volume of complex hyperbolic manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:01:36Z http://mathoverflow.net/feeds/question/84379 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84379/volume-of-complex-hyperbolic-manifolds volume of complex hyperbolic manifolds emiliocba 2011-12-27T12:00:51Z 2011-12-27T21:53:00Z <p>I would like to know if there are in the literature explicit computations of the volume of complex hyperbolic manifolds. </p> <p>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 $$\Gamma = U(n,1) \cap M(n+1,\mathcal O).$$</p> <p>How much is the volume of the complex hyperbolic manifold $M_\Gamma = H_{\mathbb C}^n / \Gamma$?</p> <p>The answer for $\mathcal O=\mathbb Z[\sqrt{-1}]$ is enough for me.</p> <p>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.</p> <p>Thank you in advance.-.</p> http://mathoverflow.net/questions/84379/volume-of-complex-hyperbolic-manifolds/84380#84380 Answer by Igor Rivin for volume of complex hyperbolic manifolds Igor Rivin 2011-12-27T12:14:06Z 2011-12-27T12:14:06Z <p>As far as I know, there is no general way to compute this. For complex surfaces, see</p> <p>MR0653917 (84i:14025) Holzapfel, R.-P. Invariants of arithmetic ball quotient surfaces. Math. Nachr. 103 (1981), 117–153. </p> <p>and </p> <p>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. </p> http://mathoverflow.net/questions/84379/volume-of-complex-hyperbolic-manifolds/84416#84416 Answer by emiliocba for volume of complex hyperbolic manifolds emiliocba 2011-12-27T21:53:00Z 2011-12-27T21:53:00Z <p>This answer belongs to the second author of this <a href="http://arxiv.org/abs/1107.5281" rel="nofollow">paper</a> (=:[ES]). </p> <p>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.</p> <p>Now, we have a finite-sheeted covering $M_0 \to M:= H_\mathbb{C}^n / \Gamma$, so $$\mathrm{vol}(M)=[PU(n, 1; \mathcal O) : PSU(n, 1; \mathcal O)]\; \mathrm{vol}(M_0).$$</p> <p>An easy computation shows that the index $[PU(n, 1; \mathcal O) : PSU(n, 1; \mathcal O)]$ is equal to $$1 \quad\text{if n is even and d_{\mathcal O}&lt;-3},$$ $$1 \quad\text{if n is even, d_{\mathcal O}=-3 and n\not\equiv 2\pmod6},$$ $$3 \quad\text{if n is even, d_{\mathcal O}=-3 and n\equiv 2\pmod6},$$ $$2 \quad\text{if n is odd and d_{\mathcal O}&lt;-3},$$ $$2 \quad\text{if n is odd, d_{\mathcal O}=-3 and n\equiv1\pmod6,}$$ $$1 \quad\text{if n is odd, d_{\mathcal O}=-3 and n\equiv3\pmod6},$$ $$6 \quad\text{if n is odd, d_{\mathcal O}=-3 and n\equiv5\pmod6}.$$</p>