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I've asked this on SE, but got no answer. So I try it here:

Let $M$ be a smooth oriented manifold and let $f \colon M \to M$ be a self-diffeomorphism with $f^p = \text{id}_M$ for some odd $p > 0$. Then $f$ is orientation preserving and it follows from the slice theorem that the fixed point set $$ F = \{ x \in M : f(x) = x\} $$ is a smooth submanifold of $M$. Is $F$ orientable?

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1 Answer 1

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I believe this is true.
Lemma
For $\mathbb{Z}_p$, $p$ odd, all nontrivial irreducible real representations are 2-dimensional. (They really come from a complex representation, forgetting the complex structure.)
Lemma
For two such real representations, there is a non trivial intertwiner if and only if the weights of the corresponding complex representations agree up to sign. (The corresponding intertwiners come from complex conjugation or the identity.)

The normal bundle of a fixpoint set as in your question is a real $\mathbb Z_p$-vector bundle $E$ over each connected component of $M^{\mathbb Z_p}$. It has the following properties:

  1. $\mathbb{Z}_p$ acts trivially on the base.
  2. $E^{\mathbb{Z}_p}_b=\{0\}$ for all $b$ in the base.

The point is now that you can put a complex structure on your bundle. Locally, your bundle splits into a direct sum of non-trivial irreducible $\mathbb{Z}_p$-representations $V_i$. They are all 2-dimensional by the first lemma. On each one of them, we can choose a complex structure. Now, because of the second lemma, we can choose in a way consistent with the transition functions.

Using the complex structure, we see that the normal bundle is orientable, and since $M$ was orientable, we get an orientation for the submanifold $M^{\mathbb Z_p}$.

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