3
$\begingroup$

Let $M$ be a connected smooth manifold such that, for every $p\in M$, the fundamental group $\pi_1(M,p)$ has no subgroup of index $2$. Prove that $M$ is orientable.

Here's what I know: there is a smooth, orientable manifold $\widetilde{M}$ and a covering map $\pi:\widetilde{M}\to M$ whose fibers have cardinality $2$, and such that $M$ is orientable $\Leftrightarrow\widetilde{M}$ is disconnected.

Supposing by contradiction that $M$ is not orientable, $\widetilde{M}$ must be connected. I suppose the "subgroup of index $2$" part has to do with the fact that the fibers have cardinality $2$, and probably $\widetilde{M}$ being connected also plays a role.

I don't know much about covering spaces except for basic definitions, so I'm pretty much stuck.

$\endgroup$
  • 1
    $\begingroup$ Do you know how covering spaces of $M$ are related to the fundamental group? $\endgroup$ – Eric Wofsey Sep 9 '18 at 21:22
  • $\begingroup$ See my comments on math.stackexchange.com/questions/2906719/… $\endgroup$ – Lord Shark the Unknown Sep 9 '18 at 21:29
  • $\begingroup$ @EricWofsey, according to Lord Shark's comment, the fact that $\widetilde{M}$ is connected means that there is a correspondence between $2$-fold coverings and subgroups of index $2$, but I don't know how to see this. $\endgroup$ – rmdmc89 Sep 9 '18 at 22:23
1
$\begingroup$

Assume $M$ is not orientable. We then have that the orientable double cover $\widetilde{M}$ of $M$ is path-connected.

For a path-connected covering $f: \widetilde{M} \to M$, it is well-known that the number of elements on a fiber is given by the index of $f_{\#}(\pi_1(\widetilde{M}) )$ on $\pi_1(M)$*. It follows that $f_{\#}(\pi_1(\widetilde{M}) )$ is a subgroup of index $2$ (since we have a double cover), a contradiction.


*Let's prove this assertion. For this, note that $$f_{\#}(\pi_1(\widetilde{M},p) )=\{[c]\mid c \text{ is a loop in } f(p) \text{ which lifts to a loop in } p\}.$$ Thus, $ f_{\#}(\pi_1(\widetilde{M},p) )$ is precisely the isotropy of the point $p$ with respect to the action of $\pi_1(M,f(p))$ on the fiber which contains $p$, which takes $\alpha \in \pi_1(M,f(p))$ and $p'$ in such fiber and takes them to where the lift of $\alpha$ which begins on $p'$ ends.

You can verify that this is indeed a group action. It is also transitive, and this is where path-connectedness is important. Now, it is an easy exercise to verify that $$\pi_1(M,f(p))/f_{\#}(\pi_1(\widetilde{M},p) ) \to F_{f(p)}$$ $$[c] \mapsto [c] \cdot p $$ is well-defined and is a bijection, where $F_{f(p)}$ is the fiber that contains $p$ (this is an algebraic exercise).

$\endgroup$
1
$\begingroup$

Consider and atlas $(U_i,f_i)_{i\in I}$ of $M$, we can suppose that $U_i\cap U_j$ is connected and therefore the sign of the Jacobian $ Jac(f_i\circ f_j^{-1})$ is constant (it takes its values in $\{-1,1\}$).Defines the $\mathbb{Z}/2$-bundle over $M$ whose coordinate changes are $g_{ij}:U_i\cap U_j\rightarrow sign(Jac(f_i\circ f_j^{-1}))$. This bundle is flat, (in fact it is a covering of $M$) thus define by a morphism $f:\pi_1(M)\rightarrow\mathbb{Z}/2$, (we can also says that since the bundle is a $2$-covering, its fundamental group is a subgroup of index $2$ of $\pi_1(M)$) the condition implies that $f$ is trivial and $Jac(f_i\circ f_j^{-1})$ is constant, we deduce that $M$ is orientable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.