Exterior differentiation and pullback

[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?

 

[lavinia]

yes. you should do the calculation

 

[center o bass]

yes. you should do the calculation

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

 

[lavinia]

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

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

 

[center o bass]

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

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.

 

[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.

 

[center o bass]

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.

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?

 

[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.

 

[center o bass]

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.

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 between 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 $M^5$?