Growth of smallest closed geodesic in congruence subgroups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:44:33Z http://mathoverflow.net/feeds/question/85651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85651/growth-of-smallest-closed-geodesic-in-congruence-subgroups Growth of smallest closed geodesic in congruence subgroups? Marc Palm 2012-01-14T11:43:29Z 2012-01-15T16:57:56Z <p>Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$.</p> <p>How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$?</p> <p>Any suggestions?</p> http://mathoverflow.net/questions/85651/growth-of-smallest-closed-geodesic-in-congruence-subgroups/85655#85655 Answer by S. Carnahan for Growth of smallest closed geodesic in congruence subgroups? S. Carnahan 2012-01-14T12:44:25Z 2012-01-14T12:44:25Z <p>I don't have an answer, but here is a suggestion. From the environs of exercise 20 in section 3.7 of Terras's <em>Harmonic analysis on symmetric spaces I</em>, 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 <code>$\left(\begin{smallmatrix} a &amp; 0 \\ 0 &amp; 1/a \end{smallmatrix} \right)$</code>. Now you just need to find how norms of hyperbolic elements grow with level.</p> http://mathoverflow.net/questions/85651/growth-of-smallest-closed-geodesic-in-congruence-subgroups/85656#85656 Answer by kassabov for Growth of smallest closed geodesic in congruence subgroups? kassabov 2012-01-14T12:45:15Z 2012-01-14T12:45:15Z <p>Let $\Gamma$ be the congruenece group $\ker SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N\mathbb{Z})$.<br> 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".</p> <p>I think that the same argument also works for $\Gamma_0$ and $\Gamma_1$</p> http://mathoverflow.net/questions/85651/growth-of-smallest-closed-geodesic-in-congruence-subgroups/85673#85673 Answer by Igor Rivin for Growth of smallest closed geodesic in congruence subgroups? Igor Rivin 2012-01-14T16:23:12Z 2012-01-14T16:23:12Z <p>For more general results and extensive references, see <a href="http://u.math.biu.ac.il/~vishne/publications/JDG-76-3-399-422.pdf" rel="nofollow">Katz, Schaps, Vishne, Logarithmic growth of systole of arithmetic riemann surfaces along congruence subgroups (JDG, 2007)</a></p> http://mathoverflow.net/questions/85651/growth-of-smallest-closed-geodesic-in-congruence-subgroups/85674#85674 Answer by Agol for Growth of smallest closed geodesic in congruence subgroups? Agol 2012-01-14T16:23:17Z 2012-01-15T16:45:47Z <p>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 </p> <p><code>$$A=\left[\begin{matrix} 1-N &amp; 1 \\ -N &amp; 1 \end{matrix}\right]$$</code></p> <p>For $\Gamma(N)$, one can obtain a better lower bound. Consider the matrix <code>$$A=\left[\begin{matrix} 1+aN &amp; bN \\ cN &amp; 1+dN \end{matrix}\right]$$</code> 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 </p> <p><code>$$A=\left[\begin{matrix} 1-N^2 &amp; N \\ -N &amp; 1 \end{matrix}\right]$$</code></p> <p><strong>Edit:</strong> (I'm modifying the answer to address Vitali's question in the comments below).</p> <p>For a matrix $A\in \Gamma_0(N)$, we have <code>$$A=\left[\begin{matrix} a &amp; b \\ cN &amp; d \end{matrix}\right]$$</code> 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 </p> <p><code>$$A=\left[\begin{matrix} a &amp; k \\ N &amp; d \end{matrix}\right]\in \Gamma_0(N)$$</code> has trace $a+d$. So the minimal trace of a hyperbolic element in $\Gamma_0(N)$ is given by $\min \{ a+d >2 | ad\equiv 1 (\mod N)\}$. </p> <p>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 $\min \{ t > 2 | \lambda^2-t\lambda+1 \equiv 0 (\mod N), some \lambda \}$. </p> <p>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 $\}$.</p> <p>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 $\infty$. </p> <p>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 $\Gamma_0(N!)$ $\to \infty$. </p>