Given $\tau$ in the upper half plane, what is a good, systematic way to find a representative in the usual fundamental domain for the $PSL_2(Z)$-orbit of $\tau$? For example, let $\tau=\frac{2}{3} + \frac{1}{2}i$. Similarly, given $\tau$ and $\tau'$ in the upper half-plane, is there a practical way to check if they lie in the same orbit?
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3$\begingroup$ This is straightforward material that can be found, say, in any introduction to modular forms. Essentially one uses the Euclidean algorithm to move a point into the fundamental domain: en.wikipedia.org/wiki/… $\endgroup$– Qiaochu YuanCommented Jan 22, 2011 at 15:36
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$\begingroup$ Thank you for the suggestion, but could you be more specific? I have looked at some books on modular forms and have not seen this question addressed. Also, the wikipedia link that you give simply describes the fundamental domain, something that I do already know where to find. $\endgroup$– A. PascalCommented Jan 22, 2011 at 15:54
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$\begingroup$ Serre gives what amounts to an algorithm for doing this in VII.1.2 of his A course in arithmetic (p. 78-79). $\endgroup$– Qiaochu YuanCommented Jan 22, 2011 at 16:13
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3$\begingroup$ @Qiaochu: actually, in most sources the proof is an existence proof and not constructive, like in Serre's cours d'arithmetique. The constructive approach moves $\tau$ to the vertical strip between $-1/2$ and $1/2$ and then reflects at the circle by sending $\tau$ to $-1/\tau$. Repeat. This algorithm is classical, since it corresponds to the reduction process for positive definite binary quadratic forms going back to Lagrange and Gauss. $\endgroup$– Franz LemmermeyerCommented Jan 22, 2011 at 16:15
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$\begingroup$ @Franz: Thank you. Indeed, the sources that I have been reading give existence not construction. This is the kind of answer that I was looking for. I'll accept it as answer if you post it as such. Could you also give a hint as to why this process converges to a point in the fundamental domain or to references for the reduction process? Did I understand correctly that Serre does describe this construction? I shall not have access to Serre's book for a while. $\endgroup$– A. PascalCommented Jan 22, 2011 at 16:33
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2 Answers
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The algorithm mentioned by Qiaochu and Franz is the following.
Let $z\in\mathbb{C}$ be arbitrary with $\Im z>0$.
Replace $z$ by $z+n$, where $n\in\mathbb{Z}$ is uniquely determined by $-1/2<\Re (z+n)\leq 1/2$.
If $|z|<1$, then replace $z$ by $-1/z$ and go back to step 2.
$z$ is in the standard fundamental domain, so stop.
To prove that the algorithm stops in a finite number of steps it suffices to show that it arrives at step 3 only finitely many times. Note that whenever the algorithm arrives at step 3, it increases the imaginary part of $z$, then in step 2 it produces another $z$ with the same imaginary part, but in the strip $-1/2<\Re z\leq 1/2$. Hence it suffices to show the following.
Proposition. Let $z\in\mathbb{C}$ be arbitrary with $\Im z>0$. Then there are only finitely many $\gamma\in\mathrm{SL}_2(\mathbb{Z})$ such that $-1/2<\Re(\gamma z)\leq 1/2$ and $\Im(\gamma z)>\Im z$.
Proof. Let $\gamma=\left(\begin{matrix}a b \\\ c d\end{matrix}\right)\in\mathrm{SL}_2(\mathbb{Z})$ be such that $-1/2<\Re(\gamma z)\leq 1/2$ and $\Im(\gamma z)>\Im z$. Then $\Im(\gamma z)=|cz+d|^{-2}\Im z$, hence $|cz+d|^2<1$. In other words, denoting $z=x+iy$, we have $(cx+d)^2+(cy)^2<1$. This shows that there are finitely many choices for $c$, and for each $c$ there are finitely many choices for $d$. On the other hand, for fixed coprime $c$ and $d$, it is straightforward to see that the matrices $\gamma=\left(\begin{matrix}a b \\\ c d\end{matrix}\right)\in\mathrm{SL}_2(\mathbb{Z})$ form a right coset of the subgroup $\left(\begin{matrix}1 \mathbb{Z} \\\ 0 1\end{matrix}\right)$, hence there is precisely one such $\gamma$ with $-1/2<\Re(\gamma z)\leq 1/2$. Done.
answered Jan 22, 2011 at 18:50
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2$\begingroup$ To justify the name "Euclidean algorithm," step 2 corresponds to replacing (a, b) with (a-nb, b) and step 3 corresponds to replacing this with (b, a-nb). $\endgroup$ Commented Jan 22, 2011 at 19:37
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Use the continued fraction algorithm: repeatedly add an integer to make the real part at most 1/2, then invert if this increases the absolute value. For real numbers this usually continues indefinitely and produces a continued fraction expansion, while for non-real numbers it terminates after a finite number of steps, producing an element in a fundamental domain.
answered Jan 22, 2011 at 19:32
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3$\begingroup$ @Richard: With all due respect, you repeated what I said (without proofs). $\endgroup$ Commented Jan 22, 2011 at 19:56