Focusing in two Riemannian examples

In this section, we give two examples (one of them new as far as we know) of focusing. Suppose that at t=0 geodesics start emanating in all possible directions from the origin. At certain times t₁, t₂, ...., we will see geodesics returning to the origin. We derive expressions for the number of geodesics returning at t_n in two cases. First, as introductory example we will discuss this for $M = \mbox{$\Bbb R$ }^2/\mbox{$\Bbb Z$ }^2$ (a more complete discussion of this example can be found in [Pe3]). Second, we will deal with a much more unusual example, namely $M=\mbox{$\Bbb H$ }^2/\Gamma(2)$ , where $\Gamma (2)$ is a subgroup of $PSL(2,\mbox{$\Bbb Z$ })$ called the principal congruence subgroup of level 2 (defined in more detail below). We note that it seems to be considerably harder to count geodesics that focus in points other than the origin. Before continuing, consider the classical problem of counting R_g(n), the number of solutions in $\mbox{$\Bbb Z$ }^2$ of

Example 4.2 Choose an origin in $M = \mbox{$\Bbb R$ }^2/\mbox{$\Bbb Z$ }^2$ and lift it to the origin in $\mbox{$\Bbb R$ }^2$ . We have that $S=\mbox{$\Bbb Z$ }^2$ . Let $\rho_x(t)$ be the number of geodesics of length t that connect the origin to the point $x \in M$ .

Proposition 4.3 In the flat torus, the number of geodesics of length t that connect the the origin to itself is given by

$\begin{displaymath}\rho_0(t) = \cases{ R_g(t^2), &if $t^2 \in \mbox{$\Bbb N$ }$\space \cr 0, &otherwise.} \end{displaymath}$

Proof: Notice that by definition geodesics of length t leaving from the origin in $\mbox{$\Bbb R$ }^2$ reach the points contained in C_t(0). Only if t²is an integer does this circle intersect points in $\mbox{$\Bbb Z$ }^2$ . $\Box$

Example 4.4 We now turn to the next example. Recall that $PSL(2,\mbox{$\Bbb Z$ })$ can be identified with the group of two by two matrices with integer entries and determinant one, and with multiplication by -1 as equivalence. The group $\Gamma (2)$ is an order 3 subgroup of $PSL(2,\mbox{$\Bbb Z$ })$ obtained as the quotient of $PSL(2,\mbox{$\Bbb Z$ })$ by the order 3 subgroup generated by $z\mapsto\frac{-1}{z-1}$ .

$\begin{displaymath}\Gamma(2) = \left\{{ \left( \begin{array}{ll} a & b\\ c & d \... ...(\bmod\, 2),\,\, b\equiv c \equiv 0\, (\bmod\,2), }\right\} . \end{displaymath}$

This group has important applications in number theory. The action of $\Gamma (2)$ on $\mbox{$\Bbb H$ }^2$ is given by the Möbius transformations

$\begin{displaymath}g(z) = \frac{az+b}{cz+d} , \end{displaymath}$

where $\left( \begin{array}{ll} a & b\\ c & d \end{array} \right) \in \Gamma(2)$ . We need a well-known result, which we reproduce here for completeness.

Lemma 4.5 The orbit S of 0 in $\mbox{$\Bbb D$ }^2$ under the fundamental group of $M=\mbox{$\Bbb H$ }^2/\Gamma(2)$ is

$\begin{displaymath}S = \left\{{ \frac{p+iq}{r+is} \,\,\vrule{}\,\,p,q,r,s \in \m... ..., p^2+q^2+1=r^2+s^2, \,\, q+s\equiv 0\, (\bmod\,2)}\right\}. \end{displaymath}$

**Figure:** The orbit of 0 under $\Gamma (2)$ in the hyperbolic disk, and the corresponding Brillouin zones.
$\begin{figure} \centerline{\psfig{figure=Figs/Gamma-2pts.ps,width=.4\hsize} \hfil \psfig{figure=Figs/Gamma-2.ps,width=.4\hsize}} { } \end{figure}$

Proof: Following the conventions in [Be], define

$\begin{displaymath}\phi : \mbox{$\Bbb D$ }^2 \rightarrow \mbox{$\Bbb H$ }^2, \quad \phi(z) = i \frac{z+1}{-z+1} . \end{displaymath}$

Push back the transformation $g \in \Gamma(2)$ from $\mbox{$\Bbb H$ }^2$ to $\mbox{$\Bbb D$ }^2$ by $g \rightarrow \phi^{-1}g\phi$ to obtain a representation of $g \in \Gamma(2)$ as a transformation in SU(1,1) acting on $\mbox{$\Bbb D$ }^2$ . The matrix representation of this transformation is given by:

$\begin{displaymath}A_g = \left( \begin{array}{rr} \frac{a+d}{2}+i\frac{b-c}{2} ... ...b+c}{2} & \frac{a+d}{2} - i\frac{b-c}{2} \end{array} \right) , \end{displaymath}$

where $\det A_g = 1$ , since this matrix is conjugated to g, whose determinant is equal to 1. Let

$\begin{displaymath}\begin{array}{rcr} p=(a-d)/2 & \qquad & q= -(b+c)/2 \\ r=(a+d)/2 & \qquad & s=-(b-c)/2 \end{array}\end{displaymath}$

and A_g now written as

$\begin{displaymath}A_g = \left( \begin{array}{lr} r-is & p+iq \\ p-iq & r+is \end{array} \right) . \end{displaymath}$

Here the numbers p,q,r,s are in $\mbox{$\Bbb Z$ }$ and must satisfy the following parity conditions:

$\begin{displaymath}\begin{array}{ccc} p+r & \equiv & 1\, (\bmod\,2) \\ q+s & \equiv & 0\, (\bmod\,2). \end{array}\end{displaymath}$

Recall that the determinant of A_g is equal to 1, therefore

p²+q²+1=r²+s².

It is easy to see that this last equation together with any one of the above parity conditions, say the one satisfied by p and r (q and s), imply the parity relation on q and s (p and r). This proves our claim. $\Box$

Minor modifications to this argument give exact counts on the number of geodesics of length t which connect the origin to itself for $\mbox{$\Bbb D$ }^2/\Gamma(3)$ and $\mbox{$\Bbb D$ }^2/\Gamma(5)$ . We shall do this now. Let

Proof: First, let us consider N=3. Since N is odd, the congruence conditions on r,p,s, and q imply that

Let $(p,q,r,s) \in \mbox{$\Bbb Z$ }^2 \times \mbox{$\Bbb Z$ }^2$ be any solution to n=p²+q² = r² + s² -1 with $p \equiv q \equiv 0 (\bmod\ 3)$ . For each choice of (p,q), we have exactly R_g(n+1) choices of (r,s). Now let R denote the product of the rotations by $\pi/2$ on each of the components of $\mbox{$\Bbb Z$ }^2 \times \mbox{$\Bbb Z$ }^2$ . Using Lemma 4.6, we see that all such solutions can be obtained from just one by applying R repeatedly. It is easy to check that each quadruple of solutions thus constructed runs exactly once through the above list. Since precisely one out of the four associated solutions is compatible with the conditions, the total number of solutions is exactly:

For N=5, the proof can be literally transcribed to obtain the second result. $\Box$