I be trying to read about them now, or is that unnecessary?
hmmm.... well, i might introduce orthogonality at some point in this thread. i m not sure. rotations and parity probably not. oh i see, you re refering to the stuff tom was saying to explain what a vector is? i don t think that stuff is too important for now, but there is some disagreement as to what is the best way to introduce vectors.
so one way to understand what a vector is, is to understand its geometric properties. that is why tom is talking about rotations and orthogonality.
by the way, these concepts are not hard. don t be put off by tom s rather advanced notations of matrices and transposes and such.
rotation means exactly what you think it means: rotation. if i have a vector that points east, and i rotate it 90 degrees clockwise, it will be pointing south. a parity transformation means reflection through a mirror. if my mirror is on the x-axis, and my vector is pointing south, i ll end up with one pointing north, right? in neither case does the magnitude of the vector change, only the direction.
the set of all such transformations is called the orthogonal group. never mind for now, what a group is, or why this group is called orthogonal. i will just mention for now that orthogonal is just another word for perpendicular.
another way to understand vectors is to understand their algebraic properties. that is what i was talking about, with the axioms and such. this approach is more general, it includes a broader class of kinds of vectors, however it is less intuitive.
more on this below.
Also, your summary of the properties that a set of vectors must have to be called a vector space is clear enough, but can you define or explain the concept of a vector space in words?
OK, so let me take another stab at is. do you know what a vector is? the starting place to understand vectors is arrows in R3. they have magnitude and direction. you can rotate them, and reflect them. you can also add them, and scale them.
well you know what? that s about all their is to a vector space. a vector space is just a bunch of guys that you can add, scale, and transform. those algebraic axioms give tell you exactly how to do arithmetic with these guys, but the axioms should look like just your regular old mathematical rules.
there is one type of arithmetic on vectors that i never mentioned, and this is important, so take note, nowhere does a vector space allow you to multiply 2 vectors. you can add vectors, or scale them (which is just multiplication by a number (scalar), like 2. multiply a vector by 2, and you get a new vector that is twice as long.) we will eventually get to some notions that are like multiplications of two vectors, but they will be kinda funny. we ll get there.
OK, so i hope this was helpful. if it was not, feel free to let me know what needs to be explained a little better. after all, this thread is here for you, not for the people who already know this stuff!! thanks for the interest.
On Radioactive Waves
Jun17-03, 07:19 AM
lethe, will this basicly be the same as that other thread (http://www.sciforums.com/showthread.php?s=&threadid=20843) or have you revised it at all?
Originally posted by On Radioactive Waves
lethe, will this basicly be the same as that other thread (http://www.sciforums.com/showthread.php?s=&threadid=20843) or have you revised it at all?
i am copy-pasting from the other forum, so yes, it will be identical. when i write new entries, i will put them on both threads as well, but obviously discussions/question-answers and such will not be the same.
Thanks, lethe. I understand your explanation of flat vs. curvy spaces, but I'm still hung up on this manifold concept. If a manifold is simply a space that either is actually flat or is curvy but up close looks flat, then a Lilliputian and a Brobdingnagian might disagree as to whether a particular space is a manifold. That's not a problem?
As to vectors, you said "nowhere does a vector space allow you to multiply 2 vectors". So, the vector cross-product doesn't apply to this discussion?
Originally posted by gnome
If a manifold is simply a space that either is actually flat or is curvy but up close looks flat, then a Lilliputian and a Brobdingnagian might disagree as to whether a particular space is a manifold. That's not a problem?
not a problem. it doesn t have to look flat to everyone, to be a manifold, it only has to look flat to someone who is really really close. and how close you would have to be might depend on how curvy the space is. a really sharply turning curve only looks flat if you re super close, whereas a very broadly turning curve, you don t have to be so close to.
this all sounds rather vague, but i assure you, these notions can be made completely precise.
As to vectors, you said "nowhere does a vector space allow you to multiply 2 vectors". So, the vector cross-product doesn't apply to this discussion?
that s right. nor do vector dot products. however, we will meet those beasties eventually. a vector dot product is an example of something called an inner product. a vector space that has this additional structure is called an inner product space. once you have an inner product, you can use words like orthogonal, to describe two perpendicular vectors (two vectors are orthogonal iff their inner product is zero), but not before then.
the vector cross product is an example of something called a Lie bracket. vector spaces with vector products like this are called algebras. if the vector product is a Lie bracket, then the algebra is a Lie algebra. there is an active discussion of this in the group theory for dummies thread, do i won t say any more about it.
the point of the story is: the definition of a vector space does not include any ways to multiply vectors, but it includes basically everything else you want to do with a vector. it does not mean that the vector space can t additionally have one, it just means that we re not talking about it.
a linear functional is simply a function that takes vectors as input, and spits out numbers as output. it should also be linear, that is to say, σ is a linear functional over a vector space if and only if
\sigma (a\mathbf{v}+b\mathbf{w}) = a\sigma(\mathbf{v})+b\sigma(\mathbf{w})
where a and b are numbers, and v and w are members of the (abstract, i.e. not necessarily arrows) vector space.
now the reason i talked at length about abstract vectors is that i want you to be comfortable with the fact that something doesn t have to be an arrow to be something i would want to call a vector. polynomials are vectors in the abstract sense, and so are linear functionals. this fact is of paramount importance for our purposes in this thread.
i will ask for a volunteer to show that the set of linear functionals on a given vector space is itself a vector space. it s not to hard, just check the vector space axioms given above.
the set of linear functionals on a vector space V is called the dual space to that vector space, and it is denoted with the symbol V*.
OK, so if you believe me that the dual space of a vector space is itself a vector space, then you should know that the dual space should have a basis. well it does. in fact, there is a special basis for the dual space called the dual basis, that i want to look at now.
suppose we are given a basis for our original vector space V, {eμ}. then this induces a natural choice of basis for the dual space V*, {σν}, determined by
\sigma^\nu(\mathbf{e}_\mu) = \delta^\nu_\mu (1)
in other words, for each basis vector, there exists exactly one linear functional that takes that basis vector to the number 1, and takes every other basis vector to the number 0. these linear functionals form a basis of the dual space, that we will have occasion to use.
the μ in {eμ} is just a label that runs from 1 to n, where n is the dimension of the vector space. so that is a set of n independent vectors. likewise, the ν in {σν} is just an index that runs from 1 to n, where n is the dimension of my dual space, which is the same as the dimension of the vector space. δνμ is the kronicker delta. it is 1 when μ = ν, and 0 when μ ≠ ν. so this just a mathematical symbol that means what i said in words: the ith basis linear functional has the value 1 when it acts on the ith basis vector, and has the value 0 when it acts on any other basis vector.
i invite anyone to try, as an exercise, to show that these linear functionals are are unique, and independent, and span the dual space, i.e. that they are a basis as i claim they are. it s not hard, everything follows from the linearity.
some examples: if your vector space is a set of column vectors, then the dual space is the set of row vectors. a row vector operates on a column vector linearly, and yields a number. if your vector space is some quantum mechanical hilbert space, then the dual to the space of kets is the space of bras.
I think it's worthwhile describing manifolds in a tiny bit more detail. So, an n-dimensional manifold is a (topological) space such that every point has a neighbourhood homeomorphic to Rn. This means that the points in such neighbourhoods may be coordinatized as if the neighbourhoods were open subsets of Rn. It also means that manifolds can't in general be covered by a single coordinate system: coordinate systems stretched too far become singular. In general, coordinate systems that together coordinatize every point on a manifold overlap each other. Manifolds come equipped with functions that allow one to change coordinate systems in these regions of overlap. If these function are differentiable, we have a differentiable manifold.
Originally posted by lethe
what do i mean by curvy (non-flat)? what is an example of a curvy 3-dimensional space? well this is a bit tricky to explain. recall that my examples of non-flat 2-dimensional spaces were surfaces drawn in R 3. it is the case that you always need to draw your space in some R n, where n is more than the dimension of your space. the reason is, that the space neds extra room to bend. like when you bend the line into a circle, you need the second dimension to bend into, even though your space (the circle) is only 1-dimensional. when you bend a plane into a sphere, you need the third dimension to bend around, even though the space (the sphere) is only 2-dimensional. so non-flat 3-dimensional space can only fit in some R nif n is 4 or more.
You're talking about "extrinsic" curvature which describes how a surface is embedded in a higher dimensional space. It is the type of curvature in terms of which we ordinarily perceive and describe shapes. However, it's really a surface's "intrinsic" curvature that's of interest here.
Intrinsic curvature is defined by using the fairly easy to understand idea of "parallel transport". Imagine some closed curve on a flat surface with the tail of a vector placed on a point of this curve. Now push the tail around the curve in such a way that in moving it between infinitessimally separated points on the curve, the vector is kept parallel to itself. When the tail returns to the starting point the vector will be pointing in the same direction as it was initially. However, in performing the same exercise on a curved surface, the final and initial orientations of the vector will in general differ.
We can use this process of parallel transport to define curvature at any given point x in a space. We simply let x be the initial point on some closed curve in the space and observe the change due to parallel transport in orientation of the vector in the limit that the loop shrinks down to x.
The extra mathematical structure needed on manifolds to define parallel transport is known as the "connection".
Originally posted by jeff
You're talking about "extrinsic" curvature which describes how a surface is embedded in a higher dimensional space. It is the type of curvature in terms of which we ordinarily perceive and describe shapes. However, it's really a surface's "intrinsic" curvature that's of interest here.
well, intrinsic curvature is certainly more important when you re doing geometry. but so far, in this thread, we have not introduced any metric, so curvature, either intrinsic or extrinsic, is not defined. really, i was just trying to give a layman description of what it means for a higher dimensional space to be flat or not flat. just an intuitive picture.
for right now, restricting ourself to intrinsic geometry would not be useful. we are going to talk about tangent vectors and 1-forms, and these would look different, for example, if you write them down for a cylinder embedded in R3 than for a plane, even though both have the same intrinsic geometry.
however, your comments are useful. a discussion of riemannian geometry would be a nice addition to the math forum here, and this thread could supplement it nicely, so if you want to talk about geometry, i encourage you to start a thread about it.
Originally posted by lethe
i was just trying to give a layman description of what it means for a higher dimensional space to be flat or not flat. just an intuitive picture.
It's simply wrong - whatever the theme of this thread - to describe "what it means for a higher dimensional space to be flat or not flat" in terms of it's embedding in a higher dimensional space. As you pointed out, surfaces can have the same intrinsic curvature but different extrinsic curvatures.
Originally posted by lethe
...we have not introduced any metric, so curvature, either intrinsic or extrinsic, is not defined.
A metric is not needed to distinguish between surfaces that are flat and curved. All one needs is the idea of parallel transport which requires only a connection to define.
Edited in:
In fact, a metric isn't needed to define the ( intrinsic)curvature.
Originally posted by jeff
It's simply wrong - whatever the theme of this thread - to describe "what it means for a higher dimensional space to be flat or not flat" in terms of it's embedding in a higher dimensional space. As you pointed out, surfaces can have the same intrinsic curvature but different extrinsic curvatures.
yes, perhaps. but you know, i think the notion of intrinsic geometry takes a little while to develop, whereas anyone can picture a "curvy" embedding. and since i m not going to do anything with metrics or connections in this thread (except possibly the hodge star operator), then i didn t think it was the right place for that.
True, but one doesn't need the metric to distinguish between surfaces that are flat and curved. All one needs is the idea of parallel transport which requires only a connection to define.
yeah, well you know what? just like there is no metric yet, there is also no connection yet.
the point that you re missing here, is that differential forms are specifically designed to be metric independent (they are also connection independent). there are a lot of things that you can do on a differentiable manifold even without a connection, even without a metric. like integration, most importantly.
a lot of times, your metric (or your connection) is a dynamic object. perhaps you want to be able to work on some space before you know what the metric is. there are still some things you can do, without knowing the geometry, and when you do introduce the metric, it is nice to explicitly know what objects depend on it.
I agree with lethe. The most interesting examples to begin studying diff forms deal with embedded manifolds -- or, I should say, embeddable manifolds. It's nice to tell everyone about the fact that a manifold need not actually be embedded, and that there exists mathematical machinery to describe curvature without reference to a higher dimension -- but it's also unnecessary.
You can learn about diff forms with embedded manifolds (the easiest and often most useful case) and make the extrinsic/intrinsic distinction a bit later. We don't need a whole class on differential geometry just to appreciate forms.
- Warren
Originally posted by lethe
...differential forms are specifically designed to be metric independent.
Yes, differential forms are simply dual to vectors with the kronecker delta allowing contraction over indices. Also, they're coordinate independent.
Originally posted by lethe
there are a lot of things that you can do on a differentiable manifold without... a metric. like integration...
Unless the domain of integration is restricted, integration over a manifold in general requires a partition of unity, which is no easier to explain than parallel transport.
Originally posted by lethe
...we have not introduced any metric, so curvature, either intrinsic or extrinsic, is not defined.
I failed to mention that a metric isn't needed to define the (intrinsic) curvature.
I do understand what you're trying to do. I apologize if I'm annoying you. My style of technical writing is very dry and sometimes comes across as obnoxious or unfriendly.
Originally posted by jeff
Unless the domain of integration is restricted, integration over a manifold in general requires a partition of unity, which is no easier to explain than parallel transport.
i hadn t thought of it that way, an interesting point, i will think about that some more.
I failed to mention that a metric isn't needed to define the (intrinsic) curvature.
whoa! is this true? i did not know that, in fact, i would be very surprised to find out that you don t need a metric (or connection) to define intrinsic curvature. what exactly do you need?
i am under the impression that curvature is a geometric quantity, and that a manifold, as we have defined it, is just a topological space. is there such a notion as a "topological curvature"?
I do understand what you're trying to do. I apologize if I'm annoying you. My style of technical writing is very dry and sometimes comes across as obnoxious or unfriendly.
jeff!
no need to apologize, i quite enjoy when there is a disagreement, it allows for fruitful debates. when i put this thread on the other forum, i got almost nothing but silence. when there is a disagreement, there is an opportunity to learn something!
and who knows, i might be wrong in lots of places. i freely admit that i am just a student myself.
but just to be clear, let me restate my stance on this issue. the lesson that has been drilled into my head over and over again is: introduce any geometry dependent structures as late as possible. it is advantageous to distinguish between those objects that are geometric and those that are topological. in the end, your geometry might be dynamic, or unknown. or you might prefer to work with a symplectic manifold, instead of a riemannian one.
i have no intention of doing any riemannian geometry in this thread, although i wouldn t mind at all if it grew into that after some time. therefore, i don t consider any geometric objects on the manifold at this stage.
of course, even from a purely topological standpoint, you can still make your argument: to homeomorphic (instead of isometric, as you were arguing) manifolds are the same. however, i think that it takes some familiarity with homeomorphisms (resp. isometries), before these spaces become transparently the same. in the same vein, it takes some practice with algebraic isomorphisms before you really learn that two isomorphic groups are really the same thing. and thus, at this stage, i view that issue as a finer point, not really important for the layman, at this stage of the game.
and to that end, i attempted to use language that was only imprecise or vague, but not actually incorrect, to just convey the idea of a manifold, not anything too rigorous. if i did say something that was actually incorrect, i do want to know about it.
I'm curious what everyone thinks the intended audience is; I got the impression at the beginning we were gonna aim at something someone fresh out of multivariable calc could understand, but the bar has been raised fairly swiftly as the thread has developed...
That's what happens when you get too many cooks in the kitchen... they start to disagree about the recipe.
- Warren
Originally posted by Hurkyl
I'm curious what everyone thinks the intended audience is; I got the impression at the beginning we were gonna aim at something someone fresh out of multivariable calc could understand, but the bar has been raised fairly swiftly as the thread has developed...
that is indeed the goal. there was a bit of a tangent here about geometry, but it is not really relevant to the differential forms discussion. right now, our discussion is up to "The Dual Space", and there is an exercise there that i m hoping someone will do.
Originally posted by lethe
i am under the impression that curvature is a geometric quantity, and that a manifold, as we have defined it, is just a topological space. is there such a notion as a "topological curvature"?...to define intrinsic curvature. what exactly do you need?
Manifolds are more than just topological spaces because with each open set Uα comes a homeomorphism φα:Uα → Rn. If the transition functions φβοφα-1:φα(Uα∩Uβ) → φβ(Uα∩Uβ) are differentiable, the manifold has a differentiable structure; if analytic, an analytic structure etc.
By a "geometrical" quantity is meant one whose properties are defined locally, usually in terms of some continuous limiting process. For example, intrinsic curvature is defined at a point p in terms of parallel transport around a loop containing p in the limit that the loop is shrunk down to p.
To define intrinsic curvature, the manifold must have at least a C2 differentiable structure and a connection. Whatever indicial contractions that need be made only require the kronecker delta that comes for free and relates the contravariant and covariant spaces as duals of each other.
Sometimes there's confusion about this in people who've studied GR because in that theory the connection determines the metric. Also, sometimes people hear that curvature is a geometric property and understand such properties as requiring a metric to define.
Originally posted by lethe
the lesson that has been drilled into my head over and over again is: introduce any geometry dependent structures as late as possible. it is advantageous to distinguish between those objects that are geometric and those that are topological. in the end, your geometry might be dynamic, or unknown. or you might prefer to work with a symplectic manifold, instead of a riemannian one.
Yes, I strongly agree, no question about it - uh oh, wait - just kidding! You're definitely right on that.
Originally posted by lethe
of course, even from a purely topological standpoint, you can still make your argument: to homeomorphic (instead of isometric, as you were arguing) manifolds are the same.
I don't quite understand what you meant by this. Yes, homeomorphisms relate topologically equivalent spaces and isometries are metric preserving diffeomorphisms, but we've yet to introduce a metric.
Originally posted by jeff
Manifolds are more than just topological spaces because with each open set Uα comes a homeomorphism φα:Uα → Rn. If the transition functions φβοφα-1:φα(Uα∩Uβ) → φβ(Uα∩Uβ) are differentiable, the manifold has a differentiable structure; if analytic, an analytic structure etc.
OK, yes, my language was sloppy there. what i meant by that was the definition of a manifold does not include any geometrical structures (metrics, connections, etc).
To define intrinsic curvature, the manifold must have at least a C2 differentiable structure and a connection. Whatever indicial contractions that need be made only require the kronecker delta that comes for free and relates the contravariant and covariant spaces as duals of each other.
wait a second. are you saying that the kronecker delta that pairs up the dual space with the tangent space can be used to build a connection? that doesn t seem right to me.... can you explain that in a little more detail?
Sometimes there's confusion about this in people who've studied GR because in that theory a metric is introduced that determines the connection, and of course contractions are made with the metric. Also, sometimes people hear that curvature is a geometric property and understand such properties as requiring a metric to define.
this is exactly the school of thought in which i find myself, at the moment. well, i am aware that the connection in gauge theory (the gauge potential) has nothing to do with a metric, but i m not very clear about the relation between the two kinds of connection.
I don't quite understand what you meant by this. Yes, homeomorphisms relate topologically equivalent spaces and isometries are metric preserving diffeomorphisms, but we've yet to introduce a metric.
what i meant was this. i started talking about different embeddings, and you objected, because different embeddings can have the same intrinsic geometry. two manifolds with the same intrinsic geometry are (locally) isometric, and are therefore, for certain purposes, the same manifold. you shouldn t distinguish manifolds as "flat" or "curved" simply by their embeddings, which contain the extrinsic geometry. this was your objection, as i understood it.
i objected to your introduction of intrinsic geometry, saying "my manifold doesn t have any geometry yet, so you can t talk about intrinsic geometry!". so what i was then saying later was, you could have still applied your argument, by weakening your isometry to a homeomorphism (well actually, a diffeomorphism, since these are differentiable manifolds). you could have said, you are distinguishing two differentiable manifolds by their embeddings, when in fact they are diffeomorphic! i guess this wouldn t make a lot of sense, though.
the problem is with the whole argument is, i have not made precise what i mean when i say "flat" or "nonflat". when i say "flat" do i mean having no curvature? if so, then your objection was a very valid one, and i countered by saying i have no geometry. when i say "flat", do i mean just "locally homeomorphic to Rn"? if so, then every manifold is "flat".
i was just trying to give an explanation of a "not necessarily flat" space, on a level of someone who just knew a bit of calculus or linear algebra, without all the mathematical apparatus.
let s forget it.
Originally posted by lethe
are you saying that the kronecker delta that pairs up the dual space with the tangent space can be used to build a connection?
Sorry, I didn't mean it to sound that way. Though their definition in the absence of a metric necessarily involves kronecker deltas, connections aren't constructed from them and are in fact assignments to vector fields at each point a certain differential operator. The reason I've made a point of mentioning kronecker deltas is that I've seen people scratching their heads when they wonder how contractions are being made without a metric.
Originally posted by lethe
manifolds with the same intrinsic geometry are (locally) isometric
As you probably know, manifolds are isometric when there exists a diffeomorphism between them that carries their metrics into each other. Since metric and connection are in general independent, that manifolds have the same geometry in terms of (intrinsic) curvature implies nothing about the relation between their metrics.
OK, so i mentioned that on general manifolds, it makes no sense to think of the points in the space as a vectors. there is simply no way to define addition of points on a sphere in such a way that the abstract vector axioms are satisfied. but what any smooth manifold will have is something called tangent vectors. but we are going to define tangent vectors without any reference to the ambient space.
let me explain what i mean by that last statement in a little more depth. consider the circle, drawn in ℜ2. we can use our well-known calculus in \mathbb{R}^2 to calculate an \mathbb{R}^2 vector that is tangent to the circle.
but what if i weren t allowed to draw my circle in \mathbb{R}^2? how would i then write down the tangent vector? this is a subtle point, that some people have trouble with, so if you don t know what i mean here, feel free to ask questions. have you heard people talk about how our 4 dimensional spacetime is curved, but you wondered "what does it curve around?" or "what direction does it curve in?", well that s what i m talking about.
well anyway, let s get on with it. the way we define tangent vectors is through directional derivatives. the formula we learned in regular calculus for a directional derivative is \mathbf{v}\cdot\nabla f. we re going to use the same concept to define a tangent vector to a manifold. remember, a manifold, by definition, is equivalent to some \mathbb{R}^n, so we can always introduce coordinates near a point, and then take derivatives along the lines of thhose coordinates, and never make reference to any ambient space.
at this point, i will stop using classical vector notation. i will write the directional derivative as v^\mu\partial_\mu f, where \partial_\mu is shorthand for \partial/\partial x^\mu, and x^\mu is one of the coordinates, and \mu is a number that ranges over the number of dimensions of the manifold, from 0 to n-1 usually. so there will be n different coordinates for an n dimensional manifold. and v^\mu is going to be associated with the \mu^{\text{th}} component of the vector, to be defined. and even though i didn t write it, i meant for that to be a summation: \mbox{$\mathbf{v}\cdot\nabla f = \sum_{\mu=0}^{n-1} v^\mu\partial_\mu f = v^\mu\partial_\mu f$}. i just leave off the ∑ from now on. every time you see an equation with the same letter as a superscript and a subscript, you should sum over that index.
OK, now the meat of the post: we are going to say that a tangent vector is an operator that takes a function, and returns a directional derivative. this is not too hard to understand. given some function on the manifold (doesn t matter what the function is), one way to characterize tangent vectors is to say they specify a direction along the manifold at an instantaneous point. and one way to characterize directions is by directional derivatives. that means, that for every direction, you will get a different value for the directional derivative, given some function. the notion of how much a function changes in any particular direction is independent of the function itself: different functions might increase at different rates in a given direction, but that value is just a linear differential operator on the function. since all the directional information is encoded in that operator, that is what we will want our vectors to be.
we define the vector to be that operator. this is how it operates on a function:
\mathbf{v}(f) = v^\mu\partial_\mu f ( 2)
since this is independent of the function that i want to operator on, let me just write the vector operator:
\mathbf{v} = v^\mu\partial_\mu (3)
and this is the point of this post. a tangent vector is defined to be/associated with/thought of as a differential operator. v is the vector, and vμ are the coordinate components of the vector, and ∂μ are the coordinate basis vectors of the tangent space. the vector itself is coordinate independent, but the components are not, and the basis vectors are not (that sounds a little redundant, eh? the basis vectors are not independent of the basis vectors. heh. fsck off.)
OK, it should be easy to show that the set of tangent vectors, thusly defined, satisy the axioms of the vector space. i will call this vector space TMp. that is, the tangent space to the manifold M at the point p is TMp. for an n dimensional manifold, the tangent space is always an n dimensional vector space.
this should make some sense, because on a curved manifold, you can only consider directions between two infinitesmally close points: the arrow pointing between to finitely seperated points on, say, a circle, is not a tangent vector to the circle, only infinitely close points determine a tangent vector. to determine tangent vectors between two infinitesimally close points, you have to take a limit, and you will end up with a derivative.
nevertheless, a lot of people have a hard time swallowing this equation, including me when i first learned it. why are coordinate derivatives vectors? well, let me just say, think carefully about what s written here, and please, ask questions. it s subtle, and if you can t really convince yourself of why, then just take it as given, so that you can procede with the rest of the thread.
Originally posted by lethe
the vector itself is coordinate independent, but the components are not, and the basis vectors are not (that sounds a little redundant, eh? the basis vectors are not independent of the basis vectors. heh. fsck off.)
OK, it should be easy to show that the set of tangent vectors, thusly defined, satisy the axioms of the vector space. i will call this vector space TMp. that is, the tangent space to the manifold M at the point p is TMp. for an n dimensional manifold, the tangent space is always an n dimensional vector space.
this should make some sense, because on a curved manifold, you can only consider directions between two infinitesmally close points: the arrow pointing between to finitely seperated points on, say, a circle, is not a tangent vector to the circle, only infinitely close points determine a tangent vector. to determine tangent vectors between two infinitesimally close points, you have to take a limit, and you will end up with a derivative.
nevertheless, a lot of people have a hard time swallowing this equation, including me when i first learned it. why are coordinate derivatives vectors? well, let me just say, think carefully about what s written here, and please, ask questions. it s subtle, and if you can t really convince yourself of why, then just take it as given, so that you can procede with the rest of the thread.
I have no difficulty seeing the tangent space as made of directional derivatives. This seems the natural way to define it. And moreover it is the standard and time-honored practice differential geometers.
You say to please assimilate this definition of the tangent space "so that you can procede with the rest of the thread". So let us procede and avoid unnecessary nit-picking. differential geometry is done for the honor of the human mind and not as an exercise in one-upmanship. procede to define the dual and the wedge and the star according to the eternal commandments of nature.
OK, at this point, we are ready to introduce the first kind of differential forms, the 1 forms. a 1-form is simply a member of the dual space to the tangent space at a point.
if M is our manifold, then TM_p is the tangent space at the point p, and T^*M_p is the dual space to that tangent space, according to the notation we introduced above for dual spaces.
i will sometimes call a member of the dual space of the tangent space a cotangent vector, and call the dual space itself the cotangent space. thus, a 1-form is simply a cotangent vector.
now, let s recall what a member of the dual space is: it s a linear functional on the vectors. that means that if i operate a member of the dual space on the vector at a point, then i get a number.
but i also recall that i defined the vector space itself as a set of operators! the vectors take functions on the manifold and return the value of the directional derivative at that point. which is also a number! in fact, since the differential operator that defines the vector is a linear operator on functions, i can make a parallel between the functions on the manifold, and the linear functionals in the dual space.
let me explain that a little further: a linear functional on the vector space says "take a vector, spit out a number". a vector says "take a function on the manifold, spit out a number". so for each function on the manifold, i can assign to it a linear functional that spits out the same number when it acts on the vector, as the vector spits out when it acts on the function. for any function f, let me call this associated linear functional df. the d will come to have a familiar meaning, but for right now, it just means find the linear functional who spits out the number required. it s just a symbol that means "the linear functional associated with the function f".
read that paragraph again, and see if you can follow it. let me write down what i said above, using the symbols we have introduced:
df(\mathbf{v}) = \mathbf{v}(f)
(4)
on the left hand side, we have a linear functional in the dual space acting on a vector in the tangent space, and on the right hand side, we have that same vector , it remembered that in addition to being a vector in the tangent space, it is also a differential operator, and as a differential operator it is acting on the function associated with my linear functional.
the linear functional takes the vector to the same number that the vector takes the associated function.
be careful of my use of the words function and functional. there isn t any deep difference between the two words, they are just usually used in different contexts. the word functional is usually reserved for mappings that act on vectors or more complicated objects, and functions usually act on numbers.
so df is a functional that acts on vectors, and f is a function, that acts on numbers. (well actually, in our case, it acts on points in our manfold M.)
sorry if i m getting repetitive here, but this is important, and i want to make it clear.
OK, so lets explore some properties of these 1-forms. first of all, let s write down the dual basis, \{\sigma^\nu\}. these are, by definition, linear functionals such that \sigma^\nu(\mathbf{e}_\mu) = \delta^\nu_\mu (eq. (1) above). where \{\mathbf{e}_\mu\} is the basis for the vector space. but remember, for the tangent space, we already chose a basis, the coordinate basis \{\partial_\mu\}. also, like we discussed above, our dual space linear functionals on the tangent space can be associated with functions on the manifold. so let s do that for each \sigma^\nu: \sigma^\nu = df^\nu, where f^\nu is some function on the manifold that we will determine.
with these changes, let s rewrite that condition for the dual basis (1):
df^\nu(\partial_\mu) = \delta^\nu_\mu
( 5)
now using (4) above, this becomes:
df^\nu(\partial_\mu) = \partial_\mu f^\nu = \frac{\partial f^\nu}{\partial x^\mu} = \delta^\nu_\mu
now, can you think of a function whose derivative with respect to x^\mu is 1, and whose derivative with respect to all other coordinates is 0? it s easy..
think about it....
got it?
it s f^\mu = x^\mu! no sweat!
OK, so then the dual basis of the 1 forms is just {dx^\nu}. now let s check what the components of a general 1-form are in terms of this basis:
df = \alpha_\nu dx^\nu
where αν are the components of the 1 form. let s solve for those components by acting this 1-form on a basis vector of the tangent space.
df(\partial_\mu) = \alpha_\nu dx^\nu(\partial_\mu)
\partial_\mu f = \alpha_\nu\partial_\mu x^\nu = \alpha_\nu\delta^\nu_\mu = \alpha_\mu
i have used (4) twice in the second equation there.
how about that! the components of the 1 form \alpha_\nu are just the partial derivatives of the function!
df = \frac{\partial f}{\partial x^\nu}d x^\nu = \partial_\nu fdx^\nu (6)
now that equation should look familiar perhaps to some of you from your calc classes. it s just the chain rule of multivariable calculus, or at least it looks like it. this explains why we used the symbol "d" to create a 1 form out of a function, because it is done by simply differentiating. at first "d" was just a symbol to associate a linear functional with a function on the manifold. but now we see that it is actually a differential operator on the functions. this "d" operator we will see again, it is called the exterior derivative. it is very important. it is the "d" that appears in the integrand of stoke s theorem.
OK, so that s 1-forms! you folks with me so far? feel free to ask some questions, or if any of you wants to clarify any points that you think i didn t make very well, feel free.
next up, higher order forms!
this is a good rough and ready writing style which shows a sensitivity to the occasional points where a reader might experience difficulty but which never seems to talk down (as to an inferior). This is a good style and not everyone can achieve it consistently, or so I think anyway. but there are some typographical boxes which I will experiment with getting rid of.
This is surprising. the boxes all went away simply by a font change. complements on the text L. as it is really quite nicely done
Originally posted by lethe
let me explain that a little further: a linear functional on the vector space says "take a vector, spit out a number". a vector says "take a function on the manifold, spit out a number". so for each function on the manifold, i can assign to it a linear functional that spits out the same number when it acts on the vector, as the vector spits out when it acts on the function. for any function ƒ, let me call this associated linear functional dƒ. the d will come to have a familiar meaning, but for right now, it just means find the linear functional who spits out the number required. it s just a symbol that means "the linear functional associated with the function ƒ".
read that paragraph again, and see if you can follow it. let me write down what i said above, using the symbols we have introduced:
dƒ(v) = v(ƒ) (4)
on the left hand side, we have a linear functional in the dual space acting on a vector in the tangent space, and on the right hand side, we have that same vector , it remembered that in addition to being a vector in the tangent space, it is also a differential operator, and as a differential operator it is acting on the function associated with my linear functional.
the linear functional takes the vector to the same number that the vector takes the associated function.
be careful of my use of the words function and functional. there isn t any deep difference between the two words, they are just usually used in different contexts. the word functional is usually reserved for mappings that act on vectors or more complicated objects, and functions usually act on numbers.
so dƒ is a functional that acts on vectors, and ƒ is a function, that acts on numbers. (well actually, in our case, it acts on points in our manfold M.)
sorry if i m getting repetitive here, but this is important, and i want to make it clear.
OK, so lets explore some properties of these 1 forms. first of all, let s write down the dual basis, {σν}. these are, by definition, linear functionals such that σν(eμ) = δνμ (eq. (1) above). where {eμ} is the basis for the vector space. but remember, for the tangent space, we already chose a basis, the coordinate basis {∂μ}. also, like we discussed above, our dual space linear functionals on the tangent space can be associated with functions on the manifold. so let s do that for each σν: σν = dƒν, where ƒν is some function on the manifold that we will determine.
with these changes, let s rewrite that condition for the dual basis (1):
dƒν(∂μ) = δνμ (5)
now using (4) above, this becomes:
dƒν(∂μ) = ∂μƒν = ∂ƒν/∂xμ = δνμ
now, can you think of a function whose derivative with respect to xμ is 1, and whose derivative with respect to all other coordinates is 0? it s easy..
think about it....
got it?
it s ƒμ = xμ! no sweat!
OK, so then the dual basis of the 1 forms is just {dxν}. now let s check what the components of a general 1 form are in terms of this basis:
dƒ = ανdxν
where αν are the components of the 1 form. let s solve for those components by acting this 1 form on a basis vector of the tangent space.
dƒ(∂μ) = ανdxν(∂μ)
∂μƒ = αν∂μxν = ανδνμ = αμ
i have used (4) twice in the second equation there.
how about that! the components of the 1 form αν are just the partial derivatives of the function!
dƒ = ∑(∂ƒ/∂xν)dxν = ∂νƒdxν (6)
now that equation should look familiar perhaps to some of you from your calc classes. it s just the chain rule of multivariable calculus, or at least it looks like it. this explains why we used the symbol "d" to create a 1 form out of a function, because it is done by simply differentiating. at first "d" was just a symbol to associate a linear functional with a function on the manifold. but now we see that it is actually a differential operator on the functions. this "d" operator we will see again, it is called the exterior derivative. it is very important. it is the "d" that appears in the integrand of stoke s theorem.
Originally posted by marcus
this is a good rough and ready writing style which shows a sensitivity to the occasional points where a reader might experience difficulty but which never seems to talk down (as to an inferior). This is a good style and not everyone can achieve it consistently, or so I think anyway. but there are some typographical boxes which I will experiment with getting rid of.
This is surprising. the boxes all went away simply by a font change. complements on the text L. as it is really quite nicely done
thanks for the compliment. i enjoy writing this a lot, and i appreciate your encouragement.
by the way, what font did you use? if it s just a matter of changing fonts, i d be more than happy to try a different font. are you using the default font? what id that, courier?
Originally posted by lethe
thanks for the compliment. i enjoy writing this a lot, and i appreciate your encouragement.
by the way, what font did you use? if it s just a matter of changing fonts, i d be more than happy to try a different font. are you using the default font? what id that, courier?
Yes, default. I didnt make a conscious selection, just erased
the "[font equals times roman]" statement in your post
I left the [size equals 3] statement and it seems that number 3 in the default is bigger than number 3 in times roman. I like big, so that is fine with me.
Greg would know what the default is.
I think you should have the freedom to choose your font, it is author's choice, and I would not want you to choose one you did not like merely on one person's account. I can cope by reposting passages in the default font---so am happy either way.
it will be very nice if both these sticky threads are active and accessible at kind of entry level-----accessible to the novice-with-gumption, and you know what I mean so I wont say it.
Originally posted by lethe
thanks for the compliment. i enjoy writing this a lot, and i appreciate your encouragement...
Lethe, I'm hoping that time permitting you will continue the exposition of diff forms.
I will tell you a personal wish motivating my interest in 1-forms. I would like to better assimilate the idea of a
(quoting Carlo Rovelli)
"a 1-form field in a principal Lorentz bundle over the spacetime manifold M whose fiber is Minkowski space M."
This is where the (classical, non-quantum) gravitational field lives. There is a shapeless manifold M, not precommitted to any particular metric or geometry. It acquires a geometry dynamically, from the gravitational field, which is a certain vector-valued ONE-FORM---valued in a 4D vectorspace which one can think of as the tangent space at each point. So at each point the gravitational field looks like a linear map T-->T
Cartan called this one-form the "moving frame", others call it the "soldering form", others call it the 'tetrad"---but as Rovelli points out it is not moving. It is just a vectorvalued 1-form
It lives in a principal G-bundle, where G is the lorentz group. So I need to understand what a principal G-bundle is about.
A differential geometry book (Bishop and Crittenden) that I happened to pick up defines a "principal bundle" as a triple (P, G, M) where P and M are smooth manifolds and G is a Lie group
(1) G acts freely on P, PxG --> P (they choose a right action, it could be left)
(2) M is the quotient space of P mod equivalence by G
the projection map is [pi]:P --> M
G acts transitively on the fiber [pi]-1(m) over any point m in M
(3) P is locally trivial. that means that around any point m in M there is a neighborhood U ( picture a disk) such that the part of P that is over U ( picture a cylinder over the disk), namely
[pi]-1(U), is diffeomorphic to the cartesian product
U x G ( picture a second cylinder U x G, with U a disk and G a vertical line).
The diffeomorphism [pi]-1(U) --> UxG takes a point p to ([pi](p), FU(p)) and this map FU: [pi]-1(U) --> G satisfies an equation FU(pg) = map FU(p)g.
The equation says you can do the group action first and then do F, or you can do F first and then do the group action, same result. In other words F "commutes with the group action."
Originally posted by marcus
Lethe, I'm hoping that time permitting you will continue the exposition of diff forms.
I will tell you a personal wish motivating my interest in 1-forms. I would like to better assimilate the idea of a
(quoting Carlo Rovelli)
"a 1-form field in a principal Lorentz bundle over the spacetime manifold M whose fiber is Minkowski space M."
my personal goal with this thread is the geometric formulation of the yang-mills equation. i think the machinery for our two goals is the same, so i think we should try to accommodate both. of course, i am trying to really pitch this thread for someone with basically just an undergrad level of calc and linear algebra, so i had to start slowly.
and also, as i guess you ve learned, if you don t occasionally remind me, i will forget to update the thread.
Now it s time to tell you what a tensor product is. basically, a (first rank) tensor can be either a tangent vector, or a linear functional, i.e. a 1 form. i will use the word as a single word that encompasses both of those notions, which i defined above, as well as certain composites that i will define now.
remember, a 1-form is a functional that takes 1 vector, and spits out a number. if i have 2 1-forms, say ω and σ, then the tensor product of these two 1-forms is a new (second rank) tensor that takes two vectors and spits out a number. it does this by feeding the first vector to the first 1-form, which gives you a single number, and feeding the second vector to the to the second 1-form, and getting another single number, and then returning the product of those two numbers. the notation usually used for this tensor product of 1-forms is ω⊗σ. so then value of the tensor product, acting on the vectors v and w i just described can be written with my symbols:
ω⊗σ(v,w) = ω(v)σ(w)
on the left hand side of the equation, i show you the tensor product of two 1-forms, acting on an ordered pair of vectors, and on the right hand side, i act the two 1-forms individually on the two vectors, and multiply the 2 numbers that come out.
and that s all there is to the tensor product! pretty simple.
Now, we must define the wedge product of two 1-forms. if you followed my definition of the tensor product, this will be easy:
σ∧ω = σ⊗ω - ω⊗σ
easy enough. let s check how that new wedge product acts on our two vectors:
σ∧ω(v,w) = σ(v)ω(w) - ω(v)σ(w)
one obvious property of this new product is that it is alternating, which means that if you switch the order of the two vectors that you feed to it, you pick up a minus sign from the original product before you switched:
σ∧ω(w,v) = σ(w)ω(v) - ω(w)σ(v) = ω(v)σ(w) - σ(v)ω(w) = -(σ(v)ω(w) - ω(v)σ(w)) = -σ∧ω(v,w)
also, the wedge product is itself antisymmetric, meaning that if you switch the order that you multiply the to 1-forms in, you again pick up a minus sign:
ω∧σ = ω⊗σ - σ⊗ω = - (σ⊗ω - ω⊗σ) = -σ∧ω
compare this with the tensor product: if you switch the order of the two vectors you input, you get a new number with no general relationship. and if you switch the order of the two 1-forms that you re taking the tensor product of, you get a new tensor that is in no general way related to the original tensor.
anyway, i define a 2-form to be the wedge product of two 1-forms. we can get to a p-form by simply taking the wedge product of p 1-forms, using these definitions.
next up, we will investigate a few more of the algebraic properties of the wedge product.
marcus: i left these posts in the default font. do the tensor product and wedge product symbols show up for you?
Hi Lethe,
as long as you dont mind my taking the liberty of reproducing your text with the ad hoc symbol /\ replacing "& and ;"
then i'm OK
the reason it works for me is that everything else comes thru except for the wedge and the tensorproduct. I can train my mind to see boxes as tensorproducts, if I rewrite the wedge.
Now, we must define the wedge product of two 1-forms. if you followed my definition of the tensor product, this will be easy:
σ/\ω = σ⊗ω - ω⊗σ
easy enough. let s check how that new wedge product acts on our two vectors:
σ/\ω(v,w) = σ(v)ω(w) - ω(v)σ(w)
one obvious property of this new product is that it is alternating, which means that if you switch the order of the two vectors that you feed to it, you pick up a minus sign from the original product before you switched:
σ/\ω(w,v) = σ(w)ω(v) - ω(w)σ(v) = ω(v)σ(w) - σ(v)ω(w) = -(σ(v)ω(w) - ω(v)σ(w)) = -σ/\ω(v,w)
also, the wedge product is itself antisymmetric, meaning that if you switch the order that you multiply the to 1-forms in, you again pick up a minus sign:
ω/\σ = ω⊗σ - σ⊗ω = - (σ⊗ω - ω⊗σ) = -σ/\ω
compare this with the tensor product: if you switch the order of the two vectors you input, you get a new number with no general relationship. and if you switch the order of the two 1-forms that you re taking the tensor product of, you get a new tensor that is in no general way related to the original tensor.
anyway, i define a 2-form to be the wedge product of two 1-forms. we can get to a p-form by simply taking the wedge product of p 1-forms, using these definitions.
next up, we will investigate a few more of the algebraic properties of the wedge product.
lethe,
I discovered this thread on that other site a while ago, then more recently discovered it on this one. I was very excited to find it, as it contains math I don't yet know and is actaully being aimed at someone with my level of math (completed multivariable calc). However, I have been real lazy (common theme in my life), and delayed a lot on getting around to reading all of it. Well, today I finally finished reading it. I am going to reread some of it to find the stuff I had most difficulty with, but for now i was wondering if you could give an example of a 1-form or some other exercises I could try to work through. I think I understand most of it...I just need to be sure I get the ideas down.
In a final note, I wanted to say I think what you are doing here is very cool, and there are people who appreciate it (me and others). I would love to see you keep going with this.
thanks
You've used 1-forms already. [:)] A simple example is just dx!
Also, for line integral you have ever performed, the integrand is a one-form. For example, you may recall that the area enclosed by any simple closed curve γ(x, y) in R2 can be computed by the line integral:
A = ∫ x dy
where the integral is taken along γ. Well, "x dy" is an example of a one-form!
I reread all the main stuff you posted lethe, and I think I am understanding it a little better. I just need to ask some questions to make sure I am on track.
In equation 4. dƒ(v) = v(ƒ) is the d, which is the linear functional, actaully the diffenrial form for which we are trying to solve?
Also, about the functions on the manifolds, suppose my manifold is just flat 2-d space, the cartesian plane. Can I say my function is x^2? IS this what is meant by function. It doesnt have to be a vector function, does it?
Also, early I was asking for an example, and i guess what i meant was something that went through the whole process. Like what would the tangent space be for the cartesian plane?
When you say "a 1-form is simply a member of the dual space to the tangent space at a point. " does that mean it is just one linear functional of the tangent space since you said member? If this is so, does this mean you can have multiple 1-forms for a given tangent space or did I misunderstand your wording? Perhaps it is that we are trying to solve for the "right" member of the dual space to the tangent space that allows equation 4 to work?
I am sure if I sat and stared at your posts some more I could come up with lots more questions, but I will get the answers to these first. I must say I am really enjoying this, tonight I felt like I actaully figured a lot more of it then the last time I read it.
Originally posted by cephas
In equation 4. dƒ(v) = v(ƒ) is the d, which is the linear functional, actaully the diffenrial form for which we are trying to solve?
dƒ is the differential form. d is the exterior derivative.
Also, about the functions on the manifolds, suppose my manifold is just flat 2-d space, the cartesian plane. Can I say my function is x^2? IS this what is meant by function. It doesnt have to be a vector function, does it?
this is a fine example of a function on the cartesian plane. can you tell me what the exterior derivative of this function is? (Hint: 2x dx)
so anyway, no, it is not a vector function.
Also, early I was asking for an example, and i guess what i meant was something that went through the whole process. Like what would the tangent space be for the cartesian plane?
well, the most convenient way for me to tell you what a vector space looks like is to choose a basis for the vector space. then i can tell you that the vector space is just the span of that basis.
furthermore, a good basis for the tangent space appears naturally once you have chosen coordinates for the manifold.
so OK, your manifold is the plane. a common choice of coordinates is x and y, the cartesian coordinates. with this choice of coordinates, the tangent space is the span of ∂/∂x and ∂/∂y. it would be similar with polar or hyperbolic coordinates.
now lets see what the cotangent space is. the natural basis of a dual of a vector space is chosen by finding those linear functionals which take each basis vector of the first space to 1. so i need a linear functional dƒ(∂/∂x)=1, dƒ(∂/∂y)=0. now let s use equation 4. dƒ(∂/∂x)=∂ƒ/∂x=1 and dƒ(∂/∂y)=∂ƒ/∂y=0. integrate those two, and you will see that ƒ=x. so the dual vector to ∂/∂x is just dx. similarly, the dual to ∂/∂y is dy. these two dual vectors (1-forms) span the vector space that is the cotangent space of the cartesian plane.
in general, it will always be the same. if i had chosen to use polar coordinates instead, the basis of the cotangent space would have been dr and dθ.
When you say "a 1-form is simply a member of the dual space to the tangent space at a point. " does that mean it is just one linear functional of the tangent space since you said member? If this is so, does this mean you can have multiple 1-forms for a given tangent space or did I misunderstand your wording? Perhaps it is that we are trying to solve for the "right" member of the dual space to the tangent space that allows equation 4 to work?
yes, a 1-form is a single linear functional. a 1-form is just one member from the set of all possible members. yes, there are many 1-forms available for a given tangent space. in fact, an infinite number. they form an n-dimensional vector space called the cotangent space.
but yes, equation 4 is the rule for finding the "right" 1-form associated with each function. i would not say that we are trying to solve for the correct differential form. we are defining the correct differential form in the only natural way that is available to us, namely that described in equation 4.
I am sure if I sat and stared at your posts some more I could come up with lots more questions, but I will get the answers to these first. I must say I am really enjoying this, tonight I felt like I actaully figured a lot more of it then the last time I read it.
hey, thanks for reading! i m happy to do it!
okay, thanks so much, that clears up a lot of stuff.
Now a few more questions.
First off, I was wondering if the 1-form or any differntial form for that matter is independent of the function on the manifold. What I mean is do you get the same 1-form for all functions on some manifold. Also, I asked something about functions being vectors, and I was wondering if you could use differntial forms on parametric equations, or on vectors functions r=fi=gj+hk where i, j, and k are unit vectors, and f,g, and h are functions that depend on t.
More so, if I can use this type of function I would I go about coming up with the 1-form for it? That's it for now since I don't have much more time. I have been doing so more research on this stuff in other areas, primarily mathworld at wolfram. From what I understand 1-forms and all differential forms are actaully just tensors. I will have more soon.
I was wondering if you would like me to copy these posts and put them on the other forum also so maybe other ppl could learn more over there too, what do you think?
let me just check my understanding real quick on the tensor product.
So dx and dy are 1-forms. They are also linear functionals. So how does dx and dy operate on a vector? Is it the same as the derivative? dx(x)=1 or dx(x^2)=2x? I am not quite sure I get that...
And then with the tensor product (which is why I asked the above question) could I go like dxdy(x,y)=1*1=1? or dxdy(y,x)=0? Do I have the right idea here?
Originally posted by cephas
So dx and dy are 1-forms. They are also linear functionals.
yes, and yes.
So how does dx and dy operate on a vector? Is it the same as the derivative? dx(x)=1 or dx(x^2)=2x? I am not quite sure I get that...
those equations are no good, because x and x^2 are not vectors in the tangent space, and therefore it is invalid to act on them with a differential form. the hardest leap to make in this thread is thinking of vectors as differential operators on functions. d/dx is a good tangent vector that a differential form might eat for breakfast. x^2 is a function that a vector might eat for lunch.
And then with the tensor product (which is why I asked the above question) could I go like dxdy(x,y)=1*1=1? or dxdy(y,x)=0? Do I have the right idea here?
assuming you replace x and y in the arguments of those equations, then yes, you have exactly the right idea.
rick1138
Oct2-03, 01:30 AM
Great thread everyone. Here is a quick example of calculating with forms. In general, the value of a form is the value of the determinant of the matrix formed by the vectors that define it. Consider the wedge sum of two one forms:
α/\β ≡ 1/2 (α⊗β - β⊗α)
Hey, does anyone know how I can create sub- and superscripts here, I noticed html is off, I can't finish this post othewise? Any info would be appreciated.
selfAdjoint
Oct2-03, 05:04 PM
Use brackets [ , ] instead of < , >. Otherwise just the same.
rick1138
Oct4-03, 04:16 PM
Thanks, I am going to start again. In general, the value of a form is the value of the determinant of the matrix formed by the vectors that define it. Consider the wedge sum of two one forms:
α/\β ≡ 1/2 (α⊗β - β⊗α)
and consider the two one-forms in expanded form:
α ≡ α1e1 + α2e2 + α3e3
β ≡ β1e1 + β2e2 + β3e3
and remember these rules for the bases of forms:
ei /\ ej = - ej /\ ei
(reversing order reverses sign)
ei /\ ei = 0
(wedge summing a basis form by itself annihilates it)
so:
α/\β = 1/2 ((α1e1 + α2e2 + α3e3) ⊗ (β1e1 + β2e2 + β3e3) - (β1e1 + β2e2 + β3e3) ⊗ (α1e1 + α2e2 + α3e3)) =
(α1β1 e1 /\ e1 + α1β2 e1 /\ e2 +α1β3 e1 /\ e3 +α2β1 e2 /\ e1 +α2β2 e2 /\ e2 +α2β3 e2 /\ e3 +α3β1 e3 /\ e1 +α3β2 e3 /\ e2 +α3β3 e3 /\ e3) - (β1α1 e1 /\ e1 + β1α2 e1 /\ e2 +β1α3 e1 /\ e3 +β2α1 e2 /\ e1 +β2α2 e2 /\ e2 +β2α3 e2 /\ e3 +β3α1 e3 /\ e1 +β3α2 e3 /\ e2 +β3α3 e3 /\ e3)
which reduces to, by virtue of the rules above:
(α1β2 - α2β1) e1 /\ e2 + (α2β3 - α3β2) e2 /\ e3 + (α3β1 - α1β3 )e3 /\ e1
Which is isomorphic to the cross product, but not quite the same thing - the wedge sum of two one forms produces a two-form, wheras the cross product produces a scalar - the cross product is the Hodge star dual of the wedge sum of two one-forms.
Originally posted by lethe
... \mathbb{R}^n is itself a manifold, albeit a flat one, but we want to extend our idea of a space to include curved spaces, so lemme just give a few examples: a parabola is a curved 1 dimensional manifold, that extends to infinity. a circle is a 1 dimensional manifold that folds back on itself.
...
a manifold is just a space that is not necessarily flat.Can you give the justification that \mathbb{R}^n is flat and that a parabola is curved (I'm assuming that you mean a parabola to be the 2-D representation of all points (x,y) that exist in the 2-D space which satisfy the coordinate values y = x2, or some scaled, translated, or rotated version thereof).
I can see the literal curvature of the parabola, and the literal flatness of the x axis (\mathbb{R}^1) if I view them in the context of the x-y plane, but I don't see how you can characterize such a thing in 1-D, and such characterization seems to be in the spirit of this thread.
Here's my problem: I can't see the fundamental difference between the parabola and the real number line as 1-D manifolds. Both have 1 dimension (another concept I don't quite understand, but I'll defer that until later). I don't see what more you can say without a metric, or at least a coordinatization. If I choose to label the points on the parabola by the arclength along the parabola from the origin (which is, IMO, the most natural way to do it), then how would I know it was curved? Alternatively, how would I know to label points using their x values to show the curvature, when, for the sake of purity, I should not be appealing to any x axis in some x-y plane? In other words, how do I know that the parabola imbeds itself in the x-y plane as a parabola instead of a straight flat line, without already knowing that it was, in fact, a parabola in the x-y plane.
Originally posted by Tom
Vectors vi (i=1,2,3,...) in Rn are independent iff
a1v1+a2v2+a3v3+...=0
implies that
a1+a2+a3+...=0Did you mean for:
"a1+a2+a3+...=0"
to be:
"a1=0, a2=0, a3=0, ...?"
Originally posted by lethe
for any vector v, -v is also a vector.I was just a little uncomfortable with this notation. Do you mean:
-v is defined as (-1)v
where (-1) is a member of the field? Isn't this rather trivial? I'm assuming you mean to require an additive inverse for your vector space (or "abelian group" or whatever you called it). IM very HO, this could be reworded to:
"for every vector, v, there is a vector, vinv, such that: v + vinv = 0. This is the existence of an inverse."
Shankar has the more indicative notation, using the kets, to include the minus sign inside the ket, to distinguish it from a literal negative sign as a multiplication by (-1).
Originally posted by lethe
suppose we are given a basis for our original vector space V, {eμ}. then this induces a natural choice of basis for the dual space V*, {σν}, determined by
\sigma^\nu(\mathbf{e}_\mu) = \delta^\nu_\mu (1)
...
these linear functionals form a basis of the dual space, that we will have occasion to use."This defines THE dual space," as in, "there is ONLY ONE WAY to do it," or, "this defines the dual space," as in, "this is the way we HAPPEN TO do it?" This looks suspiciously like you have sneeked a metric tensor into the discussion under the guise of defining the dual space. Is this the case? Is there some way to define a dual space without using this metric-ish definition? Does this Kronecker-Delta generalize to the metric tensor for general spaces?
Originally posted by jeff
Intrinsic curvature is defined by using the fairly easy to understand idea of "parallel transport". Imagine some closed curve ...Does this mean that 1-D manifolds can not have intrinsic curvature (how do you make a closed curve on the parabola)?
Originally posted by turin
Can you give the justification that \mathbb{R}^n is flat and that a parabola is curved (I'm assuming that you mean a parabola to be the 2-D representation of all points (x,y) that exist in the 2-D space which satisfy the coordinate values y = x2, or some scaled, translated, or rotated version thereof).
I can see the literal curvature of the parabola, and the literal flatness of the x axis (\mathbb{R}^1) if I view them in the context of the x-y plane, but I don't see how you can characterize such a thing in 1-D, and such characterization seems to be in the spirit of this thread.
i don t really need to be rigorous about the difference between flat and curved in those posts, since i was just mentioning it to give an intuition, and the notion is not actually well defined: the difference between the parabola and the real line is just a different embedding in R2, i.e. it is not intrinsic.
the only reason i brought it up was to convince people why the notions we learned in R3 just won t work for a general manifold. R3 is a vector space, and that is what i meant by calling it flat (no metric involved). you can add points in the manifold to each other if it is flat. tangent vectors to the manifold can also be thought of as living in the manifold itself if it is flat.
neither of these things is true if the manifold is not a vector space, and so that s all i meant.
Here's my problem: I can't see the fundamental difference between the parabola and the real number line as 1-D manifolds. Both have 1 dimension (another concept I don't quite understand, but I'll defer that until later). I don't see what more you can say without a metric, or at least a coordinatization. If I choose to label the points on the parabola by the arclength along the parabola from the origin (which is, IMO, the most natural way to do it), then how would I know it was curved? Alternatively, how would I know to label points using their x values to show the curvature, when, for the sake of purity, I should not be appealing to any x axis in some x-y plane? In other words, how do I know that the parabola imbeds itself in the x-y plane as a parabola instead of a straight flat line, without already knowing that it was, in fact, a parabola in the x-y plane.
yes, you are correct. a good observation. instrinsically, all 1D spaces have the same geometry.
Originally posted by turin
I was just a little uncomfortable with this notation. Do you mean:
-v is defined as (-1)v
where (-1) is a member of the field? Isn't this rather trivial? I'm assuming you mean to require an additive inverse for your vector space (or "abelian group" or whatever you called it). IM very HO, this could be reworded to:
"for every vector, v, there is a vector, vinv, such that: v + vinv = 0. This is the existence of an inverse."
Shankar has the more indicative notation, using the kets, to include the minus sign inside the ket, to distinguish it from a literal negative sign as a multiplication by (-1).
yes, i agree with all this. i guess i just can t be bothered with that level of formalism, but i do think that it is very important to see that kind of thing when you first do abstract algebra. it lets you divorce yourself of misconceptions or generalizations that you learned in your high school algebra class.
for abelian groups, i think its pretty harmless. it is trivially easy to show that (-1)v=vinv
Originally posted by turin
"This defines THE dual space," as in, "there is ONLY ONE WAY to do it," or, "this defines the dual space," as in, "this is the way we HAPPEN TO do it?" This looks suspiciously like you have sneeked a metric tensor into the discussion under the guise of defining the dual space. Is this the case?
note that i didn t use the kronecker delta to define the dual space, but only to choose a basis. it just happened to be on hand as a way of choosing a basis.
you can choose any basis you like, as long as you can make sure that it is actually a basis (linearly independent, etc). with the choice i made above this was easy to check.
Is there some way to define a dual space without using this metric-ish definition? Does this Kronecker-Delta generalize to the metric tensor for general spaces?
like i said above, the kronecker delta is for choosing a basis, not for defining the dual space. a dual vector acting on a vector gives me a real number. i just have to make a choice for which numbers my basis vectors will give, and i choose 1s and 0s.
this notion does not generalize to other metrics: the metric is not defined between vectors and covectors. some books use an inner product type notation, but i dislike this a lot.
Originally posted by turin
Does this mean that 1-D manifolds can not have intrinsic curvature (how do you make a closed curve on the parabola)?
you can make a closed curve on the parabola, you just have to be willing to trace back on yourself.
a 1D manifold cannot have any intrinsic curvature, but for other reasons.
Originally posted by lethe
you can add points in the manifold to each other if it is flat.Can this be done without applying coordinates to the manifold? What does it mean to add point P to point Q?
Originally posted by lethe
instrinsically, all 1D spaces have the same geometry. What does "geometry" mean? I thought we were discussing pre-geometry manifolds. Does a circle have the same geometry as the real number line?
Originally posted by lethe
... the kronecker delta is for choosing a basis, not for defining the dual space.I think I understand the distinction here, but I don't understand the significance. If you want to talk about the objects that live in your dual space, then aren't you going to need a basis? Can you give some non-trivial demonstration/identity/proof (not a definition) that does not require a basis?
Originally posted by lethe
a dual vector acting on a vector gives me a real number.Is this THE definition of a dual vector?
Originally posted by turin
Can this be done without applying coordinates to the manifold? What does it mean to add point P to point Q?
some manifolds admit algebraic structures, and some don t. linear spaces all do, since it is part of their definition. you do not have to choose coordinates on your manifold to have algebra.
What does "geometry" mean? I thought we were discussing pre-geometry manifolds. Does a circle have the same geometry as the real number line?
in this instance, geometry means curvature. any 1 dimensional manifold has no intrinsic curvature, and thus, locally, all 1D manifolds have the same geometry.
we are discussing differentiable manifolds (pregeometry manifolds, as you say) in this thread. it was you who brought up the issue about the parabola and the line being the same, and so i only mentioned that to make that discussion a little clearer.
I think I understand the distinction here, but I don't understand the significance. If you want to talk about the objects that live in your dual space, then aren't you going to need a basis?
no
Can you give some non-trivial demonstration/identity/proof (not a definition) that does not require a basis?
the dual of the dual of a vector space is canonically isomorphic to the vector space. this theorem can be proved without ever choosing a basis.
the problem with choosing a basis is that there are many equally good bases to pick from, and there is no "best" basis, so sticking to only one is unnatural. but once i have made this unnatural choice, there is a best choice for the basis of the dual space, which i describe above.
Is this THE definition of a dual vector?
yes
This all strikes me as funny for some reason. A circle is a 1D manifold, and it is curved in the sense that walking along the circle in a constant direction eventually leads you back over your own footsteps. Of course, that definition of "curved" is not mathematically sound.
At the same time, there's no way for a 1D creature who lives on the circle to do any experiments to determine if there is or is not curvature. The only figures he can draw is his 1D space are lines and points, and his lines will always have the same length, no matter which direction he draws them...
I suppose I accept the fact that a 1D curve has no intrinsic curvature, but it bugs me somehow.
- Warren
Originally posted by chroot
This all strikes me as funny for some reason. A circle is a 1D manifold, and it is curved in the sense that walking along the circle in a constant direction eventually leads you back over your own footsteps. Of course, that definition of "curved" is not mathematically sound.
this notion is mathematically sound, its just not the definition of curvature. you are talking about some global property (comes back to the beginning), and geometry talks about local properties (curvature). this kind of coming back on itself is a common subject of study in topology.
At the same time, there's no way for a 1D creature who lives on the circle to do any experiments to determine if there is or is not curvature. The only figures he can draw is his 1D space are lines and points, and his lines will always have the same length, no matter which direction he draws them...
i think this bug can do experiments to measure curvature. he just won t ever measure anything other than zero.
Originally posted by lethe
this notion is mathematically sound, its just not the definition of curvature. you are talking about some global property (comes back to the beginning), and geometry talks about local properties (curvature). this kind of coming back on itself is a common subject of study in topology.
Hmm, but I thought the definition of a Riemannian manifold was that it was locally flat at every point? Don't all manifolds have this property of being locally flat?
I suppose being locally flat just means you can introduce a Euclidean coordinate system anywhere and neglect the curvature; it's still intrinsically present, you're just neglecting it.
i think this bug can do experiments to measure curvature. he just won t ever measure anything other than zero.
What sorts of experiments can he do? Besides drawing lines and measuring their lengths?
- Warren
Originally posted by chroot
Hmm, but I thought the definition of a Riemannian manifold was that it was locally flat at every point? Don't all manifolds have this property of being locally flat?
all manifolds have the property of being locally euclidean. this is a topological property that has nothing to say about flatness.
I suppose being locally flat just means you can introduce a Euclidean coordinate system anywhere and neglect the curvature; it's still intrinsically present, you're just neglecting it.
the definition of a manifold makes no mention of curvature, no assumptions about curvature, nothing like that. the Riemannian manifold is just a differentiable manifold with a Riemannian metric on it.
the curvature is not constrained.
What sorts of experiments can he do? Besides drawing lines and measuring their lengths?
the curvature tensor tells you how a vector transforms when you go in a loop. so to measure curvature, he could draw 1dimensional loops, and see what happens. he would surely find that all vectors remain unchanged.
Hmmm but how does he draw a closed 1D loop in his 1D space? That's just a line, eh? And it can't be closed. So I guess that's why a 1D space has no intrinsic curvature. It doesn't make sense with the definition of intrinsic curvature.
- Warren
Originally posted by chroot
Hmmm but how does he draw a closed 1D loop in his 1D space? That's just a line, eh? And it can't be closed. So I guess that's why a 1D space has no intrinsic curvature. It doesn't make sense with the definition of intrinsic curvature.
- Warren
why can t the loop be closed? what about the curvature doesn t make sense?
Originally posted by lethe
why can t the loop be closed? what about the curvature doesn t make sense?
If you're an ant living in a 1D space, how can you draw a loop in the first place? I mean, how can a loop even exist in 1D? Am I missing something?
- Warren
Originally posted by chroot
If you're an ant living in a 1D space, how can you draw a loop in the first place? I mean, how can a loop even exist in 1D? Am I missing something?
- Warren
here is a loop on the real line:
\gamma(t)=\begin{cases}2t& 0\leq t\leq 1/2\\ 2-2t & 1/2\leq t\leq 1\end{cases}
Okay, I guess I can't argue that it's a loop, even though it's sort of a "degenerate" one.
So if the concept of a loop is well-defined in 1D space, why then does the curvature tensor always vanish? Sorry, I believe it, I just don't grok it.
- Warren
Originally posted by chroot
Okay, I guess I can't argue that it's a loop, even though it's sort of a "degenerate" one.
So if the concept of a loop is well-defined in 1D space, why then does the curvature tensor always vanish? Sorry, I believe it, I just don't grok it.
the tangent space is 1 dimensional. any vector has only one choice, it cannot rotate.
Aha, that makes sense now. If our ant on the 1D circular manifold pushes a vector around, there's no way it can rotate (assuming that it can't snap around 180 degrees -- that sort of wacky behavior is impossible in a smooth manifold, I guess?). And since curvature is defined by the angle between a vector and its counterpart after being parallely transported in a small loop, the angle must be zero, so there must be no (intrinsic) curvature.
Okay, got it. Thanks.
- Warren
Originally posted by lethe
some manifolds admit algebraic structures, and some don t.
...
you do not have to choose coordinates on your manifold to have algebra.If it doesn't deviate terribly from the main discussion, could you just give the basic requirements of an algebra, specifically to distinguish them from the requirements of a vector space?
Originally posted by lethe
it was you who brought up the issue about the parabola and the line being the same, and so i only mentioned that to make that discussion a little clearer.I'm just trying to distinquish between a parabola and a line in our context. I thought you were trying to make a point of it in the beginning. [*(]
Originally posted by lethe
the dual of the dual of a vector space is canonically isomorphic to the vector space.That sure isn't obvious using the stack of pancakes notion. [:D] Is this a good example of why you don't like it?
Originally posted by lethe
the problem with choosing a basis is that there are many equally good bases to pick from, and there is no "best" basis, so sticking to only one is unnatural. but once i have made this unnatural choice, ...I'm, assuming this was talking about the basis of the vector space, as opposed to the dual space?
Originally posted by turin
If it doesn't deviate terribly from the main discussion, could you just give the basic requirements of an algebra, specifically to distinguish them from the requirements of a vector space?
an algebra is a vector space that has a vector product.
in general, tangent spaces will just be vector spaces, not algebras, but if the manifold is also a group, then some of the tangent spaces will be algebras. R3 with the vector cross product is an example of this.
just to be clear though: when i said algebra above, i didn t say an algebra. an algebra is a vector space with vector product. algebra is a more general term, it just means anything having to do with addition and multiplication. so i said above something about manifolds not having an algebraic structure, i just meant that there is no way to add or multiply, in a consistent way, points on, say, a sphere, whereas there are such notions for linear spaces (by definition).
I'm just trying to distinquish between a parabola and a line in our context. I thought you were trying to make a point of it in the beginning. [*(]
its been a while, but i think the reason i wrote that in the beginning was just to show why you need vectors to live in the tangent space, and they cannot, in general, live in the manifold. that was the only point.
i didn t want to imply that the parabola had intrinsic curvature or anything like that. i m beginning to regret even mentioning the word "curved" up there, since i had not defined it yet. it was just supposed to help your intuitive picture, when thinking of vectors and manifolds.
That sure isn't obvious using the stack of pancakes notion. [:D] Is this a good example of why you don't like it?]
sure, i guess so.
I'm, assuming this was talking about the basis of the vector space, as opposed to the dual space?
yeah, i guess that is what i was talking about, but it could work the other way as well.
matt grime
Jan16-04, 06:56 PM
An algebra is not a vector space which has a vector product.
Exercise define vector product properly. Vector product is usually the term reserved for the cross product in three dimensions. there are plenty of three dimensional algebras not isomorphic to R^3 with the vector product.
An algebra is a Ring which is also a vector space.
Originally posted by matt grime
An algebra is not a vector space which has a vector product.
i suppose this is a semantic argument. i define a vector product to be a bilinear map on a vector space into the vector space. under this definition, a vector space with a vector product is an algebra.
Exercise define vector product properly. Vector product is usually the term reserved for the cross product in three dimensions.
i think you should say vector cross product (or simply cross product), when you mean the cross product in R3.
the names of objects in mathematics ought to be descriptive enough to leave no ambiguity, this is my opinion, at least. under this philosophy, vector product, scalar product, cross product, inner product, outer product and dot product are all different, and there is no ambiguity in any of the terms.
there are plenty of three dimensional algebras not isomorphic to R^3 with the vector product.
i would say that R3 with the vector cross product is an example of an algebra. certainly there are other examples.
An algebra is a Ring which is also a vector space.
this definition also works. of course, the ring is not as familiar to people as the vector space, so i prefer my definition. but it is a matter of taste.
selfAdjoint
Jan17-04, 11:31 AM
How about, an algebra is a vector space which has(completely outside its VS structure) a distributive product. Very often physicists will define an algebra just by defining the product, since the underlying vector space is "obvious".
In the case of 3D vectors, of course, the cross product is really an outer product (Grassmann style) and its result is not exactly a true (polar) vector, but an axial vector that behaves differently under parity operations. This distinction had an important role in physicsts' attempts to understnd the weak force.
The true algebra that contains the 3D vectors is the quaternions.
Originally posted by selfAdjoint
How about, an algebra is a vector space which has(completely outside its VS structure) a distributive product.
but, as i am sure you are aware, the dot product is distributive, but an inner product space is not an algebra. i think distributivity isn t worth mentioning. in my world, if its not distributive, it isn t a product. so the word "product", for me, contains the information "bilinear" and "distributive". what does need to be mentioned to distinguish it from other products is that it is vector valued
In the case of 3D vectors, of course, the cross product is really an outer product (Grassmann style)
i don t quite agree with this. there is a sense in which the cross product can be thought of as a Grassman product (which i call a "wedge product", not an "outer product". for me, "outer product" is synonymous with "tensor product"), but the Grassman algebra is certainly not isomorphic to R3 as an algebra. for example, inR3, you have (ixj)xi=j whereas in the Grassman algebra, you have (ixj)xi=0. not isomorphic. if you toss the Hodge dual in there in the appropriate place, then you can make an isomorphism
what R3 is isomorphic to is the Lie algebra \mathfrak{so}(3)
RDT2 is around, lemme see if i cant post some more on this thread. this is still taken from the thread at sciforums.
now... where were we. ah, yes. we had just finished building the 2-forms, and we re about ready to move to more general p-forms.
but before we leave the 2-forms, let s find a basis for them, and look at their coordinate representation. it should be obvious how to do that, right? we built our 2-form from two 1-forms, so we should be able to build a basis for our 2-forms from the basis for our 1-forms. let s recall what that was (6):
df=\partial_\nu f dx^\nu
let s take two of those, and wedge them together:
df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu
here the advantages of using the einstein summation notation become more clear. when you are multiplying two (or more) long summations, carrying around a lot of extra sigmas can get quite unwieldy.
this isn t quite a proof that dx^\nu\wedge dx^\mu is the basis, i.e. that any alternating second rank tensor can be written as such a sum, but it should be convincing at any rate.
now that we ve found a basis, let s count the dimension of this vector space, the space of all 2-forms. remember, the dimension of a vector space is just the number of elements in the basis. so how many independent dx^\nu\wedge dx^\mu are there? well there are N different dx^\nu and N different dx^\mu, so there should be N^2 ways to write the product, where N is the dimension of the manifold, and the tangent vector space, and the cotangent space.
so the dimension of \bigwedge^2T^*M_p is N^2. right?
not so fast, hot shot! there may be N^2 ways to write that product, but they are not all linearly independent. remember the properties of 2-forms: dx^\nu\wedge dx^\mu = -dx^\mu\wedge dx^\nu. so we don t want to count this guy twice. furthermore dx^\nu\wedge dx^\nu = 0! so we definitely don t want to count those cases when \nu=\mu. so when counting the basis elements we should only count those for which, say, \nu < \mu
if you like combinatorics, you can work out the formula. i don t really so i m just going to say the answer: there are N(N-1)/2 linearly independent 2-forms. that formula might look familiar to some of you, it is {}_NC_2 (N choose 2).
Originally posted by lethe
... we should be able to build a basis for our 2-forms from the basis for our 1-forms. let s recall what that was (6):
df=\partial_\nu f dx^\nuI still don't understand how this gives us a basis. Does this notation imply something more specific than generalized coordinates? This is just the total derivative of a multivariable function, right?
Originally posted by lethe
let s take two of those, and wedge them together:
df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu dx^\nu
Two of whats? they look like scalars to me since there is a contraction. Is there a previous post in which you explain how these would be vectors? I'm so confused. I read your post in which you introduced vectors, but they look like scalars to me.
Originally posted by lethe
this isn t quite a proof that dx^\nu\wedge dx^\mu is the basis, i.e. that any alternating second rank tensor can be written as such a sum, ...I don't see how the dx^\nu\wedge dx^\mu shows up.
Originally posted by turin
I still don't understand how this gives us a basis. Does this notation imply something more specific than generalized coordinates? This is just the total derivative of a multivariable function, right?
no, this notation implies nothing beyond the representation of a 1-form in terms of some general coordinates. your manifold has coordinates, which in turn yield a basis for the tangent space (\{\partial/\partial x^\mu\}), which in turn induces a basis for the dual space (\{dx^\mu\}). for this dual space, as for any vector space, expressing any vector in terms of the basis means finding a linear combination of the basis vectors that equals the vector in question. in this case, it is df=\partial_\mu f dx^\mu. df is the vector, \partial_\mu f are the coefficients, and dx^\mu are the basis vectors
this formula does look like the formula one learns in elementary calculus for the derivative of a function. there we have the chain rule df/dt=(\partial f/\partial x^\mu)(dx^\mu/dt). if we "multiply" both sides of this equation by dt, then we get the above formula. of course, this multiplication step is invalid, since in elementary calculus, we have no object called dt, and dx^\mu/dt is not a fraction, but a single object. but the similarity is no coincidence. it looks this way because the exterior derivative really is a kind of derivative, and so has to include the chain rule of elementary calculus.
Two of whats? they look like scalars to me since there is a contraction. Is there a previous post in which you explain how these would be vectors? I'm so confused. I read your post in which you introduced vectors, but they look like scalars to me.
this is an excellent question. you probably have learned the following rule of thumb: anything with no indices is a scalar, anything with one index is a vector, anything with more indices is a tensor.
this rule is nonsense. or at least, it is only true about the coordinate components of those objects, and not the objects themselves.
here is a better rule: any geometric object, which has a coordinate independent meaning, cannot have any indices (since indices indicate dependence on your choice of coordinates). scalars, vectors, tensors, 1-forms, and anything else worth talking about, are all geometric objects, with coordinate independent existences. therefore they cannot have indices.
in the expression \mathbf{v}=v^\mu\partial_\mu, v^\mu is the coefficient of my vector in terms of some basis. this number depends on my choice of basis (i.e. my choice of coordinates) and any coordinate dependent object should carry an index. but it is not a vector, it is only a component of a vector. you probably learned that anything with a raised index is a (contravariant) vector. part of my goal with this stuff is to teach you why that picture is misleading (and why the word contravariant is a mistake in this context).
\partial_\mu also carries an index. but this one is lowered. does that mean that it is (covariant) vector? no it doesn t!! it is still a tangent vector. the real vector is \mathbf{v}=v^\mu\mathbg{\partial_\mu}=v^1\mathbf{\ partial_1}+v^2\mathbf{\partial_2}+....+v^n\mathbf{ \partial_n} which carries no indices! to sum up, the point is: anything with an index cannot have any intrinsic meaning, since it is a coordinate dependent object.
in some other post, i will explain why the words covariant and contravariant are exactly backwards. above, i used contravariant to mean "having a raised index, like the coordinates", and covariant to mean "having a lowered index, like the derivative", since this is how it is usually taught to physicists. i will never use those words in that sense again, and for me, the words are actually switched.
I don't see how the dx^\nu\wedge dx^\mu shows up.
well, the dx^\nu\wedge dx^\mu comes from df\wedge dg by linearity on the basis vectors. i actually slipped up in one of the equations in my post, and forgot to include the \wedge. maybe it is clearer now? the point is, df and dg both have coordinate representations in terms of our basis vectors dx^\mu, so i pull the coefficients of the two 1-forms in this wedge product out front (since the wedge product is linear), and am left with a sum over the wedge products of the basis vectors.
note that the coefficients of the 2-form have two indices (\partial_\mu f\partial_\nu g), and the basis (co-)vectors also have two indices (dx^\mu\wedge dx^\nu), but the expression for the entire 2-form has both indices contracted, and so carries no indices. but it is certainly not a scalar!
I HATE ITEX!
I will type the corresponding html next to the itex.
OK, I will delete all itex crap and never use it again. It keeps changing on me.
Originally posted by lethe
... you probably have learned the following rule of thumb: anything with no indices is a scalar, anything with one index is a vector, anything with more indices is a tensor.
this rule is nonsense.I have this appreciation. My major prof adamantly declares that a tensor is defined by its transformation properties. If this is true, then it is obvious that the rule of thumb you mention here is nonsense. If it is a bad way to think of it in terms of the transformation properties (I seem to vaguely remember you discouraging this way of thinking), then please remind me.
I wasn't thinking that the index free quality indicated scalar-ness. I was more concerned with the apparent contraction of two aparent vectors.
According to my major prof:
- the contraction of two rank 1 tensors (vectors) is a rank 0 tensor (scalar)
- ∂μ and dxμ are rank 1 tensors (at least in Minkowski space-time).
From this I infer:
- the object in question still looks like a scalar to me, unless I radically change my understanding of tensors.
What about the proper time interval: dτ2 = dxμdxμ?
It seems like there are two inconsistent ways of looking at it:
- either this is a contraction and therefore a scalar
- or this is a 1-form with components dxμ.
Do the components of any 1-form form a vector basis, and do the components of any vector form a covector basis?
Originally posted by lethe
in the expression \mathbf{v}=v^\mu\partial_\mu, v^\mu is the coefficient of my vector in terms of some basis. this number depends on my choice of basis (i.e. my choice of coordinates) and any coordinate dependent object should carry an index. but it is not a vector, it is only a component of a vector.My major prof would say that it is OK to call vμ a vector because it implies all of the components (and I guess because it implies the basis?). What say you? I don't want to be picky, just trying to get a handle on the different notational formalisms.
Originally posted by lethe
i actually slipped up in one of the equations in my post, and forgot to include the \wedge. maybe it is clearer now?Ya. If you meant for that wedge to be in there, then I get it. Again, I'm not trying to be picky, but I have been given the impression lately that these kinds of notational issues are important.
Originally posted by lethe
... df and dg both have coordinate representations in terms of our basis vectors dx^\mu, ...I thought the basis vectors were ∂μ and that the 1-forms were dxμ. Did you mean ∂μ here?
Originally posted by turin
I HATE ITEX!
I will type the corresponding html next to the itex.
well, feel free to use html. i actually prefered the html, since it seems to fit more nicely with the text, however it doesn t display on some peoples browsers, and of course it can t do as much stuff as tex.
I have this appreciation. My major prof. adamantly declares that a tensor is defined by its transformation properties. If this is true, then it is obvious that the rule of thumb you mention here is nonsense. If it is a bad way to think of it in terms of the transformation properties (I seem to vaguely remember you discouraging this way of thinking), then please remind me.
physicists definition of a tensor:
an object with (r,s) raised, lowered indices (and therefore a coordinate dependent object) that transforms in such and such a way when you change coordinates
mathematicians definition of a tensor:
a tensor product of r,s vectors, covectors. the mathematicians definition of a vector and covector is such that it makes no reference to coordinates, and thus neither does the definition of a tensor. it is an exercise for the reader in most math books to check that when you look at the coordinate components of a tensor, they transform in the physicists way when you change your choice of coordinates.
you take your pick as to which definition. it is nice to understand both definitions, and then one doesn t have to adamantly adhere to one or the other. but certainly one can have a preference, mine is the mathematicians definition.
I wasn't thinking that the index free quality indicated scalar-ness. I was more concerned with the apparent contraction of two aparent vectors. According to my major prof:
- the contraction of two rank 1 tensors (vectors) is a rank 0 tensor (scalar)
you cannot contract two (1,0) rank tensors. you can only contract a (1,0) tensor with a (0,1) tensor. of course, since the metric (if you are doing Riemannian geometry, known to physicists as relativity) and the symplectic form (if you are doing symplectic geometry, known to physicists as classical mechanics) are both nondegenerate, you can always convert one of your (1,0) rank tensors into a (0,1) rank tensor and then contract them (known to physicists as raising and lowering indices). but in the absence of a metric or symplectic form, there is no canonical isomorphism between the vector space of tangent vectors ((1,0) rank tensors) and the vector space of covectors/dual vectors ((0,1) rank tensors), and therefore you cannot contract them.
if you would like any of those terms explained further, please ask.
- \partial_\mu (that is, ∂μ) and dx^\mu (that is, dxμ) are tensors (at least in Minkowski space-time)
therefore:
yes indeed, those are both tensors (they are basis tensors, and therefore coordinate dependent). also, none of this is particular to Minkowski space.
- the object in question still looks like a scalar to me, unless I radically change my understanding of tensors.
yes indeed, the contraction of dx^\mu and \mathbf{\partial_\mu} is indeed a scalar. in fact, it is 1.
What about the proper time interval:
d\tau^2 = dx_\mu dx^\mu (that is, dτ2 = dxμdxμ)?
It seems like there are two inconsistent ways of looking at it:
- either this is a contraction and therefore a scalar
contraction makes a scalar if the contracted objects are a (1,0) tensor and a (0,1) tensor. that is not the case here, so that thing is not a scalar. the fact that it doesn t carry any indices indicates that it is a geometric object, independent of coordinates.
- or this is a 1-form with components dx_\mu (that is, dxμ).
it is also not a 1-form. it is a tensor product of 2 1-forms. if it were also antisymmetric, i would call it a 2-form, but it is not antisymmetric, so i will call it a (0,2) rank tensor.
but you could have figured out that it was a (0,2) rank tensor just by looking at its coordinate components g_{\mu\nu}. 2 lowered indices on the coordinate components = (0,2) rank tensor.
Do the components of any 1-form form a vector basis, and do the components of any vector form a covector basis?
components do not form a basis, since they are not vectors. the components of a 1-form happen to transform like the basis vectors of tangent space, and components of the tangent vector happen to transform like the basis vectors of the cotangent vector space, but this does not mean that the components are themselves vectors
in fact, this point of confusion is exactly the reason that i dislike the physicists definition of a tensor. you become confused about what is a vector, what is a covector, and what are just components.
i said above "happen to transform", but it is no coincidence. recall that any vector (for any vector space. i m thinking linear algebra here) can be written like this:
\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2+...x_n\ mathbf{e}_n
here, \mathbf{x} is a vector, which exists in any basis, but has different components. the components live in some field (sometimes these guys are called scalars in math class, but i won t use that word here. for physicists, scalar means something that is invariant under coordinate transformations). thus the components are not vectors. the basis vectors are vectors, but they also depend on your choice of basis (obviously).
if you make a change of basis, you can achieve this by multiplying the basis vectors by some matrix to get a new basis. then you multiply the components by the inverse of that matrix to get the components of the vector in the new basis. the vector itself has matrix times matrix^-1, and thus doesn t change. it is independent of your choice of basis. it is only the coordinates that depend on your choice of basis, and they change in the opposite way that the basis vectors themselves change.
this is why the components of a tangent vector transform like the basis vectors of the cotangent space.
My major prof. would say that it is OK to call v^\mu (that is, vμ) a vector because it implies all of the components (and I guess because it implies the basis?). What say you? I don't want to be picky, just trying to get a handle on the different notational formalisms.
yeah, all physicists do this. it is fine to call v^\mu a vector. in fact, i do it myself whenever i am doing physics. but just keep in the back of your head that v^\mu are really the components of a vector, strictly speaking, they are not the vector itself. since, in physics, we only ever deal with components, we can replace them in our minds. but be aware that doing so will lead to confusion when you try to do this in a math class. and when it comes time to start doing non-Abelian gauge theory, you will wish you were in the math camp, instead of the physics camp.
Ya. If you meant for that wedge to be in there, then I get it. Again, I'm not trying to be picky, but I have been given the impression lately that these kinds of notational issues are important.
indeed they are (in my opinion)
I thought the basis vectors were \partial_\mu (that is, ∂μ) and that the 1-forms were dx^\mu (that is, dxμ). Did you mean \partial_\mu (that is, ∂μ) here?
no.
\mathbf{\partial}_\mu are the basis vectors for the tangent space, and dx^\mu are the basis vectors for the cotangent space. since df and dg are 1-forms (by definition, a 1-form is a member of the cotangent space), they can be written in terms of the basis of that space. of course, since the basis vectors of any vector space are themselves members of that vector space, dx^\mu is itself a 1-form, it is a basis 1-form. but this 1-form depends on your coordinates. and likewise for \mathbf{\partial}_\mu
Holy crap!
Originally posted by lethe
... and the symplectic form (if you are doing symplectic geometry, known to physicists as classical mechanics) are both nondegenerate, you can always convert one of your (1,0) rank tensors into a (0,1) rank tensor and then contract them (known to physicists as raising and lowering indices).Ya, uh, question. What is "symplectic form?"
Originally posted by lethe
... the contraction of dx^\mu and \mathbf{\partial_\mu} is indeed a scalar.This seems to contradict the definition of ∂μf dxμ as a vector. Is a vector the same thing as a scalar in math land?
Originally posted by lethe
contraction makes a scalar if the contracted objects are a (1,0) tensor and a (0,1) tensor. that is not the case here, so that thing is not a scalar. the fact that it doesn t carry any indices indicates that it is a geometric object, independent of coordinates.OK, so a (1,0) tensor is not synonymous with a contravariant vector, nor is a (0,1) tensor synonymous with a covariant vector?
I have also been told rather emphatically that the metric is a scalar because it does not get transformed by a Lorentz transformation. Not true?
Originally posted by lethe
it is also not a 1-form. it is a tensor product of 2 1-forms. if it were also antisymmetric, i would call it a 2-form, but it is not antisymmetric, so i will call it a (0,2) rank tensor.By writing one of the indices as a subscript, contraction with the metric tensor is already implied, and the dxμ is supposed to be the covariant form of dxμ. Is this just a matter of confusing termonology? I think it may be deeper than termonology and notation, because I would have sworn yesterday that dτ2 was a scalar, and that gμν was a second rank tensor. You're starting to scare me.
Originally posted by lethe
but you could have figured out that it was a (0,2) rank tensor just by looking at its coordinate components g_{\mu\nu}. 2 lowered indices on the coordinate components = (0,2) rank tensor.I understand that dxμdxν is a second rank tensor (and so, I guess that means a (0,2) tensor?). But dxβdxβ?
Originally posted by lethe
and when it comes time to start doing non-Abelian gauge theory, you will wish you were in the math camp, instead of the physics camp.I wish I was in the math camp right now. I am starting to think that physics is teaching me bad habits.
Originally posted by lethe
since df and dg are 1-forms (by definition, a 1-form is a member of the cotangent space), they can be written in terms of the basis of that space.How do you know that they are 1-forms and not vectors? What is wrong with saying that df is a vector, and in the ∂μf basis, it has components dxμ?
Originally posted by turin
Holy crap!
Ya, uh, question. What is "symplectic form?"
a symplectic form is a nondegenerate closed 2-form. in the context of classical mechanics, it is related to the Poisson bracket (which is, as you know, antisymmetric)
This seems to contradict the definition of ∂μf dxμ as a vector. Is a vector the same thing as a scalar in math land?
no, in math land, vectors and scalars are, of course, different.
i think where we are getting confused is this: dx^\mu are the basis 1-forms. they are dual to the basis vectors \mathbf{\partial}_\mu which means that you feed a basis vector to a basis dual vector and get a number dx^\mu(\mathbf{\partial}_\nu)=\delta^\mu_\nu. make special note of the fact that i made those basis vectors bold. so here, i contract a dual vector with a vector and get a scalar.
now lets recall how i defined vectors in the first place: they are derivations on the algebra of functions. in other words, they are differential operators (this is why i use the symbol \mathbf{\partial} for the basis vectors. it makes them look like differential operators.
so a bold \partial represents a tangent (basis) vector.
the exterior derivative of a function was defined by df(\mathbf{v})=\mathbf{v}(f), from which we derived that the components of df are \partial_\mu f. notice that i didn t put my \partial in bold. because \partial by itself is a tangent vector, it should be bold. but \partial_\mu f is just a number, a component of a 1-form. a scalar (in the math sense of the word, not the physics sense; it is not invariant under coordinate transformations)
so when you contract \partial_\mu f with dx^\mu, you just taking a linear combination of 1-forms, and so you end up with a 1-form (not a scalar)
when you contract dx^\mu with \mathbf{\partial}_\mu, you are letting your 1-form eat your vector, and since 1-forms (by definition) eat vectors and spit out scalars (here these are scalars in the math sense and the physics sense)
note: i am not very strict or consistent about requiring my vectors to be bold, but i m hoping in this case, it will help.
OK, so a (1,0) tensor is not synonymous with a contravariant vector, nor is a (0,1) tensor synonymous with a covariant vector?
well, that depends. are you using the words contravariant and covariant according to the physics convention? if so, then a (1,0) is a contravariant tensor, and a (0,1) tensor is a covariant tensor, as you say.
I have also been told rather emphatically that the metric is a scalar because it does not get transformed by a Lorentz transformation. Not true?
this is simply not true. not even for a physicist. whoever told you that was just wrong.
according to the math convention, the metric is a coordinate independent object, and it does not transform under coordinate transformations. but in the math convention, this does not make something a scalar. so the statement is not correct.
in the physics convention, we say something is a scalar if it is invariant under coordinate transformations, and i just finished saying the metric is coordinate independent, right?
no, not right. in this physics convention, when we say metric, we don t mean the coordinate independent (0,2) tensor, we mean its coordinate components g_{\mu\nu} and these certainly do transform under coordinate transformations.
so i can t think of any interpretation in which that statement makes any sense.
By writing one of the indices as a subscript, contraction with the metric tensor is already implied, and the dxμ is supposed to be the covariant form of dxμ. Is this just a matter of confusing termonology?
in math land, there is no object dx_\mu. that is physics shorthand for g_{\mu\nu}dx^\nu. there is also no object dx^\mu dx^\nu, that is physics shorthand for dx^\mu\otimes dx^\nu. putting the two statements together, i have g(\cdot,\cdot)=g_{\mu\nu}dx^\mu\otimes dx^\nu. this is the metric, written in coordinate components, in terms of the dx^\mu\otimes dx^\nu basis for the (0,2) tensors.
I think it may be deeper than termonology and notation, because I would have sworn yesterday that dτ2 was a scalar, and that gμν was a second rank tensor. You're starting to scare me.
but i suppose if you mean ds^2 to be the invariant distance between two very close points, then this is indeed a scalar. but here, in math land, dx^\mu does not mean distance between neighboring points (such a concept has no meaning; there is always a finite seperation, and to find the distance, you must integrate)
I understand that dxμdxν is a second rank tensor (and so, I guess that means a (0,2) tensor?). But dxβdxβ?
like i said above, that is just physics shorthand for something else.
I wish I was in the math camp right now. I am starting to think that physics is teaching me bad habits.
How do you know that they are 1-forms and not vectors? What is wrong with saying that df is a vector, and in the ∂μf basis, it has components dxμ?
well... i guess you can say that if you want, but in this thread, we are going to use my definitions, not yours. but let me point out why i think your choice of notation sucks: the notation \partial_\mu f suggests that this object depends on some function f. but basis vectors should only depend on your choice of coordinates, not on some function. also, the components of the vector associated with f should depend on f, and your notation doesn t show that.
Originally posted by lethe
a symplectic form is a nondegenerate closed 2-form.I guess I will need to know what degenerate means in this context. Does it mean that it gives a unique value for every unique pair of vectors that you feed it?
Originally posted by lethe
make special note of the fact that i made those basis vectors bold.
so a bold \partial represents a tangent (basis) vector.
notice that i didn t put my \partial in bold. because \partial by itself is a tangent vector, it should be bold. but \partial_\mu f is just a number, a component of a 1-form. a scalar (in the math sense of the word, not the physics sense; it is not invariant under coordinate transformations)
so when you contract \partial_\mu f with dx^\mu, you just taking a linear combination of 1-forms, and so you end up with a 1-form (not a scalar)
when you contract dx^\mu with \mathbf{\partial}_\mu, you are letting your 1-form eat your vector, and since 1-forms (by definition) eat vectors and spit out scalars (here these are scalars in the math sense and the physics sense)I'm not quite following the distinction between bold and not-bold. Is the difference whether or not the partial derivative operates on a function?
Originally posted by lethe
... i use the symbol \mathbf{\partial} for the basis vectors. it makes them look like differential operators.I don't understand what you mean by this. If they are in fact differential operators, then of course you would use the differential operator symbol? Are they not really differential operators, but they just act alot like them?
Originally posted by lethe
in this physics convention, when we say metric, we don t mean the coordinate independent (0,2) tensor, we mean its coordinate components g_{\mu\nu} and these certainly do transform under coordinate transformations.I thought that "metric" meant "ds2" in physics. At any rate, that's what I meant by it. So, is ds2 a (0,2) tensor? I think I'm starting to see how it could be. It returns a math_scalar value when you feed it two 1-forms, but if you don't feed it anything, then it doesn't even make sense.
Originally posted by lethe
according to the math convention, the metric is a coordinate independent object, and it does not transform under coordinate transformations. but in the math convention, this does not make something a scalar. so the statement is not correct.Can you give the math definition of scalar?
Originally posted by lethe
in math land, there is no object dx_\mu. that is physics shorthand for g_{\mu\nu}dx^\nu.Is not the gμνdxν a 1-form? Are you saying that gμνdxν is also mathematically meaningless?
Originally posted by lethe
... in this thread, we are going to use my definitions, not yours.
but let me point out why i think your choice of notation sucks:Hey, whoa. Sorry, there, partner. I didn't mean to offend you or stomp on your territory unwarranted. I just want to figure out how the notation and stuff that I have learned fits here, because some of the things that have been said seemed contradictory. I in no way intended to imply that you were contradicting yourself, and I appologize if I have admitted such interpretation.
Originally posted by lethe
the notation \partial_\mu f suggests that this object depends on some function f. but basis vectors should only depend on your choice of coordinates, not on some function.I seem to remember the definition of coordinates as functions. Are they a different kind of function or what? By choosing coordinates, have you not chosen functions?
Originally posted by lethe
also, the components of the vector associated with f should depend on f, and your notation doesn t show that. What does it mean for a vector to be associated with f? Is this related to the selection of the dual space?
Originally posted by turin
I guess I will need to know what degenerate means in this context. Does it mean that it gives a unique value for every unique pair of vectors that you feed it?
a bilinear form B(\cdot,\cdot) is degenerate if there exists some vector \mathbf{v}\neq 0 such that B(\mathbf{v},\mathbf{w})=0 for all \mathbf{w}\in V. if you think of the bilinear form as a matrix (if V is finite dimensional, then you can always do this), then this is equivalent to saying that its determinant is zero.
the existence of a nondegenerate bilinear form on a vector space gives you an isomorphism between the vector space and its dual space. it is this isomorphism that allows you to pretend that vectors with raised and lowered indices are just different names for the same thing. in the absence of this bilinear form, you cannot do this. in Riemannian geometry, the bilinear form is the metric.
I'm not quite following the distinction between bold and not-bold. Is the difference whether or not the partial derivative operates on a function?
the boldness is supposed to draw your attention away from the fact that these guys are differential operators, and remind you that they are tangent vectors. and tangent vectors can really be thought of as arrows. bold face is somehow supposed to remind you of that.
but they are still differential operators (by definition).
I don't understand what you mean by this. If they are in fact differential operators, then of course you would use the differential operator symbol? Are they not really differential operators, but they just act alot like them?
tangent vectors are differential operators, and to remind you of this, i use the symbol \partial
to recap: \mathbf{\partial}_\mu is a vector. if you feed it to a 1-form, you get a scalar. if you feed a function to this vector, you get a scalar.
\partial_\mu f is a component of a 1-form. if you contract it with basis 1-forms, you get a linear combination of those basis 1-forms, which is of course still a 1-form, not a scalar.
I thought that "metric" meant "ds2" in physics. At any rate, that's what I meant by it. So, is ds2 a (0,2) tensor? I think I'm starting to see how it could be. It returns a math_scalar value when you feed it two 1-forms, but if you don't feed it anything, then it doesn't even make sense.
feed it 2 tangent vectors, not 1-forms. things that eat vectors are type (0,s) tensors, things that eat 1-forms are type (r,0) tensors.
Can you give the math definition of scalar?
a vector space is defined to be the pair (F,V) where F is a field, and V is an abelian group, and which satisfies a few axioms (which you probably know)
in math, i call things that live in F scalars, and things that live in V vectors. since we are doing differential geometry, F will probably always be R, the real numbers. sometimes the complexes. those guys are math scalars.
Is not the gμνdxν a 1-form? Are you saying that gμνdxν is also mathematically meaningless
yes, it is a 1-form, and no it is not mathematically meaningless. recall where we said this:
Originally posted by turin
Originally posted by lethe
but you could have figured out that it was a (0,2) rank tensor just by looking at its coordinate components g_{\mu\nu}. 2 lowered indices on the coordinate components = (0,2) rank tensor.
I understand that dxμdxν is a second rank tensor (and so, I guess that means a (0,2) tensor?). But dxβdxβ?
now since you yourself have told me that dx_\mu is a 1-form (which is a (0,1) rank tensor), and we know that dx^\mu is also a 1-form, then it should be clear that dx_\mu\otimes dx^\mu must be a (0,2) rank tensor. i only said that bit about it mathematically meaningless because i thought you were getting hung up on the fact that one of those 1-forms had a lowered index. that is just physics short hand for the components of a metric. nothing more. it doesn t change the fact that you are tensoring 2 (0,1) tensors, and therefore get a (0,2) tensor. just because you are contracting an index, it does not change the fact that there are 2 tensored 1-forms up there.
Hey, whoa. Sorry, there, partner. I didn't mean to offend you or stomp on your territory unwarranted. I just want to figure out how the notation and stuff that I have learned fits here, because some of the things that have been said seemed contradictory. I in no way intended to imply that you were contradicting yourself, and I appologize if I have admitted such interpretation.
hey man, no need to apologize. you didn t offend me. i just thought your suggestion for a change of notation was silly, and i was ribbing you for it.
I seem to remember the definition of coordinates as functions. Are they a different kind of function or what? By choosing coordinates, have you not chosen functions?
coordinates are indeed a different kind of function. coordinates are mappings from the manifold to Rn (for an n dimensional manifold), whereas functions that we are dealing with here are mappings from the manifold to R.
you can consider a single coordinate as a single valued function. then i guess you could call this a choice of a function. but f is a different (arbitrary) function, and your coordinates shouldn t depend on it.
What does it mean for a vector to be associated with f? Is this related to the selection of the dual space?
well, in this context, it means that there is a unique differential form df for each function f on the manifold that satisfies the following equation:
df(\mathbf{v})=\mathbf{v}(f)
It's "soak" time for me. I think you've answered all of my questions. As always, thanks.
Please proceed with your original exposition.
Peterdevis
Feb3-04, 08:50 AM
Excuse me for my rather bad english (I'm dutch speaking)
first:
In many replies people are making a distinguis between physists and maths views about tensors ( p-forms and vectors).
I think that both agree that tensors are coordinate independent objects (because thats the whole idea behind differential geometry)
Only when you have to do some real calculation you have to choose a coördinate system and base vectors and p forms. The tensor components you get then are of cours coordinate dependent.
second:
It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)
Originally posted by Peterdevis
Excuse me for my rather bad english (I'm dutch speaking)
first:
In many replies people are making a distinguis between physists and maths views about tensors ( p-forms and vectors).
I think that both agree that tensors are coordinate independent objects (because thats the whole idea behind differential geometry)
Only when you have to do some real calculation you have to choose a coördinate system and base vectors and p forms. The tensor components you get then are of cours coordinate dependent.
there are two reasons why i outline the distinction between the maths convention and the physics convention.
1. while, as you say, both the mathematician and the physicist know that a tensor is a coordinate independent object, i think this fact is completely obscure for the physics student who sees only the coordinate transformations of the components which follow a prescribed but somewhat mysterious rule.
2. math and physics have the use of the words covariant and contravariant exactly switched, so there is a tangible difference that needs to be made clear.
second:
It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)
yeah, sure. i don t know who said the coordinates of the metric are coordinate independent. they are not. the components of any tensor (except a (0,0) tensor, otherwise known as a scalar) are not coordinate independent. i agree with you, but i don t know why you say "it seems that way". in fact, it doesn t even seem like the coordinates of the metric are invariant under coordinate transformations. the transformation looks like this:
g_{\mu\nu}'=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}g_{\rho\sigma}
so i don t know why anyone would think that the components are invariant, or even seem invariant.
although.... now that i think of it, what did turin actually say before?
Originally posted by turin
I have also been told rather emphatically that the metric is a scalar because it does not get transformed by a Lorentz transformation. Not true?
you know, when i answered this before, i answered it talking about general coordinate transformations, for which there is no sense in which this statement can me made to even resemble the truth. For some reason, i didn t even notice that you were asking about Lorentz transformations, instead of coordinate transformations. i guess it is because when i am in the differential geometry subforum, i am never thinking about flat manifolds.
when the manifold in question has a large number of symmetries, we may consider those coordinate transformations that respect those symmetries. in the case of your manifold being flat Minkowski space, the set of coordinate transformations that respect its symmetries are Lorentz transformations.
in other words, by construction, a Lorentz transformation leaves the metric invariant. even the components.
i guess you would say that the components of the metric are scalars, but i think only a sick person would do this. i think it is much better to keep the metric in your head as a (0,2) tensor, whose components are therefore neither tensors or scalars, but rather components of a tensor, and just view the Lorentz transformation as a specific coordinate transformation that leaves those components invariant. not by luck, but because this is what we want, this is how we define the Lorentz transformation.
Peterdevis
Mar2-04, 03:55 PM
It is a mistake to believe that the components of the metric are coördinate independent. It seems only that way because when we change coördianete we change directly the base vectors too (when we use basevectors in the direction of te coördinates)
I wrote a mistake independent must be dependent. So the components of the metric (and any tensor) can be seen as coördinate independent.
The whole transformation rule of the metric is:
g= g_{\mu^\prime\upsilon^\prime}(dx^{\mu^\prime}\otim es dx^{\upsilon^\prime}) =\frac{\partial x^\mu}{\partial x^{\mu^\prime}}\frac{\partial x^\upsilon}{\partial x^{\upsilon^\prime}}g_{\mu \upsilon }(\frac{\partial x^{\mu^\prime}}{\partial x^\mu}dx^{\mu }\otimes \frac{\partial x^{\upsilon^\prime}}{\partial x^\upsilon}dx^{\upsilon })
whow it takes me 20 minutes with latex
So the componenets are changing not because of te coordinate transformation but because when we change of coordinates, we automaticaly change the basic tensors of the vector space. By changing the basic tensors, the components of a tensor must change.
But it is not necessary to change basic tensors when you change coordinates!
...but before we leave the 2-forms, let s find a basis for them, and look at their coordinate representation. it should be obvious how to do that, right? we built our 2-form from two 1-forms, so we should be able to build a basis for our 2-forms from the basis for our 1-forms. let s recall what that was (6):
df=\partial_\nu f dx^\nu
let s take two of those, and wedge them together:
df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu
lethe, are you going to continue with this thread further later i.e. more on exterior derivatives and Hodge star operators? Thanks to your thread, am getting a lot of confusions cleared up esp the difference between a vector and the vector components and the basis vectors.
I'd like to ask some questions related to the above. Can we write d\omega^\mu=df\wedge dg?
Let the {\omega^\mu} be the basis of 1-forms dual to a basis {X_\mu}.
Can we relate the structure coefficients (\partial_\mu f)(\partial_\nu g) in df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu
with the structure coefficients of the expansion of the Lie bracket of 2 basis vectors? (am trying to work out an equation in the section dealing with exterior differentiation in Ryan and Shepley's Homogeneous Relativistic Cosmologies)
lethe, are you going to continue with this thread further later i.e. more on exterior derivatives and Hodge star operators?
well, when i started this thread, i had planned on going all the way to the Yang-Mills lagrangian, which means i would include more on exterior derivatives, covariant derivatives, and exterior covariant derivatives, as well as the Hodge dual. and i had ideas on where i could go after that. higher form gauge fields, Palatini formalism of GR, Chern-Simmons forms, i dunno, a lot of things, some of which i am still learning myself, but the best way to learn a subject is to try to teach it, right?
however, i haven't written a new addition to the main exposition of this thread in almost a year. i seem to have run out of steam. i need something to motivate me to write some more of these, but they are a lot of work.
i have gotten a lot of encouraging private messages about this thread, and i like to think that i have made a difficult subject accesible to people. but i have abandoned it halfway through.
i will make you a deal, maddy. throughout my exposition (this thread is long, it has almost 100 replies. but there are only about 10 or so posts, written by me only, that i consider to be the main kernel of the exposition. i wouldn't mind if every post other than those that make up my exposition were deleted), so throughout these 10 posts or so by me, there have been a couple of homework exercises. no one ever did my homework exercises. i think they should have been very easy for anyone following along. so my deal for you is: do my homework exercises, and, in exchange, i will write 5 more posts.
what do you think?
Thanks to your thread, am getting a lot of confusions cleared up esp the difference between a vector and the vector components and the basis vectors.
I am glad you liked it! i think it is an exciting subject, I really liked learning it, and this thread is a way for me to communicate that excitement to others. i am glad someone is listening
I'd like to ask some questions related to the above. Can we write d\omega^\mu=df\wedge dg?
hmm... well, you can write whatever you want, and if you define things appropriately, it may even make sense. i am not sure what \omega is in your equation above, so i am having trouble making sense of it. usually \omega^\mu is a function, in which case, the left-hand side of your equation is a 1-form, and the right-hand side is a 2-form. this would make the equation meaningless. furthermore, it violates the einstein summation convention, where the index content of both sides of the equation should match. your right-hand side contains a raised \mu index, whereas your left-hand side contains no index at all. in short, i would say this is an invalid equation.
Let the {\omega^\mu} be the basis of 1-forms dual to a basis {X_\mu}.
these symbols don't agree with the spirit of my thread here, in that you are using coefficients of vectors with respect to some basis as the basis itself. but, OK, i can live with that. it is the convention that most physicists follow.
Can we relate the structure coefficients (\partial_\mu f)(\partial_\nu g) in df\wedge dg=(\partial_\mu fdx^\mu)\wedge(\partial_\nu gdx^\nu)=(\partial_\mu f)(\partial_\nu g) dx^\mu\wedge dx^\nu
with the structure coefficients of the expansion of the Lie bracket of 2 basis vectors? (am trying to work out an equation in the section dealing with exterior differentiation in Ryan and Shepley's Homogeneous Relativistic Cosmologies)
the equation above has pretty much nothing to do with the structure constants of my basis.
why don't you post the equation here that you are trying to make sense of (i don't have the book you refer to), and let's see if we can help you
well, when i started this thread, i had planned on going all the way to the Yang-Mills lagrangian, which means i would include more on exterior derivatives, covariant derivatives, and exterior covariant derivatives, as well as the Hodge dual......
however, i haven't written a new addition to the main exposition of this thread in almost a year. i seem to have run out of steam. i need something to motivate me to write some more of these, but they are a lot of work.....
i will make you a deal, maddy. throughout my exposition (this thread is long, it has almost 100 replies. but there are only about 10 or so posts, written by me only, that i consider to be the main kernel of the exposition....
The original thread on http://www.sciforums.com/showthread.php?t=20843&page=2&pp=20 is still fairly clean. Over there I started a new thread for comments/questions/side-issues about Lethe's notes, leaving the main thread for the notes themselves. Then I was off-line for some time and lost the thread, so to speak. But perhaps that's what's needed here.
RDT2.
Still around - and still trying to integrate this into mech eng teaching. So many notations, so little time!
i will make you a deal, maddy.....my deal for you is: do my homework exercises, and, in exchange, i will write 5 more posts.
what do you think?
It's a deal.
Homework no 1 is:-
i will ask for a volunteer to show that the set of linear functionals on a given vector space is itself a vector space. it s not to hard, just check the vector space axioms given above.
Let \sigma and \omega are members of a set of linear functionals, p^\alpha be the basis of the linear functional \sigma, \nu be a vector, {e_\alpha} be its basis, and a and b be arbitrary members of R.
(\sigma+\omega)\nu
=(\sigma+\omega)(\nu^\alpha e_\alpha)
=\sigma(\nu^\alpha e_\alpha)+\omega(\nu^\alpha e_\alpha)
=\sigma(\nu)+\omega(\nu)
=(\omega+\sigma)\nu
So, the addition of linear functionals is commutative.
Because \sigma=\sigma_\alpha p^\alpha,
-\sigma=-(\sigma_\alpha p^\alpha)
=(-\sigma_\alpha)p^\alpha
So, given a linear functional \sigma, -\sigma is also a linear functional.
(a+b)\sigma(\nu)=a\sigma(\nu)+b\sigma(\nu)
So, the addition of the real numbers is distributive.
a(\sigma+\omega)(\nu)
=a(\sigma(\nu)+\omega(\nu))
=a\sigma(\nu)+a\omega(\nu)
So, the addition of linear functionals is distributive.
a(b\sigma(\nu))
=a(b\sigma(\nu^\alpha e_\alpha))
=a(b \nu^\alpha \sigma_\alpha)
=(ab)\nu^\alpha \sigma_\alpha
=(ab)\sigma(\nu^\alpha e_\alpha)
=(ab)\sigma(\nu)
So, scalar multiplication of the linear functionals is associative.
The set of linear functionals thus form a vector space.
Homework no 2 is:-
OK, it should be easy to show that the set of tangent vectors, thusly defined, satisy the axioms of the vector space. i will call this vector space TMp. that is, the tangent space to the manifold M at the point p is TMp. for an n dimensional manifold, the tangent space is always an n dimensional vector space.
Let f and g be functions, and e_\alpha=\partial_\alpha.
Likewise,
(\partial_\alpha+\partial_\mu)f
=(\partial_\alpha)f+(\partial_\mu)f
=(\partial_\mu+\partial_\alpha)f
Addition of tangent vectors is commutative.
(a+b)\partial_\alpha f
=a\partial_\alpha f+b\partial_\alpha f
Addition of the real numbers is distributive.
a(\partial_\alpha+\partial_\mu)f
=a((\partial_\alpha f)+(\partial_\mu f))
=a(\partial_\alpha f)+a(\partial_\mu f)
Addition of tangent vectors is distributive.
a(b\partial_\alpha f)
=ab(\partial_\alpha f)
=(ab)(\partial_\alpha f)
Scalar multiplication of the tangent vectors is associative.
The set of tangent vectors thus form a vector space.
i am not sure what \omega is in your equation above, so i am having trouble making sense of it. usually \omega^\mu is a function, in which case, the left-hand side of your equation is a 1-form, and the right-hand side is a 2-form. this would make the equation meaningless. furthermore, it violates the einstein summation convention, where the index content of both sides of the equation should match. your right-hand side contains a raised \mu index, whereas your left-hand side contains no index at all. in short, i would say this is an invalid equation.
Oops, sorry, yes, I made a lethal careless mistake!
The equation I was referring to is
d\omega^\mu=D^\mu_{\alpha\beta} \omega^\alpha \wedge \omega^\beta.
(which at a glance, I saw the right-hand side wrongly as the wedge product of d\omega^\alpha and d\omega^\beta)
Ok, so the above equation shows that the exterior derivative operator turns the 1-form \omega^\muinto a 2-form \omega^\alpha \wedge \omega^\beta.
these symbols don't agree with the spirit of my thread here, in that you are using coefficients of vectors with respect to some basis as the basis itself. but, OK, i can live with that. it is the convention that most physicists follow.
The {\omega^\mu} is the basis of a 1-form dual to a basis {X_\mu} of a vector.
the equation above has pretty much nothing to do with the structure constants of my basis.
why don't you post the equation here that you are trying to make sense of (i don't have the book you refer to), and let's see if we can help you
Here is the Lie bracket of the basis of a vector.
\left[ X_\mu , X_\nu \right] = C^\lambda_{\mu\nu} X_\lambda
I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.
I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.
Is there any concrete way that I can prove this?
Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?
Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?
Mmm, as far as I learnt, exterior calculus applies only to functions and forms to generate forms of higher rank. An arbitrary tensor is a product of an arbitrary number of forms and vectors. We use another type of calculus for tensors i.e. Lie derivatives, covariant derivatives. (I stand corrected)
I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.
I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.
Is there any concrete way that I can prove this?
Oops, didn't know that the Maurer-Cartan machinery does the trick.
Some discussions on Maurer-Cartan forms at SPR (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&threadm=bdlr17%24epj%241%40glue.ucr.edu&rnum=1&prev=/groups%3Fq%3Dmaurer-cartan%2Bform%2Bgroup:sci.physics.research%26hl%3D en%26lr%3D%26ie%3DUTF-8%26group%3Dsci.physics.research%26selm%3Dbdlr17%2 524epj%25241%2540glue.ucr.edu%26rnum%3D1) :redface:
Any other recommendations for an excellent book on Lie algebras that's not too hard for a physics student? (sorry if veering out of topic here)
The set of linear functionals thus form a vector space.
OK, this is all correct. one comment though, as Matt Grime would say "gentlemen never choose a basis". none of these proofs require you to specify a basis, and you can just delete those steps where you expand in terms of a basis, and the proofs still go through correctly
The set of tangent vectors thus form a vector space.
yes, very good. there is a basis independent way to prove this too, but since the definition of a tangent vector was in terms of a basis, the proof must be as well. thus proving that i am not a gentleman.
i think these are too easy for you!
Oops, sorry, yes, I made a lethal careless mistake!
The equation I was referring to is
d\omega^\mu=D^\mu_{\alpha\beta} \omega^\alpha \wedge \omega^\beta.
(which at a glance, I saw the right-hand side wrongly as the wedge product of d\omega^\alpha and d\omega^\beta)
OK, now i see. this is Cartan's first structure equation for a torsion free connection. if there is torsion, then you have T=d\omega+D\wedge\omega
and that D is your connection coefficient, right? not a differential operator?
Ok, so the above equation shows that the exterior derivative operator turns the 1-form \omega^\muinto a 2-form \omega^\alpha \wedge \omega^\beta.
yes. but of course that is no surprise. an exterior derivative, by construction, always turns a p-form into a (p+1)-form.
[quote]The {\omega^\mu} is the basis of a 1-form dual to a basis {X_\mu} of a vector.
OK, i apologize. I saw X_\mu and just assumed that meant coefficients of a vector. which is what i usually use that symbol for, with raised index. but i do use \omega^\mu for basis 1-forms, so you are all good.
Here is the Lie bracket of the basis of a vector.
\left[ X_\mu , X_\nu \right] = C^\lambda_{\mu\nu} X_\lambda
I'm supposed to connect D^\mu_{\alpha\beta} with the structure coefficients C^\mu_{\alpha\beta}.
I should be getting D^\mu_{\alpha\beta}=-\frac{1}{2}C^\mu_{\alpha\beta},
and this will result in
d\omega^i=-\frac{1}{2}C^i_{st}\omega^s \wedge \omega^t.
Is there any concrete way that I can prove this?
sure:
by definition, the torsion tensor acts on a pair of vectors like this:
T(v,w)=D_vw-D_wv-[v,w]
since we are dealing with a torsion-free connection, we can set that equal to zero, and this gives you your required result.
Hi, is my understanding that you can take the exterior derivarive of a differential form, but my question is, can you take the exterior derivative of some other object (e.g., can you take the exterior derivative of a tensor that is not a differential form?)?
you can take the exterior covariant derivative of a tensor, but not a plain old exterior derivative.
mathwonk
Aug9-04, 11:54 PM
How fascinating. The popularity of this thread shows that many people would prefer to exchange posts than to just read a book on the topic, where all of these questions would be answered. That is not shocking, as science too is a social activity, and it is fun to share it. Still if these questions are really of interest, one could just read a book about it and learn most of the answers, I am guessing. Sometimes it seems as if the questions here are asked by people who do not read, and the answers are provided by those who do. No disrespect intended, just curious about the learning habits of the younger generation. I also like to read these posts to see what I can learn. Books are so dry. Keep it going.
mathwonk,
Too bad you didn't get to see this thread when it was active.. the starter of this thread, lethe, decided to leave PF after our membership voted to keep our profanity filter in place (he felt it was unfair censorship: http://www.physicsforums.com/showthread.php?t=24373). It's a shame he decided to delete all his posts before leaving out of some sort of vindictiveness. The posts are still physically present in our database, however, and I might be legally able to reinstate them -- I'll have to look into it.
- Warren
mathwonk
Sep18-04, 12:21 PM
a riemannian manifold is locally flat if and only if the curvature tensor is zero. this is what spivak class the "test case" for the relevance of riemann's curvature tensor.
mathwonk
Sep18-04, 12:26 PM
trying again after the post did not appear:
a riemannian manifold is locally eucldean in the sense of isometric, if and only if the riemann curvature tensor is zero. that is the whole point of the curvature tensor.
VASILSIS
Oct20-05, 10:30 PM
[QUOTE=lethe]i am going to assume that you are a little familiar with euclidean vectors. a euclidean vector is an arrow between two points. it has direction and magnitude1. mathematically, we can specify a vector in euclidean space with a pair of points in the space, and let the vector be the arrow directed from one point to the other. or you can assume that the first point is always the origin, and specify the vector with just a single point. by doing this, you are essentially moving the vector from it s basepoint, to the origin. this is possible because euclidean space is both a manifold and a vector space.
this won t be true when we move to noneuclidean manifolds. for example, there is no sensible way to make points on a sphere into a vector space. there is no sensible way to define addition on these points.
1Well, the vectors don t have magnitude or direction until we endow the space with a metric. almost everything we are going to talk about here is independent of metric, and we will not need to specify a metric on this space. when using metric dependent quantities, this is the differential geometry, and when dealing with the more general metric independent quantities, this is differential topology. if you don t know what any of this means, ignore it.[/QUO
I may say that all familiar curced shapes can be expressed in a Eucledian space using one extra dimension.
A line can be expressed in a 2D eucl.. space .A sphere or other 2D shape can be expressed in an 3D eucl. space .
a) Is this correct that all shapes no matter how many dimension have can be expressed in D+1 eucledian space?
b) If we want to express them using a space with different metric can we always manage it with a tranformation equation ?
I may say that all familiar curced shapes can be expressed in a Eucledian space using one extra dimension.
A line can be expressed in a 2D eucl.. space .A sphere or other 2D shape can be expressed in an 3D eucl. space .
a) Is this correct that all shapes no matter how many dimension have can be expressed in D+1 eucledian space?
I'm probably misunderstanding your question, but doesn't a (flat) 2-torus (for example) only stay flat in 4D (not in 3D)? Sure someone with more knowledge will answer fully.
The differential forms webbook is great by the way for someone just learning differential geometry (e.g. me). Clears up 4 weeks lectures straightaway.
mathwonk
Nov1-05, 02:56 PM
even smooth manifolds without metric amy not be embeddable as hypersurfaces - e.g. klein bottle in 4 space.
the general theorem (whitney embedding) says it takes 2n dimensions to embed a smooth n manifold, and this is best possible in general. e.g. the 8 dimensional projective spacedoes not embed in 15 dimensional euclidean space projective space cannot be smoothl;y embedded in 15 dimensional
Ratzinger
Nov3-05, 05:53 AM
Reading this thread I must say very sad that Lethe left PF.
PLEASE LETHE, COME BACK!
fisico30
Sep14-08, 05:06 PM
Hi Lethe,
can I bug you with some super elementary questions about differentials, in order to catch up with the rest of you?
thanks
mathwonk
Sep14-08, 09:08 PM
sure, anyone can ask anything about them at all.
fisico30
Sep14-08, 09:55 PM
Thanks mathwonk!
I am used to the idea of differential as an infinitesimal delta of something: dx is the infinitesimal distance, dq the infinitesimal amount of charge....
Infinitesimal intended as something infinitesimally small, but always bigger than zero.
Q1: In differential geometry, the differential changes meaning:
is is true that it represents a unit tangent vector?
is it a functional (operates on a function and outputs a number)?
why its symbol is dx? how is it related to the infinitesimal, differential in calculus?
Q2:p-forms. I am not sure I understand what they are. Could you provide a baisc explanation/example of 1-form and 2 form?
Q3: Manifold: is it correct to say that a manifold is a space, flat or curved?Lethe explains that a surface (a sphere) is a curved 2-manifold. how do we figure out if a space is curved or flat? Is there a curvature function related to the metric distance?
But, what is a space, first of all? For instance, could the electric field, which is a vector field be called a manifold? If so, why?
It seems that a manifold is a general, abstract concept applicable to many things which satisfy a certain criterion of membership But what is that criterion?
(like anything can be a vector, as long as it satisfies those 10 rules that make it be a vector...)
thanks
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