What Are Differential Forms and How Do They Apply in Differential Geometry?

lethe · Jun 11, 2003

don't want my words on PF

lethe · Jun 11, 2003

a couple of weeks ago there was a thread over there about Stoke s theorems. 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 let me 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.
[/size]

lethe · Jun 11, 2003

Euclidean Vectors

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.[/size]

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.[/size]

lethe · Jun 11, 2003

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.[/size]

Brad_Ad23 · Jun 12, 2003

This should be interesting! Keep it up!

quantumdude · Jun 12, 2003

Prerequisite Review Sheet

Originally posted by lethe
oh and i assume that you know what it means for vectors to be linearly independent,

Vectors v_i (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 a_i in R}

edit: fixed subscript bracket

quantumdude · Jun 12, 2003

A note on Euclidean vectors.

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 (R^TR=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 (v^T and v, respectively):

v'^Tv'=v^TR^TRv

For the equality of the inner products to hold, we can see that we must have R^TR=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

lethe · Jun 12, 2003

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 (R^TR=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 (v^T and v, respectively):

v'^Tv'=v^TR^TRv

For the equality of the inner products to hold, we can see that we must have R^TR=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.

Hurkyl · Jun 12, 2003

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.

quantumdude · Jun 13, 2003

Examples and Exercises

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

v^T₁=[1 2 3]
v^T₂=[2 -1 4]
v^T₃=[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:

Code:

      [1  2  0|0] 
[V|[b]0[/b]]=[2 -1  5|0]
      [3  4  2|0]

This reduces to:

Code:

       [1  2  0|0]
[V|[b]0[/b]]=[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[/color].

Spanning Sets
Example:
In R³ (regular Euclidean 3-space) let S={u₁,u₂,u₃}, where

u^T₁=[1 -1 0]
u^T₂=[-2 3 1]
u^T₃=[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 u_i. In other words, we must determine whether the equation

a₁u₁+a₂u₂+a₃u₃=v

always[/color] 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:

Code:

      [1  -2  1| a] 
[A|[b]v[/b]]=[-1  3  2| b]
      [0  1  4| c]

Solving this system yields:

Code:

      [1  0  0|10a+9b-7c] 
[A|[b]v[/b]]=[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

lethe · Jun 13, 2003

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?

quantumdude · Jun 13, 2003

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.

lethe · Jun 13, 2003

Abstract Vector Spaces

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

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
1. (a+b)v=av+bv
2. a(v+w)=av+aw
3. 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.
  [/size]

1. 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.[/size]

gnome · Jun 17, 2003

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 · Jun 17, 2003

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!

lethe · Jun 17, 2003

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 referring 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 · Jun 17, 2003

lethe, will this basicly be the same as that other thread or have you revised it at all?

lethe · Jun 17, 2003

Originally posted by On Radioactive Waves
lethe, will this basicly be the same as that other thread 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.

gnome · Jun 18, 2003

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?

lethe · Jun 18, 2003

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.

lethe · Jun 18, 2003

The Dual Space

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, {e_μ}. then this induces a natural choice of basis for the dual space V*, {σ^ν}, 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 {e_μ} 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 {σ^ν} 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. δ^ν_μ 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 i^th basis linear functional has the value 1 when it acts on the i^th 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.

jeff · Jun 18, 2003

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".

lethe · Jun 19, 2003

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.

jeff · Jun 19, 2003

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.

lethe · Jun 19, 2003

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.

chroot · Jun 19, 2003

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

jeff · Jun 19, 2003

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.

lethe · Jun 19, 2003

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.

Hurkyl · Jun 19, 2003

I'm curious what everyone thinks the intended audience is; I got the impression at the beginning we were going to aim at something someone fresh out of multivariable calc could understand, but the bar has been raised fairly swiftly as the thread has developed...

chroot · Jun 19, 2003

That's what happens when you get too many cooks in the kitchen... they start to disagree about the recipe.

- Warren

lethe · Jun 19, 2003

Originally posted by Hurkyl
I'm curious what everyone thinks the intended audience is; I got the impression at the beginning we were going to 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.

jeff · Jun 19, 2003

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 U_α comes a homeomorphism φ_α:U_α → Rⁿ. If the transition functions φ_βοφ_α^-1:φ_α(U_α∩U_β) → φ_β(U_α∩U_β) 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.

lethe · Jun 19, 2003

Originally posted by jeff
Manifolds are more than just topological spaces because with each open set U_α comes a homeomorphism φ_α:U_α → Rⁿ. If the transition functions φ_βοφ_α^-1:φ_α(U_α∩U_β) → φ_β(U_α∩U_β) 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.

jeff · Jun 20, 2003

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.

lethe · Jun 27, 2003

Tangent Vectors

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 &real;². 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 v^μ 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 TM_p. that is, the tangent space to the manifold M at the point p is TM_p. 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 separated 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.

marcus · Jul 27, 2003

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 TM_p. that is, the tangent space to the manifold M at the point p is TM_p. 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 separated 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.

lethe · Jul 27, 2003

1-Forms

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 let's 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 α_ν 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!

marcus · Jul 27, 2003

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 &fnof;, let me call this associated linear functional d&fnof;. 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 &fnof;".

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:

d&fnof;(v) = v(&fnof;)                                               (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 d&fnof; is a functional that acts on vectors, and &fnof; 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 let's explore some properties of these 1 forms. first of all, let s write down the dual basis, {σ^ν}. these are, by definition, linear functionals such that σ^ν(e_μ) = δ^ν_μ (eq. (1) above). where {e_μ} 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 σ^ν: σ^ν = d&fnof;^ν, where &fnof;^ν is some function on the manifold that we will determine.

with these changes, let s rewrite that condition for the dual basis (1):
d&fnof;^ν(∂_μ) = δ^ν_μ                                      (5)

now using (4) above, this becomes:

d&fnof;^ν(∂_μ) = ∂_μ&fnof;^ν = ∂&fnof;^ν/∂x^μ = δ^ν_μ

now, can you think of a function whose derivative with respect to x^μ is 1, and whose derivative with respect to all other coordinates is 0? it s easy..

think about it...

got it?

it s &fnof;^μ = x^μ! no sweat!

OK, so then the dual basis of the 1 forms is just {dx^ν}. now let s check what the components of a general 1 form are in terms of this basis:

d&fnof; = α_νdx^ν

where α_ν 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.

d&fnof;(∂_μ) = α_νdx^ν(∂_μ)
∂_μ&fnof; = α_ν∂_μx^ν = α_νδ^ν_μ = α_μ

i have used (4) twice in the second equation there.

how about that! the components of the 1 form α_ν are just the partial derivatives of the function!
d&fnof; = ∑(∂&fnof;/∂x^ν)dx^ν = ∂_ν&fnof;dx^ν                                  (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.

lethe · Jul 27, 2003

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?

marcus · Jul 28, 2003

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 won't say it.

marcus · Aug 10, 2003

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[/size]."

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), F_U(p)) and this map F_U: [pi]^-1(U) --> G satisfies an equation F_U(pg) = map F_U(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."

lethe · Aug 11, 2003

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[/size]."

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.

lethe · Aug 11, 2003

Tensor Products of 1-forms

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&otimes;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&otimes;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.[/size]

lethe · Aug 11, 2003

Wedge Product

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&and;ω = σ&otimes;ω - ω&otimes;σ

easy enough. let s check how that new wedge product acts on our two vectors:

σ&and;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&and;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&and;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&and;σ = ω&otimes;σ - σ&otimes;ω = - (σ&otimes;ω - ω&otimes;σ) = -σ&and;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.
[/size]

lethe · Aug 11, 2003

marcus: i left these posts in the default font. do the tensor product and wedge product symbols show up for you?

marcus · Aug 11, 2003

Hi Lethe,
as long as you don't 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&otimes;ω - ω&otimes;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&otimes;σ - σ&otimes;ω = - (σ&otimes;ω - ω&otimes;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.

cephas · Aug 19, 2003

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

Hurkyl · Aug 19, 2003

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!

cephas · Aug 22, 2003

a few questions on 1-forms

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 doesn't 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.

lethe · Aug 22, 2003

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?

d&fnof; 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 doesn't 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 let's 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 d&fnof;(∂/∂x)=1, d&fnof;(∂/∂y)=0. now let s use equation 4. d&fnof;(∂/∂x)=∂&fnof;/∂x=1 and d&fnof;(∂/∂y)=∂&fnof;/∂y=0. integrate those two, and you will see that &fnof;=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!

What Are Differential Forms and How Do They Apply in Differential Geometry?

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