Question of deck transformation on double cover $\tilde{M}$ of non-orientable manifold $M$.

Question

Suppose $(M,g)$ is a non-orientable, compact, connected Riemannian manifold with positive sectional curvature, $\tilde{M}$ is its orientable double cover. $\varphi$ is deck transformation of $\tilde{M}$ with $\varphi \not= id$. $\varphi$ is deck transformation of $\tilde{M}$ means for smooth covering $\pi:\tilde{M}\to M$, $\varphi$ is diffeomorphism s.t. $\pi=\pi\ \circ \varphi$.

Question 1:

Prove $\varphi$ reverses the orientation of $\tilde{M}$.

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Background:

1. Weinstein's theorem:

Suppose $N$ is a compact, orientable $n$-dim Riemannian manifold with positive sectional curvature, $f:N\to N$ is an isometry. If $f$ preserves orientation if $n$ is even and reverses orientation if $n$ is odd, then $f$ has a fixed point.

2. Synge's theorem for odd-dim:

$(M,g)$ is a compact, connected odd-dim Riemannian manifold with positive sectional curvature, then $M$ is orientable.

I want to use Weinstein's theorem to prove Synge's theorem for odd-dim.

Suppose on the contrary $M$ is non-orientable, $\tilde{M}$ is its orientable double cover, $\pi:\tilde{M}\to M$ is smooth covering and is isometry, so $(\tilde{M},\pi^*g)$ is compact, connected, orientable Riemannian manifold with positive sectional curvature. Let $\varphi$ be deck transformation of $\tilde{M}$ with $\varphi \not=$ Id, if I can prove $\varphi$ reverses the orientation of $\tilde{M}$, then from Weinstein's theorem, $\varphi$ has a fixed point. From uniqueness of lifting or property of deck transformation, $\varphi$ = Id, contradiction. Thus $M$ is orientable.

The only problem is to prove $\varphi$ reverses the orientation of $\tilde{M}$.

Double cover: Chapter 15 in Introduction to Smooth Manifolds by John.Lee.

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Question 2

Is that possible to use Weinstein's theorem to prove Synge's theorem for even-dim?

$(M,g)$ is a compact, even-dim, orientable, connected Riemannian manifold with positive sectional curvature, then $M$ is simply connected.

Thank you.

Answer 1

You do not need a Riemannian structure on $M$. However, you have to assume that $M$ is connected (if it has more than one component, you will find a non-identity deck transformation which this is the identity on all but one component).

$\tilde{M}$ can be defined as in Understanding the orientable double cover. $\pi : \tilde{M} \to M$ is a two-sheeted covering projection with $\tilde{M}$ connected. It has exactly two deck transformations: The identity and the map $r : \tilde{M} \to \tilde{M}, r(p,o) = (p,-o)$, where $-o$ denotes the reverse orientation for $o$. For each $(p,o) \in \tilde{M}$ we get an isomorphism $d_{(p,o)}\pi : T_{(p,o)}\tilde{M} \to T_pM$, and an orientation of $\tilde{M}$ is given by taking on $T_{(p,o)}\tilde{M}$ the unique orientation $\omega_{(p,o)}$ such $d_{(p,o)}\pi (\omega_{(p,o)}) = o$. We have $$d_{(p,o)} \pi = d_{(p,o)} (\pi \circ r) = d_{r(p,o)} \pi \circ d_{(p,o)}r = d_{(p,-o)} \pi \circ d_{(p,o)}r,$$ hence $$o= d_{(p,o)} \pi (\omega_{(p,o)}) = d_{(p,-o)}(d_{(p,o)}r(\omega_{(p,o)})).$$ Since $d_{(p,-o)} \pi (\omega_{(p,-o)}) = -o$, we must have $d_{(p,o)}r(\omega_{(p,o)}) = -\omega_{(p,-o)} = -\omega_{r(p,o)} $. This means that $r$ reverses the orientation of $\tilde{M}$.

Answer 2

Late to the party but for OP's question $1$, can't we just argue like this: $\widehat M$ is connected, so $\pi$ is a covering map. Now, any deck transformation $\varphi$ is uniquely determined by its action on a single element $x\in \widehat M$ and as $\varphi$ must also satisfy $\pi\circ\varphi=\pi,$ it follows immediately that either $\varphi(p,\mathscr O)=(p,\mathscr O)$ or $\varphi(p,\mathscr O)=(p,-\mathscr O).$