Principal bundles, connection forms and fundamental vector fields

Question

Suppose $\pi:P\rightarrow M$ is a principal bundle, $\omega\in \Omega^1(P;\mathfrak{g})$ is the connection one form and $\sigma(\cdot)$ is the fundamental vector field associated to some vector field in $\mathfrak{g}$. That is, for $X\in \mathfrak{g}$, the value of $\sigma(X)$ at $p\in P$ is given by

$ \begin{align*} \sigma(X)_p=\frac{d}{dt}\bigg|_{t=0}p\exp(tX). \end{align*}$

I am reading through some lecture notes at the moment and the author decomposes a vector field $u\in \mathfrak{X}(P)$ in to its horizontal and vertical components. So

$u=u_v + u_h$.

The author then takes the derivative $u_v f$ of some function $f$ defined on $P$. They make the argument that since this derivative at a point $p\in P$ depends only on the vector $u_v$ at $p$, they may assume $u_v = \sigma(\omega(u))$. I understand the idea of the derivative only depending on the vector at $p$. However, I am stuck on how they can assume that $u_v=\sigma(\omega(u))$. I don't understand how $u_v$ at $p$ is equal to $\sigma(\omega(u))$ at $p$. If anyone could help that would be greatly appreciated. Thanks!

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EDIT:

Thanks very much for your detailed response. I have a few questions

1) Why have you written that map as $\rho_p$ instead of $L_p$? Isn't it just left multiplication? On that note, is it kosher to define left multiplication on $P$?

2)Is the reason $\sigma(X)_p=(\rho_p)_{*,e}X$

because $X=\frac{d}{dt}\big|_{t=0}\exp(tX)$,

so

$(\rho_p)_{*,e}X=(\rho_p)_{*,e}\frac{d}{dt}\big|_{t=0}\exp(tX)=\frac{d}{dt}\big|_{t=0}p\exp(tX)=\sigma(X)_p$??

3) Why does $R_g$ and $\delta_g$ being diffeomorphisms imply $\rho_p$ has constant rank? Why did you go to the effort of adding $g$ and $g^{-1}$?

4) Why does dim$V_p$=dim$\mathfrak{g}$ imply the map is surjective? I'm thinking that's just something from linear algebra I'm forgetting?

Thank you again!

Answer 1

It is true that $u_v(p) = \sigma(\omega(u_p))_p$. This is because since $u_v(p)$ is vertical, it is equal to $\sigma(X)_p$ for some $X\in \mathfrak g$. But $X = \omega(\sigma(X)_p) = \omega(u_v(p))$.

Answer 2

Here is my understanding (with more details than needed) : The author wants to prove a formula $F(u_vf)$ expressing the fact that the derivative $u_vf$ is equal to something on $P$. To show that $F(u_vf)$ is valid on all of $P$, he chooses to do it one point $p \in P$ at a time.

So, choose a point $p \in P$. Since $(u_vf)(p)$ depends only on the value of $u_v$ at the point $p$, you can change the value of $u_v$ however you want at other points of $P$. You just have to keep in mind that the calculations you do will only be valid at the chosen point $p \in P$. But then if you can prove the formula at any arbitrary points $p \in P$, it will be true globally.

Now, to simplify the expressions, the author chooses to replace $u_v$ by a fundamental vector field $u' = \sigma(X)$ (for $X \in \mathfrak{g}$) such that $u_v|_p = u'|_p$. Like Eric O. Korman says, given a vertical $u_v|_p \in T_pP$ you can always find an $X \in \mathfrak{g}$ such that $\sigma(X)|_p = u_v|_p$ and you end up with a vector field $u' = \sigma(X)$ of the desired form $u' = \sigma(\omega(u'))$ by the defining properties of $\omega$.

The reason why we can find such an $X$ was not obvious to me, here's a way to see it: For $p \in P$, you can rewrite the map $\mathfrak{g} \ni X \mapsto \sigma(X)|_p \in T_pP$ as the differential at the identity $e \in G$ of the map $\rho_p : G \to P$ defined by $\rho_p(g) = p.g$ where the dot is the action on $P$. i.e. $\sigma(X)|_p = (\rho_p)_{*,e}(X)$. But by rewriting things, we can show that $(\rho_p)_{*,e}$ gives an isomorphism between $\mathfrak{g}$ and the space $V_p$ of vertical vectors at $p$. To do that, write $\delta_g : G \to G$ and $R_g : P \to P$ for the diffeomorphisms $\delta_g(h) = hg^{-1}$ and $R_g(p) = p.g$. For $$ X_g = \left.\frac{d}{dt}\right|_{t=0}\gamma(t) \in T_gG, $$ we find $$ (\rho_p)_{*,g}(X_g) = \left.\frac{d}{dt}\right|_{t=0}p.\gamma(t) = \left.\frac{d}{dt}\right|_{t=0}p.\gamma(t) g^{-1}g = (R_g)_{*,p} \circ (\rho_p)_{*,e} \circ (\delta_g)_{*,g}(X_g). $$ Since $R_g$ and $\delta_g$ are diffeomorphisms, this shows that $\rho_p : G \to P$ has constant rank. But since the action of $G$ on $P$ is free, $\rho_p$ is injective and recall that an injective map with constant rank is necessarily an immersion (for a general action, we would mod out $G$ by the stabilizer at $p$ to get an immersion). In other words, the map $$ (\rho_p)_{*,e} : \mathfrak{g} \to T_pP $$ which is $(\rho_p)_{*,e}(X) = \sigma(X)|_p$ is an isomorphism on its image, which is contained in the vertical subspace $V_p \subset T_pP$. But since $\dim V_p = \dim G = \dim \mathfrak{g} < \infty$, this map is also surjective and in fact $$ (\rho_p)_{*,e} : \mathfrak{g} \to V_p $$ is an isomorphism.

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EDIT ANSWERING YOUR QUESTIONS:

1) Well the action on $P$ is usually on the right hence $\rho$, but it’s just a notation I’ve seen on some Wikipedia article. Also since $\rho_p$ goes from $G$ to $P$, it’s not really just right multiplication on $P$ or on $G$. See that thread https://mathoverflow.net/questions/50473/why-does-the-group-act-on-the-right-on-the-principal-bundle for a discussion of why the action is usually on the right.

2) Yes exactly.

3) The fact that $R_g$ and $\delta_g$ are diffeomorphims implies that their differential is an isomorphism at every point. Hence since $(\rho_p)_{*,g} = (R_g)_{*,p} \circ (\rho_p)_{*,e} \circ (\delta_g)_{*,g}$, the rank of $(\rho_p)_{*,g}$ is constant and is equal to the rank of $(\rho_p)_{*,e}$ for any $g \in G$. Indeed, adding $g$ and $g^{-1}$ and rewriting everything like that let us see that the differential of $\rho_p$ at any element $g \in G$ is "conjugated" to the same map $(\rho_p)_{*,e}$, hence that linear properties invariant under composition with isomorphisms of $(\rho_p)_{*,g}$ (like its rank) will not depend on $g$.

4) Yes it is just the fact that an injective linear map $f : W_1 \to W_2$ between two vector spaces of the same (finite) dimension is also surjective.