I would like to know if there are in the literature explicit computations of the volume of complex hyperbolic manifolds. 


More precisely, let $\mathcal O$ be an imaginary quadratic number field, and let $\Gamma$ be the set of "integer" elements in $U(n,1)$, that is
$$
\Gamma = U(n,1) \cap M(n+1,\mathcal O).
$$


How much is the volume of the complex hyperbolic manifold $M\_\Gamma = H\_{\mathbb C}^n / \Gamma$?


The answer for $\mathcal O=\mathbb Z[\sqrt{-1}]$ is enough for me.


It's known that 
$$
\mathrm{vol}(M\_\Gamma) = \frac{(-\pi)^n 2^{2n}}{(n+1)!}\; \chi(M\_\Gamma),
$$
where $\chi(M\_\Gamma)$ denotes the Euler characteristic of $M\_\Gamma$, but I couldn't find computed the term $\chi(M\_\Gamma)$ in the literature.


Thank you in advance.-.



volume of complex hyperbolic manifolds
This answer belongs to the second author of this [paper](http://arxiv.org/abs/1107.5281) (=:[ES]). 


First, let $\Gamma\_0 = SU(n, 1) \cap M(n + 1, \mathcal O)$ and $M\_0 = H\_\mathbb{C}^n / \Gamma\_0$. By (1) and (28) in [ES], we obtain that
$$
\mathrm{vol}(M\_0) = \frac{(4 \pi)^n}{(n + 1)!} |d\_{\mathcal O}|^s \left( \prod\_{j = 1}^n \frac{j!}{(2 \pi)^{j + 1}} \right) \zeta(2) L\_{\mathcal O}(3) \zeta(4) L\_{\mathcal O}(5) \cdots F(n + 1),
$$
where $d\_{\mathcal O}$ is the discriminant of $\mathcal O$, $s$ is $\frac{n (n + 3)}{4}$ when $n$ is even and $\frac{(n - 1)(n - 2)}{4}$ when $n$ is odd, and $F(n + 1)$ is $\zeta(n + 1)$ when $n$ is odd and $L\_{\mathcal O}(n + 1)$ when $n$ is even.


Now, we have a finite-sheeted covering $M\_0 \to M:= H\_\mathbb{C}^n / \Gamma$, so 
$$
\mathrm{vol}(M)=[PU(n, 1; \mathcal O) : PSU(n, 1; \mathcal O)]\; \mathrm{vol}(M\_0).
$$


An easy computation shows that the index $[PU(n, 1; \mathcal O) : PSU(n, 1; \mathcal O)]$ is equal to 
$$
1 \quad\text{if $n$ is even and $d\_{\mathcal O}<-3$},
$$
$$
1 \quad\text{if $n$ is even, $d\_{\mathcal O}=-3$ and $n\not\equiv 2\pmod6$},
$$
$$
3 \quad\text{if $n$ is even, $d\_{\mathcal O}=-3$ and $n\equiv 2\pmod6$},
$$
$$
2 \quad\text{if $n$ is odd and $d\_{\mathcal O}<-3$},
$$
$$
2 \quad\text{if $n$ is odd, $d\_{\mathcal O}=-3$ and $n\equiv1\pmod6$,}
$$
$$
1 \quad\text{if $n$ is odd, $d\_{\mathcal O}=-3$ and $n\equiv3\pmod6$},
$$
$$
6 \quad\text{if $n$ is odd, $d\_{\mathcal O}=-3$ and $n\equiv5\pmod6$}.
$$


