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Consider the complex 3-fold $SL(2,\mathbb C)/SL(2,\mathbb Z)$ (just for clarity: note that $SL(2,\mathbb Z)$ acts without stabilizers, so this is a complex manifold, not a complex orbifold).

Is $SL(2,\mathbb C)/SL(2,\mathbb Z)$ a quasi-projective variety?

The natural generalization of this question seems to be the following. Let $G$ be a semisimple linear algebraic group over $\mathbb Q$. Then $G(\mathbb Z)$ is well-defined up to taking a finite-index subgroup. Thus we can ask the same question: is the complex manifold $G(\mathbb C)/G(\mathbb Z)$ a quasi-projective variety?

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    $\begingroup$ One observation: this can be regarded as the moduli space of rank 2 lattices in $\mathbb{C}^2$, up to rescaling. But I'm not sure if this can be made naturally into a variety. $\endgroup$
    – Ian Agol
    Commented Aug 29, 2012 at 0:20

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No, the quotient is not quasi-projective. In the paper Invariant meromorphic functions on complex semisimple Lie groups by D. N. Ahiezer you can find the following result.

Theorem. Let $G$ be a connected semisimple linear algebraic group defined over the rationals and $\Gamma$ be a subgroup of $G(\mathbb > Q)$ which is Zariski dense in $G$. Then there are no invariant analytic hypersurfaces in $G$ invariant by the action of $\Gamma$. In particular, if the quotient $G/ \Gamma$ exists as a complex variety then every meromorphic function on it is constant.

This implies that the quotient is not quasi-projective and even more: it cannot be holomorphically embedded in any algebraic variety.

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  • $\begingroup$ Ahiezer writes "no", not "non". $\endgroup$
    – Wilberd van der Kallen
    Commented Aug 31, 2012 at 8:33
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The space is not an affine variety. A smooth affine variety of dimension $n$ is a Stein manifold, and thus must be homotopy equivalent to an $n$-dimensional CW complex (see Theorem 7.2 of Milnor's Morse Theory). However, $SL_2(\mathbb{C})/SL_2(\mathbb{Z})$ is not homotopy equivalent to a 3-complex.

Notice $SL_2(\mathbb{C})/SL_2(\mathbb{Z})$ is finitely covered by e.g. $SL_2(\mathbb{C})/\Gamma(4)$, which therefore has fundamental group $\Gamma(4)$ which is a free group. There is a fibration $S^3=SU(2) \to SL_2(\mathbb{C})/\Gamma(4) \to \mathbb{H}^3/\Gamma(4)$. Take any homologically non-trivial simple closed loop in $\mathbb{H}^3/\Gamma(4)$, then the induced $S^3$ bundle over $S^1$ is trivial since it is orientable, so we see an essential $S^3\times S^1\subset SL_2(\mathbb{C})/\Gamma(4)$ (in fact, being a bit more careful, there is a retract $SL_2(\mathbb{C})/\Gamma(4)\to S^3\times S^1$). Therefore the cohomological dimension of $SL_2(\mathbb{C})/\Gamma(4)$ is 4 (one also sees it is homotopy equivalent to a 4-complex since $\mathbb{H}^3/\Gamma(4)$ is homotopy equivalent to a wedge of loops), so it is not homotopy equivalent to a 3-complex, and thus neither is $SL_2(\mathbb{C})/SL_2(\mathbb{Z})$.

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If I understand correctly, it is proved by J. Winkelmann in On complex analytic compactifications of parallelizable manifolds (manuscripta math, 2000) that the answer is always NO.

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    $\begingroup$ This paper seems to deal with $SL(2,\mathbb C)/\Gamma$ where $\Gamma$ is a discrete subgroup so that the quotient has finite volume. I'd guess that in my case $\Gamma=SL(2,\mathbb Z)$, the quotient has infinite volume. $\endgroup$
    – John Pardon
    Commented Aug 28, 2012 at 21:07
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    $\begingroup$ @unknown: The quotient is certainly not compact, but does have finite volume relative to a natural measure. $\endgroup$
    – Jim Humphreys
    Commented Aug 28, 2012 at 23:27
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    $\begingroup$ Unless by ${\mathbb Z}$ you mean Gaussian integers, the quotient obviously has infinite volume (note that you are using complex, not real, coefficients). The reason is that the quotient $H^3/SL(2,Z)$ has infinite volume as the limit set of $SL(2,Z)$ is a circle, not the entire 2-sphere. $\endgroup$
    – Misha
    Commented Aug 29, 2012 at 15:20
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    $\begingroup$ Another argument that $SL(2,\mathbb C)/SL(2,\mathbb Z)$ has infinite volume is the following. There is an infinite covering map $SL(2,\mathbb C)/SL(2,\mathbb Z)\to SL(2,\mathbb C)/SL(2,\mathbb Z[i])$, so the former must have infinite volume. $\endgroup$
    – John Pardon
    Commented Aug 29, 2012 at 22:22
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    $\begingroup$ Also, Winkelmann proves nonexistence of G-equivariant complex-analytical compactification for quotients by lattices. He conjectures that there is no analytical compactification (equivariant or not), but he does not prove it. Also, for some discrete (infinite covolume) subgroups, compactification of course exists, just take trivial subgroup and compactify $SL(2,C)$ as a subvariety in the space of all 2-by-2 matrices. Thus, one has to use something specific about the modular group to disprove existence of a compactification. If may be easier to consider a cocompact Fuchsian subgroup first. $\endgroup$
    – Misha
    Commented Aug 30, 2012 at 5:37

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