# Simple Tamagawa number calculations

As is well known, Euler proved the Basel identity $\displaystyle\sum\limits_{i=0}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}$. By far the most illuminating explanation of this fact that I've seen is as follows:

By another theorem of Euler we can rewrite the identity as $\displaystyle \frac{{\pi}^2}{6}\cdot \prod_{p} \left(1 - \frac{1}{p^2}\right) = 1$, where $p$ ranges over all primes. The first term is the normalized volume of

$\displaystyle SL(2,\mathbb{R})/SL(2, \mathbb{Z})$

and the term corresponding to $p$ in the product is the normalized volume of

$SL(2, \mathbb{Z}_p)$.

With these replacements, the right hand side can be written as the normalized volume of

$\displaystyle SL(2,\mathbb{\mathbb{A}_{\mathbb{Q}}})/SL(2, \mathbb{Q})$.

where $\mathbb{A}_{\mathbb{Q}}$ denotes the adeles of $\mathbb{Q}$ But this last volume is equal to 1: this is a special case of the Weil Conjecture on Tamagawa Numbers.

I've been fascinated by this result for years, but have never been able to understand a proof of it (from the adelic perspective) even in cases as simple the one above. In this way, my question contrasts with that of Ben Weiland who asked about the theorem in more general settings.

I tried reading André Weil "Adeles and algebraic groups" with a view toward learning a proof but found the book unintelliglbe. I gathered that the idea of the proof is to show that the nonzero volume of some object is equal to the Tamagawa number multiplied by the original volume but beyond that understood nothing. My impression is that Marie-France Vigneras' book titled Arithmetique des algebres de quaternions has this material, but I don't read French.

What are some lucid sources that you would recommend for learning proofs of the some of the first few cases (including the case above) of the Weil Tamagawa Number conjecture from an adelic perspective?

1 I learned this material from Yuri Manin's "Reflections on Arithmetical Physics" and Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds"

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I'm afraid that if you want to be an algebraic number theorist, you have to learn to read (mathematical) French. It's inescapable. – David Loeffler Nov 3 '12 at 9:06
@Jonah: Note that your question contains errors, e.g. the volume of $SL(2,\mathbb{\mathbb{Q}_p})/SL(2,\mathbb{Z}_p)$ is infinite. – GH from MO Nov 3 '12 at 16:22
@ GH - thanks, corrected (I think, unless there are other errors!) BTW, who are you? You can email me - I'd keep your identity confidential if you'd like. – Jonah Sinick Nov 3 '12 at 16:43
@Jonah: I extended my response, have a look. I dropped you an email as well. – GH from MO Nov 3 '12 at 16:56

It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this space can be identified with $$(\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})) \times \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots$$ so the volume in question is the product of the volumes of the various factors. The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of $\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$. Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.

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The book of Colmez seems a good choice for learning mathematical French, too (besides it being a fantastic book!). It starts with a 200pages chapter reviewing a large part of the standard math notions. The chapter is even called Vocabulaire Mathématique (Mathematical vocabulary). Here is a table of content editions.polytechnique.fr/files/pdf/TDMD_1587_9.pdf – user9072 Nov 3 '12 at 12:48
Regarding your addendum: (1) You're reiterating what I wrote in my original question! I was asking for a proof that the Tamagawa number is 1 that doesn't rely on resolution into components. (2) the volume of SL(2,R)/SL(2,Z) can be computed by regarding: PSL( 2, R) is a fiber bundle over the upper half plane with fibers that are circles, computing the orbifold Euler characteristic of the modular domain in the upper half plane, and applying Gauss-Bonnet to its 24 fold genus 2 constant negative curvature cover. – Jonah Sinick Nov 3 '12 at 15:55
@Jonah: That sounds like a very complicated way to calculate the volume of SL(2,R)/SL(2,Z)! There are simpler ways as in the mentioned books by Platonov-Rapinchuk or Goldfeld. I can imagine that via adelic Eisenstein series one can prove directly that the Tamagawa number is 1. – GH from MO Nov 3 '12 at 16:19
@Jonah: In general, you should consider yourself fortunate whenever you can resolve a global question into local questions at each place. The first reflex upon encountering a global problem should be to solve it locally everywhere. I don't see what anyone can have against this reflex. – Chandan Singh Dalawat Nov 3 '12 at 16:40
@GH: All your comments and answers are very pertinent. Is it fair that only Jonah knows who you are ? – Chandan Singh Dalawat Nov 3 '12 at 17:14

In order to talk about the volume you need to fix the measure first. Weil observed that there is a canonical choice for the Haar measure on $G(\mathbb{A})$. For a connected $n$-dimensional semisimple algebraic group $G$ over a global field $K$, there is a nonzero $n$-dimensional rational differential $K$-form which is invariant under left-translation. The differential form induces a Haar measure on $G(K_v)$ for each completion $K_v$ of $K$, hence it determines a Haar measure on the restricted direct product $G(\mathbb{A})$ as well. The differential form is unique up to scaling by $K^\times$, hence the obtained Haar measure on $G(\mathbb{A})$ is unique. When $G$ is simply connected and semisimple, Weil conjectured that $\mathrm{vol}(G(\mathbb{A})/G(K))=1$ for this particular measure.

In the case of $G=\mathrm{SL}_2$ and $K=\mathbb{Q}$, the invariant differential form is given by $\frac{dx\wedge dy\wedge dz}{x}$, where $\begin{pmatrix} x & y \\ z & t \end{pmatrix}$ are the usual coordinates on the group. For this form one obtains the numbers mentioned by Chandan Singh Dalawat (and also by Colmez's Un autre monde est possible), hence a proof of Weil's conjecture for $G=\mathrm{SL}_2$ and $K=\mathbb{Q}$. For details I recommend Platonov-Rapinchuk: Algebraic groups and number theory (Academic Press, 1994). See especially Example 3 on pp. 166-167, the Example on pp. 222-223, and the Example on p. 262.

Added. The OP is looking for a direct (global) proof of $\mathrm{vol}(\mathrm{SL}_2(\mathbb{A})/\mathrm{SL}_2(\mathbb{R}))=1$. For the analogous (local) statement that $\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\zeta(2)$ there are nice accounts via Eisenstein series, see for example here. No doubt this proof can be modified to yield $\mathrm{vol}(\mathrm{SL}_2(\mathbb{A})/\mathrm{SL}_2(\mathbb{Q}))=1$ directly as well, but I don't know of any simple written account. Hopefully someone knows a good reference. In general, Langlands proved Weil's conjecture for simply connected Chevalley groups with the help of his general theory of Eisenstein series.

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Of courses none of the responses so far answer the original question: why is the adelic volume equal to 1, without assuming that $\zeta(2)=\pi^2/6$. – GH from MO Nov 3 '12 at 14:59
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 $\zeta(2)^{−1}$. – Chandan Singh Dalawat Nov 3 '12 at 15:35
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. – GH from MO Nov 3 '12 at 15:57
@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 $\zeta(2)$). – GH from MO Nov 3 '12 at 16:06
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. – Asaf Nov 3 '12 at 18:38

I have only read Paul Garrett's paper. I find the use of Poisson summation rather mysterious, but it seems to me that it applies mutatis mutandis to the adelic case.

Very short version: the generic orbit of $SL(V)$ on $V$ is open. If we can compute volumes for $V$ and the stabilizer $N$, then we can do so for $SL(V)$.

It seems to me that most of the argument works for pretty general locally compact ring $A$ containing a lattice $B$. I see three steps:

1. We must choose Haar measures. First, scale Haar measure on $A$ so that $\mathop{vol}(A/B)=1$; this is important for Poisson summation. Choose invariant differential forms defined over $\mathbb Z$ for the three groups; these are unique up to sign. The differential forms lift Haar measure from $A$ to $G(A)$. Because the product of the differential forms on $V$ and $N$ is that of $SL(V)$, so are the Haar measures multiplicative. This gives the best normalization for understanding the large group in terms of the small groups, but if you care about other measures (eg, the measure on hyperbolic space), you may need a correction factor. For number rings, often a power of the discriminant is needed.

2. We need to understand the orbits of $SL(V)(R)$ on $V(R)$. In particular, if $R$ is a field, the action is transitive away from the origin. For any ring, $SL(V)(R)$ is transitive on the Zariski open $(V-0)(R)$. This is not true for general quotients. For example, $PGL_2$ is the quotient variety of $SL_2$, but $SL_2(\mathbb Q)$ does not surject onto $PGL_2(\mathbb Q)$. Thus for any representation of $SL_2$ that factors through $PGL_2$, such as the adjoint representation, $SL_2(\mathbb Q)$ is not transitive on the $\mathbb Q$-points of the algebraic orbit.

3. We need to understand the other orbits. They reflect the ideals of $R$. There is a tradeoff between the cases $\mathbb Z\subset\mathbb R$ and $\mathbb Q\subset\mathbb A$, in that we can't have both rings be fields. However, the complement for the locally compact ring has measure zero and is unimportant, while the counting measure on the discrete ring means that every point counts. This is an advantage of the adelic case and explains its simpler answer.

Finally, we have the argument. We take a Schwartz class function $f:V(A)\to\mathbb R$, and compute $$\int_{G(A)/G(B)}\sum_{x\in V(B)}f(gx)\,dg$$ One way to compute it is to break $V(B)$ into pieces according to the orbits of $SL(V)$. The zero orbit contributes $\mathop{vol}(G(A)/G(B))f(0)$. In the PID case, the other orbits are all scaled versions of the open orbit. Summing the scaling factors yields the zeta function of $B$ evaluated at the dimension of $V$. If $N$ is the stabilizer of $v$, then using point (1), the contribution of the open orbit is $$\int_{G(A)/N(B)}f(gv)\,dg=\mathop{vol}(N(A)/N(B))\int_{(V-0)(A)}f(x)\,dx =\mathop{vol}(N(A)/N(B))\hat f(0)$$

Thus the original integral is $\mathop{vol}(G(A)/G(B))f(0)+\zeta \mathop{vol}(N(A)/N(B))\hat f(0)$. Using Poisson summation for $V(B)\subset V(A)$ allows us to interchange $f$ with its Fourier transform and conclude that $\mathop{vol}(G(A)/G(B))=\zeta \mathop{vol}(N(A)/N(B))$. In the adelic case, the factor of $\zeta$ is always $1$ and this shows by induction that $\tau(SL(V))=1$. But the argument also explains where the factor of $\zeta(s)$ comes from in the case of $\mathbb Z\subset \mathbb R$ and extends to other number rings.

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