Is SL(2,C)/SL(2,Z) a quasi-projective variety? 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?
 A: 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. 
A: 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  $SL_2(\mathbb{C})/SL_2(\mathbb{Z})$. 
A: 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.
