# Exercise: show that two definitions of vertical subspace are equivalent

+ 2 like - 0 dislike
261 views

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.

edited May 22, 2014

+ 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
edited May 23, 2014

@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...

@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?

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

\begin{align*} 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)) \end{align*}

@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.

@Christoph thanks that is really helpful! I will try to spend some time on it this weekend if I have time.

@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 $u \exp(t A) \in G_p$ ?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.