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Marc Palm
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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? Please also provide either a reference or a short argument.

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? Please also provide either a reference or a short argument.

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?

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Marc Palm
  • 11.2k
  • 2
  • 35
  • 92

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 $\Gamma$ grow$N \rightarrow \infty$?

Any suggestions? Please also provide either a reference or a short argument.

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 $\Gamma$ grow?

Any suggestions? Please also provide either a reference or a short argument.

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? Please also provide either a reference or a short argument.

Source Link
Marc Palm
  • 11.2k
  • 2
  • 35
  • 92
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