This question arose from an attempt of using the Equivariant Rank Theorem to prove question 7.1 of Lee's Introduction to Smooth Manifolds.
The Equivariant Rank Theorem is stated:
Let M and N be smooth manifolds and let G be a Lie group. Suppose $F:M\rightarrow N$ is a smooth map that is equivariant with respect to a transitive smooth G-action on M and any smooth G-action on N. Then F has constant rank. Thus, if F is surjective, it is a smooth submersion; if it is injective, it is a smooth immersion; and if it is bijective, it is a diffeomorphism.
Question 7.1 is stated:
Show that for any Lie group G, the multiplication map $m: G\times G \rightarrow G$ is a smooth submersion. [Hint: use local sections.]
I was thus attempting to prove 7.1 through use of the equivariant rank theorem. I will outline my progress; however, the question of this post can be viewed disjoint from the context of this problem.
My attempt:
We know that m is surjective, as for any h in G, consider $(e,h) \in G\times G$. Then $m(e,h)=e*h=h$. So, if we can show the preliminary conditions in the theorem, we are done, as surjectivity implies smooth submersion.
Given that G is a Lie group, it is of course a smooth manifold by definition. Therefore $G\times G$ is as well, so m is a map between smooth manifolds. Since G is a Lie group, we also know that m is a smooth map.
The equivariant part is where my question emerges:
We need a transitive smooth G-action on $G\times G$, and any smooth action on G, where m is equivariant. For the smooth action on G, I was considering the trivial action, say $\theta$, which is smooth. However, m just multiplies the two coordinates of $G\times G$, so if we were to find a transitive smooth G-action from $G\times G$ to $G\times G$, say $\phi$, it would necessarily need to be trivial in order to commute with $m \circ \theta$ (as $m\circ\theta(p_1,p_2)=m(p_1,p_2)=(p_1p_2$). But the trivial action is not transitive, so we need another path.
My issue is that no matter which smooth G-action from G to G I pick, it seems that the overarching issue is finding a transitive smooth G-action from $G\times G$ to $G\times G$ at all. Thus, my gut is saying there does not exist one, but how can I be sure?
Your gut feeling is correct. Assume there is a smooth, transitive action $\theta\colon G \times G^2 \to G^2$. Pick some $p \in G^2$. As $\theta$ is transitive, the orbit map $\theta^{(p)}\colon G \to G^2$, $g \mapsto g . p$ is surjective, and since it is equivariant with regards to the left-multiplication action of $G$ on itself and $\theta$ on the codomain (!), the equivariant rank theorem tells you it is a smooth submersion. But unless $G$ is discrete this is absurd, seeing as $\dim G^2 = 2 \dim G$.