The tangent space of $G \times _H M$

Question

I've posted this question Tangent space of $G \times_H M$ in MSE, but didn't get any answer. My question is the following:

Let $G$ be a Lie group and let $H$ be a Lie subgroup of $H$. Let $M$ be a smooth manifold on which $H$ acts from the left.

Let's consider the action of $H$ on $G \times M$: $$h((g,m)):= (gh,h^{-1}m), \quad h \in H , g \in G , m \in M, $$ and define the manifold $Z$ to be the quotient $G \times_H M$.

If we fix $(g,m) \in G \times M$, what is the induced equivalence relation on $T_gG \times T_mM$?

(My background is not so good in differential geometry, so please be patient with me.)

Answer 1

When moving to the tangent space level, the general rule is to take the derivative (of everything you can). Just from this I'd expect something like the equivalence on $T_gG\oplus T_mM$ should be such that $gX - Xm = 0$ for every $X\in \mathfrak h$.

(As a side note, it may be better to write $(g,m).h$ because it's a right action. Or at least change it to $h.(g,m):=(gh^{-1},hm)$. I will stick with the way you have it for now.)

In detail, we want to explicitly describe the kernel of $\mathrm d \pi$, since this is exactly the $H$ relation you seek.

Let $h(t) = \exp(tX)$ be a path in $H$ with tangent vector $X\in \mathfrak h$, and consider the corresponding path in $G\times M$, $h(t).(g,m) = (gh(t),h(t)^{-1}m)$. The tangent vector is $gX - Xm\in T_gG\oplus T_mM$.

This whole path is projected to the same point $[(g,m)]\in Z$, so its tangent vector is in the kernel of $\mathrm d\pi$. Also since you are just quotienting by $H$ to get to $Z$, it is exactly these types of paths which make up the whole kernel. You can also convince yourself by counting dimensions. You should have exactly $\dim(H)$ conditions for the equivalence.