Just some simple facts which have come up from my research.

Let \(M\) be a (smooth) manifold with a distinguished point \(x_0\in M\). Suppose \(K\) is a (Lie) group acting (smoothly) on \(M\) on the left. So

\[\begin{align} K\times M &\to M \\ (k,x) &\mapsto kx. \end{align}\]

Each point defines a stabilizer subgroup. Let \(H\) denote the stabilizer for \(x_0\). So

\[H:= \text{Stab}_K(x_0)=\{h\in K~:~hx_0=x_0\} \leq K.\]

If we pick a different point \(kx_0\) in the orbit of \(x_0\), the stabilizer \(H\) changes to a conjugate subgroup in \(K\). That is,

\[H_{kx_0} = kH_{x_0}k^{-1} = \{khk^{-1}~:~h\in H_{x_0}\}\cong H_{x_0}\]

because \((khk^{-1})(kx_0) = khx_0 = kx_0\).

Now suppose \(P\to M\) is a principal \(G\)-bundle. That is, it is a bundle with fiber diffeomorphic to \(G\), equipped with a right \(G\) action

\[\begin{align} P\times G &\to P \\ (p,g) &\mapsto pg \end{align}\]

which only moves along the fibers of \(P\), and further is free and transitive along those fibers, so that \(M= P/G\). Also we have local \(G\)-equivariant trivializations from \(U\subset M\) to the bundle \(U\times G\to U\), which are equivariant with respect to right \(G\) action on \(U\times G\) of just right multiplying on the \(G\) component.

Suppose further that \(P\) is \(K\)-homogeneous, which means that we have a lift of the left \(K\) action to \(P\) which commutes with the right \(G\) action. By lift, we mean that we have a \(K\) action on \(P\), still on the left, which “covers” the \(K\) action on \(M\), (so \(k\) sends the fiber over \(x\) to the fiber over \(kx\)).

If we distinguish a point \(p_0\in P\) which lies in the fiber over the point \(x_0\), then this defines an “isotropy homomorphism” \(\lambda:H\to G\) by the following: Let \(h\in H\). Since \(H\) fixes \(x_0\), when we look at the lifted action to \(P\), it can only possibly jumble up the fiber over \(x_0\). Since the \(G\) action on the fiber is free and transitive, there is a unique element \(\lambda(h)\) for which

\[hp_0 = p_0\lambda(h).\]

Note this is actually a homomorphism by

\[p_0\lambda(hh') = (hh')p_0 = hp_0\lambda(h')=p_0\lambda(h)\lambda(h'),\]

which, since the \(G\) action is free on the \(p_0\) fiber, says that \(\lambda(hh') = \lambda(h)\lambda(h')\).

If we pick a different point \(p_0g\) in the fiber over \(x_0\), then \(\lambda\) changes to a conjugate homomorphism. That is,

\[\lambda_{p_0g}= g^{-1}\lambda_{p_0}g\]

because

\[p_0\lambda_{p_0g}(h) = hp_0g = p_0\lambda_{p_0}(h)g = p_0g (g^{-1}\lambda_{p_0}(h)g).\]

So \(H\) depends conjugately on \(x_0\) (in its \(K\)-orbit), and \(\lambda\) depends conjugately on \(p_0\) (in its \(G\)-orbit / fiber over \(x_0\)).

There is another point of view of the lifted action on \(P\). Since \(K\) acts on the left, and \(G\) acts on the right, and these actions commute, we have a (left) action by the group \(\hat K := K\times G\) (using a classic trick to turn a right action into a left action):

\[\begin{align} \hat K\times P &\to P \\ (k,g).p &\mapsto kpg^{-1} \end{align}\]

Now given \(x_0,p_0,H\) and \(\lambda\) as above, I ask “what is the stabilizer of \(p_0\)?”

Well, note that by the defining equation for \(\lambda\), the element \((h,\lambda(h))\) fixes \(p_0\).

\[hp_0\lambda(h)^{-1} = p_0\]

Conversely, if any \((k,g)\) fixes \(p_0\), note that \(k\) must lie in \(H\), as it sends the fiber over \(x_0\) to itself (the \(G\)-action doesn’t change fiber). Then

\[p_0 = (h,g).p_0 = hp_0g^{-1} = p_0\lambda(h)g^{-1}\]

which implies that \(g=\lambda(h)\). Thus the stabilizer of \(p_0\) is the graph of \(H\) in \(\hat K\), and is isomorphic to \(H\).

\[\hat H:= \text{Stab}_{\hat K}(p_0) = \{(h,\lambda(h))~:~h\in H\} \cong H.\]