The corresponding Lie subalgebra of the isotropy subgroup

Question

Consider the Lie group action $G$ on the manifold $M$. How to find the Lie subalgebra of the isotropy subgroup $ G_m:=\{g\in G\,|\,g.m=m\}$?

Answer 1

Regarding the continued question in the comments of José's answer: if we differentiate the action $\rho : G \times M \to M$ in the $G$ direction at the identity $e \in G$, we get the linearized action $D_e \rho : \mathfrak{g} \times M \to TM$. Evaluating this at an element $v \in \mathfrak{g}$ gives the corresponding fundamental vector field $v^\sharp : M \to TM$, which is the vector field whose flow coincides with the action of the subgroup $\exp(\mathbb R v)$. When the author writes $v|_m = 0$ what they really mean is $v^\sharp|_m = 0$, i.e. that $m$ is a zero of the fundamental vector field $v^\sharp$, or equivalently that the flow $t \mapsto \exp(tv)$ fixes $m$.

Answer 2

If $\mathfrak g$ is the Lie algebra of $G$, then the Lie algebra of $G_m$ is$$\left\{X\in\mathfrak{g}\,\middle|\,(\forall t\in\mathbb{R}):\exp(tX).m=m\right\}.$$