# Exterior differentiation and pullback

1. Oct 3, 2013

### center o bass

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. Oct 3, 2013

### lavinia

yes. you should do the calculation

3. Oct 4, 2013

### center o bass

Is'nt the above really enough calculationwise as long as we assume the commutation? What calculation would you have me do?

4. Oct 4, 2013

### lavinia

It is enough but you should write down some examples for your self.

5. Oct 4, 2013

### center o bass

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.

6. Oct 4, 2013

### WannabeNewton

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.

7. Oct 4, 2013

### center o bass

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?

8. Oct 4, 2013

### WannabeNewton

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.

9. Oct 4, 2013

### center o bass

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 $M^5$?