Let $\mathbb{Z}_2=\{1,g\},T^2=\{(e^{i\theta_1},e^{i\theta_2})\}$ and place $T^2$ in $\mathbb{R}^3$ as the locus of the rotation of $2\pi$ rads of the circle$\{(y,z)|(y-2)^2+z^2=1\}$ around $z$ axis.

It is known that there are 5 nonequivalent smooth involutions on torus,and they are:

1.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1+\pi)},e^{i\theta_2})$ (rotation$\pi$ rads around $z$ axis) with null fixed point set and orbit space $T^2$

2.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{i\theta_2})$(reflection along $x=0$) with fixed point set $S^1\times S^0$ and orbit space an annulus

3.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i\theta_2},e^{i\theta_1})$(switch the two coordinates) with fixed point set the diagonal circle and orbit space Mobius band

4.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1 +\pi)},e^{-i\theta_2})$(restriction of the involution $(x,y,z,\mapsto (-x,-y,-z)$ of $\mathbb{R}^3$ to torus)with null fixed point set and orbit space klein bottle

5.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{-i\theta_2})$(reflection along $x=0$ plus reflection along $z=0$) with fixed point set 4 points and orbit space $S^2$

i want to know how to derive the result above.for the free case it seems easy.since the action is free,the orbit space must be a manifold also,and has euler char 0,hence must be torus or klein bottle. for the nonfree case,the orbit is not manifold,but "orbifold". and we have Riemann-Hurwitz Formula:

$\chi(O)=\chi(X_O)-\sum_{i=1}^n (1-\frac{1}{q_i})-\frac{1}{2}\sum_{j=1}^m (1-\frac{1}{r_j})$

here$\chi(O)$ is the orbifold euler char and $\chi(X_o)$ is the euler char of the underlying space associated to the orbifold $O$,and $q_i$and $r_j$ denote the angles for sigular points(cone points and reflector corners can we determine the remaining 3 involutions by using this formula?Thank you!

  • $\begingroup$ What equivalence are you talking about? Conjugacy under a diffeo? $\endgroup$ Apr 13, 2010 at 4:19
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    $\begingroup$ You're getting pretty close to duplicating this thread: mathoverflow.net/questions/7746/… $\endgroup$ Apr 13, 2010 at 5:40
  • $\begingroup$ to Mariano Suárez-Alvarez:yes,two involutions are defined to be equivalent if they are conjugate in the group $Diff(T^2)$ $\endgroup$
    – student
    Apr 13, 2010 at 6:24
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    $\begingroup$ Do you really care how the torus is embedded in $R^3$? If not, perhaps edit the question to simplify the statement. $\endgroup$
    – Sam Nead
    Apr 13, 2010 at 14:54

1 Answer 1


Here is a sketch -- some of the details are a bit hazy:

Suppose that $\iota$ is a smooth involution of $T^2$. Show that the fixed point set of $\iota$ is a submanifold. Show that the orbit space of $\iota$ is an orbifold with orbifold Euler characteristic zero. Using the orbifold Euler characteristic you can enumerate all 17 compact, connected, 2-dimensional orbifolds of orbifold Euler characteristic zero. Now rule out 12 of these for topological reasons.

The second to last step is a nice exercise that everybody should do once, after learning about the orbifold Euler characteristic. The non-trivial part in the last step is eliminating $D(2,2;)$ and $P(2,2)$. Getting rid of the others is easy.


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