Timeline for Simple Tamagawa number calculations
Current License: CC BY-SA 3.0
12 events
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| Nov 3, 2012 at 18:38 | comment | added | Asaf | To just compute the volume of the homogeneous space (wrt the measure induced from the regular differential form etc), after one say constructed the usual Dirichlet domain, just use some hyperbolic geometry, for example the Gauss-Bonnet formula. | |
| Nov 3, 2012 at 18:08 | history | edited | Jonah Sinick | CC BY-SA 3.0 |
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| Nov 3, 2012 at 17:06 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 3, 2012 at 16:46 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 3, 2012 at 16:06 | comment | added | GH from MO | @Jonah: You can prove $\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\zeta(2)$ directly, via Eisenstein series. See for example Section 1.6 in Goldfeld: Automorphic forms and L-functions for the group GL(n,R). Note, however, that Goldfeld uses a different invariant measure, hence his answer is also different (but it contains $\zeta(2)$). | |
| Nov 3, 2012 at 16:03 | comment | added | GH from MO | Chandan: Alternately, and this is what the OP probably had in mind, there might exist a direct proof that $\mathrm{vol}(\mathrm{SL}_2(\mathbb{A})/\mathrm{SL}_2(\mathbb{Q}))=1$. This fact combined with $\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\pi^2/6$ and $\mathrm{vol}(\mathrm{SL}_2(\mathbb{Z}_p))=1-p^{-2}$ would prove that $\zeta(2)=\pi^2/6$. | |
| Nov 3, 2012 at 15:58 | comment | added | Jonah Sinick | @ GH - yes. @ Chandan - doesn't the integral that you referenced work out to being (pi)^2/6? How do you get zeta(2) directly? | |
| Nov 3, 2012 at 15:57 | comment | added | GH from MO | Chandan: Fair enough, for the proof that the adelic volume equals 1, it suffices to show that $V:=\mathrm{vol}(\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z}))=\zeta(2)$ for the measure above. It is possible to do this calculation by Eisenstein series. But if we want to explain Euler's formula along these lines, then we also need to show, by a more direct calculation as in Platonov-Rapinchuk, that $V=\pi^2/6$. In the end we can forget about adeles: the proof that the OP is looking for boils down to calculating $V$ in two ways: $V=\zeta(2)$ via Eisenstein series, and $V=\pi^2/6$ directly. | |
| Nov 3, 2012 at 15:35 | comment | added | Chandan Singh Dalawat | Where is the fact that $\zeta(2)=\pi^2/6$ being used ? All we seem to be using is that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$ and that of $$ \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots $$ is $\zeta(2)^{−1}$. | |
| Nov 3, 2012 at 15:11 | history | edited | Chandan Singh Dalawat | CC BY-SA 3.0 |
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| Nov 3, 2012 at 14:59 | comment | added | GH from MO | Of courses none of the responses so far answer the original question: why is the adelic volume equal to 1, without assuming that $\zeta(2)=\pi^2/6$. | |
| Nov 3, 2012 at 14:41 | history | answered | GH from MO | CC BY-SA 3.0 |