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I am reading a paper at the moment and I have come across two statements which I want to understand. Here is the setup:

Suppose that $G$ is a Lie group which acts on a manifold $E$ differentiably and transitively. Fix a point $x_0\in E$ and define a map $p:G\rightarrow E$ by \begin{align*} p(g)=g(x_0). \end{align*}

Claim 1: The mapping $p$ is differentiable. I can convince myself of this as $g$ acts differentiably on $E$, but I was wondering if there is any slightly more rigorous way to prove this?

My second question is in relation to their proposition:

Let $V$ be a real vector space over $\mathbb{R}$. Suppose $G$ acts transitively on $E$ and that $\tilde{\omega}$ is a $V$-valued form over $G$ satisfying some properties that aren't important to my question. Then there exists a unique $V$-valued differential form $\omega$ over $E$ such that $\tilde{\omega}=p^*\omega$.

What I want to get my head around first is the uniqueness. They say that the uniqueness follows from the fact that $p$ is an onto mapping. Now, in this paper, they use $p$ to denote the map $p:G\rightarrow E$ and also the pushforward $p_*:TG\rightarrow TE$. So I don't know which they are referring to here. A few things I have noted.

a) the map $p:G\rightarrow E$ is certainly surjective since $G$ acts transitively.

b) surjectivity of $p:G\rightarrow E$ does not guarantee it is a submersion (I am just recalling this general fact from geometry)

c) if $p:G\rightarrow E$ is a submersion then I think the uniqueness argument follows quite easily? Since for arbitrary $X\in T_x E$, there exists $g\in G$ and $\overrightarrow{g}\in T_g G$ such that $p_*(\overrightarrow{g})=X$. Hence, if there exists $\omega_1$ and $\omega_2$ satisfying $\tilde{\omega}=p^*\omega_i$ \begin{align} (\omega_1-\omega_2)(X)&=0\\ \iff\omega_1(X)&=\omega_2(X)\\ \iff\omega_1(p_*\overrightarrow{g})&=\omega_2(p_*\overrightarrow{g})\\ \iff\tilde{\omega}&=\tilde{\omega} \end{align}

So is the argument in c) correct? And if it is, given these assumptions, is there any way to show that the map $p$ is a submersion? Any help would be much appreciated!

NOTE: They also say later on that upon restricting $p_*$ to $T_e G$, then 'from the theory of Lie groups' the map $(p_*)_e:T_e G\rightarrow T_{x_0}(E)$ is onto. I don't know if this is relevant.


EDIT: Here is a theorem from Lee's Introduction to Smooth Manifolds:

enter image description here

Are we able to use this theorem to claim that the map $\rho$ is a submersion?

In our context, let $G$ be the Lie group, $M=G$ and $N=E$. Then $\rho:G\rightarrow E$ is a smooth map. The map $\rho$ is equivariant since the transitive $G$-action on $G$ would just be left multiplication. Thus, for $g,a\in G$, \begin{align} \rho(ga)=ga(x_0)=g(a(x_0))=g(\rho(a)). \end{align} So our situation satisfies all assumptions in the theorem and thus indeed $\rho$ is a submersion?

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    $\begingroup$ The "properties that aren't important to [your] question" must indeed be important here, because without some extra hypotheses, the claim is false. For example, consider the transitive action of $\mathbb R^2$ on $\mathbb R$ by $(y,z)\centerdot x = x+y$. Take $x_0=0$, so that $p(y,z)=y$. Let $\tilde\omega =dz$ (an $\mathbb R$-valued $1$-form). Then there is no $1$-form $\omega$ on $\mathbb R$ such that $\tilde \omega = p^*\omega$. $\endgroup$ – Jack Lee Jul 15 '15 at 14:36
  • $\begingroup$ Thanks for your reply. Obviously the properties are required for the claim - but are they needed for the uniqueness part? I'm having trouble trying to show that $(p_*)_g$ is onto at each $g\in G$. $\endgroup$ – beedge89 Jul 15 '15 at 22:30
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About point c): $p$ being a submersion means exactly $p_{*,g}:T_gG\to T_{g\cdot x_o}E$ being surjective for every $g\in G$; but $p_{*,g}= p_{*,e}\circ L_{*,g^{-1}}$ from the action property, where $L_{g^{-1}}:G\to G$ is left translation, hence the claim.

About claim 1) The compostion of differentiable maps is differentiable, right? And projection of a cartesian product onto one of its components is differentiable, right? Try to use this.

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  • $\begingroup$ Thanks for your reply. Do you mean $p_{*,g}=p_{*,e}\circ (L_{g^{-1}})_{*,g}$? I don't understand how you get your equality in your first paragraph. What do you mean 'from the action property'? $\endgroup$ – beedge89 Jul 15 '15 at 11:30

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