Growth of smallest closed geodesic in congruence subgroups? - MathOverflow

Growth of smallest closed geodesic in congruence subgroups?

2012-01-14T11:43:29Z

Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$.

How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$?

Any suggestions?

Answer by S. Carnahan for Growth of smallest closed geodesic in congruence subgroups?

2012-01-14T12:44:25Z

I don't have an answer, but here is a suggestion. From the environs of exercise 20 in section 3.7 of Terras's Harmonic analysis on symmetric spaces I, we find that a primitive geodesic whose beginning and end are joined by the hyperbolic transformation $\gamma$ has length $\log N(\gamma)$, where $N(\gamma) = a^2$ and $a$ is a real number satisfying $|a| > 1$, such that the Jordan form of $\gamma$ is $\left(\begin{smallmatrix} a & 0 \\ 0 & 1/a \end{smallmatrix} \right)$ . Now you just need to find how norms of hyperbolic elements grow with level.

Answer by kassabov for Growth of smallest closed geodesic in congruence subgroups?

2012-01-14T12:45:15Z

Let $\Gamma$ be the congruenece group $\ker SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N\mathbb{Z})$.
The length of the shortest geodesic is of the order of $\log N$. The upper bound follows from a counting argument, and the lower bound comes from the observation that a product of $c\log N$ generators of $SL_2(\mathbb{Z})$ can not be a nontrivial element $\Gamma$ because all entries are "small".

I think that the same argument also works for $\Gamma_0$ and $\Gamma_1$

Answer by Igor Rivin for Growth of smallest closed geodesic in congruence subgroups?

2012-01-14T16:23:12Z

For more general results and extensive references, see Katz, Schaps, Vishne, Logarithmic growth of systole of arithmetic riemann surfaces along congruence subgroups (JDG, 2007)

Answer by Agol for Growth of smallest closed geodesic in congruence subgroups?

2012-01-14T16:23:17Z

For a hyperbolic element $A\in SL(2,\mathbb{Z})$, we have the length of the closed geodesic is given by $\ln[(tr^2(A)-2+\sqrt{tr^4(A)-4tr^2(A)})/4]$, and this is monotonic in $|tr(A)|$ for $|tr(A)|>2$. For $A\in \Gamma(N),\Gamma_1(N)$, we have $tr(A)\equiv 2 (\mod N)$, and $tr(A)\neq \pm 2$, so the smallest that $|tr(A)|$ can be is when $|tr(A)|=N-2$ (I suppose for $N>4$). This gives a lower bound on the shortest geodesic for $\Gamma_1(N)$. This is realized by the matrix

$$A=\left[\begin{matrix} 1-N & 1 \\ -N & 1 \end{matrix}\right]$$

For $\Gamma(N)$, one can obtain a better lower bound. Consider the matrix $$A=\left[\begin{matrix} 1+aN & bN \\ cN & 1+dN \end{matrix}\right]$$ with $\det(A)=1$. Then we have $(1+aN)(1+dN)-bcN^2=1$, so $a+d+(ad-bc)N=0$. This implies that $a+d\equiv 0(\mod N)$, so $tr(A)=2+(a+d)N \equiv 2 (\mod N^2)$. Thus, we get a lower bound of $|tr(A)|\geq N^2-2$. This is realized by the matrix

$$A=\left[\begin{matrix} 1-N^2 & N \\ -N & 1 \end{matrix}\right]$$

Edit: (I'm modifying the answer to address Vitali's question in the comments below).

For a matrix $A\in \Gamma_0(N)$, we have $$A=\left[\begin{matrix} a & b \\ cN & d \end{matrix}\right]$$ with $\det(A)=1$. Then $ad-bcN=1$ implies $ad \equiv 1(\mod N)$. We want to minimize $tr(A)=a+d$ subject to the constraint $ad\equiv 1(\mod N)$. Conversely, if $ad=1+kN$ for some $k$, then the matrix

$$A=\left[\begin{matrix} a & k \\ N & d \end{matrix}\right]\in \Gamma_0(N)$$ has trace $a+d$. So the minimal trace of a hyperbolic element in $\Gamma_0(N)$ is given by $\min \{ a+d >2 | ad\equiv 1 (\mod N)\}$.

Let's reformulate this problem. $ad\equiv 1(\mod N)$ is equivalent to the characteristic polynomial $\lambda^2-tr(A)\lambda+1\equiv(\lambda-a)(\lambda-d) (\mod N)$, i.e. the characteristic polynomial of $A$ reduces $(\mod N)$. So we want to minimize $\min \{ t > 2 | \lambda^2-t\lambda+1 \equiv 0 (\mod N), some \lambda \}$.

If $t$ is even, then we complete the square to get $(\lambda-t/2)^2 \equiv t^2/4-1 (\mod N)$, that is $t^2/4-1$ is a quadratic residue $(\mod N)$. If $t$ is odd, then $N$ must be odd if $\lambda^2-t\lambda+1\equiv 0 (\mod N)$, so multiplying by $4$, this is equivalent to $(2\lambda-t)^2\equiv t^2-4 (\mod N)$. Thus, the minimal trace is given by $\min \{ t>2 | t^2-4$ is a quadratic residue $(\mod N)$, $N$ odd, or $t^2/4-1$ is a quadratic residue $(\mod N)$, $N$ even $\}$.

Thus, since there are infinitely many $N$ such that $3^2-4=5$ is a quadratic residue $(\mod N)$ (e.g. the sequence $N=a^2-3a+1$), we have that the systole does not approach $\infty$.

Also, the systoles are unbounded from above. To see this, note that if $j$ is not a quadratic residue $(\mod N)$, then it is not a quadratic residue $(\mod kN)$ for any $k$. For $t>2$, choose $n(t)$ such that $t^2-4$ is not a quadratic residue $(\mod n(t))$. Then the number $N(T)=n(3)n(4)\cdots n(T)$ has the property that the minimal trace of $\Gamma_0(N(T))$ is bigger than $T$. In particular, the systole of $\Gamma_0(N!)$ $\to \infty$.