I'm reading Michèle Audin's book, I'm confused by the Remark 1.2.7, where she said:
An orbit $G.x$ is principal if and only if the representation in $V_x$ (the normal space of $G.x$ at $x$) of its stablizer $G_x$ is trivial.
Here are some explanations of the statement:
Let $G$ be a connected compact Lie group, acts smoothly on a compact manifold $M$, for a fixed $x\in M$, choose an $g\in G_x$, the isotropy group (stablizer) at $x$, then the tangent map $(dg)_x: T_xM\longrightarrow T_xM$ is an isomorphism, and it is an identity map restricts on the subspace $T_xG.x$, thus it induces a representation of $G_x$:
$$G_x\longrightarrow GL(T_xM/T_xG.x):=GL(V_x)$$
I don't know how to show this "obviously" thing.
Best Answer
Because $G$ is compact, $M$ admits a $G$-invariant Riemannian metric, so we can assume that the $G$ action is isometric. Thus, we can replace $T_x M/T_x(G\cdot x)$ with the orthogonal complement of $(T_x (G\cdot x))^\bot\subseteq T_x M$. (I find it easier to think about vector subspaces than vector quotients).
Again, because $G\cdot x$ is compact, the normal exponential map $\exp^\bot$ gives a diffeomorphism $$\exp^\bot|_V:V:=\{v: v\in T(G\cdot x)^\bot \subseteq TM \text{ and } \|v \| < \epsilon\} \rightarrow U:=\{p\in M: d(p,G\cdot x) <\epsilon\}$$ for some small enough $\epsilon$. A subset of the form $\exp((T_x(G\cdot x))^{\bot})\cap U$ is called a slice at $x$. Notice that for $x\neq x'\in G\cdot x$, the slices at $x$ and $x'$ are disjoint since $\exp^{\bot}|_V$ is injective.
Proposition: An orbit $G\cdot x$ is principal if and only if the isotropy group $G_x$ acts on $(T_x (G\cdot x))^\bot$ trivially.
Proof. Assume first that $G\cdot x$ is principal and let $g\in G_x$.
Let $v\in (T_x(G\cdot x))^\bot$. Note that the $G_x$ action on $(T_x (G\cdot x))^\bot$ is linear, so the isotropy group at $v$ is the same as at $\lambda v$ for any $\lambda \neq 0$. Thus, by shrinking $v$, we may assume that $\exp(v)\in U$, so $\exp(v)$ is in a slice at $x$.
We claim that $G_{\exp(v)}\subseteq G_x$. Indeed, for $g\in G_{\exp{v}}$, if $gx\neq x$, then the slice at $x$ and the slice at $gx$ intersect at $\exp(v)$, giving a contradiction. Since $x$ is principal, $G_x$ is minimal, so we must have $G_x = G_{\exp(v)}$. In particular, $g\in G_{\exp(v)}$. But now, since $g$ fixes the end points of the geodesic $\exp(tv)$ and since $\exp^\bot|_V$ is injective, it follows that $g$ fixes $\exp(tv)$ pointwise, from which it follows that $g_\ast v = v$. That is, $g_\ast$ acts as the identity on $(T_x (G\cdot x))^\bot$.
Conversely, now assume that the action of $G_x$ on $(T_x (G\cdot x))^\bot$ is trivial. Choose a principal orbit $G\cdot y$ and let $\gamma$ be a minimizing geodesic connecting $G\cdot x$ and $G\cdot y$. Applying an appropriate element of $G$, we may assume $\gamma(0) = x$.
Now, let $g\in G_x$. Because $g$ fixes $\gamma(0)$ and $g_\ast$ acts trivially, $g$ fixes $\gamma(t)$ for all time. In particular, $G_x\subseteq G_{\gamma(1)}$. Since $G_{\gamma(1)}$ is principal, $G_\gamma(1) = G_x$, so $G_x$ must also be principal. $\square$