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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.-.

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Googling "volume + complex hyperbolic manifold" may help; it will direct you e.g. to papers by John Parker: projecteuclid.org/… or to papers by Saar Hersonsky and Frederic Paulin: math.uga.edu/~saarh/Papers/VolCom.pdf –  Alain Valette Dec 27 '11 at 12:09
    
I did it, but I couldn't find the explicit answer, including in these papers. –  emiliocba Dec 27 '11 at 21:54
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2 Answers

As far as I know, there is no general way to compute this. For complex surfaces, see

MR0653917 (84i:14025) Holzapfel, R.-P. Invariants of arithmetic ball quotient surfaces. Math. Nachr. 103 (1981), 117–153.

and

R. Langlands Langlands, R. P. The volume of the fundamental domain for some arithmetical subgroups of Chevalley groups. 1966 Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965) pp. 143–148 Amer. Math. Soc., Providence, R.I.

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I will consider the Langlands' paper for a general case. Thank you.-. –  emiliocba Dec 27 '11 at 21:57
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up vote 2 down vote accepted

This answer belongs to the second author of this paper (=:[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 $$ 1 \quad\text{if $n$ is even and $d_{\mathcal O}<-3$}, $$ $$ 1 \quad\text{if $n$ is even, $d_{\mathcal O}=-3$ and $n\not\equiv 2\pmod6$}, $$ $$ 3 \quad\text{if $n$ is even, $d_{\mathcal O}=-3$ and $n\equiv 2\pmod6$}, $$ $$ 2 \quad\text{if $n$ is odd and $d_{\mathcal O}<-3$}, $$ $$ 2 \quad\text{if $n$ is odd, $d_{\mathcal O}=-3$ and $n\equiv1\pmod6$,} $$ $$ 1 \quad\text{if $n$ is odd, $d_{\mathcal O}=-3$ and $n\equiv3\pmod6$}, $$ $$ 6 \quad\text{if $n$ is odd, $d_{\mathcal O}=-3$ and $n\equiv5\pmod6$}. $$

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