Let $H$ be a lie group, $\Phi : H \times M \to M$ be a differentiable group action on a Riemannian manifold $(M,g)$, and $H_p:=\{ h \in H | \Phi(h,p)=\Phi_h(p)=p \} $ be it's stabilizer/isotropy group. The claim now is that for $h \in H_p$
$$ d(\Phi_h)_p : T_pM \to T_{\Phi_h(p)=p}M $$ is an isometry, i.e.
$$ g_p(v,w)=: \langle v,w \rangle_p = \langle d(\Phi_h)_p v, d(\Phi_h)_p w \rangle_p.$$
Since we have $\Phi_{h_1}\circ \Phi_{h_2}=\Phi_{h_1h_2}$, we have $d(\Phi_{h_1})_p \circ d(\Phi_{h_2})_p=d(\Phi_{h_1h_2})_p$ and $d(\Phi_e)_p=id_{T_pM}$ and thus the isotropy representation goes to
$$ H_p \to GL(T_pM,g_p), h \mapsto d(\Phi_h)_p $$ But why does the image actually lie in $O(T_pM,g_p)$?
I'm sure the answer is rather trivial, but I don't see it currently.
Best Answer
Let $(V, \langle\cdot, \cdot\rangle)$ be an inner product space. A vector space isomorphism $L : V \to V$ is orthogonal if $\langle L(v_1), L(v_2)\rangle = \langle v_1, v_2\rangle$. The collection of all orthogonal isomorphisms of $(V, \langle\cdot, \cdot\rangle)$ is denoted by $O(V, \langle\cdot, \cdot\rangle)$.
With this in mind, the equation $\langle v, w\rangle_p = \langle d(\Phi_h)_pv, d(\Phi_h)_pw\rangle_p$ shows that $d(\Phi_h)_p \in O(T_pM, \langle\cdot, \cdot\rangle_p) = O(T_pM, g_p)$.