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  Exercise: show that two definitions of vertical subspace are equivalent

+ 2 like - 0 dislike

Let us consider a principal bundle \(P \overset{\pi }{\to} M\), let \(u \in P\) and let \(G_p\) be the fibre at \(p = \pi(u)\). According to my book, the vertical subspace \(V_uP\) is defined as the subspace of \(T_uP\)  which is tangent to \(G_p\) at \(u\). Now, one of the exercises is to show that this definition is equivalent to:

\(V_u P = \mathrm{ker} \; \pi_* = \{ X \in T_uP \mid \pi_*X = 0 \} \)

where \(\pi_* : T_u P \to T_{\pi(u)}M \) is the induced map.

In order to try this exercise, I have mainly looked at how one usually derives the pushforward between manifolds:

\(\pi_* X[g] = X[g \circ \pi] = \frac{\mathrm{d}g(\pi(u(t)))}{\mathrm{d}t}\Bigl|_{t=0} = \frac{\partial g}{\partial \pi} \frac{\mathrm{d} \pi(u(t))}{\mathrm{d}t}\Bigl|_{t=0} \)

But I'm not really sure how to continue from here on. I've tried to Google it, but most sources I found only give the second definition without even discussing the first definition. Any help is much appreciated.

asked May 22, 2014 in Mathematics by Hunter (520 points) [ revision history ]
edited May 22, 2014 by Hunter

1 Answer

+ 4 like - 0 dislike

As $\pi_*$ is surjective,

$$\dim \ker \pi_* = \dim TG_p$$

and we only need to show $TG_p\subset \ker \pi_*$.

For any vector $X\in TG_p$, choose a curve $c$ in $G_p$ representing $X$, ie $\dot c(0) =X$. By definition, $\pi_*X=(\pi\circ c)^\cdot(0)$ and thus $\pi_*X=0$ as $\pi\circ c=p$ is constant.

Written in terms of derivations as was done in the question, this reads

$$ \pi_* X[g] = X[g \circ \pi] = \frac{\mathrm{d}g(\pi(u(t)))}{\mathrm{d}t}\Bigl|_{t=0} = \frac{\mathrm{d}g(p)}{\mathrm{d}t}\Bigl|_{t=0} = 0 $$

If you do not want to argue by dimensionality, you'd still have to show $\ker \pi_*\subset TG_p$, ie find a curve in $G_p$ representing any arbitrary vector from the kernel, which is easy after choice of trivialization:

Just take an associated chart $x^\mu,g^k$. Then, the local expression of $\pi_*$ is just

$$ X^\mu\frac\partial{\partial x^\mu} + X^k\frac\partial{\partial g^k} \mapsto X^\mu\frac\partial{\partial x^\mu} $$

so that any $X\in\ker\pi_*$ can be written as

$$ X = X^k\frac\partial{\partial g^k} $$

A corresponding vertical curve is parametrized by

$$ t \mapsto (x^\mu, g^k + t\cdot X^k) $$

answered May 22, 2014 by Christoph [ revision history ]
edited May 23, 2014
Most voted comments show all comments

@Hunter, I added some comments to the answer. The dimensions are equal due to $\dim TP=\dim\mathrm{im}\pi_*+\dim\ker\pi_*$. Hope that clarifies things a bit...

@Hunter, correct. This is a standard argument. Formally, it goes like this:

TG_p=\ker\pi_* &\Leftrightarrow TG_P\subset\ker\pi_*\wedge\ker\pi_*\subset TG_p
\\&\Leftrightarrow (X\in TG_p\Rightarrow X\in\ker\pi_*)\wedge(X\in\ker\pi_*\Rightarrow X\in TG_p)
\\&\Leftrightarrow (c:(-\epsilon,\epsilon)\to G_p\Rightarrow\pi_*\dot c(0)=0)\\&\quad\quad\wedge(\pi_*X=0\Rightarrow\exists c:(-\epsilon,\epsilon)\to G_p:X=\dot c(0))

@Christoph I've thought it about, and I think/hope I may have the answer. I am posting here, because I hope you can let me know if you think it is correct (and maybe it will useful for someone else).

Since we are considering a principal bundle, we have for \(A \in \mathfrak{g}\):

\(\pi(u) = \pi(u \exp (t A)) = p\)

and so \(u \exp (t A) \in G_p\). Therefore:

\(\pi_* X[g] = \frac{\partial g}{\partial \pi}\frac{\mathrm{d}}{\mathrm{d}t} \pi(u \exp(tA)) |_{t=0} = \frac{\partial g}{\partial \pi}\frac{\mathrm{d}}{\mathrm{d}t} \pi(u ) |_{t=0} = 0\)

and so for \(\pi_* X = 0\) we can always find a curve \(u \exp(t A) \in G_p\) ?

@Hunter, this doesn't help as you'd still have to construct $A$ (or at least prove its existence); I added the missing pieces to the answer...

@Christoph thank you for your help! I get it now.

Most recent comments show all comments

Thanks for your answer, but I'm not sure if I have fully understood it. I'll think about it some more and maybe ask you some questions in a couple of days if that is ok. Either way, I've voted it up, because it shows me that the method I was trying to use did not work at all.

@Christoph I feel really silly for asking this, but please bear with me (I'm not very trained in providing mathematical proofs). You say that if we don't want to argue with dimensionality then we still have to show that \(\mathrm{ker} \; \pi_* \subset TG_p\). But the way I see it is that you have shown that \(X \in TG_p \implies \pi_* X = 0\), and we need to show the converse \(\pi_* X = 0 \implies X \in TG_p\). Is this true?

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