Let $K$ be a number field and consider the Arithmentic complex $\Gamma_{Ar}(1)^\bullet$ be defined by

$$\begin{array} A\Bbb R^{r_1+r_2} & \stackrel{\Sigma}{\longrightarrow} & \Bbb R \\ \uparrow{l(\cdot)=\Pi\log |\sigma_i(\cdot)|} & & \uparrow{-\Sigma \log|\mathfrak p_i|^{-z_i}} \\ K^* & \stackrel{\operatorname{val}}{\longrightarrow} & \oplus_\mathfrak p\Bbb Z \end{array} $$

The left arrow going up is just the same map that appears in the proof of the unit theorem. The lower arrow takes the valuation in all non-archimedian places and the upper arrow is the sum of the coordinates. Finally the right arrow is the negative of the sum of the logarithm of the norm of the primes to the power equal to the negative of coordinate in $\oplus_\mathfrak p\Bbb Z$. For example, for $K=\Bbb Q$, an element $(1,2,-3,0,0,0,0,...)\in\oplus_\mathfrak p\Bbb Z$ would be mapped to $-(log(2) +log(3^2)+log(5^{-3}))$.

This diagram anticommutes and gives us a complex

$$K^*\to\Bbb R^{r_1+r_2} \times \oplus_\mathfrak p\Bbb Z \to \Bbb R$$

Its second cohomology is zero, because the second map is surjective, and its zeroth cohomology is the group of units (by the unit theorem).

On the other hand, $H^1(\Gamma_{Ar}(1))=\frac {\Bbb R^{r_1+r_2-1}}{l(\mathcal O_K^{\,*})}\times Cl_K$ because we can quotient the $\oplus_\mathfrak p\Bbb Z$ factor by $K^*$ to obtain $Cl_K$ and then we notice that to which each element $i$ of the ideal class group corresponds a hyperplane in $\Bbb R^{r_1+r_2}$ such that $\Bbb R^{r_1+r_2}\times i \mapsto 0$. Finally, remark that the kernel of this map is $O_K^{\,*}$.

Therefore, we get that $\operatorname{vol}(H^\bullet(\Gamma_{Ar}(1))):=\frac{\det H^1}{\det H^0 \otimes \det H^2}=\frac{h_K\cdot R}{w_K}$, where $R$ is the regulator. Since the volume of a complex (as just defined) is the same as the volume of its cohomology, we get that $\frac{h_K\cdot R}{w_K}$ is also the volume of the arithmetic complex.

Now the expression $\frac{\det H^1}{\det H^0 \otimes \det H^2}$ is very akin to $\mu \left(K^* \backslash \Bbb A^{(1)}_K\right)$ since $\Bbb A_K^{(1)}$ the kernel of the map $\Gamma_{Ar}(1)^1\to\Gamma_{Ar}(1)^2$. What I am after is how we get the factors $2^{r_1+r_2}$ in the following

$$\mu \left(K^* \backslash \Bbb A^{(1)}_K\right) = 2^{r_1+r_2}\frac {h_K\cdot R}{w_K}$$

Well, that's what I should get from playing around with equations below, although in my notes I have that it should be $\mu \left(K^* \backslash \Bbb A^{(1)}_K\right) = 2^{r_1}(2\pi)^{r_2}\frac {h_K\cdot R}{w_k\sqrt{|D_K|}}$.

Other relevant formulae: $$\begin{aligned}\xi_K(s) &= \zeta_K(s) |D_K|^{s/2}\left(\pi^{-s/2}\Gamma(s/2)\right)^{r_1}\left(2(2\pi)^{-s}\Gamma(s)\right)^{r_2}\\ &=\Psi(s)+\Psi(1-s)+\mu\left(K^* \backslash \Bbb A^{(1)}_K\right)\left(\frac{1}{s-1}-\frac 1 s\right)\end{aligned}$$ where $\Psi$ is analytic in $s$.


1 Answer 1


Let $U:=\prod_{\mathfrak p\in \Sigma_{K,0}} \mathcal O_\mathfrak p$ be the product of the valuation rings for all non-archimedean valuations. Then $\Bbb A^*/U = \Bbb R_*^{r_1} \times \Bbb C_*^{r_2}\times \bigoplus_\mathfrak p \Bbb Z$. The multiplicative measure on $\Bbb R$ and $\Bbb C$ is $\frac {dx}{x}$ and $\frac {1}{\pi}d\theta\frac{dr}{r}$. (The $\frac 1 \pi$ comes from the fact that I have to be consistent with the measure I used on $\Bbb C_*$ when I computed $\Gamma_\Bbb C=\int \phi_\Bbb C(t) ||t||^sd\mu(t)$.

Now consider the map $\Psi: \Bbb A^{(1)}_K \to \Gamma_{Ar}(1)^1 = \Bbb R^{r_1+r_2} \times \bigoplus_\mathfrak p \Bbb Z$, which is just $\log |x_i|$ in the first $r_1+r_2$ arguments. The $\log$ converts the measure from $\frac {dr}{r}$ to the usual measure on $\Bbb R$, which is $dx$. But you fold the measure by two every time you take the absolute value. So you pick up the factor of $2^{r_1+r_2}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .