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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$One can compute the (group cohomological) Euler characteristic of $\SL_2(\mathbb{Z})$ via $$ \chi(\SL_2(\mathbb{Z})) = \chi(\mathbb{Z}/2) \cdot \chi(\PSL_2(\mathbb{Z})) = \frac{1}{2}\cdot \left(\frac{1}{2} + \frac{1}{3} - 1\right) = -\frac{1}{12} = \zeta(-1) $$ (since $ \PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2 * \mathbb{Z}/3 $). Alternatively it follows as $\SL_2(\mathbb{Z})$ has $F_2$ as an index-$12$ subgroup.

I was wondering whether the connection with $\zeta$ is coincidental.

The only (far fetched) connection I could come up with is that the (representation theoretic) zeta function of $\SL_2(\mathbb{C})$ is the usual $\zeta$, as its irreducible representations have dimensions $1, 2, 3, \dotsc$. (Here $\zeta_G(s) = \sum_{V\in \operatorname{Irr}(G)} \dim(V)^{-s}$.)

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    $\begingroup$ This is certainly no coincidence, it is a theorem by Harder, which associates the zeta function with the Euler characteristic of various arithmetic groups. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 8, 2022 at 21:16
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    $\begingroup$ The relevant result is proved in Harder, G., A Gauss-Bonnet formula for discrete arithmetically defined groups, Ann. Sci. Éc. Norm. Supér. (4) 4, 409-455 (1971). You can also find it expanded on in Chapter IX.8 of Brown's "Cohomology of Groups". $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 8, 2022 at 21:20
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    $\begingroup$ Somewhat roundabout, but: let Y(1) = SO(2)\SL_2(R)/SL_2(Z) = H/SL_2(Z) where H = upper-half plane. Then vol(Y(1)) * vol(SO(2)) = vol(SL_2(R)/SL_2(Z)). A very special case of the Tamagawa number formula tells us that vol(SL_2(R)/SL_2(Z)) = prod_p (1-p^{-2})^{-1} = zeta(2). Using the SL_2(Z)-action on H, one can compute the orbifold Euler characteristic χ(Y(1)); using Gauss-Bonnet, one sees that vol(Y(1)) = pi/12. So we conclude that zeta(2) = pi^2/6, as expected [contd] $\endgroup$
    – skd
    Commented Jan 8, 2022 at 21:21
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    $\begingroup$ [contd] Now using the functional equation for zeta(s), we see zeta(-1) = 1/12. (Replacing Y(1) with the moduli space of abelian varieties of dim g, you can similarly inductively compute zeta(1-2g).) $\endgroup$
    – skd
    Commented Jan 8, 2022 at 21:21
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    $\begingroup$ Can either of you expand the comments into an answer? $\endgroup$
    – Daniel Loughran
    Commented Jan 8, 2022 at 21:48

1 Answer 1

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(Expanding my comment into an answer)

It is not a coincidence. Relating the Euler characteristic of certain arithmetic groups to the Zeta function is a theorem due to Harder [1] from 1971. It is expanded on in Brown's "Cohomology of Groups", Chapter IX.8.

Taken from Gruenberg's AMS review of Brown's book is the following overview of the idea:

Let $G$ be an algebraic subgroup of $\operatorname{GL}_n$ defined over $\mathbb{Q}$ and $\Gamma$ an arithmetic subgroup of $G(\mathbb{Q})$. Then $\Gamma$ is a discrete subgroup of the Lie group $G(\mathbb{R})$. If $K$ is a maximal compact subgroup of $G(\mathbb{R})$, then $X = G(\mathbb{R})/K$ is diffeomorphic to a euclidean space of dimension say $d$.
[...]
Number theory enters through the work of Harder (1971). The Gauss-Bonnet measure on $X$ lifts to a unique invariant measure $\mu$ on $G(\mathbb{R})$. Harder proved the deep theorem that $\chi(\Gamma) = \mu(G(\mathbb{R})/\Gamma)$. This leads to an explicit fromula [sic.] for $\chi(G(\mathbb{Z}))$ in terms of values of the zeta-function. For example, $\chi(\operatorname{SL}_2(\mathbb{Z})) = \zeta(-1)$ and since $\zeta(-1) = -1/12$, this gives a third way of arriving at the Euler characteristic of $\operatorname{SL}_2(\mathbb{Z})$.

Edit: One can find more values using this method, of course. Some are given in Brown's book (p. 255-256). For example, we have $$ \chi(\operatorname{SL}_n(\mathbb{Z})) = \prod_{k=2}^n \zeta(1-k) $$ and $$ \chi(\operatorname{Sp}_{2n}(\mathbb{Z})) = \prod_{k=1}^n \zeta(1-2k). $$ Thus for example, we find $\chi(\operatorname{SL}_n(\mathbb{Z})) = 0$ for $n \geq 3$ and $\chi(\operatorname{Sp}_{4}(\mathbb{Z})) = \zeta(-1)\zeta(-3) = -\frac{1}{1440}$.

[1] Harder, G., A Gauss-Bonnet formula for discrete arithmetically defined groups, Ann. Sci. Éc. Norm. Supér. (4) 4, 409-455 (1971). ZBL0232.20088.

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    $\begingroup$ I think zeta $\zeta$ specifically enters via Langlands computation of the integral of the Euler-Poincare form $\omega$ over the fundamental domain, but that is a paper i never could understand. "R. P. LANGLANDS, The volume of the fundamental domain for some arithmetical subgroups of Cheualley groups (Proc. of Symp. Math., Amer. Math. Soc., Providence, 1966. p. 143-148)." $\endgroup$
    – JHM
    Commented Jan 9, 2022 at 1:12
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    $\begingroup$ @JHM Langlands' method extends to Chevalley groups; the Gauss-Bonnet measure $\mu$ on $G(\mathbb{R})/ K$ (as above) is explicitly calculated by Harder in the case of Chevalley groups $G$. Indeed, in this case $\mu = c\mu_a$ where $\mu_a$ is the arithmetic measure on $G(\mathbb{R})$, for some scalar $c$ (which is computed by Harder). Since $\mu_a(G(\mathbb{R})/G(\mathbb{Z}))$ is expressible using $\zeta$, this yields the connection in the cases considered by Langlands (and circumvents his to-me-seemingly more complicated methods). Perhaps the two papers are worth reading simultaneously? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 9, 2022 at 13:17

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