Let M be a smooth manifold. Suppose that $f : M \to M$ is a smooth involution without fixed points, i.e. for all points $x \in M$ the conditions $f^2(x) = x$ and $f(x) \ne x$ are satisfied. Define an equivalence relation on M by $x ∼ y \Leftrightarrow x = f^k(y)$ for some $k \in \mathbb{Z}$. I want to show that the associated quotient space $M/∼$ naturally inherits the structure of a smooth manifold such that the projection $ \pi : M \to M/∼ $ is smooth.

Given an atlas $\{U_\alpha, \phi_\alpha\}_{\alpha \in \mathcal{I}}$ of $M$, choosing symmetrized neighborhoods (that is restricting the size of each $U_\alpha$ such that $U_\alpha \cap f(U_\alpha)= \emptyset$, assuming that can be done), one can construct an atlas for $M/∼$ by considering the covering $ \{ V_\alpha = \pi(U_\alpha), \psi_\alpha = \phi_\alpha \circ (\pi |_{U_\alpha})^{-1} \}_{\alpha \in \mathcal{I}} $ which is obviously a smooth atlas on $M/∼$.

$M/∼$ is compact and connected because M is compact and connected. But I am not sure how to prove Hausdorffness. Moreover, I am not totally convinced that I can always have a choice of symmetrized neighborhoods.


First, pick an arbitrary Riemannian metric $\langle \cdot ,\cdot \rangle _1$ and define a new Riemainnian metric $\langle \cdot ,\cdot \rangle = \langle \cdot, \cdot\rangle_1 + f^\ast \langle \cdot,\cdot \rangle_1$. Then by construction, $\langle \cdot ,\cdot \rangle$ is $f$ invariant. Said another way, $f$ is an isometry with respect to this metric.

Now, let's show you can always find a symmetrizing neighborhood. So, let $x\in M$ and set $d = d(x,f(x)) > 0$. Consider the ball $B = B(x, d/2)$ of radius $d/2$ around $x$. Since $f$ is an isometry, $f(B) = B(f(x), d/2)$. We claim that $B$ is a symmetrizing neighborhood. (If it's not a chart, intersect it with a chart, giving a symmetrizing neighborhood).

To see this, assume for a contradiction that there is a $y\in B\cap f(B)$. Then $$d = d(x,f(x)) \leq d(x,y) + d(y, f(x) < d/2 + d/2 = d,$$ so $d < d$.

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I'll leave the Hausdorff condition to you, but as a hint, for $x$ distinct from both $y$ and $f(y)$, let $d = \min\{d(x,y), d(x,f(y)\}$, and consider $U = B(x,d/2)\cup B(f(x),d/2)$ and $V = B(y,d/2)\cup B(f(y), d/2)$.

  • $\begingroup$ What do you mean by $d(x,y)$? M is not a metric space. Moreover, a Riemannian metric cannot compute distances between points in the manifold. $\endgroup$ – Nanashi No Gombe Feb 14 '17 at 16:06
  • $\begingroup$ A Riemannian metric allows you to compute the distance between two points (on a connected manifold) as $d(x,y) = \inf \int_0^1 |\gamma| dt$ where the infimum is taken over all (piecewise) smooth $\gamma:[0,1]\rightarrow M$ with $\gamma(0) = x$ and $\gamma(1) = y$. (If $M$ is compact, as it seems you assume later on, then the infimum is always achieved. This is one consequence of the Hopf-Rinow Theorem). $\endgroup$ – Jason DeVito Feb 14 '17 at 18:34
  • $\begingroup$ I see. But, you mean $d(x,y) = \inf \int_0^1 |\dot{\gamma}| dt$, right? $\endgroup$ – Nanashi No Gombe Feb 14 '17 at 18:39
  • $\begingroup$ Yes - I forgot my derivative. $\endgroup$ – Jason DeVito Feb 14 '17 at 18:40
  • $\begingroup$ Okay, I understand your proof. However, can we have an alternative proof where we do not have to assume a Riemannian structure, because the claim holds regardless? $\endgroup$ – Nanashi No Gombe Feb 14 '17 at 18:47

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