2
$\begingroup$

What is known about the fundamental domain of an involution on a manifold? Eg., if the involution is free is it true that there exists a fundamental domain which is a smooth manifold with boundary?

$\endgroup$
1
  • 1
    $\begingroup$ It's the question of representing the mod 2 homology class of the quotient manifold (corresponding to the 2-fold cover) by a smooth hypersurface. Cutting along this hypersurface and lifting to the covering space will give you a fundamental domain. $\endgroup$ Commented Nov 17 at 5:45

1 Answer 1

5
$\begingroup$

I will assume that you are working in the DIFF category, i.e. your manifold $X$ and involution $\tau$ of $X$ are smooth. It suffices t consider the case when $X$ is connected. Let $Y:=X/\tau$, $p: X\to Y$ is a 2-fold covering map. Then the image of $\pi_1(X)$ has index 2 in $\pi_1(Y)$, thus, is the kernel of an epimorphism $h: \pi_1(Y)\to \mathbb Z_2$. There exists a smooth map $f: Y\to P^n$ (the real-projective space of sufficiently large dimension) inducing the homomorphism $$ h: \pi_1(Y)\to \mathbb Z_2= \pi_1(P^n). $$ The map $h$ desceds to an isomorphism $\phi: \langle \tau\rangle\to \pi_1(P^n)$. Take a generic hyperplane $P^{n-1}\subset P^n$, such that $f$ is transversal to $P^{n-1}$. (Then $Z:= f^{-1}(P^{n-1})$ is a smooth properly embedded hypersurface in $Y$ whose mod 2 locally finite homology class is the Poincare-dual to the cohomology class $c\in H^1(Y; \mathbb Z_2)$ corresponding to $h$.) The map $f$ lifts to an equivariant map $F: X\to S^n$ (equivariant with respect to the homomorphism $\phi: \langle \tau\rangle \to \mathbb Z_2\cong \pi_1(P^n)$, where the latter is generated by the antipodal map of $S^n$). Let $H$ be the great sphere in $S^n$ which is the preimage of $P^{n-1}$ above. The sphere $H$ divides $S^n$ in two hemispheres $H^\pm$. Lastly, take $\Phi:= F^{-1}(H^+)$. This is your fundamental domain for the $\tau$-action on $X$. It is bounded by the smooth hypersurface $F^{-1}(H)$.

Note that this $\Phi$ need not be connected (I do not know if your definition of a fundamental domain assumes connectivity). Getting a connected fundamental domain would require more work.

$\endgroup$
1
  • $\begingroup$ Thank you for this very detail reply $\endgroup$ Commented Nov 19 at 13:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .