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Is Every Diff. Form on a Submanifold the Restriction of a Form in R^n?

  1. Jul 18, 2014 #1

    WWGD

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    (From another site)
    I think the answer is no, for ##i: S \rightarrow \mathbb R^n ## the inclusion/restriction, and ##w## any form, we have that ##i_{*}dw=d(i_{*}w )## , but the homology of ##\mathbb R^n## is trivial(i.e., every closed form is exact), so that we can write ##w= d(\alpha)##, for some form ##\alpha##. Substituting back in above: ##i_{*}(d(d\alpha))=0 ##, since ##d^2=0##, so that any EDIT closed form in ##\mathbb R^n## restricts to a trivial form.

    So as a counterexample, take any nonzero form on a submanifold. Is this correct ?

    A problem seems to be that ##w## may not be closed. So this only shows closed forms restrict to closed forms.

    (Any examples of non-closed forms in ##\mathbb R^n##?)

    Just curious as to how to answer this. Maybe using the inclusion/restriction map? What if we showed the restriction/inclusion map is/ (is not) onto?

    We have: ##i^{*}: T_p \mathbb R^n → T_p^* S## as a map between the respective cotangent bundles, given by:

    ##i_{*}w(X_p)= w(di)(X_p)##

    Where di is the pushforward/tangent map of the inclusion. How do I take it from here?

    (Let me use R^n instead of ##\mathbb R^n## to simplify )

    I think we need to use properties of the inclusion map of a submanifold S into the ambient manifold re the tangent space. Doesn't then T_p S inject into T_p R^n (meaning T_pS is a subspace of T_p R^n as (del/delX_1,...,del/delX_n)→ (del/delX_1,....,del/delX_n,0,0,...0))?

    I think I am missing some basic linear algebra here, but not sure on what.
     
    Last edited: Jul 18, 2014
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  3. Jul 18, 2014 #2

    Ben Niehoff

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    An obvious counterexample is any volume form on the submanifold. For example, take the arc length differential on a 1-dimensional submanifold. There is no linear functional on (the tangent bundle of) ##\mathbb{R}^n## whose restriction (pullback) to a 1-dimensional submanifold gives arc length.
     
  4. Jul 18, 2014 #3

    WWGD

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    Thanks, Ben, I agree, but I a also trying to practice my technique/knowledge of bundles, pullbacks, etc. by using the inclusion map. Would you please check my argument?
     
  5. Jul 18, 2014 #4

    WWGD

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    Sorry, let me see if I can write this more clearly:

    We have a submanifold S of ## \mathbb R^n ##

    I want to show that the dual of the differential of the inclusion map ## i: S → \mathbb R^n ##(as a map between the cotangent bundles of ## \mathbb R^n , S ## respectively) is an onto map.

    Say we have an inclusion## i : S → \mathbb R^n ## , so this is a parametrization of the submanifold. We then want to compute the differential ## di : T_p S →T_p R^n ## . But I think if S is a submanifold, we can find coordinates for i so that ## di :T_p S →T_p \mathbb R^n ## is given by:

    ##di: ( ∂/∂x_1, ∂ /∂x_2,....,∂/∂x_n) →(∂/∂x_1, ∂ /∂x_2,....,∂/∂x_n,0,0,...,0) ##




    Then we must only show that ## di^* : T^*_i(p) \mathbb R^n → T^*_p S ## is an onto map.


    But maybe I'm off here somewhere.
     
    Last edited: Jul 18, 2014
  6. Oct 6, 2014 #5

    lavinia

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    Start with the 1 form on R^2-{0} that is the length form of circles centered at the origin and zero on vectors that are perpendicular to these circles. Along each radius line from the origin multiply this form by a smooth function defined on the positive real numbers that is identically 1 for numbers greater than equal to 1 and is identically zero for numbers less than 1/2. This produces a 1 form on the plane that pulls back to the volume element of the unit circle.

    This form is not closed and thus, is not exact.

    This construction might generalize to any form on a smoothly embedded compact submanifold of another manifold. I think it follows from the tubular neighborhood theorem.
     
    Last edited: Oct 6, 2014
  7. Oct 6, 2014 #6

    mathwonk

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    people seem to be assuming that continuity is a requirement although that was not stated. presumably anything is a restriction of a discontinuous object.

    what about dtheta on the plane minus the origin? i.e. in this example thew submanifold is open and dense.
     
  8. Oct 6, 2014 #7

    WWGD

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    You're right, wonk, we need smoothness, otherwise, like you said, discontinuous objects can be trivially extended/restricted to discontinuous objects. And I guess my knowledge is still introductory at this point, since I am still using big machinery, and I cannot think of counterexamples.
     
    Last edited: Oct 6, 2014
  9. Oct 7, 2014 #8

    lavinia

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    I think it instructive to think this through for smooth forms.

    For example when a compact oriented manifold is smoothly embedded in another oriented manifold, it has an open neighborhood - called a tubular neighborhood - that is diffeomorphic to the normal bundle of the embedding. Any differential form on the normal bundle with compact supports along the fibers can be extended by zero to the entire ambient manifold. So any form with compact supports along the fibers comes from a form on the ambient manifold. E.g. the Thom class of the normal bundle when extended to the entire ambient manifold is the Poincare dual to the embedded manifold.
     
    Last edited: Oct 7, 2014
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