Let $G$ be a compact Lie group that acts on a smooth, finite dimensional, oriented manifold $M$, and suppose that such action preserves orientation, i.e., for each $g\in G$, the diffeomorphism $\mu_g$ of $M$ induced by the action preserves orientation. Consider the stratification $$M=\bigcup M_j$$ by orbit types, i.e., $M_j$ is the set of points that have the same isotropy, up to conjugation.

My question is, are there "simple" conditions under which the quotient manifold $M_j/G$ is orientable? Is it automatic true with the orientation hypothesis on $M$ and the action?


1 Answer 1


In your question, you may as well take $M=M_j=$ a ($G$-invariant) tubular neighborhood of a single orbit $G/S$. Near the point $S/S$, the space looks like $\mathfrak g/\mathfrak s \times N$, where $N$ is the normal space. Indeed, using a $G$-invariant metric, we can make $S$ act on $N$ and make this identification $S$-equivariant. By your condition that the stabilizers are all conjugate, the $S$-action on $N$ should be trivial.

You say that $\mathfrak g/\mathfrak s \times N$ is oriented, and ask whether $N$ (the tangent space in the quotient) is naturally oriented. This is therefore equivalent to $G/S$ being oriented. (Which is easy to violate, e.g. $SO(3)$ acting on $\mathbb{RP}^2$.) If we assume $G$ to be simply connected (which we don't get for free, since taking the universal cover might spoil its compactness), then for $G/S$ to be nonorientable you'd need $S$ to be disconnected. (Here's it's $S(O(2)\times O(1))$.)

For a counterexample, we want a $G/S$-bundle over a non-orientable manifold $M$. These are easy to cook up but I haven't yet seen how to guarantee that the total space is oriented.


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