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Exterior differentiation and pullback

  1. Oct 3, 2013 #1
    Say one one have a projection map ##\pi : M^5 \to M^4## which in adapted coordinates are of the form

    $$ \pi(x^\mu, x^4) = x^\mu$$

    where ##\mu = 0,1,2,3##. Now if one ##M^4## introduce an orthonormal frame ##\left\{ e_\mu, e_4\right\}## where ##e_\mu## are tangential to ##M^4## and ##e_4## orthogonal to it. The corresponding dual basis is ##\left\{\omega^\mu, \omega^4\right\}##. If one now considers ##\omega^\nu## to be part of the cotangent space ##T^*_pM^4## and one take the exterior derivative

    $$d ^{^{(4)}}\omega^\nu$$

    Where ##^{^{(4)}}\omega^\nu## denotes that i consider the covector a part of ##T_p^*M^4##. Now if one considers the same covector as a covector in ##T^*_pM^5## and we take the exterior derivative

    $$d ^{^{(5)}}\omega^\nu$$

    would it then be correct to say that

    $$\pi^* \left( d ^{^{(4)}}\omega^\nu\right) = d \pi^* \left(^{^{(4)}}\omega^\nu\right) = d^{^{(5)}}\omega^\nu$$

    due to the commutation of the pullback ##\pi^*## with the exterior derivative?
     
  2. jcsd
  3. Oct 3, 2013 #2

    lavinia

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    yes. you should do the calculation
     
  4. Oct 4, 2013 #3
    Is'nt the above really enough calculationwise as long as we assume the commutation? What calculation would you have me do?
     
  5. Oct 4, 2013 #4

    lavinia

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    It is enough but you should write down some examples for your self.
     
  6. Oct 4, 2013 #5
    I will try to do that :)
    By the way; the reason I asked was because I've seen this be applied on the problem of relating curvatures. By using Cartan's first structural equation

    $$d \omega^\nu = - \Omega^\nu_\alpha \wedge \omega^\alpha$$

    in both the lower dimensional and higher dimensional space, and pulling back the lower dimensional one, one can obtain a relation between the connection one-forms ##\Omega^\nu_\alpha## in the two spaces. I'm wondering what the assumptions behind this method are? Can one always assume that Cartan's first equation holds in any space with a metric? Or do one define it to be hold and then find the connection from that? In the case I am lookig at a 4-dimensional space (which is not a hypersurface) inherits a metric for a 5-dimensional one.
     
  7. Oct 4, 2013 #6

    WannabeNewton

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    You don't need a metric for Cartan's structure equations to hold, all you need is an affine connection. They are proven not assumed.

    Let ##M## be a smooth manifold and ##\nabla## an affine connection on ##M##. Furthermore, let ##\{e_{\alpha}\}## be a basis field defined on some open subset ##U## of ##M## and let ##\{\theta^{\alpha}\}## be the corresponding dual basis field. We have of course that ##\nabla \theta^{\alpha} = -\theta^{\beta}\otimes \omega^{\alpha}{}{}_{\beta}## where ##\nabla_{X}e_{\alpha} = \omega^{\beta}{}{}_{\alpha}(X)e_{\beta}## define the connection coefficients.

    Then for any 1-form ##\alpha##, ##\nabla \alpha = \theta^{\beta}\otimes (d\alpha_{\beta} - \omega^{\gamma}{}{}_{\beta}\alpha_{\gamma})## and for any vector field ##X##, ##\nabla X = e_{\beta}\otimes(dX^{\beta} + \omega^{\beta}{}{}_{\gamma}X^{\gamma})##. Finally, the torsion forms ##\Theta ^{\alpha}## and curvature forms ##\Omega ^{\alpha}{}{}_{\beta}## are defined in terms of the Torsion ##T## and Riemann curvature ##R## in the usual way.

    Then we can easily show for example that ##\Theta^{\alpha} = d\theta^{\alpha}+ \omega^{\alpha}{}{}_{\beta}\wedge \theta^{\beta}## which is the first of the structure equations. We have ##\Theta^{\alpha}(X,Y)e_{i} = \nabla_{X}Y - \nabla_{Y}X - [X,Y] \\= \nabla_{X}(\theta^{\alpha}(Y)e_{\alpha})- \nabla_{Y}(\theta^{\alpha}(X)e_{\alpha}) - \theta^{\alpha}([X,Y])e_{\alpha}\\ = d\theta^{\alpha}(X,Y)e_{\alpha}+ (\omega^{\alpha}{}{}_{\beta}\wedge \theta^{\beta})(X,Y)e_{\alpha}##

    This gives us the desired result. The calculation for the second structure equation is similar.
     
  8. Oct 4, 2013 #7
    That's very interesting WbN! Thanks for for that! So Cartan's equation only require the existence only off an affine connection on the manifold. I know that a Levi-Civita connection arises naturally if a manifold is a hypersurface of a bigger ambient space with a metric. But say one generally have a m-dimensional manifold with a metric; how can one be sure that one has an affine-connection on a lowerdimensional submanifold? I guess what one would be doing by going about the process of relating the connection one-forms as described above is to find the (unique) affine connection consistent with the given metric on the ambient space?
     
  9. Oct 4, 2013 #8

    WannabeNewton

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    I'm not sure I'm understanding your question correctly. If ##(M,g)## is a (pseudo)Riemannian manifold and ##\Sigma \subset M## is a submanifold of some codimension (e.g. a space-like hypersurface if ##M## is a Lorentz manifold) then the embedding map ##i:\Sigma \rightarrow M## induces a metric ##\bar{g} = i^{*}g## on ##\Sigma##. It follows that there exists a unique affine connection ##\bar{\nabla}## on ##\Sigma## such that ##\bar{\nabla}## is metric and symmetric i.e. ##\bar{\nabla}\bar{g} = 0## and ##\bar{\nabla}_{X}Y = \bar{\nabla}_{Y}X + [X,Y]##; this is a fundamental result of (pseudo)Riemannian geometry.
     
  10. Oct 4, 2013 #9
    I agree. Hmm, I'm a bit confused myself. The case I'm thinking of is Kaluza-Klein theory which supposedly (http://ptp.oxfordjournals.org/content/128/3/541.full.pdf+html) can be thought of as a trivial bundle of a four-dimensional manifold ##M^4## with a one-dimensional manifold ##S^1## and a projection ##\pi:M^5 = M^4 \times S^1 \to M^4## (##S^1## is often taken to be compact and thought of as a circle). ##M^5## comes with a metric ##g##. The method I outlined for finding the relation between the curvature tensor in ##M^5## and ##M^4## is applied the book "Einstein's general theory of Relativity" by Grøn and Hervik where goes about by taking the exterior derivative of a basis form in ##T_pM^5##. By then using Cartan's first equation in both ##M^4## and ##M^5## one can relate the curvature-forms ##\omega^\mu_\nu## in the respective spaces (by pull-back/inclusion) from which follows the relation between the curvature tensors.

    There are several things here which is not clear to me: Is there any difference betwen a product manifold and a trivial bundle except for the projection mapping? And if ##M^5## were either a product manifold or trivial bundle, would not ##M^4## then automatically be a hypersurface in ##M^5##?
     
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