I've posted this question [Tangent space of $G \times\_H M$](https://math.stackexchange.com/questions/4365925/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.)



The tangent space of $G \times _H M$
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.


