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

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)<br />                                               (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):
<br /> 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)<br /> \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!

cephas · Aug 23, 2003

more on 1-forms

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 differential 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 differential 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?

cephas · Aug 24, 2003

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?

lethe · Aug 27, 2003

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 · Oct 2, 2003

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 &equiv; 1/2 (α&otimes;β - β&otimes;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 · Oct 2, 2003

Use brackets [ , ] instead of < , >. Otherwise just the same.

rick1138 · Oct 4, 2003

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 &equiv; 1/2 (α&otimes;β - β&otimes;α)

and consider the two one-forms in expanded form:

α &equiv; α₁e₁ + α₂e₂ + α₃e₃

β &equiv; β₁e₁ + β₂e₂ + β₃e₃

and remember these rules for the bases of forms:

e_i /\ e_j = - e_j /\ e_i

(reversing order reverses sign)

e_i /\ e_i = 0

(wedge summing a basis form by itself annihilates it)

so:

α/\β = 1/2 ((α₁e₁ + α₂e₂ + α₃e₃) &otimes; (β₁e₁ + β₂e₂ + β₃e₃) - (β₁e₁ + β₂e₂ + β₃e₃) &otimes; (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.

turin · Dec 15, 2003

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 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 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.

turin · Dec 15, 2003

Originally posted by Tom
Vectors v_i (i=1,2,3,...) in Rⁿ are independent iff

a₁v₁+a₂v₂+a₃v₃+...=0

implies that

a₁+a₂+a₃+...=0

Did you mean for:

"a₁+a₂+a₃+...=0"

to be:

"a₁=0, a₂=0, a₃=0, ...?"

turin · Dec 15, 2003

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, v^inv, such that: v + v^inv = 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).

turin · Dec 15, 2003

Originally posted by lethe
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)
...
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?

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

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