Let $\pi: M\to B$ be a fiber bundle of smooth manifolds with $B$ connected and each fiber of $\pi$ is a compact manifold. Let $G$ be a compact Lie group acting smoothly on $M$ such that $\pi(g\cdot m)=\pi(m)$. It is clear that $G$ acts smoothly on each fiber $M_b$ for $b\in B$.

Noe fix a $g\in G$. For each $b\in B$ we consider the fixed point submanifold $M_b^g\subset M_b$.

My question is: when $b$ varies, does the diffeomorphic type of $M_b^g$ unchanged?


1 Answer 1


I think the answer is yes.

Since $G$ is compact, there is a $G$-invariant Riemannian metric on $M$ (by averaging any metric). The orthogonal distribution to the fiber for this metric is a $G$-invariant Ehresmann connection, and the parallel transport for this connection thus commutes with the $G$-action.

This shows that the action of $G$ on any two fibers are conjugate. In particular, the fixed loci are diffeomorphic.

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    $\begingroup$ More generally, two nearby smooth actions of a compact Lie group on a closed manifold are equivariantly diffeomorphic. This is proved in [Palais, Equivalence of nearby differentiable actions of a compact group. Bull. Amer. Math. Soc. 67 (1961), 362–364], projecteuclid.org/journals/…. $\endgroup$ Jul 5 at 19:18

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