12
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ What equivalence are you talking about? Conjugacy under a diffeo? $\endgroup$ Apr 13, 2010 at 4:19
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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