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Let $\Gamma(2)$ be the subgroup of $\mathrm{SL}_2(\mathbf{Z})$ satisfying the usual congruence conditions. It acts on the complex upper half-plane.

Does it have a fundamental domain contained in the strip $$\{x+iy: -1\leq x \leq 1\}?$$

Is the following a correct argument?

The matrix $$\left( \begin{matrix} 1 & \pm 2 \\ 0 & 1 \end{matrix}\right)$$ is in $\Gamma(2)$. If $\tau$ is not in the above strip, we can translate $\tau$ into this strip by multiplying with the above matrix a couple of times.

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the argument is certainly valid. If you want a particular fund. domain in your strip then your strip minus the two half-discs with centers $\pm1/2$ and radii $1/2$ is an example. –  user8268 Apr 17 '12 at 8:55
    
Ow that reminds me a lot of the standard fundamental domain of $\mathrm{SL}_2(\mathbf{Z})$. –  John M Apr 17 '12 at 15:45

1 Answer 1

Here is a visual method to see that $\Gamma(2)$ has a fundamental domain contained in the strip $\{x+iy : 0\leq x\leq 1\}\subset \{x+iy : -1\leq x\leq 1\}$. You can use Helena Verrill's fundamental domain drawer applet to draw the usual fundamental domain for $\Gamma(2)$ and then modify it to fit other conditions. First select "Gamma(N)" below, choose $N=2$, click draw, and then click edit. By clicking on the little yellow dots you can move different triangular components of the fundamental domain to other regions. By following these instructions, I got the following picture of a fundamental domain for $\Gamma(2)$:

A fundamental domain for $\Gamma(2)$

Now, in order to prove that this is in fact a fundamental domain for $\Gamma(2)$, you can start by showing that the fundamental domain originally depicted by the app is indeed a fundamental domain for $\Gamma(2)$, and then just carefully write down what matrices bring the original domain to this one. As long as these matrices are in $\Gamma(2)$, the two domains are equivalent.

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