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Parameterized p-subset of a manifold M

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data

    A parameterized ##p##-subset ##(U,F)## of a manifold ##M^{n}## is ''irregular'' at ##u_0## if rank ##F<p## at ##u_0##.

    Show that if ##\alpha^{p}## is a form at such a ##u_0## then ##F^{*}\alpha^{p}=0##.

    2. Relevant equations

    3. The attempt at a solution

    A parameterized ##p##-subset of a manifold ##M^{n}## is a pair ##(U,F)## consisting of a region ##U## in ##\mathbb{R}^{p}## and a differentiable map ##F:U(u)\rightarrow M^{n}(x)##.

    If ##\alpha^{p}## is a form at a ##u_0## where rank ##F<p##, then

    ##\displaystyle{F^{*}\alpha^{p}=(F^{*}\alpha^{p})\bigg[\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{1}}}, \dots , \frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}##

    ##\displaystyle{F^{*}\alpha^{p}=\alpha^{p}\bigg[F_{*}\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{1}}}, \dots , F_{*}\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}##

    ##\displaystyle{F^{*}\alpha^{p}=\alpha^{p}\bigg[\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}}\frac{{\bf{\partial}}}{{\bf{\partial}} y^{j_{p}}}, \dots , \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}\frac{{\bf{\partial}}}{{\bf{\partial}} y^{j_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}##

    ##\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}##

    Next, I need to use the fact that the matrix of partial derivatives has rank less than ##p##. Can you suggest a trick to show that ##F^{*}\alpha^{p}## is ##0##?
     
    Last edited: Oct 17, 2016
  2. jcsd
  3. Oct 17, 2016 #2

    Orodruin

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    What is the determinant of a square matrix with rank lower than its size?

    Alternatively you can try acting with the form on p vectors and use the fact that at least one of the pushforwards is linearly dependent on the others.
     
  4. Oct 17, 2016 #3
    The determinant is surely zero, but in this case, we have ##\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}##, which is clearly not the determinant of the matrix of partial derivatives.
     
  5. Oct 17, 2016 #4

    Orodruin

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    You are contracting it with something antisymmetric and hence ... (fill in the dots)

    On the other hand, the alternative approach may be more aesthetic ...
     
  6. Oct 17, 2016 #5
    I get it.

    Thank you so very much!
     
  7. Oct 17, 2016 #6
    Oh, just one thing.

    I have not really used the fact that the rank of ##F## is less than ##p##.
     
  8. Oct 17, 2016 #7

    Orodruin

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    This would be much easier to address if you told me what you did.
     
  9. Oct 17, 2016 #8
    ##\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}## is symmetric but ##\displaystyle{du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}## is antisymmetric. This seems to imply that the product of the two is ##0##, regardless of the rank of ##F##.

    Oh wait, ##\alpha_{j_{1}\dots j_{p}}## is also antisymmetric, so that the total product is symmetric and not necessarily ##0##.

    So, I am not really sure how to approach the problem.
     
  10. Oct 17, 2016 #9

    Orodruin

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    No, this is incorrect. The matrix of partial derivatives is not symmetric.
     
  11. Oct 17, 2016 #10

    Orodruin

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    You still have both my suggestions left. Either use that the determinant is zero or that the vector pushforwards are linearly dependent.
     
  12. Oct 17, 2016 #11
    Well, the matrix of partial derivatives is not necessarily symmetric - I agree.

    But the product of partial derivatives ##\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}## is surely symmetric, is it not?

    Well, the determinant of the matrix of partial derivatives is ##0##, since the matrix has rank less than its number of dimensions. Also, the product ##\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}## is being contracted on both sides, on the left with the antisymmetric ##\alpha_{j_{1}\dots j_{p}}## and on the right with the antisymmetric ##du^{i_{1}}\wedge \dots \wedge du^{i_{p}}##.

    I know that the determinant of the matrix of partial derivatives is given by ##\displaystyle{\epsilon_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}##, where the epsilon tensor is antisymmetric, but I am not sure how this relates to the above.
     
  13. Oct 17, 2016 #12

    Orodruin

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    How many linearly independent p-forms exist in a p-dimensional space?
     
  14. Oct 17, 2016 #13
    In a ##p##-dimensional space, there is ##1## linearly independent ##p##-form.
     
  15. Oct 17, 2016 #14

    Orodruin

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    And therefore ...
     
  16. Oct 17, 2016 #15
    I give up!

    I understand that this means that ##\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}=\alpha_{1\dots p}\frac{\partial y^{1}}{\partial u^{1}} \dots \frac{\partial y^{p}}{\partial u^{p}}du^{1}\wedge \dots \wedge du^{p}}##,

    but I can't carry this idea any further.
     
  17. Oct 17, 2016 #16

    Orodruin

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    No it does not mean that. You missed several (most) terms of the sum. Can you write down a particular p-form using the permutation symbol?
     
  18. Oct 17, 2016 #17
    ##\displaystyle{{\alpha}_{{\mu}_{\sigma(1)}\dots{\mu}_{\sigma(p)}}}=\text{sgn}(\sigma)\alpha_{\mu_{1}\dots\mu_{p}}##,

    where the signature ##\text{sgn}(\sigma)## of a permutation ##\sigma=\sigma(1)\dots\sigma(p)## is defined to be ##+1## if ##\sigma## is even and ##−1## if ##\sigma## is odd.
     
  19. Oct 17, 2016 #18

    Orodruin

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    No, I am asking if you can write down a p-form using ##\epsilon_{i_1\ldots i_p}##.
     
  20. Oct 17, 2016 #19
    So, do you mean the following:

    ##\alpha_{j_{1}\dots j_{p}}=\epsilon_{j_{1}\dots j_{p}}\alpha_{1\dots p}##?
     
  21. Oct 17, 2016 #20

    Orodruin

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    Yes. Now what does this tell you?
     
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