In this question it is noted that $SL(2, \mathbb{R})/SL(2, \mathbb{Z})$ is homeomorphic to the trefoil complement in $S^3.$ Is there a similarly nice interpretation of $SL(2, \mathbb{R})/\Gamma(N)$ for various $N?$
-
1$\begingroup$ Igor, for large $N$, the Seifert manifold in question will have high genus base and would not embed in the 3-sphere. Thus, it is unclear what nice interpretation would even mean. You can say that it is a Seifert link complement in a compact Seifert manifold, but this is not much of a description. $\endgroup$– MishaCommented Apr 13, 2012 at 21:03
1 Answer
$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $Y(N)$, which is a Riemann surface of a certain genus with a certain number of punctures. The action on the left is free, so $SL(2,\mathbb R)/SL(2,\mathbb Z)$ is a circle bundle on a punctured Riemann surface. All such bundles are trivial, so it is a circle cross a punctured Riemann surface.
The circle cross the sphere with $k$ points is a link complement, because the circle cross a disc is the unknot complement and removing additional points removes circles. This gets you $\Gamma(2)$ through $\Gamma(5)$.
Specifically, $\Gamma(2)$ gives a chain of three loops, $\Gamma(3)$ gives 3 loops each linked once to a central loop, $\Gamma(4)$ has 5 loops around 1, and $\Gamma(5)$ has 11.
The quotient by $\Gamma(6)$ is the complement of twelve circles in $S^1 \times S^1 \times S^1$, and $\Gamma(7)$ and above are quite strange.
If you want the tangent bundle to an orbifold rather than a manifold, you need to choose one of the less well-behaved subgroups, probably $\Gamma_0(N)$. In that case I do not know.
-
1$\begingroup$ I'm not sure if this convention is universal, but I have seen $Y(N)$ used to refer to the non-compact modular curve, and $X(N)$ to refer to the compactified curve that parametrizes generalized elliptic curves with level structure. $\endgroup$– S. Carnahan ♦Commented Apr 14, 2012 at 11:27
-
$\begingroup$ Even if it's not universal, it seems like a good convention. I'll change it. $\endgroup$ Commented Apr 14, 2012 at 18:39