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Exact definition of differential forms

  1. Sep 8, 2007 #1
    First off, I'm no geometer. I've jumped from looking into QFT from an operator algebra perspective to one looking at it from a differential geometry perspective. It's been a fairly nice ride...modulo the fact that I know very little differential geometry. Thus I have been going through a bit of a crash course lately. I've spent the last couple days learning the classical theorems of differential forms (i.e. worked my way up through Poincare's Lemma, Stokes theorem, etc.) and that hasn't really been a problem (makes me feel stupid for not learning it earlier). But I am still often confused by some basic aspects of it all.

    This is from "Formal Definition" in the wikipedia article on differential forms:

    I'm not quite understanding how the first sentence and the second are meant to be equivalent. As I've understood it, a differential form assigns each point of a manifold an alternating tensor defined on the tangent space to that point (i.e. what I've read in books fits more with the second definition). The first sentence seems to say the same thing except now it's cotangent space instead of tangent space. In finite dimensions the spaces would be canonically isomorphic, but I don't think that's the point of it (especially since that's no longer true in infinite dimensions). It seems to me like using the cotangent bundle just sends us to the base field twice unnecessarily.

    Am I missing something obvious here? Which of the following definitions is true for a k-form on a manifold?

    [tex]\omega(p) \in \Lambda^k(T_p(M))[/tex]

    [tex]\omega(p) \in \Lambda^k(T_p^*(M))[/tex]
     
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  3. Sep 8, 2007 #2
    Looking at a little harder, it seems like my confusion may just be a result of using different definitions of the exterior algebra of a vector space. Some books seemed to define the tensor product by by using multilinear maps and then modding out by vv=0. It seems that the wikipedia definition of exterior algebras is just a base algebraic definition (which is isomorphic to the multilinear maps approach) and thus needs the extra "functional aspect" provided by using the cotangent space instead of the tangent space at each point.

    Is that it? Does anybody understand what I'm saying. (If so, I'm impressed. I barely understand myself...)
     
  4. Sep 8, 2007 #3

    Hurkyl

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    One of the wonderful things about multilinear algebra (and sometimes confusing) is the sheer number of ways you can view an object.

    The wedge product of two vector spaces comes equipped with an embedding into their tensor product. In particular, we have the natural injection
    [tex]
    \Lambda^k \left( V^* \right) \to \otimes^k V^*
    [/tex]

    Now, the space [itex]\otimes^k V^*[/itex] is naturally isomorphic to several interesting spaces, such as the space of k-linear maps [itex]V^k \to \mathbb{R}[/itex], or the space of linear functionals on [itex]\otimes^k V[/itex].

    So, by the above injection, we can view elements of [itex]\Lambda^k \left( V^* \right)[/itex] in these ways, and we get alternating k-linear maps, or alternating linear functionals.

    Because elements of [itex]\Lambda^k \left( V^* \right)[/itex] naturally correspond to alternating functions [itex]\otimes^k V \to \mathbb{R}[/itex], that means they can naturally be viewed as functions [itex]\Lambda^k (V) \to \mathbb{R}[/itex]. In fact, there is a natural isomorphism
    [tex]\Lambda^k \left( V^* \right) \cong \left( \Lambda^k \left( V \right) \right)^*[/tex]

    (The wikipedia article misspoke; it should have said linear functionals on the exterior power of V)
     
  5. Sep 8, 2007 #4
    Do you know what the cotangent bundle is? One can view the cotangent bundle as the tangent bundle with the tangent spaces replaced by their dual spaces. So an element of a cotangent space acts on the tangent space, i.e. you feed it a tangent vector and out comes a number. That's precisely what the second sentence expresses, right?
     
  6. Sep 8, 2007 #5

    mathwonk

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    a differential one form is a section of the cotangent bundle. a differential p form is a section of the pth exterior power of the cotangent bundle.
     
  7. Sep 11, 2007 #6
    Thanks guys. I read your posts a couple days ago, but wanted to think a bit before replying. My confusion seemed to stem from this fact:

    ...and from the fact that I was thinking of the exterior algebra as alternating multilinear maps instead of its algebraic definition of tensors (mod vv)...and I was unknowingly interchanging the definitions. Thus at times I was seeing "too many" duals (i.e. two when I needed one) and sometimes "too few" (i.e. none when I needed one). So really it just came down to the fact that I was using slightly different definitions that were basically the same, but the way I was thinking about them confused me.

    Thanks though. I've got it cleared up now.
     
  8. Sep 11, 2007 #7

    Hurkyl

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    I have a strong dislike for thinking about any of these things as actual maps. (Except, for some reason, I'm able to think about elements in the dual space as linear functionals without trouble)

    Instead, I think of evaluation maps. e.g. you could think of [itex]V^*[/itex] as a set of linear functionals on V... or, you could just treat [itex]V^*[/itex] as an abstract vector space, and when appropriate, invoke the bilinear map
    [tex]V^* \times V \to \mathbb{R}[/tex]
    as appropriate. Actually, I even dislike thinking about bilinear maps; I actually think in terms of the corresponding evaluation map on the tensor product
    [tex]V^* \otimes V \to \mathbb{R}.[/tex]

    The same goes for the space [itex]\hom(V, W)[/itex] of linear maps from V to W; I tend to treat it as an abstract space.
     
    Last edited: Sep 11, 2007
  9. Sep 13, 2007 #8
    I've always been more of an analyst. My functional analysis background has built in certain habits. Thinking of the dual space as anything but functionals is screwy to me.
     
  10. Sep 18, 2007 #9

    mathwonk

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    to the op, i agree the probem is in definitions. with stabndard definitions, your second version is the correct one, without the asterisk, but somebody somewhere might define the symbol wedge to include the asterisk.

    the little book by david bachman studied here once seemed helpful to all concerned, even if at the time i picked on its tiny flaws, visible only to experts.
     
  11. Sep 18, 2007 #10

    mathwonk

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    well you said exact, but i rofer intuitive. you know how einstein had systems of clocks distributed throiught his universe, all synchronized?


    well we have systems of k dimensional volume measures distributed over our manifolds, one at each point, and all synchronized by varying smoothly. at each point you have a k form, i.e. a gadget for measuring k dimensional volumes of parallelpipeds. thats all they are. (think systems of kbyk subdeterminants.)
     
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