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$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $X(N)$, 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.

$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ 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.

$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $X(N)$, 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 $\Gamma(5)$. 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.

Edit: 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.

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 $11$.

$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $X(N)$, 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.

$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $X(N)$, 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 $\Gamma(5)$. 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.

Edit: 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.

$SO(1)\backslash SL(2,\mathbb R)/\Gamma(N)$ is the modular curve $X(N)$, 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 $\Gamma(5)$. 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.

Edit: 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.

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 $11$.