On characterisation of smooth $G$ equivariant morphisms between Product manifolds with $G$ action

Question

In particular I am interested in the following!

Let $M$ be a smooth manifold and $G$ be a Lie Group. Let $\rho: (M \times G)\times G \rightarrow M \times G$ be the smooth action of $G$ on $M \times G$ given by $(m,g).g' \mapsto (m,g.g')$. Let $\phi: M \times G \rightarrow M \times G$ be a $G-$ equivariant smooth map (that is $\phi(m,g).g'=\phi(m,g.g')$) such that it is identity on the first component.

Then can we say that $\phi$ is always of the form $(x,g) \mapsto (x, \psi(x)^{-1}.g)$ for some map $\psi:M \rightarrow G$? .If such map $\psi$ exists then can we say $\psi$ is smooth?

Answer 1

By your assumption about $\phi$ being identity on the first component we have

$$\phi(m,g)=\big(m,\beta(m, g)\big)$$

for some $G$-equivariant $\beta:M\times G\to G$. Note that $\beta$ is smooth since $\phi$ is.

Now $\beta(m,g)=\beta(m,1)g$ and therefore the map you are looking for is given by $\psi(m)=\beta(m,1)^{-1}$ which is smooth as well.