The following question is stated on an exercise sheet of Riemannian Geometry.
We look at the pseudo Riemannian metric, defined on $M = \mathbb{R}^2 \ 0 $ by
\begin{align*}
< \partial_x, \partial_x > &= < \partial_y, \partial_y> = 0 \\
< \partial_x, \partial_y > &= \frac{2}{x^2 + y^2}
\end{align*}
Show that the map $F(x,y) = 2(x,y)$ is an isometry and that the group of isometries $\Gamma$ generated by $F$ acts in a proper and discontinuous way on $M$. Hence $\frac{M}{\Gamma}$, which can be identified with the annulus between radius 1 and 2 (with both curves identified), is a torus and therefore compact.
I do not know
- what the group of isometries generated by a certain isometry is
- How to take the quotient group of a manifold and an isometry group
- what to interprete as a proper and discontinuous action.
I do have knowledge of group theory, quotient groups and isometries, but their use in this particular context confuses me and the course material nor the internet is helping me any further.
Could someone help me out with those definitions/interpretations? Any help is duly appreciated.
Regards.
Best Answer
The group of isometries generated by $F$ is simply the set of powers of $F$ (i.e. $\{1, F, F^2, F^3, \dots\}$) and their inverses. Since $F$ is just rescaling by $2$, the group is just all rescalings by powers of $2$.
The quotient manifold $M \, / \, \Gamma$ is the set of equivalence classes of points in $M$ under the equivalence relation of being in the same orbit of $\Gamma$. By the description above, this means that $(x_1,y_1) \sim (x_2,y_2)$ if there is some integer $n$ so that $(x_2,y_2) = 2^n (x_1,y_1)$. So in the quotient, the points $(x,y)$, $(2x,2y)$, $(4x,4y)$, $(8x,8y)$, etc... all represent the same point. It can be identified by the annulus because any point can be multiplied by some $2^n$ to bring it to a point with a radius between 1 and 2. In other words, any equivalence class has a representative in this annulus.
The definition of properly discontinuous action can be found in a topology textbook. For example, see the wikipedia page: here.