Tamagawa numbers of CM tori]On Tamagawa numbers of CM tori

Pei-Xin Liang
(Liang) Department of Mathematics, National Taiwan Normal University

Taipei, Taiwan,

Hsin-Yi Yang
Fakultät für Mathematik,
Universität Duisburg-Essen,
45117 Essen, Germany

Chia-Fu Yu
(Yu) Institute of Mathematics, Academia Sinica and NCTS,
6F Astronomy Mathematics Building, No. 1, Roosevelt Rd. Sec. 4
Taipei, Taiwan, 10617

[2010]14K22, 11R29.

In this article we investigate the problem of computing Tamagawa numbers of CM tori, which play an important role in counting polarized CM abelian varieties, polarized abelian varieties with commutative endomorphism algebras over finite fields, and components of unitary Shimura varieties. Using Ono's formula and extending the methods of Li-Rüd and Guo-Sheu-Yu, we make a systematic study on Galois cohomological groups in question and compute the Tamagawa numbers of CM tori associated to various Galois CM fields. We also show that every (positive or negative) power of $2$ is the Tamagawa number of a CM tori provided that a generalized Landau conjecture holds true, and show an evidence of our result unconditionally.

§ INTRODUCTION

An algebraic torus $T$ over a number field $k$ is a connected linear algebraic group over $k$ such that its base change $T\otimes_k \bar k$ over the algebraic closure $\bar k$ of $k$ is isomorphic to a product of copies of the split multiplicative group $\Gmkbar$. In his two fundamental papers [12, 14] Takashi Ono investigated the arithmetic of algebraic tori. He introduced and explored the class number and Tamagawa number of $T$, which will be denoted by $h(T)$ and $\tau(T)$ respectively (also see Section 2 for the definitions). One arithmetic significant of these invariants is that
the class number $h(\G_{{\rm m},k})$ is equal to the class number $h_k$ of the number field $k$, and the analytic class number formula for $k$ can be reformulated by the simple statement
$\tau(\G_{{\rm m},k})=1$. Thus,  the class numbers of algebraic tori can be viewed as generalizations of class numbers of number fields, while Tamagawa numbers play a key role in the extension of analytic class number formulas.

Ono showed [14] that the Tamagawa number $\tau(T)$, can be calculated by the Galois cohomology groups $H^1(L/k, X(T))$ and  $\Sha^1(L/k,T)$, where $L$ is a splitting field of $T$, $X(T)$ is the group of characters of $T$ and $\Sha^1(L/k,T)$ is the Tate-Shafarevich group for $H^1(L/k, T)$. As an immediate corollary of Ono's formula, the Tamagawa number of any algebraic torus is a positive rational number. It is natural
to ask the precise range of Tamagawa numbers of all algebraic tori. For this, Ono constructed a $15$-dimensional algebraic torus over $\Q$ with Tamagawa number $1/4$, showing that the Tamagawa number of an algebraic torus can be a non-integral rational number.
This question was answered by S. Katayama [7]
who proved Ono's conjecture that every positive rational number can be realized as the Tamagawa number of an algebraic torus (and the ground field $k$ can be fixed and independent of the given rational number).
For other studies of class numbers and Tamagawa numbers of algebraic tori we refer to J.-M. Shyr <cit.>,
S. Katayama [8], M. Morishita [10], C. González-Avilés [3, 4] and M.-H. Tran [21].

In this article we are mainly concerned with the problem of computing the Tamagawa numbers of complex multiplication (CM) algebraic tori.
Achter, Altug and Gordon [1] computed the numbers of principally polarized abelian varieties in an isogeny class over finite fields
with commutative endomorphism algebra. They utilize the Langlands-Kottwitz counting method and express the number of abelian varieties in terms of discriminants and a product of certain local density factors.
In [5] J. Guo, N. Sheu and the third named author gave formulas for the class numbers and Tamagawa numbers of CM tori. As applications they gave explicit formulas for the numbers of CM abelian varieties with additional structures and certain conditions, and the numbers of certain polarized abelian varieties over finite fields with commutative endomorphism algebras. Using the class number formula for CM tori, they also computed the numbers of connected components of complex unitary Shimura varieties.
The most complicated term appearing in the class number formulas in [1] or in [5] is the Tamagawa number of a CM algebraic torus associated to the data. To make the counting formula explicitly, one needs to solve the problem of computing the Tamagawa number $\tau(T)$ of a CM torus. In the appendix of [1], W.-W. Li and Rüd have already obtained a few results of the values of $\tau(T)$. Our goal is to prove more cases of CM tori and to determine the range of the values of Tamagawa numbers of all CM tori.
With a similar goal but using different methods, T. Rüd obtains several results along this direction. He makes an algorithm, among others, for giving precise lower bounds and determining possible Tamagawa numbers, and obtains the values $\tau(T)$ for CM tori of lower dimension.

We shall describe our results towards computing Tamagawa numbers for a class of algebraic tori which include CM tori. We then give more detailed results of CM tori. Let $k$ be a number field and $K:=\prod_{i=1}^rK_{i}$ be the product of finite field extensions $K_i$ of $k$. Let $E:=\prod_{i=1}^r E_i$, where each $E_i\subset K_i$ is a subextension of $K_i$. Denote by $T^K=\prod_i T^{K_i}$ and $T^E=\prod_i T^{E_i}$ the algebraic $k$-tori associated to the multiplicative groups of $K$ and $E$, respectively, and let $N_{K/E}=\prod_i N_{K_i/E_i}: T^K\to T^E$ be the norm map. We write $T^{K/E,1}$ for the kernel of $N_{K/E}$ and $T^{K/E,k}:=N_{K/E}^{-1}(\Gmk)$ for the preimage of the subtorus $\Gmk \embed T^E$ via the diagonal embedding. Let $L$ be the smallest splitting field of $T^{K/E,k}$ and let $G=\Gal(L/k)$. We let $\Lambda:=X(T^{K/E,k})$ and $\Lambda^1:=X(T^{K/E,1})$ be the character groups of $T^{K/E,k}$ and $T^{K/E,1}$, respectively. They fit in the following short exact sequence of $G$-lattices:

\begin{equation}
\label{eq:I.1}
\begin{CD}
0 @>>> \Z @>>> \Lambda @>>> \Lambda^1 @>>> 0.
\end{CD}
\end{equation}

Write $H_i:=\Gal(L/K_i)$,  $\wt N_i:=\Gal(L/E_i)$, and $N_i^{\rm ab}:=\wt N_i/D(\wt N_i)H_i$, where $D(\wt N_i)$ denotes the commutator group of $\wt N_i$.
If $K_i/E_i$ is Galois with group $N_i=\Gal(K_i/E_i)=\wt N_i/H_i$, then $N_i^{\rm ab}$ coincides with the abelianization of $N_i$.
When there is no confusion, for brevity we write $H^q(A)$ for $H^q(G,A)$ for any $G$-module $A$.

Let $\Ver_{G,N_i}:G\to N_i^{\rm ab}$ denote the transfer from $G$ into $N_i^{\rm ab}$; see Definition <ref>.
For any torsion abelian group $H$, the Pontryagin dual of $H$ is denoted by $H^\vee$. By Ono's formula (cf. (<ref>)) and
the Poitou-Tate duality <cit.>, we have
$\tau(T)=|H^1(G,X(T))|/|\Sha^2(G,X(T))|$, where $\Sha^{i}(G,X(T))$ is the $i$th Tate-Shafarevich group of $X( T)$.

Let $\scrD$ be the set of all decomposition groups of $G$.
Denote by

\[ r_{\scrD,A}^i: H^i(G,A) \to \bigoplus_{D\in \scrD} H^i(D,A)\]

the restriction map for each $G$-module $A$. Put

\begin{equation}
\label{eq:I.1.5}
%  \begin{split}
H^2(\Z)':=\{x\in H^2(G, \Z) : r^2_{\scrD,\Z} (x) \in {\rm Im} (\delta_\scrD) \},
%  \end{split}
\end{equation}

where $\delta_\scrD: \bigoplus_{D\in \scrD} H^1(D, \Lambda^1) \to    \bigoplus_{D\in \scrD} H^2(D, \Z)$ is the connecting homomorphism induced from (<ref>). The group $H^2(\Z)'$ plays a similar role of a Selmer group.

Let the notation be as above.

(1) There is a canonical isomorphism $H^1(G,\Lambda^1)\simeq \bigoplus_{i} N_{i}^{\rm ab,\vee}$.

(2) There is a canonical isomorphism

\begin{equation}
\label{eq:I.2}
H^1(G,\Lambda)\simeq \Ker\left (\sum \Ver_{G,N_i}^{\vee}: \bigoplus_i N_i^{\rm ab,\vee} \to G^{\rm ab,\vee} \right ),
\end{equation}

where $\Ver_{G,N_i}:G\to N_i^{\rm ab}$ is the transfer map.

(3)  Assume that $K_i/E_i$ is cyclic with Galois group $N_i$ for all $i$. Then there is an exact sequence of abelian groups

\begin{equation}
\label{eq:I.3}
\begin{CD}
0 @>>> H^1(\Lambda) @>>>  H^1(\Lambda^1) @>{\delta}>> H^2( \Z)' @>{\wh N}>> \Sha^2(\Lambda) @>{}>> 0.
\end{CD}
\end{equation}

In particular, we have
$\Sha^2(\Lambda)\simeq H^2(\Z)'/{\rm Im}(\delta)$ and

\begin{equation}
\label{eq:I.}
\tau(T^{K/E,k})=\frac{\prod_{i=1}^r |N_i|}{|H^2(\Z)'|}.
\end{equation}

(4) If we further assume that each subgroup $\wt N_i$ is normal in $G$, and

\[ \Ver_{G, \bfN}=(\Ver_{G, N_i})_i : G^{\rm ab} \to \prod_i N_i \text{ and }
\Ver_{D, \ol \bfD}=(\Ver_{D, \ol D_i})_i: D^{\rm ab}\to \prod_{i} \ol D_i \]

denote the corresponding transfer maps, where $\ol D_i$ is defined in (<ref>), respectively, then

\begin{equation}
\label{eq:I.6}
H^2(\Z)'=\{f \in G^{{\rm ab}, \vee}: f {\big |}_{D^{\rm ab}}
\in {\rm Im}(\Ver_{D, \ol \bfD}^\vee)\  \forall\, D\in \scrD \},
\end{equation}

\begin{equation}
\label{eq:I.7}
\Sha^2(\Lambda) \simeq \frac{\{f \in G^{{\rm ab}, \vee}: f {\big |}_{D^{\rm ab}} \in {\rm Im}(\Ver_{D, \ol \bfD}^\vee) \ \forall\, D\in \scrD \}}{{\rm Im} (\Ver^{\vee}_{G,\bfN})}.
\end{equation}

The assumption on the normality of $\wt N_i$ in Theorem <ref>(4) is used in the step of the description of $H^1(D, \Lambda^1)$ by Mackey's formula. This assumption is not really necessary but it simplifies our description of (<ref>) and (<ref>) significantly. Using Theorem <ref>(2), we give a different proof of a result of Li and Rüd <cit.> which does not reply on Kottwitz's isomorphism (cf. <cit.>).

Suppose the following conditions hold:
(a) for each $1\le i \le r$, the extension $K_i/k$ is Galois with group $G_i$, $N_i=\Gal(K_i/E_i)$ is cyclic and every decomposition group of $G_i$ is cyclic; and (b) $G\simeq G_1\times \dots \times G_r$. Then we have $\tau(T^{K/E,k})=\prod_{i=1}^r \tau(T^{K_i/E_i,k})$.

Now let $K=\prod_{i=1}^r K_i$ be a CM algebra and $E:=K^+$ the $\Q$-subalgebra fixed by the canonical involution $\iota$, and let $T=T^{K/E,\Q}$ be the associated CM torus over $\Q$. We call a pair of odd primes $(p,q)$ a Landau pair if $p=1+a^2$
and $q=1+pb^2$ for some integers $a$ and $b$.
Two Landau pairs $(p,q)$ and $(p',q')$ are said to be disjoint if $\gcd (pq,p'q')=1$.

If there exists infinitely many disjoint Landau pairs of primes, then for any integer $n$, there exists a CM torus $T$ over $\Q$ such that $\tau(T)=2^n$.

The condition in Theorem <ref> is satisfied if a generalized conjecture of Landau (cf. Conjecture <ref>) holds true. We also obtain Theorem <ref> unconditionally for $n>-7 \times 10^5$ by a simple computer search. Finally, we give a numerous of results for $\tau(T)$ when $K$ is a Galois CM field.

Suppose that $K$ is a Galois CM field with Galois group $G=\Gal(K/\Q)$. Let $g:=[K^+:\Q]$, $G^+=\Gal(K^+/\Q)$ and $T$ be the associated CM torus.

(1) If $K=\Q(\zeta_n)\neq \Q$ is the $n$th cyclotomic field with either $4|n$ or odd $n$, then

$(a)$ If $n$ is either a power of an odd prime $p$ or $n=4$, then $\tau(T) = 1$;

$(b)$ In other cases, we have $\tau(T) = 2$.

(2) If $G$ is abelian, then $\tau(T) \in \{1,2\}$.
Moreover, the following statements hold:

$(a)$ If $g$ is odd, then $\tau(T) = 1$.

$(b)$ If $g$ is even and the exact sequence

\[1 \rightarrow \langle\iota\rangle \rightarrow G \rightarrow G^{+} \rightarrow 1 \tag{$*$}\]

splits, then $\tau(T) = 2$.

(3) Suppose that the short exact sequence $(*)$ splits. Then $\tau(T)\in \{1,2\}$ and the following statements hold:

(a) When $g$ is odd, $\tau(T) = 1$;

(b) Suppose $g$ is even.

(i) $\tau(T) = 2$ if $g^{ab}:=|G^{+{\rm ab}}|$ is even;

(ii) If $g^{\rm ab}$ is odd, then there is a unique nonzero
element $\xi$ in the $2$-torsion subgroup $H^{2}(\Lambda)[2]$
of $H^{2}(\Lambda)$. Moreover, $\tau(T) = 1$ if and only if

its restriction $r_D(\xi) = 0$ in $H^2(D,\Lambda)$ for all $D\in \scrD$.

(4) Let $P$ and $Q$ be two odd non-square positive integers such that $P-1=a^2$ and $Q-1=Pb^2$
for some integers $a,b\in  \bbN$. Let $K:=\Q(\sqrt{\alpha})$ with $\alpha:=-(P+\sqrt{P})(Q+\sqrt{Q})$. Then
$K$ is a  Galois CM field with Galois group $Q_8$ and

\begin{equation}
\label{eq:I.8}
\tau(T)=\begin{cases}
1/2, & \text{if $\left ( \frac{P}{q} \right) =1$ for all prime  $q|Q$;} \\
2, & \text{otherwise}.
\end{cases}
\end{equation}

(5) Suppose the Galois extension $K/\Q$ has Galois group $D_n$ of order $2n$. Then $n$ is even and $\tau(T)=2$.

This article is organized as follows. Section <ref> includes preliminaries and background on Tamagawa numbers of algebraic tori, and some known results of those of CM tori due to Li-Rüd and Guo-Sheu-Yu. Section <ref> discusses transfer maps, their extensions and connection with class field theory. In Sections <ref> and <ref>
we compute the Galois cohomology groups of character groups of a class of algebraic tori $T^{K/E,k}$ and $T^{K/E,1}$. Section <ref> treats a numerous of Galois CM tori and in Section <ref> we determine the precise ranges of Tamagawa numbers of CM tori under a generalized Landau hypothesis.

Let $K/k$ be a finite extension and let $R_{K/k}$ denote the Weil restriction of scalars form $K$ to $k$, then we have the following  exact sequence of tori defined over $k$

1\longrightarrow R^{(1)}_{K/k}(\G_{{\rm m},K})\longrightarrow R_{K/k}(\G_{{\rm m},K})\longrightarrow \G_{{\rm m},k}\longrightarrow 1,

where $R^{(1)}_{K/k}(\G_{{\rm m},K})$ is the kernel of the norm map $N: R_{K/k}(\G_{{\rm m},K})\longrightarrow\G_{{\rm m},k}$. It is easy to see that $h(R_{K/k}(\G_{{\rm m},K}))$ and $h(\G_{{\rm m},k})$ coincide with the class numbers $h_K$ and $h_k$ of $K$ and $k$, respectively. In order to compute the class number $h(R^{(1)}_{K/k}(\G_{{\rm m},K}))$, Ono <cit.> introduced the arithmetic invariant

\[ E(K/k):= \frac{h_K}{h_k\cdot h(R^{(1)}_{K/k}(\G_{{\rm m},K}))} \]

and expressed it in terms of certain cohomological invariants when $K/k$ is Galois. In [8], S. Katayama proved a formula for $E(K/k)$ for any finite extension $K/k$. He also studied its dual arithmetic invariant $E'(K/k)$ and gave a similar formula. The latter gives a formula for the class number of the quotient torus $R_{K/k}(\G_{{\rm m},K})/\G_{{\rm m},k}$. The class numbers of general tori have been investigated by J.-M. Shyr <cit.>, M. Morishita [10], C. González-Avilés [3, 4] and M.-H. Tran [21].

§ PRELIMINARIES, BACKGROUND AND SOME KNOWN RESULTS

§.§ Class numbers and Tamagawa numbers of algebraic tori

The cardinality of a set $S$ will be denoted by $\abs{S}$ or $\# S$.
For any field $k$, let $\bar k$ be a fixed algebraic closure of $k$,
let $k^{\sep}$ be the separable closure of $k$ in $\bar k$ and denote by $\Gamma_k:=\Gal(k^{\sep}/k)$ the Galois group of $k$.
Let $\G_{{\rm m},k}:=\Spec k[X,X^{-1}]$ denote the

multiplicative group associated to $k$ with the usual multiplicative group law.

(1) A connected linear algebraic group $T$ over a field $k$ is said to be an algebraic torus over  $k$ if there exists a finite field extension
$K/k$ such that there exists an isomorphism
$ T \otimes_k K \simeq \GmK^d$ of algebraic groups over $K$ for some positive integer $d$. Then $d$ is
the dimension of $T$. If $T$ is an algebraic $k$-torus and $K$ is a field extension of $k$ that satisfies the  above property,
then $K$ is called a splitting field
of $T$. The smallest splitting field (which is unique and Galois, see below) is
called the minimal splitting field of $T$.

(2) For any algebraic torus $T$ over $k$, denote by $X(T):=\Hom_{\bar k}(T\otimes_{k} \bar k, \Gmkbar)$ the group of characters of $T$.
It is a finite free $\Z$-module of rank $d$ together with a continuous action of the Galois group $\Gamma_k$ of $k$.

It is shown in <cit.>
(also see [23] for other proofs) that every algebraic $k$-torus splits over a finite separable field extension $K/k$. The action of $\Gamma_k$ on $X(T)$ gives a continuous representation

\begin{equation}
\label{eq:rT}
r_T: \Gamma_k=\Gal(k^{\sep}/k) \to \Aut(X(T))
\end{equation}

which factors through a faithful representation of a finite quotient $\Gal(L/k)$ of $\Gamma_k$.
Here $L$ is the fixed field of the kernel of $r_T$ and is the smallest splitting field of $T$. In particular, $L$ is  a finite Galois extension of $k$ and $X(T)$ can be also regarded as a $\Gal(L/k)$-module.

From the remainder of this article,
$k$ denotes a number field unless otherwise stated.
For each place $v$ of $k$, denote by
$k_v$ the completion of $k$ at $v$, and $O_{v}$ the ring of integers of $k_v$ if $v$ is finite. For each finite place $v$, the group $T(k_v)$ of $k_v$-valued points of $T$ contains a unique maximal open compact subgroup, which is denoted by $T(O_v)$. Let $\A_{k,f}$ and $\A_k$ be the finite adele ring and the adele ring of $k$, respectively. Then the locally compact topological group $T(\A_{k,f})$ contains a
unique maximal open compact subgroup $T(\wh O_k)=\prod_{v \nmid \infty} T(O_v)$, where $v$ runs through all finite places of $k$.
By a finiteness theorem of Borel [2], the double coset space $T(k)\backslash T(\A_{k,f})/T(\wh O_k)$
is a finite set.

Let $T$ be an algebraic torus over $k$. The class number $h(T)$ of $T$ is defined as

\begin{equation}
\label{eq:hT}
h(T):=\abs{T(k)\backslash T(\A_{k,f})/T(\wh O_k)}.
\end{equation}

It follows immediately from the definition that if $K/k$ is a finite extension and $T_K$ is an algebraic $K$-torus, then $h(R_{K/k}T_K)=h(T_K)$, where $R_{K/k}$ denotes the Weil restriction of scalars from $K$ to $k$. Note that
$\dim R_{K/k} T_K = \dim T_K \cdot [K:k]$. It is well known (cf. [12, 14]) that

\begin{equation}
\label{eq:ind_char}
X(R_{K/k}T_K)\simeq \Ind_{\Gamma_K}^{\Gamma_k} X(T_K)=\Ind_{\Gal(L/K)}^{\Gal(L/k)} X(T_K),
\end{equation}

as $\Gal(L/k)$-modules,  where $L$ is any finite Galois extension of $k$ over which the algebraic $k$-torus $R_{K/k}T_K$ splits.

We recall the definition of the Tamagawa number of an algebraic $k$-torus $T$.
Fix a finite Galois splitting field extension $K/k$ of $T$ with Galois group $G$.
Let $ \chi _T $ be the character of representation $ \big( X(T)\otimes {\C} , r_T \big) $ of $G$ over $\mathbb{C}$, that is,  $ \chi _T : G \rightarrow \mathbb{C} $, $ \chi _T(g) := \tr ( r_T ( g ) ) $ for all $g\in G$.

\begin{equation}
\label{eq:artinL}
L( s , K/k , \chi _T ) := \prod_{ v \nmid \infty } L_v( s , K/k , \chi _T )
\end{equation}

be the Artin $L$-function of the character $\chi_T$, where
$L_v(s,K/k,\chi_T)$ is the local Artin $L$-factor at $v$.
It is well known (cf. [18])
$L( s , K/k , \chi _T )$ has a pole of order $a$ at $s=1$, where $a$ is the rank of the $G$-invariant sublattice $X(T)^G$.

Let $\omega$ be a nonzero invariant differential form on $T$ of highest degree defined over $k$.
To each place $v$, one associates a Haar measure $\omega_v$ on
$T(k_v)$. Then the product of the Haar measures

\[  \prod_{ v \mid \infty } \omega_v \cdot \prod_{ v \nmid \infty }
\left ( L_v( 1 , K/k , \chi _T ) \cdot \omega_v \right ) \]

converges absolutely and defines a Haar measure on $T(\A_k)$.

Let $T$ be an algebraic torus over a number field $k$ and $\omega$ be a nonzero invariant differential form on $T$ defined over $k$ of highest degree. Then

\[
\omega_{\mathbb{A} , can } := \frac{ \prod_{ v \mid \infty } \omega_v \cdot
\prod_{ v \nmid \infty } \big( L_v( 1 , K/k , \chi _T )\cdot \omega_v \big) }{ | d_k |^{ (\dim{T}) / 2 } \cdot \rho_T },
\]

defines a Haar measure on $T(\A_k)$, which is called
the Tamagawa measure on $T(\mathbb{A}_k)$,
where $d_k$ is the discriminant of $k$,

\[ \rho_T := \lim_{ s \to 1 } ( s - 1 )^a L( s , K/k , \chi _T ), \]

and $a$ is the order of the pole of the Artin $L$-function
$L( s , K/k , \chi _T )$ at $ s = 1 $.

Let $\xi_1, \dots ,\xi_a$ be a basis of $X(T)^G$. Define

\[ \xi: T(\A_k) \to (\R_{+}^\times)^{a}, \quad x\mapsto
(\vert\vert \xi_1(x)||,\dots, \vert \vert\xi_{a}(x)\vert\vert), \]

where $\R_+^\times:=\{x\in \R: x>0 \}$.
Let $T(\A_k)^1$ denote the kernel of $\xi$; one has an isomorphism $T(\A_k)/T(\A_k)^1 \simeq (\R^\times_{+})^a \subset (\R^\X)^{a}$. Let $d^\times t:=\prod_{i=1}^{a} dt_i/t_i$ be the canonical measure on $(\R^{\X}_{+})^a$.
Let $\omega^1_{\A,can}$ be the unique Haar measure on $T(\A_k)^1$ such that

\begin{equation}
\label{eq:omega1}
\omega_{\A,can}=\omega^1_{\A,can}\cdot d^\times t,
\end{equation}

that is, for any measurable function $F$
on $T(\A_k)$ one has

\[ \int_{T(\A_k)/T(\A_k)^1} \int_{T(\A_k)^1} F(xt) \,
\omega^1_{\A,can}\cdot d^\times t=\int_{T(\A_k)} F(x)\,
\omega_{\A,can}. \]

By a well-known theorem of Borel and Harish-Chandra <cit.>, the quotient
space $T(\A_k)^1/T(k)$ has finite volume with respect to every Haar
measure. In fact $T(\A_k)^1/T(k)$ is the unique maximal
compact subgroup of $T(\A_k)/T(k)$, because $(\R^{\X}_{+})^a$
has no nontrivial compact subgroup.

Let $T$ be an algebraic torus over a number field $k$.
The  Tamagawa number $\tau_k(T)$ of $T$
is defined by

\begin{equation}\label{eq:tau}
\tau_k(T):=\int_{T(\A_k)^1/T(k)}\ \omega^1_{\A,can},
\end{equation}

the volume of $T(\A_k)^1/T(k)$ with respect to $\omega^1_{\A,can}$, where
$\omega^1_{\A,can}$ is the Haar measure on $T(\A_k)^1$ defined in (<ref>).

One has the following properties <cit.>:

(i) For any two algebraic $k$-tori $T$ and $T'$, one has $ \tau_k(T \times T') = \tau_k(T) \cdot \tau_k(T') $.

(ii) For any finite extension $K/k$ and any algebraic $K$-torus $T_K$, one has  $\tau_k(R_{K/k}T_K) = \tau_K(T_K) $.

(iii) One has $\tau_k({\Gm}_{,k})=1 $.

Note that the last statement (iii) is equivalent to the analytic class number formula <cit.>.

§.§ Values of Tamagawa numbers

Let $K/k $ be a finite Galois extension with Galois group $\Gamma$, and $T$ be an algebraic torus over $k$ with splitting field $K$. Then

\begin{equation}
\label{eq:ono-1}
\tau_k (T) = \frac{| H^1 \big( \Gamma , X(T) \big) |}{| \Sha^1(\Gamma, T)|},
\end{equation}

\begin{equation}
\label{eq:ShaT}
\Sha^i(\Gamma, T):=\Ker \big( H^i (K/k , T) \to \prod_v H^i (K_w/k_v , T) \big)
\end{equation}

is the Tate-Shafarevich group associated to $H^i (\Gamma , T)$ for $i\ge 0$ and $w$ is a place of $K$ over $v$.

See <cit.>.

The cohomology groups $H^1 \big( \Gamma , X(T) \big)$ and
$\Sha^1(\Gamma, T)$ are independent of the choice of the splitting
field $K$;  see <cit.>.

According to Ono's formula, the Tamagawa number of any algebraic $k$-torus  is a positive rational number. Ono constructed an infinite family of algebraic $\Q$-tori $T$ with $\tau(T)=\tau_{\Q}(T)=1/4$,  which particularly shows that the Tamagawa number $\tau(T)$ need not to be an integer. More precisely,
take an abelian extension $K=\Q(\sqrt{p_1}, \sqrt{p_2}, \sqrt{p_3},\sqrt{p_4})$ of type $(2,2,2,2)$ for a $4$-tuple $(p_1,p_2,p_3,p_4)$ of distinct primes which satisfy the conditions:

\begin{equation}
\label{eq:pppp}
p_i\equiv 1 \pmod 4, \quad \text{and}\quad \left (\frac{p_i}{p_j}\right )=1\quad \text{for all $i,j$,}
\end{equation}

where $(p_i/p_j)$ is the Legendre symbol. For example, the tuple $(5,29,109,281)$ is one of them.

Chebotarev's density theorem shows that there are infinity many of such $4$-tuple of primes. Let

\[ T=R_{K/\Q}^{(1)} \GmK :=
\Ker (R_{K/\Q} \GmK \xrightarrow{N_{K/\Q}} {\Gm}_{,\Q}), \]

and one has $\tau(T)=\tau_\Q(T)=1/4$. Ono conjectured [13]
that every positive rational number can be realized as $\tau_k(T)$ for some algebraic $k$-torus $T$.
Ono's conjecture was proved by S. Katayama [7].
Katayama's result tells us the precise range of values of Tamagawa numbers of
algebraic tori.

Let $k$ be a number field. For any positive rational number $r$, there exists an algebraic torus $T$ over $k$ such that

\[
\tau_k(T) = r.
\]

The proof is to reduce the statement to the construction of an abelian extension $K/k$ of type $(p,p,p,p)$ such that all decomposition groups of $K/k$ are cyclic.

For any torsion abelian group $G$, the Pontryagin dual of $G$ is defined to be

\begin{equation}
\label{eq:Pdual}
\end{equation}

The Poitou-Tate duality <cit.>
says that there is a natural isomorphism

\begin{equation}
\label{eq:Sha}
\Sha^1(\Gamma, T)^\vee\simeq  \Sha^2(\Gamma, X(T)),
\end{equation}

where $\Sha^2(\Gamma, X(T))$ is the Tate-Shafarevich group associated to
$H^2(\Gamma, X(T))$ as in (<ref>). Thus, one can compute $|\Sha^2(\Gamma, X(T))|$ instead of $|\Sha^1(\Gamma, T)|$ in Ono's formula, that is,

\begin{equation}
\label{eq:ono-f-2}
\tau_k (T) = \frac{| H^1 \big( \Gamma , X(T) \big) |}{| \Sha^2(\Gamma, X(T))|}.
\end{equation}

To simply the notation we shall often
suppress the Galois group from Galois
cohomology groups and write $H^1(X(T))$ and $\Sha^2(X(T))$ for
$H^1(\Gamma, X(T))$ and $\Sha^2(\Gamma, X(T))$ etc.,
if there is no risk of confusion.

§.§ Some known results for CM tori

Let $K=\prod_{i=1}^r K_i$ be a CM algebra, where each $K_i$ is a CM field, with the canonical involution $\iota$.
The subalgebra $K^+$ of elements in $K$ fixed by $\iota$ is the product $K^+=\prod_{i=1}^rK_i^+$ of the maximal totally real subfields $K_i^+$ of $K_i$.
Denote $N_{K_i/K_i^+}$ as the norm map from $K_i$ to $K^+_i$ and let $N_{K/K^+}=\prod_{i=1}^r N_{K_i/K_i^+}:K\to K^+$ be the norm map.

Denote by $T^K$ (resp. $T^{K^+}$) the $\Q$-torus associated to the multiplicative group of $K$ (resp. $K^+$): for any commutative $\Q$-algebra $R$, the group of its $R$-valued points is given by $(K\otimes_\Q R)^\X$ (resp. $(K^+\otimes_\Q R)^\X$).  Then $T^{K_i}=R_{K_i/\Q}(\G_{{\rm
m},K_i})$ and $T^K=\prod_{i=1}^r T^{K_i}$; similarly $T^{K^+}=\prod_{i=1}^r T^{K_i^+}=\prod_{i=1}^r R_{K^+_i/\Q}(\G_{{\rm m},K_i^+})$, and we have the norm map $N_{K/K^+}: T^K\to T^{K^+}$.

Consider the following exact sequence of algebraic tori over $\Q$

\begin{equation}\label{exact sequence of T^K}
\begin{tikzcd}
1 \arrow{r} & T^{K,1}\arrow{r} & T^K\arrow{r}{N_{K\slash K^+}} & T^{K^+}\arrow{r} & 1,
\end{tikzcd}
\end{equation}

$T^{K,1}:=\Ker (N_{K/K^+})=\prod_{i=1}^r T^{K_i,1}$ and
each $T^{K_i,1}:=\left\{x\in T^{K_i}\mid N_{K_i/K_i^+}(x)=1\right\}$ is a norm one subtorus.
We regard $\Gm$ as a $\Q$-subtorus of $T^{K^+}$ via the diagonal embedding. Let $T^{K,\Q}$ denote the preimage of $\Gm$ in $T^K$ under the map $N_{K\slash K^+}$.
So we have the following commutative diagram of algebraic $\Q$-tori with each row exact

\begin{equation}\label{diagram for exact sequence of T^K}
\begin{tikzcd}
\qquad 0\arrow{r} & T^{K,1}\arrow{r}{\iota}\arrow[equal]{d} & T^K\arrow{r}{N}& T^{K^+}\arrow{r} & 0 \\
\qquad 0\arrow{r} & T^{K,1}\arrow{r}{\iota} & T^{K,\Q}\arrow{r}{N}\arrow[hook]{u} & \G_{m}\arrow[hook]{u}{}\arrow{r} & 0.
\end{tikzcd}
\end{equation}

Here, for brevity, we write $N$ for $N_{K\slash K^+}$.

Let $K$ be a CM algebra and $T=T^{K,\Q}$ the associated CM torus over $\Q$.

(1) We have

\begin{equation}
\label{eq:CMformula}
\tau(T)=\frac{2^r}{n_K},
\end{equation}

where $r$ is the number of components of $K$ and

\begin{equation}
\label{eq:nK}
n_K:=[\A^\times: N(T(\A))\cdot \Q^\times]
\end{equation}

is the global norm index associated to $T$.

(2) We have

\begin{equation}
\label{eq:bound_nK}
n_K\, \Big{|}\!  \prod_{p\in S_{K/K^+}}\!\!\!\! e_{T,p},
\end{equation}

where $S_{K/K^+}$ is the finite set of rational primes $p$
with some place $v|p$ of $K^+$ ramified in $K$, and
$e_{T, p}:=[\Z^\times_p: N(\,T(\Z_p)\,)]$. Here $T(\Zp)$ denotes the unique maximal open compact subgroup of $T(\Qp)$.

(1) This is <cit.>. (2) This is <cit.>.

Let $K$ and $T$ be as in Proposition <ref>.

(1) If $K$ contains an imaginary quadratic field, then $n_K\in \{1,2\}$ and $\tau(T)\in \{2^r, 2^{r-1}\}$.

(2) If $K$ contains two distinct imaginary quadratic fields, then $n_K=1$ and $\tau(T)=2^r$.

This is proved in <cit.> (cf. <cit.>) in the case where $K$ is a CM field.
The same proof using class field theory also proves the CM algebra case.

Let $K$ be a CM field and $T$ the associated CM torus over $\Q$. Put $g=[K^+:\Q]$ and let $K^{\Gal}$ be the Galois closure of $K$ over $\Q$ with Galois group $\Gal(K^{\Gal}/\Q)=G$.

(1) If $g$ is odd, then $H^1(G,X(T))=0$.

(2) If $K/\Q$ is Galois and $g$ is odd, then $\tau(T)=1$.

(3) If $K/\Q$ is cyclic, then $\tau(T)=1$.

(4) If $K/\Q$ is Galois and $g=2$, then $\tau(T)=1,2$, and furthermore $\tau(T)=2$ if and only if $G\simeq (\Z/2\Z)^2$.

See Propositions A.2 and A.12 of [1].

Note that if $K/\Q$ is Galois with Galois group $G=(\Z/2\Z)^2$, then $K$ contains two distinct imaginary quadratic fields. Therefore, Proposition <ref>(4) also follows from Proposition <ref>(3) and Lemma <ref>(2).

§ TRANSFER MAPS,  CORESTRICTION MAPS AND EXTENSIONS

§.§ Transfer maps

Let $G$ be a group and let $H$ be a subgroup of $G$ of finite index. Let $X:=G/H$, and let $\varphi: X\to G$ be a section. If $g\in G$ and $x\in X$, the elements $\varphi(gx)$ and $g\varphi(x)$ belong to the same class of mod $H$; hence there exists a unique element $h^\varphi_{g,x}\in H$ such that  $g\varphi(x)=\varphi(gx)h^\varphi_{g,x}$. Let $\Ver_{G,H}(g)\in H^{\rm ab}$ be defined by:

\[ \Ver_{G,H}(g):=\prod_{x\in X} h^\varphi_{g,x} \mod D(H), \]

where $D(H)=[H,H]$ is the commutator group of $H$ and the product is computed in $H^{\rm ab}=H/D(H)$.

By <cit.>, the map $\Ver_{G,H}: G\to H^{\rm ab}$ is a group homomorphism and it does not depend on the choice of the section $\varphi$. This homomorphism is called the transfer of $G$ into $H^{\rm ab}$ (originally from the term  “Verlagerung” in German). One may also view it as a homomorphism $\Ver_{G,H}:G^{\rm ab}\to H^{\rm ab}$.

In the literature one also uses the right coset space $X':=H\backslash G$ but this does not effect the result.

Let $\varphi':X'\to G$ be a section, and define for each $g\in G$

\[ \Ver_{G,H}'(g):=\prod_{x'\in X'} h^{\varphi'}_{x',g} \mod D(H), \]

where $h^{\varphi'}_{x',g}\in H$ is the unique element such that $\varphi'(x')g= h^{\varphi'}_{x',g} \varphi'(x'g)$. Then $\Ver_{G,H}'=\Ver_{G,H}$.

Let $S=\{\varphi(x): x\in X\}$ be the set of representatives of $G/H$ defined by $\varphi$. Then $S':=\{\varphi(x)^{-1}: x\in X\}$ is a set of representatives of $H\backslash G$. Let $c':G\to X'$ be the natural projection. Define $\varphi'(x'):=\varphi(x)^{-1}$ if $x'=c'( \varphi(x)^{-1})$.
Taking the inverse of $g\varphi(x)=\varphi(gx)h^\varphi_{g,x}$, one has
the relation $\varphi(x)^{-1} g^{-1}=(h^\varphi_{g,x})^{-1} \varphi(gx)^{-1}$. Put $x'=c'(\varphi(x)^{-1})$, and applying $c'$ to the previous relation   one gets $\varphi(gx)^{-1}=\varphi'(x'g^{-1})$ and the relation
$\varphi'(x') g^{-1}=(h^\varphi_{g,x})^{-1} \varphi'(x' g^{-1})$. This shows
$h^{\varphi'}_{x',g^{-1}}=(h^\varphi_{g,x})^{-1}$. Then

\[ \Ver_{G,H}'(g^{-1}) = \prod_{x'\in X'}  h^{\varphi'}_{x',g^{-1}}
% \prod_{x'\in X'}
= \prod_{x\in X} (h^\varphi_{g,x})^{-1} =\Ver_{G,H}(g)^{-1}=\Ver_{G,H}(g^{-1}) \quad \text{in $H^{\rm ab}$}. \]

This shows $\Ver_{G,H}'=\Ver_{G,H}$.

One has the following functorial property (see <cit.>).

Let $H\subset G$ and $H'\subset G'$ be subgroups of finite index.
If $\sigma$ be a group homomorphism from the pair $(G,H)$ to $(G',H')$ which induces a bijection $G/H\isoto G'/H'$, then the following diagram

\begin{equation} \label{eq:func}
\begin{tikzcd}
\qquad  & G^{\rm ab} \arrow{r}{\sigma} \arrow{d}{\Ver_{G,H}}
& G^{' \rm ab} \arrow{d}{\Ver_{G'.H'}} \\
\qquad & H^{\rm ab} \arrow{r}{\sigma} \arrow{r} & H^{' \rm ab}
\end{tikzcd}
\end{equation}

Let $G_1$ and $G_2$ be groups, and let $H_i\subset G_i$ be a subgroup of finite index for $i=1,2$. Then

\[  \Ver_{G_1\times G_2, H_1\times H_1}(g_1,g_2)=\Ver_{G_1,H_1}(g_1)^{[G_2:H_2]} \cdot \Ver_{G_2,H_2}(g_2)^{[G_1:H_1]}. \]

Put $G=G_1\times G_2$, $H=H_1\times H_2$, $X_i:=G_i/H_i$ and $X=G/H=X_1\times X_2$. Fix a section $\varphi_i:X_i \to G_i$ for each $i$ and let $\varphi=(\varphi_1,\varphi_2):X\to G$. For $g=(g_1,g_2)$ and $x=(x_1,x_2)$, we have $h^\varphi_{g,x}=(h^{\varphi_1}_{g_1,x_1},h^{\varphi_2}_{g_2,x_2})$. Then in $H^{\rm ab}=H_1^{\rm ab} \times H_2^{\rm ab}$ we have

\[
\begin{split}
\Ver_{G,H}(g)&=\prod_{x\in X} h^\varphi_{g,x}=\prod_{x_1\in X_1} \prod_{x_2\in X_2} (h^{\varphi_1}_{g_1,x_1}, h^{\varphi_2}_{g_2,x_2})\\
&=\prod_{x_1\in X_1} ((h^{\varphi_1}_{g_1,x_1})^{|X_2|}, \Ver_{G_2,H_2}(g_2))= (\Ver_{G_1,H_1}(g_1)^{|X_2|}, \Ver_{G_2,H_2}(g_2)^{|X_1|}). \quad \text{\qed}
\end{split}  \]

If $G$ is a finite group and $p$ is a prime, $G_p$ denotes a $p$-Sylow subgroup of $G$.

Let $G$ be a finite group and $N\triangleleft G$ be a cyclic normal subgroup of prime order $p$. Then the transfer $\Ver_{G,N}:G^{\rm ab}\to N^{\rm ab}$ is surjective if and only if $G_p$ is cyclic.

See Proposition 4.8 of [16].
Note that the author assumes that the normal subgroup $N$ is central in $G$ throughout the whole
paper; however, this assumption is not needed for
this proposition in his proof.

§.§ Connection with corestriction maps

Let $H \subset G$ be a subgroup of finite index, and let $A$ be a $G$-module.

\[ f: \Ind_H^G A=\Z[G]\otimes_{\Z[H]} A \to A, \quad g\otimes a \mapsto ga \]

be the natural map of $G$-modules. Applying Galois cohomology $H^i(G, -)$ to the map $f$ and by Shapiro's lemma, we obtain  for each $i\ge 0$ a morphism

\[ \Cor: H^i(H, A) \to H^i(G,A), \]

called the corestriction form $H$ to $G$.

Applying $H^i(G,-)$ to the short exact sequence

\[ 0 \longrightarrow \Z  \longrightarrow \Q  \longrightarrow \Q/\Z  \longrightarrow 0, \]

one obtains an isomorphism

\[ H^i(G,\Z) \simeq H^{i-1}(G, \Q/\Z) \]

for all $i\ge 1$. When $i=2$, this gives the isomorphism

\begin{equation}
\label{eq:H2Z}
H^2(G,\Z)\simeq \Hom(G^{\rm ab},\Q/\Z).
\end{equation}

Let $H\subset G$ be a subgroup of finite index. Through the isomorphism (<ref>) we have the following commutative diagram

\begin{equation} \label{eq:cor-ver}
\begin{tikzcd}
\qquad  & H^2(H,\Z) \arrow{r}{\sim} \arrow{d}{\Cor}
& \Hom(H^{\rm ab},\Q/\Z) \arrow{d}{\Ver_{G,H}^\vee} \\
\qquad & H^2(G,\Z) \arrow{r}{\sim} \arrow{r} & \Hom(G^{\rm ab},\Q/\Z),
\end{tikzcd}
\end{equation}

where $\Ver_{G,H}^{\vee}$ is dual of the transfer $\Ver_{G,H}: G^{\rm ab} \to H^{\rm ab}$.
Moreover, if $H$ is normal in $G$, the composition $H^{\rm ab}\to G^{\rm ab}\to H^{\rm ab}$ is the norm $N_{G/H}$. Here $H^{\rm ab}$ is viewed as a $G$-module by conjugation and also as a $G/H$-module, since $H$ acts trivially on
$H^{\rm ab}$.

See <cit.>.

§.§ Connection with class field theory

Let $k\subset K \subset L$ be three number fields such that the extension $L/k$ is Galois.
Put $G=\Gal(L/k)$ and $H=\Gal(L/K)\subset G$.
Denote by $C_k$ and $C_K$ the idele class groups of $k$ and $K$, respectively.
The Artin map is a surjective homomorphism $\Art_{L/k}: C_k \to G^{\rm ab}$; similarly we have $\Art_{L/K}: C_K \to H^{\rm ab}$.
By class field theory we have the following commutative diagrams

\begin{equation}
\label{eq:cft}
\begin{CD}
C_k @>\Art_{L/k}>> G^{\rm ab} \\
@V{\rm func}VV @VV{\Ver_{G,H}}V \\
C_K @>\Art_{L/K}>> H^{\rm ab} \\
\end{CD} \qquad \qquad
\begin{CD}
C_k @>\Art_{L/k}>> G^{\rm ab} \\
@A{N_{K/k}}AA @AA{\rm func}A \\
C_K @>\Art_{L/K}>> H^{\rm ab}, \\
\end{CD}
\end{equation}

where ${\rm func}$ denotes the natural map induced from the inclusion $\A_k^\times \embed \A_K^\times$ or $H\embed G$. When $K/k$ is Galois and let
the second diagram of (<ref>) induces a homomorphism

\[ \Art_{K/k}: C_k/N_{K/k}(C_K) \to \Gal(K/k)^{\rm ab}, \]
which is an isomorphism by class field theory.

§.§ Relative transfer maps

Let $H\subset N \subset G$ be two subgroups of $G$ of finite index.
Let $\wt X:= G/H$ and $X:=G/N$ with natural $G$-equivariant projections
$\wt c: G\to X$ and $c: \wt X\to X$. Let $\wt \varphi: X\to G$ be a section, which induces a section $\varphi: X\to \wt X$. For each $g\in G$ and $x\in X$, let $n^{\wt \varphi}_{g,x}$ be the unique element in $N$ such that $g \wt \varphi (x)=\wt \varphi (gx) n^{\wt \varphi}_{g,x}$.
Let $(N/H)^{\rm ab}=N/D(N)H$ denote the maximal abelian quotient of $N$ which maps $H$ to zero.
Let $\Ver_{G,N/H}(g) \in (N/H)^{\rm ab}$ be the element defined by

\begin{equation}
\label{eq:rt}
\Ver_{G,N/H}(g):=\prod_{x\in X} n^{\wt \varphi}_{g,x} \mod D(N)H.
\end{equation}

(1) The map $\Ver_{G,N/H}: G\to (N/H)^{\rm ab}$ does not depend on the choice of the section $\wt \varphi$ and it is a group homomorphism.

(2) One has $\Ver_{G,N/H}=\pi_H \circ \Ver$, where $\pi_H: N^{\rm ab} \to (N/H)^{\rm ab}$ is the morphism mod $H$.

Clearly, the statement (1) follows from (2), because $\Ver$ does not depend on the choice of $\wt \varphi$ and is a group homomorphism .
(2) By definition
$\Ver(g)=\prod_{x\in X} n^{\wt \varphi}_{g,x} \mod D(N)$, thus $\Ver_{G,N/H}=\pi_H \circ
\Ver$.

Let $H\subset N$ be two subgroups of $G$ of finite index. The group homomorphism $\Ver_{G,N/H}: G\to (N/H)^{\rm ab}$ defined in (<ref>) is called the transfer of $G$ into $(N/H)^{\rm ab}$ relative to $H$. By abuse of notation, we denote the induced map by $\Ver_{G,N/H}: G^{\rm ab} \to (N/H)^{\rm ab}$.

One can check directly that the map $\Ver_{G,N/H}: G^{\rm ab} \to (N/H)^{\rm ab}$ factors through $\pi_H: G^{\rm ab} \to (G/H)^{\rm ab}$, the map modulo $H$.
We denote the induced map by

\begin{equation}\label{eq:rV} % reduced transfer
\Ver_{G/H,N/H}: (G/H)^{\rm ab} \to (N/H)^{\rm ab}.
\end{equation}

If $H\triangleleft G$ is a normal subgroup, then the induced map $\Ver_{G/H,N/H}$ is the transfer map from $G/H$ to $N/H$ associated to the subgroup $N/H\subset G/H$ of finite index.

Let $H\triangleleft \wt  N$ be two subgroups of finite index in $G$ with $H$ normal in $\wt N$ and cyclic quotient $N=\wt N/H=\< \sigma \>$ of order $n$. Fix a lift $\wt \sigma\in \wt N$ of $\sigma$.

(1) Let $g\in G$ and $\{x_1,\dots x_r\}$ be double coset representatives of $\<g\>\backslash G/ \wt N$. Then

\[ \Ver_{G,N}(g)=\sigma^{\sum_{i=1}^r m(g,x_i)}, \]

where $f_i:=|\<g\> x_i H/H|$ and $0\le m(g,x_i)\le n-1$ is the unique integer such that $g^{f_i} x_iH=x_i \wt \sigma^{m(g,x_i)} H$.

(2) Let $\sigma^*\in N^{\vee}:=\Hom(N,\Q/\Z)$ be the element defined by $\sigma^*(\sigma)=1/n \mod \Z$,  $\Ver_{G,N}^\vee: N^\vee\to \Hom(G,\Q/\Z)$ the induced map, and $f:=\Ver_{G,N}^\vee(\sigma^*)$. Then $f(g)=\left (\sum_{i=1}^r m(g,x_i) \right )/n \mod \Z$.

(1) We choose the set of  representatives $S=\{g^j x_i: i=1,\dots, r,\  j=0,\dots, f_i-1\}$ of $X=G/\wt N$, which defines a section $\wt \varphi$ of the natural projection $G\to G/\wt N$. Then the image of the element $n^{\wt \varphi}_{g, x}$ in $H$ is given by

\[ n^{\wt \varphi}_{g, x} \mod H=\begin{cases} \sigma^{m(g,x_i)} & \text{if $x=g^{f_i-1} x_i H$ for some $1\le i\le r$;} \\
1 & \text{otherwise.}  \
\end{cases} \]

(2) By definition, $f(g)=\sigma^*(\Ver_{G,N}(g))=\sigma^*(\sigma^{\sum_i m(g,x_i)})=\left (\sum_i m(g,x_i)\right )/n \mod \Z$.

One can show that the integer $m(g,x_i)$ is independent of the choice of double coset representative in $\<g\> x_i H$.

Let $G=\prod_{i=1}^r G_i$ be a product of groups $G_i$ and let $N_i\subset G_i$, for each $1\le i \le r$, be a subgroup of finite index. Put

\[ H_i:=G_1\times \cdots \times G_{i-1}\times \{1\} \times G_{i+1}\times \cdots \times G_r  , \quad \wt N_i:=G_1\times \cdots  \times G_{i-1}\times N_i \times G_{i+1}\times \cdots \times G_r. \]

Then the map

\begin{equation}\label{eq:prod_ver-1}
\prod_{i=1}^r \Ver_{G, \wt N_i/H_i}: G \longrightarrow (\wt N_1/H_1)^{\rm ab}\times \cdots \times (\wt N_r/H_r)^{\rm ab}=N_1^{\rm ab} \times \cdots \times N_r^{\rm ab}
\end{equation}

is given by the product of the maps

\begin{equation}\label{eq:prod_ver-2}
\prod_{i=1}^r \Ver_{G_i, N_i}: \prod_{i=1}^r G_i \longrightarrow \prod_{i=1}^r N_i^{\rm ab}.
\end{equation}

Let $\pr_i: G\to G_i$ be the $i$-th projection. Then

\[ \Ver_{G, \wt N_i/H_i}=\Ver_{G/H_i,\wt N_i/H_i} \circ \,\pr_i =\Ver_{G_i,N_i} \circ \,\pr_i,  \]

cf. Remark <ref>. Thus, the map (<ref>) is equal to $\prod_i \Ver_{G_i, N_i} \circ \, (\pr_i)_i$. Since the map $(\pr_i)_i:G\to \prod_i G_i$ is the identity, we show that the map (<ref>) is equal to $\prod_i \Ver_{G_i, N_i}$.

§.§ Connection with class field theory

Let $k\subset K\subset L$ be three number fields as in Section <ref> and let $H=\Gal(L/K)\subset G=\Gal(L/k)$. The right commutative diagram of (<ref>) induces an isomorphism

\begin{equation}
\label{eq:art}
\Art: C_k/N_{K/k}(C_K) \to (G/H)^{\rm ab}.
\end{equation}

Note that $K/k$ is any finite extension (not necessarily Galois) and $(G/H)^{\rm ab}=\Gal(K_{ab}/k)$ does not depend on the choice of the Galois extension
$L/k$ containing $K$, where $K_{\rm ab}$ is the maximal abelian extension of $k$ in $K$.

Let $k\subset E\subset K$ be three number fields. Let $L/k$ be a finite Galois extension containing $K$ with Galois group $G=\Gal(L/k)$.
Let $H=\Gal(L/K)\subset N=\Gal(L/E)$ be subgroups of $G$.

The Artin map produces the following isomorphisms

\[ C_k/N_{L/k}(C_L)\simeq G^{\rm ab}, \quad C_k/N_{K/k}(C_K)\simeq (G/H)^{\rm ab}, \]

\[ \quad C_E/N_{L/E}(C_L)\simeq N^{\rm ab}, \quad
C_E/N_{K/E}(C_K)\simeq (N/H)^{\rm ab}.
\]

We obtain the following commutative diagrams

\begin{equation}
\label{eq:cft_ver}
\begin{CD}
C_k/N_{L/k}(C_L) @>>> C_k/N_{K/k}(C_K)\\
@VVV @VVV \\
C_E/N_{L/E}(C_L)  @>>>  C_E/N_{K/E}(C_K)\
\end{CD} \qquad \qquad
\begin{CD}
G^{\rm ab}  @>\pi_H>> (G/H)^{\rm ab} \\
@V{\Ver}VV @VV{\Ver_{G/H,N/H}}V \\
N^{\rm ab} @>\pi_H>> (N/H)^{\rm ab}. \\
\end{CD}
\end{equation}

From this we see that $\Ver_{G/H,N/H}: (G/H)^{\rm ab} \to (N/H)^{\rm ab}$ does not depend on the choice of the Galois extension $L/k$.

§ COHOMOLOGY GROUPS OF ALGEBRAIC TORI

§.§ The problem

The problem of computing the Tamagawa number of CM algebraic tori can
be recasted in the following more general problem.

Let $k$ be a number field. For every finite commutative
etale $k$-algebra $K$,
denote by $T^K$ the algebraic $k$-torus whose group of $R$-valued points of $T^K$
for any commutative $k$-algebra $R$, is

\[ T^K(R)=(K \otimes_k R)^\times. \]

Explicitly, if $K=\prod_{i=1}^r K_i$ is a product of finite separable field
extensions $K_i$ of $k$,

\[ T^K=\prod_{i=1}^r T^{K_i}=\prod_{i=1}^r R_{K_i/k} (\G_{m,K_i}), \]

where $R_{K_i/k}$
is the Weil restriction of scalars from $K_i$ to $k$.
Let $E=\prod_{i=1}^r E_i$ be a product of finite subfield extensions
$E_i\subset K_i$ over $k$. Let $N_{K_i/E_i}: T^{K_i} \to T^{E_i}$
be the norm map and put $N=\prod_{i=1}^r N_{K_i/E_i}: T^K \to T^E$. Define
$T^{K/E,1}:=\Ker N$, the kernel of the norm map $N$, and

\[ T^{K/E,k}:=\{x\in T^K \, |\, N(x)\in \G_{m,k} \}, \]

the preimage of $\G_{m,k}$ in $T^K$ under $N$,
where $\G_{m,k}\embed T^E$ is viewed as a subtorus of $T^E$ via the diagonal
embedding. Thus, we have the following commutative diagram of algebraic $k$-tori in which each row is an exact sequence:

\begin{equation}\label{eq:diagram_tori}
\begin{tikzcd}
\qquad 1 \arrow{r} & T^{K/E,1}\arrow{r}{j}\arrow[equal]{d}
& T^K\arrow{r}{N}& T^{E}\arrow{r} & 1 \\
\qquad 1 \arrow{r} & T^{K/E,1} \arrow{r}{j} & T^{K/E,k}\arrow{r}{N}
\arrow[hook]{u} & \G_{m,k}\arrow[hook]{u}{}\arrow{r} & 1.
\end{tikzcd}
\end{equation}

We investigate the problem of computing the Tamagawa
number $\tau_k(T^{K/E,k})$ of the algebraic $k$-torus $T^{K/E,k}$.
When $k=\Q$, $K$
is a CM algebra and $E=K^+$ is the maximal totally real subalgebra of
$K$, this reduces to the CM case in Section <ref> and this is the original problem we are mainly interested in.

\[ \Lambda:=X(T^{K/E,k})\quad \text{and}\quad \Lambda^1:=X(T^{K/E,1}), \]

$X(T)$ denotes the group of characters of an algebraic torus $T$ over
$k$. Let $L$ be a finite Galois extension of $k$ which contains all
Then $L$ is a splitting field for all tori in
(<ref>). Let $G=\Gal(L/k)$ be the Galois group of
$L/k$. By Ono's formula and the duality $\Sha^1(G,T)\simeq
\Sha^2(G,\Lambda)^\vee$, it suffices to compute
the cardinalities of the groups $H^1(G,\Lambda)$ and $\Sha^2(G,\Lambda)$.
To shorten our notation, we shall often omit
the symbol $G$ from the Galois cohomology concerned
if there is no risk of confusion, that is,
$H^i(\Lambda)=H^i(G,\Lambda)$, and also write $H^i(L/K_j, \Lambda)$ for $H^i(\Gal(L/K_j), \Lambda)$ as convention.

§.§ $H^1(\Lambda^1)$ and $H^1(\Lambda)$

Let $G=\Gal(L/k)$, and $\Lambda$ and $\Lambda^1$ be $G$-modules as above. Put $H_i:=\Gal(L/K_i)$, $\wt N_i:=\Gal(L/E_i)$ and $N_i^{\rm ab}=(\wt N_i/H_i)^{\rm ab}:=\wt N_i/H_i D(\wt N_i)$ for each $1\le i\le
r$. In case that $K_i$ is Galois over $E_i$, we put $N_i:=\wt N_i/H_i=\Gal(K_i/E_i)$ and then $N_i^{\rm ab}$ is the abelianization of the group $N_i$.

Taking the character groups of the diagram (<ref>), we obtain the commutative diagram of $G$-modules:

\begin{equation}\label{eq:diagram_char}
\begin{tikzcd}
\qquad 0 \arrow{r} & X(T^{E}) \arrow{r}{\wh N}\arrow{d}{\wh \Delta}
& X(T^K) \arrow{d}{\wh j_T} \arrow{r}{\wh j}& \Lambda^1 \arrow{r} \arrow[equal]{d} & 0 \\
\qquad 0 \arrow{r} & \Z \arrow{r}{\wh N} & \Lambda \arrow{r}{\wh j}
& \Lambda^1 \arrow{r} & 0.
\end{tikzcd}
\end{equation}

Taking the character groups of the upper row of (<ref>), we get a short exact sequence of $G$-modules:

\begin{equation}\label{eq:ses_char}
\begin{tikzcd}
\qquad 0 \arrow{r} & X(T^{E}) \arrow{r}{\wh N}
& X(T^K) \arrow{r}{\wh j}& \Lambda^1 \arrow{r} & 0. \\
\end{tikzcd}
\end{equation}

We have

\begin{equation}
\label{eq:XTK}
X(T^K)=\bigoplus_{i=1}^r X(T^{K_i})=\bigoplus_{i=1}^r \Ind_{H_i}^G X({\Gm}_{,K_i})=
\bigoplus_{i=1}^r \Ind_{H_i}^G \Z
\end{equation}

\begin{equation}
\label{eq:XTE}
X(T^E)=\bigoplus_{i=1}^r X(T^{E_i})=\bigoplus_{i=1}^r \Ind_{\wt N_i}^G X({\Gm}_{,E_i})=\bigoplus_{i=1}^r \Ind_{\wt N_i}^G \Z
\end{equation}

as $G$-modules.
The torus $T^{K/E,1}=\prod_{i=1}^r T^{K_i/E_i,1}=\prod_{i=1}^r R_{E_i/k} R^{(1)}_{K_i/E_i} {\Gm}_{,K_i} $. Thus,

\begin{equation}
\label{eq:XKE1}
\Lambda^1=\bigoplus_{i=1}^r X(T^{K_i/E_i,1})=\bigoplus_{i=1}^r \Ind_{\wt N_i}^G \Lambda^1_{E_i}, \quad \text{where  $\Lambda^1_{E_i}:=X(R^{(1)}_{K_i/E_i} {\Gm}_{,K_i})$.}
\end{equation}

By Shapiro's lemma, we have

\begin{equation}
\label{eq:HqL1}
H^q(G,\Lambda^1)=\bigoplus_{i=1}^r H^q(G, \Ind_{\wt N_i}^G \Lambda^1_{E_i})=\bigoplus_{i=1}^r H^q(\wt N_i,\Lambda^1_{E_i}), \quad \forall\, q\ge 0.
\end{equation}

Note that we have

\begin{equation}
\label{eq:H1L1E}
H^1(\wt N_i,\Lambda^1_{E_i})=H^1(K_i^{\Gal}/E_i, \Lambda^1_{E_i}),
\end{equation}

where $K_i^{\Gal}$ denotes the Galois closure of $K_i$ over $E_i$, because the Galois cohomology $H^1(X(T))$ does not depend on the choice of the splitting field of an algebraic torus $T$, cf. Remark <ref>. In particular, if $K_i/E_i$ is Galois, then
$H^1(\wt N_i,\Lambda^1_{E_i})=H^1(N_i,\Lambda^1_{E_i})$

Since the torus $R^{(1)}_{K_i/E_i} {\Gm}_{,K_i}$ is anisotropic, it has no non-trivial characters defined over $E_i$ and therefore,

\begin{equation}
\label{eq:H0L1}
H^0(G,\Lambda^1)=\bigoplus_{i=1}^r H^0(\wt N_i,\Lambda^1_{E_i})=0.
\end{equation}

Consider the following exact sequence of algebraic $E_i$-tori

\begin{equation}
\label{eq:R1Ei}
\begin{CD}
1 @>>> R^{(1)}_{K_i/E_i} (\G_{{\rm m}, K_i})
@>j>>  R_{K_i/E_i} (\G_{{\rm m}, K_i}) @>N_{K_i/E_i}>> \G_{{\rm m}, E_i} \to 1.
\end{CD}
\end{equation}
Taking the character groups,
we have an exact sequence of $\wt N_i$-modules

\begin{equation}
\label{eq:L1Ei}
\begin{CD}
0 @>>> \Z @>\wh N >>  X(R_{K_i/E_i} (\G_{{\rm m}, K_i})) @>\wh j >> \Lambda^1_{E_i} @>>> 0.
\end{CD}
\end{equation}

When $K_i/E_i$ is Galois, all $E_i$-tori in (<ref>) split over $K_i$ and hence (<ref>) becomes an exact sequence of $N_i$-modules and $X(R_{K_i/E_i} (\G_{{\rm m}, K_i}))\simeq \Ind^{N_i}_1 \Z$ is an induced module.

Let the notation be as above.

(1) There is a canonical isomorphism $H^1(G,\Lambda^1)\simeq \bigoplus_{i} N_{i}^{\rm ab,\vee}$.

(2) Under the canonical isomorphisms $H^1(G,\Lambda^1)\simeq \bigoplus_{i} N_{i}^{\rm ab,\vee}$ and $H^2(G,\Z)\simeq G^{\rm ab, \vee}$, the connecting homomorphism $\delta: H^1(G, \Lambda^1)\to H^2(G, \Z)$ obtained from the lower exact sequence of (<ref>) expresses as

\begin{equation}
\label{eq:delta}
\sum \Ver_{G,N_i}^{\vee}: \bigoplus_i N_i^{\rm ab,\vee} \to G^{\rm ab,\vee},
\end{equation}

where $\Ver_{G,N_i}:G\to N_i^{\rm ab}$ is the transfer map.
In particular, $H^1(G,\Lambda)\simeq \Ker (\sum \Ver_{G,N_i}^{\vee})$.
Furthermore, the map in (<ref>) factors as

\begin{equation} \label{eq:dec}
\begin{CD}
\bigoplus_i N_i^{\rm ab,\vee} @>{(\Ver_{G/H_i, N_i}^{\vee})}>>  &
\bigoplus_{i} (G/H_i)^{\rm ab,\vee} @>{\sum_i \pi_{H_i}^\vee}>> &
\ G^{\rm ab,\vee},
\end{CD}
%  \qquad \bigoplus_i N_i^{\rm ab,\vee} \arrow{r}{\oplus \Ver_{G/H_i,
% %  N_i}^{\vee}}\  &
%  \bigoplus_{i} (G/H_i)^{\rm ab,\vee} \arrow{r}{\sum_i \pi_{H_i}^\vee} &  \ %G^{\rm ab,\vee},
\end{equation}

where $\pi_{H_i}: G^{\rm ab} \to (G/H_i)^{\rm ab}$ the map mod $H_i$.

There is an exact sequence of abelian groups

\begin{equation} \label{eq:H1L}
\begin{tikzcd}
\qquad 0 \arrow{r} & H^1(G,\Lambda) \arrow{r}  & \bigoplus_i N_i^{\rm ab,\vee} \arrow{r}{\sum \Ver_{G,N_i}^{\vee}}\  & \ G^{\rm ab,\vee}.
\end{tikzcd}
\end{equation}

Taking Galois cohomology of the upper exact sequence of (<ref>), we have a long exact sequence of abelian groups:

\begin{equation} \label{eq:1strow1}
\begin{tikzcd}
\qquad H^1(G, X(T^K)) \arrow{r} & H^1(G,\Lambda^1) \arrow{r}{\delta}
& H^2(G, X(T^E)) \arrow{r}{\wh N^2}  &  H^2(G, X(T^K)).
\end{tikzcd}
\end{equation}

By (<ref>), the first term

\[ H^1(G,X(T^K))=\bigoplus_i H^1(H_i,\Z)=\bigoplus_i \Hom(H_i,\Z)=0. \]

Using the relations (<ref>) and (<ref>) and by Shapiro's lemma, $H^1(G,\Lambda^1)=\bigoplus_i H^1(\wt N_i, \Lambda^1_{E_i})$ and the exact sequence (<ref>) becomes the following:

\begin{equation} \label{eq:1row}
\begin{tikzcd}
\qquad 0 \arrow{r} & \bigoplus_i H^1(\wt N_i, \Lambda^1_{E_i}) \arrow{r}{\wt \delta}
& \bigoplus_i H^2(\wt N_i,\Z) \arrow{r}{\rm Res}  & \bigoplus_i H^2(H_i,\Z).
\end{tikzcd}
\end{equation}

The group $N_i^{\rm ab}$ is the maximal abelian quotient of $\wt N_i$ which maps $H_i$ to zero. It follows that
the following sequence

\begin{equation} \label{eq:1row2}
\begin{tikzcd}
\qquad 0 \arrow{r} &  \Hom(N_i^{\rm ab}, \Q/\Z)  \arrow{r}{\rm Inf}
&  \Hom(\wt N_i, \Q/\Z) \arrow{r}{\rm Res}  &  \Hom(H_i,\Q/\Z)
\end{tikzcd}
\end{equation}

is exact.

Using the canonical isomorphism
$H^2(H,\Z)\simeq \Hom(H,\Q/\Z)$ (<ref>) for any group $H$, we replace the group $\Hom(H, \Q/\Z)$ in (<ref>) by $H^2(H,\Z)$ for $H=N_i^{\rm ab}$, $H=\wt N_i$ and $H=H_i$, respectively and obtain the exact sequence

\begin{equation} \label{eq:1row3}
\begin{tikzcd}
\qquad 0 \arrow{r} &  H^2(N_i^{\rm ab}, \Z)  \arrow{r}{\rm Inf}
&  H^2(\wt N_i, \Z) \arrow{r}{\rm Res}  &  H^2(H_i,\Z).
\end{tikzcd}
\end{equation}

Comparing (<ref>) and (<ref>), there is a unique isomorphism

\begin{equation}
\label{eq:H1L1}
H^1(G,\Lambda^1)=\bigoplus_{i=1}^r H^1(\wt N_i, \Lambda^1_{E_i}) \simeq \bigoplus_{i=1}^r H^2(N_i^{\rm ab},\Z)
\end{equation}

which fits the following commutative diagram

\begin{equation} \label{eq:1row4}
\begin{tikzcd}
\qquad 0 \arrow{r} &  \bigoplus_i H^2(N_i^{\rm ab}, \Z)  \arrow{r}{\rm Inf}  \arrow[equal]{d}
&  \bigoplus_i H^2(\wt N_i, \Z) \arrow{r}{\rm Res}  \arrow[equal]{d} &  \bigoplus_i H^2(H_i,\Z) \arrow[equal]{d} \\
\qquad 0 \arrow{r} & \bigoplus_i H^1(\wt N_i, \Lambda^1_{E_i}) \arrow{r}{\wt \delta}
& \bigoplus_i H^2(\wt N_i,\Z) \arrow{r}{\rm Res}  & \bigoplus_i H^2(H_i,\Z).
\end{tikzcd}
\end{equation}

This shows the first statement.

(2) The map $\wh \Delta$ in (<ref>) is induced from the restriction of $X(T^E)$ to the subtorus $\Gmk$ and therefore is given by

\begin{equation}
\label{eq:hatD}
\wh \Delta=\sum_i \wh \Delta_i : \bigoplus_{i=1}^r \Ind_{\wt N_i}^G \Z\to \Z, \quad \wh \Delta_i(g\otimes n)=n, \quad \forall\, g\in G, n\in\Z.
\end{equation}

Thus, by the definition, the induced map $\wh \Delta_i^2$ on $H^2(G,-)$ is nothing but the corestriction $\Cor: H^2(\wt N_i, \Z) \to H^2(G,\Z)$; see Section <ref>.

Taking Galois cohomology to upper row of (<ref>), we have an exact sequence of abelian groups:

\begin{equation} \label{eq:1strow1}
\begin{tikzcd}
\qquad H^1(G, X(T^K)) \arrow{r} & H^1(G,\Lambda^1) \arrow{r}{\delta}
& H^2(G, X(T^E)) \arrow{r}{\wh N^2}  &  H^2(G, X(T^K)).
\end{tikzcd}
\end{equation}

The first term $H^1(G,X(T^K))=\bigoplus_i H^1(H_i,\Z)=0$. Using the relations (<ref>) and (<ref>) and by Shapiro's lemma, (<ref>) gives an exact sequence of abelian groups

\begin{equation} \label{eq:1row}
\begin{tikzcd}
\qquad 0 \arrow{r} & \bigoplus_i H^1(\wt N_i, \Lambda^1_{E_i}) \arrow{r}{\wt \delta}
& \bigoplus_i H^2(\wt N_i,\Z) \arrow{r}{\rm Res}  & \bigoplus_i H^2(H_i,\Z).
\end{tikzcd}
\end{equation}

Taking Galois cohomology of the diagram (<ref>), we then obtain the following commutative diagram:

\begin{equation} \label{eq:fund_diagram}
\begin{tikzcd}
\qquad  & 0 \arrow{r} \arrow{d} & \bigoplus_i H^1(\wt N_i, \Lambda^1_{E_i}) \arrow{r}{\wt \delta} \arrow[equal]{d}
& \bigoplus_i H^2(\wt N_i,\Z) \arrow{r}{\rm Res} \arrow{d}{\wh \Delta^2=\Cor} & \bigoplus_i H^2(H_i,\Z) \arrow{d}    \\
\qquad 0 \arrow{r} & H^1(G,\Lambda) \arrow{r}  &  H^1(G,\Lambda^1)  \arrow{r}{\delta} & H^2(G,\Z)  \arrow{r}{}
& H^2(G,\Lambda).
\end{tikzcd}
\end{equation}

With the identification (<ref>), we have
$\delta= \Cor \circ \wt \delta= \Cor\circ \Inf $ and
the lower long exact sequence of (<ref>) becomes as follows:

\begin{equation} \label{eq:4term_es}
\begin{tikzcd}
\qquad 0 \arrow{r} & H^1(G,\Lambda) \arrow{r}  & \bigoplus_i H^2(N_i^{\rm ab}, \Z) \arrow{r}{\Cor\circ \Inf} & H^2(G,\Z)  \arrow{r}{\wh N^2}
& H^2(G,\Lambda).  &
\end{tikzcd}
\end{equation}

By Propositions <ref> and <ref>,
under the isomorphism (<ref>)

the map $\Cor \circ \Inf: H^2(N_i^{\rm ab},\Z) \to H^2(G,\Z)$ corresponds to
$\Ver^{\vee}_{G,N_i}: N_i^{\rm ab,\vee} \to G^{\rm ab,\vee}$, where $\Ver^{\vee}_{G,N_i}$ is the dual of the transfer $\Ver_{G, N_i}: G^{\rm ab}\to N_i^{\rm ab}$ relative to $H_i$.
This proves (<ref>). Then it follows from  (<ref>) that $H^1(\Lambda)\simeq \Ker (\sum_i \Ver^{\vee}_{G,N_i})$. As the map $\Ver_{G, N_i}: G^{\rm ab}\to N_i^{\rm ab}$ factors as

\[ \begin{CD}
G @>\pi_{H_i}>> (G/H_i)^{\rm ab} @>\Ver_{G/H_i, \wt N_i/H_i}>> N_i^{\rm ab},
\end{CD}
\]

the last assertion (<ref>) follows.

In terms of class field theory, we have $C_{E_i}/N_{K_i/E_i}(C_{K_i})\simeq N_i^{\rm ab}$, $C_k/N_{K_i/k}(C_{K_i})\simeq (G/H_i)^{\rm ab}$ and $C_k/N_{L/k}(C_L)\simeq G^{\rm ab}$. Let $L_0$ be the compositum of all Galois closures of $K_i$ over $k$; this is the minimal splitting field of the algebraic torus $T^{K/E,k}$. One has $L_0\subset L$ and its Galois group $G_0:=\Gal(L_0/k)$ is a quotient of $G$. Thus, we have the following commutative diagram:

\begin{equation}
\label{eq:indep}
\begin{CD}
\frac{C_k}{N_{L/k}(C_L)} @>>> \frac{ C_k}{N_{L_0/k}(C_{L_0})} @>>> \prod_i \frac{C_k}{N_{K_i/k}(C_{K_i})} @>>>   \prod_i \frac{C_{E_i}}{N_{K_i/E_i}(C_{K_i})} \\
@VVV @VVV @VVV @VVV \\
G^{\rm ab} @>>> G_0^{\rm ab} @>>> \prod_{i} (G/H_i)^{\rm ab} @>>> \prod_{i} N_i^{\rm ab}.
\end{CD}
\end{equation}

Taking the Pontryagin dual, the lower row gives

\begin{equation}
\label{eq:indp-dual}
\begin{CD}
\prod_{i} N_i^{\rm ab,\vee}  @>>> \prod_{i} (G/H_i)^{\rm ab,\vee} @>>> G_0^{\rm ab,\vee}
\subset  G^{\rm ab,\vee}.
\end{CD}
\end{equation}

From this, we see that the map $\sum \Ver_{G,N_i}^{\vee}$ in (<ref>) has image contained in $G_0^{{\rm ab},\vee}$. It follows from (<ref>) that this map is independent of the choice of the splitting field $L$. As a result, its kernel $H^1(G,\Lambda)$ is also independent of the choice of $L$, which is a known fact from remarks in <cit.>, also cf. Remark <ref>.

(Here we add the computation of relative transfer map)
We compute the relative transfer map in the case when $K_i/E_i$ is cyclic with group $N_i=\<\sigma_i\>$ generated by an element $\sigma_i$. It suffices to describe $\Ver_{G, N_i}: G\to N_i$ for each $i$ and we drop $i$ from the notation for simplicity.

Since $\wt N/H$ is abelian, $H\supset D(\wt N)$ and $H=D(\wt N)H$. Put $X:=G/ \wt N$, $\wt X:=G/H$ and $\wt \varphi:X\to G$ a section and $\varphi: X\to \wt X$ the induced section.

(Here we add the computation o

§.§ On $\Sha^2(\Lambda)$

Let $\scrD$ be a set of representatives of the conjugacy classes of
all decomposition groups of $G$. For any $G$-module $A$, denote by

$$H^i_\scrD(A):=\prod_{D\in \scrD} H^i_D(A), \quad H^i_D(A):=H^i(D,A)$$

\[ r^i_{\scrD,A}=(r^i_{D,A})_{D\in \scrD}: H^i(G,A) \to H^i_\scrD(A) \]

the restriction map to all possible decomposition groups up to conjugate.
If $D$ and $D'$ are conjugate under $G$, then $\Ker r^i_{D,A}=\Ker r^i_{D',A}$. Thus, $\Sha^i(A)= \Ker r^i_{\scrD,A}$. We shall write $r_D$ and $r_{\scrD}$ for $r^i_{D,A}$ and $r^i_{\scrD,A}$, respectively, if it is clear from the content.

For the remainder of this section, we assume that the extension $K_i/E_i$ is cyclic with Galois group $N_i$ for each $i$. Consider the following commutative diagram

\begin{equation}
\label{eq:lgd} % local-global diagram
\begin{CD}
H^1(\Lambda^1) @>{\delta}>> H^2( \Z) @>{\wh N}>> H^2(\Lambda) @>{\wh j}>> H^2(\Lambda^1) \\
@VV{r^1_{\scrD,\Lambda^1}}V @VV{r^2_{\scrD,\Z}}V @VV{r^2_{\scrD,\Lambda}}V @VV{r^2_{\scrD,\Lambda^1}}V \\
H^1_{\scrD}(\Lambda^1) @>\delta_\scrD>>   H^2_{\scrD}( \Z) @>{\wh N}>> H^2_{\scrD}(\Lambda) @>{\wh j}>> H^2_{\scrD}(\Lambda^1). \\
\end{CD}
\end{equation}

\begin{equation}
\label{eq:H2Zp}
\begin{split}
H^2(\Z)':&=\{x\in  H^2(\Z) : \wh N (x) \in \Sha^2(\Lambda) \} \\
&=\{x\in H^2(\Z) : r^2_{\scrD,\Z} (x) \in {\rm Im} (\delta_\scrD) \} .
\end{split}
\end{equation}

We shall call $H^2(\Z)'$ the primitive part of $H^2(\Z)$, which plays an important role in computing Tamagawa numbers.

Assume that $K_i/E_i$ is cyclic with Galois group $N_i$ for all $i$. Then
$\Sha^2(\Lambda)\simeq H^2(\Z)'/{\rm Im}(\delta)$ and

\begin{equation}
\label{eq:tauT}
\tau_k(T^{K/E,k})=\frac{\prod_{i=1}^r |N_i|}{|H^2(\Z)'|}.
\end{equation}

It is obvious that $\wh j(\Sha^2(\Lambda))\subset \Sha^2(\Lambda^1)$ and then we have a long exact sequence

\begin{equation}
\label{eq:formula}
\begin{CD}
0 @>>> H^1(\Lambda) @>>>  H^1(\Lambda^1) @>{\delta}>> H^2( \Z)' @>{\wh N}>> \Sha^2(\Lambda) @>{\wh j}>> \Sha^2(\Lambda^1).
\end{CD}
\end{equation}

One has

\[ \Sha^2(\Lambda^1)=\bigoplus_{i} \Sha^2(G,\Ind_{\wt N_i}^G \Lambda_{E_i}^1)
=\bigoplus_{i} \Sha^2(\wt N_i, \Lambda_{E_i}^1)=\bigoplus_{i} \Sha^2(N_i, \Lambda_{E_i}^1)
\]

because $\Sha^2(X(T))$ does not depend on the choice of the splitting field.
Since $K_i/E_i$ is cyclic, its Galois group $N_i$ is a decomposition group of $K_i/E_i$ by Chebotarev's
density theorem. Thus, $\Sha^2(N_i, \Lambda^1_{E_i})=0$ for all $i$ and $\Sha^2(\Lambda^1)=0$.

Therefore, one gets the 4-term
exact sequence

\begin{equation}
\label{eq:formula-2}
\begin{CD}
0 @>>> H^1(\Lambda) @>>>  H^1(\Lambda^1) @>{\delta}>> H^2( \Z)' @>{\wh N}>> \Sha^2(\Lambda) @>{}>> 0.
\end{CD}
\end{equation}

From this we obtain $\Sha^2(\Lambda)\simeq H^2(\Z)'/{\rm Im}(\delta)$ and

\[ \tau_k(T^{K/E,k})=\frac{|H^1(\Lambda)|}{|\Sha^2(\Lambda)|}=
\frac{|H^1(\Lambda^1)|}{|H^2(\Z)'|}=\frac{\prod_{i=1}^r |N_i|}{|H^2(\Z)'|}. \quad \text{\qed}\]

Ono showed [14] that $\tau(R^{(1)}_{K/k} \GmK)=[K:k]$ for any cyclic extension $K/k$. Since $|H^1(K/k, X(R^{(1)}_{K/k} \GmK))|=[K:k]$, it follows that $\Sha^2(K/k, X(R^{(1)}_{K/k} \GmK))=0.$ This gives an alternative proof of the first statement $\Sha^2(\Lambda^1)=0$ of Proposition <ref>.

In order to compute the groups $H^2(\Z)'$ and $\Sha^2(\Lambda)$, we describe the maps $\delta$ and $\delta_\scrD$ in the first commutative diagram of (<ref>).

Since $\Lambda^1=\oplus_i \Lambda^1_i$, where $\Lambda^1_i=X(T^{K_i/E_i,1})=\Ind^G_{\wt N_i} \Lambda^1_{E_i}$, it suffices to describe the following commutative diagram:

\begin{equation}
\label{eq:lgd-1}
\begin{CD}
H^1(\Lambda^1_i) @>{\delta}>> H^2( \Z) \\
@VV{r_{D}}V @VV{r_{D}}V \\
H^1_{D}(\Lambda^1_i) @>\delta_D>>   H^2_{D}( \Z).  \\
\end{CD}
\end{equation}

Using the commutative diagram (<ref>), it factors as the following one:

\begin{equation}
\label{eq:lgd-2}
\begin{CD}
H^1(\Ind_{\wt N_i}^G \Lambda^1_{E_i}) @>{\wt \delta}>> H^2(\Ind_{\wt N_i}^G \Z)
@>{\wh \Delta^2}>> H^2( \Z) \\
@VV{r_{D}}V @VV{r_{D}}V @VV{r_{D}}V \\
H^1_{D}(\Ind_{\wt N_i}^G \Lambda^1_{E_i}) @>\wt \delta_D>>   H^2_{D}(\Ind_{\wt N_i}^G \Z)@>{\wh \Delta^2_D} >>   H^2_{D}( \Z).  \\
\end{CD}
\end{equation}

For any element $g\in G$ and any $\wt N_i$-module $X$, let $X^g=X$ be the $g \wt N_ig^{-1}$-module defined by

\[ h' x= (g^{-1} h' g) x, \quad \text{for $x\in X^g$ and $h'\in g \wt N_ig^{-1}$}. \]

Mackey's formula <cit.> says that as $D$-modules one has

\begin{equation}
\label{eq:M} %Mackey's formula
\Ind_{\wt N_i}^G X=\bigoplus_{g\in D\backslash G/\wt N_i} \Ind_{\wt N_{i,g}}^D X^g,
\end{equation}

where $g$ runs through double coset representatives for $D\backslash G/\wt N_i$ and

\[ \wt N_{i,g}:=D \cap g \wt N_ig^{-1} \subset g \wt N_ig^{-1}. \]

By Mackay's formula (<ref>) and Shapiro's lemma, we obtain

\begin{equation}
\label{eq:lgd-3}
\begin{CD}
H^1(\wt N_i, \Lambda^1_{E_i}) @>{\wt \delta}>> H^2({\wt N_i}, \Z)
@>{\Cor}>> H^2( \Z) \\
@VV{r_{D}}V @VV{r_{D}}V @VV{r_{D}}V \\
\bigoplus_{g} H^1(\wt N_{i,g}, (\Lambda^{1}_{E_i})^g)
@>\wt \delta_D>>  \bigoplus_{g} H^2(\wt N_{i,g}, \Z)@>{\Cor} >>
H^2_{D}( \Z).
\end{CD}
\end{equation}

If $\wt N_i$ is normal in $G$, then

\[ D\backslash G/\wt N_i=G/D \wt N_i, \quad \wt N_{i,g}=D\cap \wt N_i=:D_i,\quad \text{and} \quad X^g=X. \]

In this case, the bottom row of (<ref>) expresses as

\begin{equation}
\label{eq:lgd-4}
\begin{CD}
H^1(D_i, \Lambda^{1}_{E_i})^{[G:D\wt N_i]}
@>\wt \delta_D>>  H^2(D_i, \Z)^{[G:D \wt N_i]} @>{\Cor} >>   H^2_{D}( \Z).
\end{CD}
\end{equation}

Recall from (<ref>) we have an exact sequence of $\wt N_i$-modules

\begin{equation}
\label{eq:L1Ei-2}
\begin{CD}
0 @>>> \Z @>\wh N >>  \Ind^{\wt N_i}_{H_i} \Z @>\wh j >> \Lambda^1_{E_i} @>>> 0
\end{CD}
\end{equation}

on which $H_i$ acts trivially.
Note that as an $N_i$-module,  $\Ind^{\wt N_i}_{H_i} \Z=\Ind^{N_i}_1 \Z$. Taking the Galois cohomology $H^{*}(D_i,-)$, we get an exact sequence

\begin{equation}
\label{eq:lgd-5}
\begin{CD}
0 @>>> H^1(D_i, \Lambda^{1}_{E_i})
@>\wt \delta_{D_i}>>  H^2(D_i, \Z) @>\wh N >>   H^2(D_i, \Ind_{H_i}^{\wt N_i} \Z)\simeq H^2(D_i\cap H_i, \Z)^{[\wt N_i: D_i H_i]}
\end{CD}
\end{equation}

as one has

\[ H^2(D_i,\Ind_{H_i}^{\wt N_i} \Z)=H^2(D_i, \bigoplus_{\wt N_i/D_i H_i} \Ind_{D_i\cap H_i}^{D_i} \Z)\simeq H^2(D_i\cap H_i, \Z)^{[\wt N_i: D_i H_i]} \]

by Mackey's formula and Shapiro's lemma.

Similar to (<ref>) and (<ref>), using the Inflation-Restriction exact sequence, we make the following identification

\begin{equation}\label{eq:Dbari}
H^1(D_i, \Lambda^{1}_{E_i})=H^2(\ol D_i, \Z), \quad \text{where} \ \ol D_i:=D_i/D_i\cap H_i \subset N_i
\end{equation}

and (<ref>) becomes

\begin{equation}
\label{eq:lgd-6}
\begin{CD}
H^2(\ol D_i, \Z)^{[G:D\wt N_i]}
@>\Inf>>  H^2(D_i, \Z)^{[G:D \wt N_i]} @>{\Cor} >>   H^2_{D}( \Z).
\end{CD}
\end{equation}

We obtain the following proposition.

Assume that the extension $K_i/E_i$ is cyclic and the subgroup $\wt N_i=\Gal(L/E_i)  \triangleleft G$ is normal (i.e. $E_i/k$ is Galois) for all $i$. There are natural isomorphisms $H^1(\Lambda^1)\simeq \bigoplus_iH^2(N_i,\Z)$ and $H^1_{\scrD}(\Lambda^1)=\bigoplus_{D} \bigoplus_{i}  H^2(\ol D_i,\Z)^{[G:D\wt N_i]}$. Under these identifications the first commutative diagram in (<ref>) decomposes as the following:

\begin{equation}
\label{eq:lgd-7}
\begin{CD}
\bigoplus_{i=1}^r H^2(N_i,\Z)  @>{\Inf}>> \bigoplus_{i=1}^r H^2({\wt N_i}, \Z)
@>{\Cor}>> H^2( \Z) \\
@VV{r_{\scrD}}V @VV{r_{\scrD}}V @VV{r_{\scrD}}V \\
\bigoplus_{D} \bigoplus_{i=1}^rH^2(\ol D_i,\Z)^{[G:D\wt N_i]}
@>\Inf>>  \bigoplus_{D} \bigoplus_{i=1}^r H^2(D_i,\Z)^{[G:D\wt N_i]} @>{\Cor} >>   \bigoplus_{D} H^2_{D}( \Z).
\end{CD}
\end{equation}

The following result gives a group-theoretic description of $\Sha^2(\Lambda)$.

Let the notation and assumptions be as in Proposition <ref>.

\[ \Ver_{G, \bfN}=(\Ver_{G, N_i})_i : G^{\rm ab} \to \prod_i N_i \text{ and }
\Ver_{D, \ol \bfD}=(\Ver_{D, \ol D_i})_i: D^{\rm ab}\to \prod_{i} \ol D_i \]

denote the corresponding transfer maps, respectively. Then

\begin{equation}
\label{eq:H2Zp-1}
H^2(G,\Z)'=\{f \in G^{{\rm ab}, \vee}: f {\big |}_{D^{\rm ab}}
\in {\rm Im}(\Ver_{D, \ol \bfD}^\vee)\  \forall\, D\in \scrD \},
\end{equation}

\begin{equation}
\label{eq:Sha2L-gt}
\Sha^2(\Lambda) \simeq \frac{\{f \in G^{{\rm ab}, \vee}: f {\big |}_{D^{\rm ab}} \in {\rm Im}(\Ver_{D, \ol \bfD}^\vee) \ \forall\, D\in \scrD \}}{{\rm Im} (\Ver^{\vee}_{G,\bfN})}.
\end{equation}

We translate the commutative diagram (<ref>)
in terms of group theory. For each decomposition group $D\in \scrD$, we have the following corresponding  commutative diagram:

\begin{equation}
\label{eq:lgd-8}
\begin{CD}
\prod_{i=1}^r N_i  @<{\pi_{\bf H}=(\pi_{H_i})}<< \prod_{i=1}^r \wt N_i^{\rm ab}
@<{\Ver_{G,\bfN}=(\Ver_{G,N_i})}<< G^{\rm ab}\\
@AA{\rm func}A @AA{\rm func}A @AA{\rm func}A \\
\prod_{i=1}^r \ol D_i  @<{\pi_{{\bf D \cap  H}}=(\pi_{D_i\cap H_i})}<<  \prod_{i=1}^r D_i^{\rm ab}  @<{\Ver_{D,\bfD}=(\Ver_{D,D_i})}<<  D^{\rm ab}.
\end{CD}
\end{equation}

Here we ignore the multiplicity $[G:D\wt N_i]$ because we are only concerned with the image of the map $\Cor\circ \Inf$ in (<ref>) and this does not effect the result.
Then $\Ver_{G, \bfN}: G^{\rm ab} \to \prod_i N_i$ and $\Ver_{D, \ol \bfD}: D^{\rm ab}\to \prod_{i} \ol D_i$ are the respective compositions. By Proposition <ref>, the map $\delta_D$ corresponds to
$\Ver_{D,\ol \bfD}^{\vee}$. From the second description of $H^2(\Z)'$ in  (<ref>), we obtain (<ref>).

By Propositions <ref> and <ref>, the map $\delta: H^1(\Lambda^1) \to H^2(\Z)'\subset H^2(\Z)$ (<ref>) corresponds to $\Ver_{G, \bfN}^\vee: \prod_i N_i^\vee \to G^{{\rm ab},\vee}$.
Thus, by Proposition <ref>, we obtain (<ref>). This proves the proposition.

§ MORE COMPUTATIONS OF SOME PRODUCT CASES

We keep the notation of the previous section. In this section we consider the case where the extensions $K_i/k$ are all Galois with group $G_i=\Gal(K_i/k)$. Let $L=K_1K_2\cdots K_r$ be the compositum of all $K_i$ over $k$ with Galois group $G=\Gal(L/k)$. Assume that the canonical injection $G\to G_1\times \dots \times G_r$  is an isomorphism and that
the Galois group $N_i:=\Gal(K_i/E_i)$ is cyclic.
As before, we put $T=T^{K/E,k}$, $\Lambda:=X(T)$, $T^1=T^{K/E,1}$ and $\Lambda^1:=X(T^{1})$.

§.§ $H^1(\Lambda)$ and $H^2(\Lambda^1)$

We discuss two cases of algebraic tori in Section <ref>.

First we consider the case where the extensions $K_i/k$ are all Galois with Galois group $G_i=\Gal(K_i/k)$. We keep the assumption previously that every extension $K_i/E_i$ is cyclic with Galois group $N_i$ and that subgroups $N_i$ are all normal in $G_i$ (i.e. the extensions $E_i/k$'s are Galois), or equivalently, $\wt N_i$ are all normal in $G$.

Let $L=K_1K_2\cdots K_r$ be the composite of all $K_i$ over $k$ with Galois group $G=\Gal(L/k)$. Assume that the canonical injection $G\to G_1\times \dots \times G_r$  is an isomorphism. Put

$T^1:=T^{K/E,1}$ and $\Lambda^1:=X(T^1)$.

Let $\Lambda_i:=X(T^{K_i/E_i,k})$. We have $H^1(\Lambda)\simeq \oplus_{i=1}^r H^1(\Lambda_i)$.

By Proposition <ref>, $H^1(\Lambda)$ is isomorphic to the kernel of the dual of the map $\prod_i \Ver_{G,N_i} : G \to \prod_i N_i^{\rm ab}$. By Lemma <ref>, $\prod_i \Ver_{G,N_i}=\prod_i \Ver_{G_i, N_i}: \prod_{i=1}^r G_i \to \prod_i N_i^{\rm ab}$,
and therefore the kernel of its dual is equal to the product of abelian groups $H^1(\Lambda_i)$ for $i=1,\dots, r$, by Proposition <ref>. This proves the lemma.

Note that in the proof of Lemma <ref> the cyclicity assumption of $N_i$ is actually not needed.
Now we compute $H^2(\Lambda^1)=H^2(G,\Lambda^1)$. Note that unlike the first cohomology group $H^1(X(T))$, the second cohomology group $H^2(G, X(T))$ does depend on the choice of the splitting field. For example, if we take $T=\Gmk$ and let $L/k$ be any finite Galois extension with group $G$, then  $H^2(G,X(T))=H^2(G,\Z)\simeq \Hom(G,\Q/\Z)$, which clearly depends on the Galois group $G$.

By (<ref>), we have

\begin{equation}
\label{eq:HqL1-1}
H^q(G,\Lambda^1)=\bigoplus_{i=1}^r H^q(\wt N_i,\Lambda^1_{E_i}), \quad \forall\, q\ge 0.
\end{equation}

Taking the Galois cohomology $H^{*}(N_i,-)$ of the exact sequence (<ref>), we get the long exact sequence

\begin{equation}
\label{eq:HqNiL1}
\begin{CD}
H^q(N_i, \Ind_{H_i}^{\wt N_i} \Z) @>>>
H^q(N_i, \Lambda^1_{E_i}) @>>>
H^{q+1}(N_i,  \Z) @>>>
H^{q+1}(N_i, \Ind_{H_i}^{\wt N_i} \Z).
\end{CD}
\end{equation}

Since $\Ind_{H_i}^{\wt N_i} \Z=\Ind_1^{N_i} \Z$  is an induced module, we get

\begin{equation}\label{eq:H2L1Ei}
H^2(N_i, \Lambda^1_{E_i})\isoto H^3(N_i, \Z)=H^1(N_i, \Z)=0,
\end{equation}

as $N_i$ is cyclic. Recall $H_i:=\Gal(L/K_i)$. Since $K_i/k$ is Galois and the algebraic torus $T^{K_i}$ splits over $K_i$, $\Lambda^1_{E_i}\simeq \Z^{a_i}$ is a trivial $H_i$-module with $a_i=\dim R^{(1)}_{K_i/E_i} (\G_{{\rm m}, K_i})=[K_i:E_i]-1$. It follows that

\begin{equation}\label{eq:H1HiL1Ei}
H^1(H_i, \Lambda^1_{E_i})=H^1(H_i, \Z^{a_i})=0.
\end{equation}

We will need the following result.

Let $H\subset G$ be a normal subgroup of a finite group $G$, let $A$ be a $G$-module, and let $n\ge 1$ be a positive integer. If $H^q(H, A)=0$ for all $0 < q < n$, then we have a 5-term long exact sequence

\begin{equation} \label{eq:HS5n}
0 \to H^n(G/H, A^H) \to H^n(G, A) \to H^n(H, A)^{G/H} \xrightarrow{d} H^{n+1}(G/H, A^H) \to H^{n+1}(G, A).
\end{equation}

We show that this follows from the Hochschild-Serre spectral sequence:

\[ E_2^{p, q}:=H^p(G/H, H^q(H, A^H)) \implies H^{p+q}(G,A), \]

cf. <cit.>.
Indeed, since $H^q(H, A)=0$ for all $0 < q < n$, we have $E_2^{p,q}=H^p(G/H, H^q(H,A))=0$ for all $0 < q < n$ and all $p$. By <cit.> we get the $5$-term long exact sequence

\[
0 \to E_2^{n,0} \to E^n \to E^{0,n}_2 \xrightarrow{d} E^{n+1,0}_2 \to E^{n+1}_2
\]

and this gives the 5-term exact sequence

By (<ref>) and Proposition <ref> with $n=2$ and $(G,H)=(\wt N_i, H_i)$,
we have the exact sequence (noting $H^2(N_i,\Lambda^1_{E_i})=0$ by (<ref>))

\begin{equation}
\begin{CD}
0   @>>> H^2(\wt N_i, \Lambda^1_{E_i}) @>>> H^2(H_i, \Lambda^1_{E_i})^{N_i} @>d>> H^3(N_i, \Lambda^1_{E_i}).
\end{CD}
\end{equation}

Using the same argument as (<ref>), we get

\begin{equation}
\label{eq:H3L1Ei}
H^3(N_i,\Lambda^1_{E_i})\simeq H^4(N_i,\Z)\simeq H^2(N_i,\Z)\simeq \Hom(N_i,\Q/\Z).
\end{equation}

On the other hand,

\begin{equation}\label{eq:H2HiL1Ei}
H^2(H_i, \Lambda^1_{E_i})\simeq H^2(H_i, \Z)^{a_i}\simeq \Hom(H_i,\Q/\Z)^{a_i}.
\end{equation}

Note that the action of $N_i$ on $H^2(H_i, \Lambda^1_{E_i})$ is induced by the conjugation by $\wt N_i$ on $H_i$ and by the fact that the action of $H_i$ on $H^2(H_i, \Lambda^1_{E_i})$ is trivial. Since the isomorphisms in (<ref>) are functorial, the action on $N_i$ on $\Hom(H_i, \Q/\Z)$ is simply the one induced by the conjugation of $\wt N_i$ on $H_i$. Note that

\begin{equation}\label{eq:HiwtNi}
\begin{split}
H_i&=\, G_1\times \dots \times G_{i-1} \times \{1 \} \times G_{i+1} \times \dots \times G_r, \\
\wt N_i&=\, G_1\times \dots \times G_{i-1} \times N_i \times G_{i+1} \times \dots \times G_r.
\end{split}
\end{equation}

The conjugation of $\wt N_i$ on $H_i^{\rm ab}$ is trivial.
Therefore, $N_i$ acts trivially on $\Hom(H_i,\Q/\Z)^{a_i}$ and $H^2(H_i, \Lambda^1_{E_i})^{N_i}\simeq \Hom(H_i,\Q/\Z)^{a_i}$.
So, we obtain an exact sequence

\begin{equation}\label{eq:di}
\begin{CD}
0   @>>> H^2(\wt N_i, \Lambda^1_{E_i}) @>>> \Hom(H_i, \Q/\Z)^{a_i} @>d_i>> \Hom(N_i, \Q/\Z).
\end{CD}
\end{equation}

We conclude the following:

Let $K=\prod_{i=1}^r K_i$ and $E=\prod_{i=1}^r E_i$ be as before. Suppose that each $K_i/k$ is Galois with group $G_i$. Assume $G\simeq G_1\times \dots \times G_r$ and $N_i=\Gal(K_i/E_i)$ is cyclic for all $i$. Then $H^2(\Lambda^1)\simeq \oplus_{i=1}^r H^2(\wt N_i, \Lambda_{E_i}^1)$ and

\begin{equation}
H^2(\wt N_i, \Lambda_{E_i}^1)\simeq \Ker \left (d_i: \Hom(H_i, \Q/\Z)^{[K_i:E_i]-1} \longrightarrow \Hom(N_i, \Q/\Z) \right ),
\end{equation}

where $d_i$ is the map in (<ref>).
If $r=1$, that is $K/k$ is a Galois field extension, then $H^2(\Lambda^1)=0$.

Let the notation and assumptions be as in Proposition <ref>. Assume further that the orders of groups $G_i$ are mutually relatively prime.  Then

\begin{equation}\label{eq:H2L1-abel}
H^2(\Lambda^1)\simeq \bigoplus_{i=1}^r \Hom(H_i, \Q/\Z)^{[K_i:E_i]-1}.
\end{equation}

(Note that in our notation the group $H_1=\{1\}$ if $r=1$.)

The assumption implies that $|H_i|$ and $|N_i|$ are relatively prime. Thus, the maps $d_i$'s are all zero.

From Corollary <ref>, we see that the second cohomology group $H^2(\Lambda^1)$ may not vanish, which is different from the case $r=1$.
It is desirable to describe the map $d_i$ in terms of group theory. Since the map $d_i$ induces a natural map $N_i \to (H_i^{\rm ab})^{[K_i/E_i]-1}$ and such a functorial map we could only think of is the trivial one, it seems that the map $d_i$ is simply the trivial map.

§.§ $\tau(T)$
In this subsection we shall further assume that each subgroup $N_i$, besides being cyclic, is also normal in $G_i$. This assumption simplifies the description of the group $H^2(\Z)'$ through Mackey's formula.
For each
$1 \le i \le r$
let $T_i:=T^{K_i/E_i, k}$, and let $\Lambda_i:=X(T_i)$ be the character group of $T_i$.

Our goal is to reduce the problem of computing $\tau(T)$ into  that of $\tau(T_i)$. Recall by Proposition <ref>, we have

\begin{equation}\label{eq:tauTi}
\tau(T)=\frac{\prod_{i=1}^r |N_i|}{|H^2(\Z)'|} \quad \text{and} \quad \tau(T_i)=\frac{|N_i|}{|H^2(G_i,\Z)'|}.
\end{equation}

Thus, it reduces to compare the groups $H^2(\Z)'$ and $\prod_{i=1}^r H^2(G_i,\Z)'$.
The following lemma gives a partial result.

Let $K=\prod_{i=1}^r K_i$ and $E=\prod_{i=1}^r {E_i}$ be $k$-algebra, and
let $T=T^{K/E,k}$ and $T_i=T^{K_i/E_i,k}$ for each $1\le i \le r$ as in Section <ref>.
Assume the following conditions:

(a) each $K_i/k$ is a Galois extension with Galois group $G_i=\Gal(K_i/k)$;

(b) each $K_i/E_i$ is a cyclic extension with Galois group $N_i=\Gal(K_i/E_i)$ normal in $G_i$;

(c) the Galois extensions $K_i/k$ are mutually linearly disjoint over $k$. That is, the Galois group $G:=\Gal(L/k)$ of the the composite $L=K_1K_2\cdots K_r$ over $k$ of $K_i$'s is isomorphic to the product $G_1\times \dots \times G_r$ of the Galois groups $G_i$.

Then $\tau_k(T)=\prod_{i=1}^r \tau_k(T_i)$.

Let $L/k$ be a Galois extension and $K/k$ a Galois subextension of $L/k$. Then the natural projection $\pr:\Gal(L/k)\to \Gal(K/k)$ maps the decomposition group $D_\grP$ of a prime $\grP$ of $L$ onto $D_{\grP_K}$, where $\grP_K$ is the prime of $K$ below $\grP$.

Let $\grp$ be the prime of $k$ below the prime $\grP$.
If $\sigma\in \Gal(L/k)$ and $\sigma(\grP)=\grP$, then $\sigma(\grP_K)=\grP_K$ and $\pr(D_\grP)\subset D_{\grP_K}$. Note that the fixed subfield $K^{D_{\grP_K}}$ is characterized as the maximal subfield extension $K'$ of $K/k$ such that $e(\grP'|\grp)=f(\grP'|\grp)=1$, where $\grP'$ is the prime of $K'$ below $\grP_K$. Thus, if $\grP_1$ is the prime of $L^{D_\grP}$ below $\grP$, then $e(\grP_1|\grp)=f(\grP_1|\grp)=1$. Since $K^{D_\grP}\subset L^{D_\grP}$, the prime $\grP_2$ of $K^{D_\grP}$ below $\grP_1$ satisfies  $e(\grP_2|\grp)=f(\grP_2|\grp)=1$. Therefore, $K^{D_\grP}$ is contained in $K^{D_{\grP_K}}$ and $D_{\grP_K} \subset \pr (D_\grP)$. This proves the lemma.

Note that $D_\grP$ may not contain $\Gal(L/K)$ and hence it is not the preimage of $D_{\grP_K}$ of the map $\pr$ in general.

Let $\scrD$ be the set of all decomposition subgroups of $G$, and let $\scrD_i$ be that of $G_i$ for each $i$.
Let $f\in \Hom(G, \Q/\Z)=\prod_{i=1}^r \Hom(G_i, \Q/\Z)$ be an element and write $f=(f_1, \dots, f_r)$ with $f_i\in \Hom(G_i, \Q/\Z)$. Consider the following two conditions:

(a) For each $D\in \scrD$, the restriction $f|_{D_{}^{}}$ of $f$ to $D^{}$ lies in the image of the map $\sum_{i=1}^r \Ver_{D,\ol D_i}^\vee$.

(b) For each $1 \le i\le r$ and $D'_i\in \scrD_i$,
the restriction $f_i|_{D_{i}'^{}}$ to $D_{i}'^{}$ lies in the image of the map
$\Ver_{D_{i}', D_{i}' \cap N_i}^\vee$.

Then the condition (b) implies (a). In particular, one has the inclusion $\prod_{i=1}^r H^2(G_i,\Z)'\subset H^2(\Z)'$.

We have $G=\prod_{i=1}^r G_i$, and $H_i$ and $\wt N_i$ are as in (<ref>).
Let $D=D_\grP\in \scrD$ be a decomposition group associated to a prime $\grP$ of $L$.
By definition $D_i=D\cap \wt N_i$ and $\ol D_i=D_i/D_i \cap H_i$. The projection map $\pr_i: G\to G_i$ sends each element $\sigma$ to its restriction $\sigma|_{K_i}$ to $K_i$. Let $\grP_i$ denote the prime of $K_i$ below $\grP$ and we have $\pr_i(D_{\grP})= D_{\grP_i}$ by Lemma <ref>.

We show that $\pr_i(D_\grP \cap \wt N_i)=D_{\grP_i} \cap N_i$. The inclusion $\subseteq$ is obvious because $\pr_i(D_\grP \cap \wt N_i)\subset \pr_i(D_{\grP_i}) \cap \pr_i(\wt N_i)=D_{\grP_i} \cap N_i$. For the other inclusion, since $\pr_i: D_\grP\to D_{\grP_i}$ is surjective, for each $y\in D_{\grP_i} \cap N_i$ there exists an element $x=(x_i)\in D_\grP$ such that $x_i=y$. Since $x_i\in N_i$, $x$ also lies in $\wt N_i$. This proves the other inclusion. Therefore, $\ol D_i=D_{\grP_i} \cap N_i$.

The map $\sum_{i=1}^r \Ver_{D,\ol D_i}^\vee$ is dual to the map

\[ \prod_{i=1}^r \Ver_{D, D_{\grP_i} \cap N_i}: D \to \prod_{i=1}^r (D_{\grP_i} \cap N_i). \]

Since each $\Ver_{D, D_{\grP_i} \cap N_i}= \Ver_{D_{\grP_i}, D_{\grP_i} \cap N_i}\circ \, \pr_i$ (cf. (<ref>)),
the above map factorizes into the following composition

\[
\begin{CD}
D @>(\pr_i)_i>> \prod_{i=1}^r D_{\grP_i} @>
\prod_{i} {\Ver_{D_{\grP_i}, D_{\grP_i} \cap N_i}}>> \prod_{i=1}^r (D_{\grP_i} \cap N_i).
\end{CD}
\]

Consider the restriction $f|_{\prod_i D_{\grP_i}}=(f_i |_{D_{\grP_i}})$ of $f$ to $\prod_i D_{\grP_i}$, and put $D_i':=D_{\grP_i}$.
By condition (b), there exists an element $h_i\in \Hom(D_{\grP_i} \cap N_i, \Q/\Z)$ such that $f_i |_{D_{\grP_i}} = \Ver^\vee_{D_{\grP_i}, D_{\grP_i} \cap N_i} (h_i)$. Then $f|_{\prod_i D_{\grP_i}}=\left (\Ver^\vee_{D_{\grP_i}, D_{\grP_i} \cap N_i} (h_i) \right )$.
Restrict this function to $D$, we obtain $f|_D=\sum_{i=1}^r \Ver_{D,\ol D_i}^\vee (h_i)$. This shows that $f|_D$ lies in the image of $\sum_{i=1}^r \Ver_{D,\ol D_i}^\vee$ and hence proves the lemma.
Conversely, if $f|_{D^{\rm ab}}$ lies in the image of $\sum_{i=1}^r \Ver_{D,\ol D_i}^\vee$. Since $D^{\rm ab} \to D_{\grP_i}^{\rm ab}$ is surjective, the function $f_i|_{D_{\grP_i}}$ is determined by $f|_D$

Let the notation and assumptions be that as in Lemma <ref>. Assume that for each $1\le i \le r$ and each $D_i'\in \scrD_i$, there exists a member $D\in \scrD$ such that $\pr_i(D)=D'_i$ and a section $s_i: D'_i\to D$ of $\pr_i:D\twoheadrightarrow D_i'$ such that
the composition $\pr_k \circ s_i: D_i'\to D \to D_k'$ is trivial for $k\neq i$. Then $\prod_{i=1}^r H^2(G_i,\Z)'= H^2(\Z)'$.

Let $f\in \Hom(G, \Q/\Z)=\prod_{i=1}^r \Hom(G_i, \Q/\Z)$ be an element and write $f=(f_1, \dots, f_r)$ with $f_i\in \Hom(G_i, \Q/\Z)$. By Lemma <ref>, we have $\prod_{i=1}^r H^2(G_i,\Z)'\subset H^2(\Z)'$. To prove the other inclusion, we must show that each $f_i$ satisfies condition (b) of Lemma <ref> provided $f$ satisfies condition (a), which we assume from now on. For each $D'_i\in \scrD_i$, let $D\in \scrD$ be a decomposition group of $G$ over $D_i'$ and $s_i:D_i'\to D$ be a section such that $\pr_k \circ s_i$ is trivial for $k\neq i$. Then

\[ f|_D=\sum_{k=1}^r \Ver_{D,\ol D_k}^\vee (h_k) \]
for some $h_k\in \Hom(\ol D_k, \Q/\Z)$. Pulling back to $D_i'$ via $s_i$, we have

\[ f_i|_{D_i'}=s_i^* f|_D=\sum_{k=1}^r s_i^* \Ver_{D,\ol D_k}^\vee (h_k)=\sum_{k=1}^r s_i^* \pr_k^* \Ver_{D'_k,\ol D_k}^\vee (h_k). \]

Since $s_i^*\pr_k^*=1$ if $k=i$ and  $s_i^*\pr_k^*=0$ otherwise, we get $f_i|_{D_i'}=\Ver_{D'_k,\ol D_k}^\vee (h_k)$. This proves the lemma.

Let the notation and assumptions be that as in Lemma <ref>, and let $T:=T^{K/E,k}$ and $T_i:=T^{K_i/E_i,k}$. Suppose further that for each $1\le i\le r$, every decomposition group of $G_i$ is cyclic. Then $\tau(T)=\prod_{i=1}^r \tau(T_i)$.

We shall show that the assumption in Lemma <ref> holds. Let $D'_i\in \scrD_i$ be a decomposition group of $G_i$; by our assumption $D'_i$ is cyclic and let $a_i$ be a generator.
Let $D$ be a cyclic subgroup of $G$ generated by $\tilde a_i=(1, \dots, 1,a_i, 1,\dots, 1)$ with $a_i$ at the $i$th place.
Clearly $D$ is the decomposition group of some prime and let $s_i:D_i' \to D$ be the section sending $a_i$ to $\tilde a_i$. Clearly for $k\neq i$, $\pr_k \circ s_i:D_i' \to D_k'$ is trivial. By Lemma <ref>, we have $H^2(G,\Z)'=\prod_{i=1}^r H^2(G_i,\Z)'$. It follows from (<ref>) that $\tau(T)=\prod_{i=1}^r \tau(T_i)$.

We remark that results in Sections <ref> and <ref> are valid for a global field $k$. The Tamagawa number of an algebraic $k$-torus for a global function field $k$ is defined slightly different in <cit.> and Ono's formula is also true <cit.>.

(Perhaps formulate a group theoretic lemma to let the argument clear)

Now we consider the special case where $K/k$ is a finite Galois extension with Galois group $G$, and assume $N:=\Gal(K/E)$ is a normal cyclic subgroup of $G$. By Proposition

\[ H^2(\Z)'=\{ f\in G^{\rm ab , \vee}: f|_{D^{\rm ab}} \in {\rm Im} \Ver^\vee_{D, D\cap N}, \, \forall\, D\in \scrD \}. \]

§ GALOIS CM FIELDS

We return to the case of CM tori and keep the notation of Section <ref>.
In this section, we assume that $K$ is a Galois CM field. Let $G=\Gal(K/\Q)$ and $G^+:=\Gal(K^+/\Q)$.
Then one has the short exact sequence

\[1 \rightarrow \langle\iota\rangle \rightarrow G \rightarrow G^{+} \rightarrow 1. \tag{$*$}\]

Let $T=T^{K,\Q}$ be the associated CM torus over $\Q$.

§.§ Cyclotomic extensions

We first consider cyclotomic fields $K=\Q(\zeta_n)$ for positive integer $n>1$, where
$\zeta_n$ is a primitive $n$th root of unity. We may and will assume that either $n>1$ is odd or $4|n$

Let $K=\Q(\zeta_n)$ be the $n$th cyclotomic field.

If $n$ is not a power of a prime, then the CM extension
$K/K^{+}$ is unramified at all finite places of $K^+$.

(2) If $n=p^m$ is a power of a
prime $p$, then $K/K^{+}$ is ramified precisely at the unique place $v$ lying over $p$ and all real places of $K^+$.

See <cit.>.

Let $K=\mathbb{Q}(\zeta_{n})\neq \Q$ be the $n$th cyclotomic field and $T$ the associated CM torus over $\Q$.

$(1)$ If $n$ is either a power of an odd prime $p$ or $n=4$, then $\tau(T) = 1.$

$(2)$ In other cases, $\tau(T) = 2$.

$(1)$  In this case the Galois group $G=\Gal(K/\Q)$ is cyclic and Proposition <ref>(3), $\tau(T) = 1$.

$(2)$ Suppose first that $n$ is not a power of a prime. By Proposition <ref>(2) we have
$n_{K} | \prod_{p \in S_{K/K^{+}}}e_{T,p}$. By
Lemma <ref> we have $S_{K/K^{+}} = \emptyset$, so $\prod_{p \in
S_{K/K^{+}}}e_{T,p}~=~1$. As a consequence, $n_{K} = 1$ and $\tau(T)=2$ by Proposition <ref>(1). Now suppose
$n=2^v$ with $v\ge 3$, then $K$ contains $\mathbb{Q}(\zeta_{8})=\Q(\sqrt{2},\sqrt{-1})$,
which contains two distinct imaginary quadratic fields. Thus,
by Lemma <ref>, we have $n_{K} = 1$ and hence $\tau(T)=2$ by Proposition <ref>(1).

This completes the proof.

§.§ Abelian extensions

If $\,G=\Gal(K/\Q)$ is abelian, then $\tau(T) \in \{1,2\}$.

Write $G = \prod_{i=1}^{\ell} G_{i}$ as a product of cyclic subgroups $G_i$.

Then there exists $i$ such that the image of $\iota$ in the projection
$G \to G_{i}$ is nontrivial, say $i=1$. Let $K_{1}$ be a subfield of $K$ with $\Gal(K_{1}/\mathbb{Q}) = G_{1}$, which is a cyclic CM field.

Then we have $\tau(T^{K_{1}, \mathbb{Q}})=1$ by Proposition <ref>(3), that is, $n_{K_1} = 2$. Since $K\supset K_1$, we have $n_{K} \mid n_{K_{1}}$ and $n_{K} \in \{1,2\}$. Therefore, $\tau(T) \in \{2,1\}$.

Let $g=[K^+:\Q]$. Assume $G$ is abelian. Then the following statements hold:

$(1)$ If $g$ is odd, then $\tau(T) = 1$.

$(2)$ If $g$ is even and $(*)$ splits, then $\tau(T) = 2$.

$(1)$ This is proved by Proposition <ref>.

(2) Since $(*)$ splits, $G \simeq \langle\iota\rangle \times G^{+}$. As $g$ is even, there is an epimorphism

\begin{equation}\label{eq:biquad}
\langle\iota\rangle \times G^{+} \twoheadrightarrow
\langle\iota\rangle \times \mathbb{Z}/2\mathbb{Z}.
\end{equation}

In particular, $K$ contains two distinct imaginary quadratic fields; thereby
$n_K=1$ and $\tau(T)=2$ by Lemma <ref>.

Assume $G$ is abelian and $(*)$ is non-split and $G^{2}$ be a maximal $2$-torsion subgroup such that $G/G^{2}$ isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{r}$ for some $r$. Then

\[
|\Sha ^{2}(X(T))| \; \mathlarger{\mid} \; 2^{r+1}.
\]

Apply Kronecker's theorem, there exists $\mathbb{Q}(\zeta_{m})$ such
that $K \subseteq \mathbb{Q}(\zeta_{m})$. In particular, $K=
\mathbb{Q}(\zeta_{m})^{H}$, where $H \leq
\Gal(\mathbb{Q}(\zeta_{m})/\mathbb{Q})$ satisfies $\iota \notin H$. It
is clear that $\Gal(\mathbb{Q}(\zeta_{m})/\q) \cong
(\mathbb{Z}/m\mathbb{Z})^{\times}$, so $G =
(\mathbb{Z}/m\mathbb{Z})^{\times}/H$. Let $K_{2} = K^{G^{2}}$. We
therefore obtain a split exact sequence

$$1 \rightarrow G^{2} \rightarrow G \xrightarrow{pr} G/
G^{2} \rightarrow 1,$$

where $G / G^{2} = \Gal(K_{2}/\mathbb{Q}) \cong (\mathbb{Z}/
2\mathbb{Z})^r$. Then we have an imaginary quadratic field extension
$E$ over $\mathbb{Q}$ in $K_{2}$. Consequently, we have the following
equivalence relation:

\begin{align*}\vspace{-1cm}
\iota \notin G^{2} &\Leftrightarrow \Im_{pr}(\iota)\neq
0.\\&\Leftrightarrow \exists \text{ imaginary quadratic field $E$ over
$\q$ in $K_{2}$ }.  \\&\Leftrightarrow (*) \text{  splits}.
\end{align*}

Since $(*)$ is non-split, $\langle\iota\rangle \subset G^{2}$, we
may assume that $K_{2}$ is a totally real field of $K$. And we notice
that $g= [K^{+}:\mathbb{Q}] = [K_{2}:\mathbb{Q}] =
|\mathbb{Z}/2\mathbb{Z}|^{r}, r \geq 1$, in particular, $g$ is
even. We consider the following exact sequence:

$$H^{1}(\mathbb{Z}) \rightarrow H^{1}(X(T)) \rightarrow
H^{1}(X(T_{1}^{K})) \rightarrow H^{2}(\mathbb{Z})\rightarrow
H^{2}(X(T)) \rightarrow 0 .$$

In our case, the exact sequence turns out to be

$$0 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow
\mathbb{Z}/2\mathbb{Z} \rightarrow H^{2}(\mathbb{Z})\rightarrow
H^{2}(X(T)) \rightarrow 0 ,$$

$$0 \rightarrow 0 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow
H^{2}(\mathbb{Z})\rightarrow H^{2}(X(T)) \rightarrow 0 .$$

Hence $H^{2}(\mathbb{Z}) \cong H^{2}(X(T))$ or $H^{2}/
(\mathbb{Z}/2\mathbb{Z}) \cong H^{2}(X(T))$.

Also, we know that $\Sha^{2}(X(T)) \subseteq H^{2}(X(T))[2]$ and
$H^{2}(\mathbb{Z})[2] = G^{\vee}[2] = (G/G^{2})^{\vee}
\cong (\mathbb{Z}/2\mathbb{Z})^{r}$. Therefore, in either case,

$$ |\Sha^{2}(X(T))| \; \mathlarger{\mid} \; 2^{r+1}.$$

§.§ Certain Galois extensions

Assume the short exact sequence $(*)$ splits. Let $g^{ab} := |G^{+ {\rm ab}}|$ and $\Lambda:=X(T)$. Then $\tau(T)\in \{1,2\}$. Moreover, the following statements hold:

(1) When $g$ is odd, $\tau(T) = 1$.

(2) When $g$ is even,

(i) $\tau(T) = 2$ if $g^{ab}$ is even;

(ii) In the other case, there is a unique nonzero
element $\xi$ in the $2$-torsion subgroup $H^{2}(\Lambda)[2]$
of $H^{2}(\Lambda)$. Moreover, $\tau(T) = 1$ if and only if

its restriction $r_D(\xi) = 0$ in $H^2(D,\Lambda)$ for all $D\in \scrD$.

$(1)$$\tau(T)\in \{1,2\}$.

$(2)$If $g$ is odd, then $\tau(T) = 1$.

$(3)$Assume $g$ is even.

$(\romannumeral1)$ If $g^{ab}$ is even, then $\tau(T) =

$(\romannumeral2)$ If $g^{ab}$ is odd, then there is a
unique nonzero element $\xi \in H^{2}(X(T))[2]$,

$2$-torsion subgroup of $H^{2}(X(T))$.
$\tau(T) = 1 \Leftrightarrow \xi_{v} = 0$ for all places $v$
of $\Q$.

Since $(*)$ splits, $G \simeq \langle\iota\rangle \times G^{+}$. It follows that $K$ contains an imaginary quadratic field $E$. Since $n_E=2$, one has $n_K\in\{1,2\}$ and $\tau(T)\in \{2,1\}$.
The statement (1) follows from Proposition <ref>(2).

(2) We have $G \simeq \langle\iota\rangle \times G^{+}$ and $G^{\rm ab}\simeq \<\iota\>\times G^{+ {\rm ab}}$.
(i) Suppose $g^{ab}$ is even. As (<ref>), $K$ contains
two distinct imaginary quadratic fields. Thus, $n_{K} = 1$ and $\tau(T) = 2$.
(ii) Suppose $g^{ab}$ is odd. Put $\Lambda^1=X(T^{K,1})$ and we have the following exact sequence

$$0 \rightarrow
H^{1}(\Lambda) \rightarrow H^{1}(\Lambda^1) \rightarrow
H^{2}(\mathbb{Z}) \rightarrow H^{2}(\Lambda) \rightarrow H^2(\Lambda^1). $$

We have $H^{1}(\Lambda^1)\simeq \Z/2\Z$ by Proposition <ref>(1), $H^2(\Lambda^1)=0$ by Proposition <ref>, and $H^{1}(\Lambda)\simeq \Z/2\Z$ by <cit.>.
This gives $H^{2}(\mathbb{Z}) \cong H^{2}(\Lambda).$
The proof of <cit.> shows that $H^1(G,T)$ is a $2$-torsion group. It follows that $\Sha^2(\Lambda)$ is also a $2$-torsion group. Therefore, $\Sha^2(\Lambda)\subset H^2(\Lambda)[2]$.
Since $g^{\rm ab}$ is odd,

$$ H^2(\Lambda)[2]\simeq H^2(\Z)[2]=\Hom(G^{\rm ab}, \Q/\Z)[2]=\Hom(\langle\iota\rangle \times {G^{+{\rm ab}}},
\mathbb{Q}/\mathbb{Z})[2]\cong
\mathbb{Z}/2\mathbb{Z}.$$

Let $\xi \in H^{2}(X(T))[2]$ be the unique nonzero
element. We obtain the following equivalence relations:

\begin{align*}
\tau(T)=1 & \iff \Sha^2(\Lambda)\simeq \Z/2\Z \iff \xi \in \Sha^{2}(\Lambda) \\
& \iff r_D(\xi)=0 \quad \text{for all $D\in \scrD$. }
\end{align*}

The last equivalence follows from the definition of the Tate-Shafarevich group. This completes the proof of the proposition.

When $G$ is non-abelian and the short exact sequence $(*)$ does not split,
we don't know the value of $\tau(T)$ in general. However,
if the involution $\iota$ is non-trivial on the maximal abelian extension
$K_{ab}$ over $\Q$ in $K$,
then $K_{ab}$ is a CM abelian field. So from Proposition <ref>, the value of $n_{K_{ab}}$ is in $\{1,2\}$. Also,
since $n_{K}\mid n_{K_{ab}} $, we have $n_{K} =
\{1,2\}$ and $\tau(T) \in \{2,1\}$.

It remains to determine the Tamagawa number when $G$ is non-abelian, $(*)$ is non-split, and $K_{ab}$ is totally real. This includes the case $G = Q_{8}$ or $G=D_{4}$, which will be treated in the next two subsections.

§.§ $Q_8$-extensions

The quaternion group $Q_8=\{\pm 1, \pm i, \pm j, \pm k\}$ is the group of $8$ elements generated by $i, j$ with usual relations $i^2=j^2=-1$ and $ij=-ji=k$.
We have the following properties of $Q_8$, whose proofs are omitted.
We have the following properties:

(i) $Q_8$ contains $6$ elements of order $4$, one element of order $2$, and the identity. It does not has a subgroup isomorphic to $\Z/2\Z\times \Z/2\Z$. Thus, every proper subgroup is cyclic.

(ii) Every non-trivial subgroup contains $\{\pm 1\}$, and only subgroup which maps onto $Q_8/\{\pm 1\}$ is $Q_8$.

(iii) The center $Z(Q_8)=\{\pm 1\}=D(Q_8)$, where $D(Q_8)=[Q_8,Q_8]$.

Let $P$ and $Q$ be two odd positive integers such that

\begin{equation}\label{eq:PQ}
P-1=a^2,\quad Q-1=Pb^2
\end{equation}

for some integers $a,b\in  \bbN$. Assume that $Q$ is not a square. Let $K:=\Q(\beta)$ be the simple extension of $\Q$ generated by $\beta$ which satisfies $\beta^2=\alpha:=-(P+\sqrt{P})(Q+\sqrt{Q})$. Then
$K$ is a  Galois CM field with group $Q_8$ with maximal totally real field $K^+=\Q(\sqrt{P},\sqrt{Q})$.  The Galois group $\Gal(K/\Q)$ is generated by $\tau_1$ and $\tau_2$ given by

\begin{equation}
\label{eq:tau12}
\tau_1(\beta)=\frac{(\sqrt{P}-1)}{a} \beta, \quad  \tau_2(\beta)=\frac{(\sqrt{Q}-1)}{\sqrt{P} a} \beta.
\end{equation}

Moreover, for each prime $\ell$, the decomposition group $D_\ell$ is cyclic except when $\ell|Q$. For $\ell|Q$, one has

\begin{equation}
\label{eq:DlQ}
\Z/4\Z, & \text{if $\left(\frac{P}{\ell} \right)=1$;} \\
Q_8 ,  & \text{if $\left(\frac{P}{\ell} \right)=-1$.}
\end{cases}
\end{equation}

Note that $\gcd(P,Q)=1$ and $P$ is not a square. Then $\Q(\sqrt{P},\sqrt{Q})$ is a totally real  biquadratic field and its Galois group is generated by $\sigma_1$ and $\sigma_2$, where

\[ \sigma_1: \begin{cases} \sqrt{P} \mapsto -\sqrt{P} \\
\sqrt{Q} \mapsto \sqrt{Q}\end{cases}, \quad
\sigma_2: \begin{cases} \sqrt{P} \mapsto \sqrt{P} \\
\sqrt{Q} \mapsto -\sqrt{Q}. \end{cases} \]

It is clear that $\alpha$ is a totally negative, and hence $K$ is a CM field. The maximal totally real subfield $K^+=\Q(\alpha)\subset \Q(\sqrt{P},\sqrt{Q})$.
Since the minimal polynomial of $\alpha$ is of degree $4$, one has $K^+=\Q(\sqrt{P},\sqrt{Q})$.
Let $\tau_i\in \Gal(\Qbar/\Q)$, $i=1,2$, be elements such that $\tau_i|_{K^+}=\sigma_i$. One computes

\[ \frac{\tau_1(\beta)^2}{\beta^2}= \frac{\tau_1(\alpha)}{\alpha}=\frac{-\sigma_1(P+\sqrt{P})\sigma_1(Q+\sqrt{Q})}{-(P+\sqrt{P})(Q+\sqrt{Q})}
=\frac{(P-\sqrt{P})^2}{P^2-P}=\frac{(\sqrt{P}-1)^2}{P-1}=\frac{(\sqrt{P}-1)^2}{a^2}, \]

and obtains

\[ \tau_1(\beta)=\pm \frac{(\sqrt{P}-1)}{a} \beta,\quad  \text{and} \quad \tau_1(K)\subset K. \]

Similarly, one computes

\[ \frac{\tau_2(\beta)^2}{\beta^2}=\frac{(\sqrt{Q}-1)^2}{Pb^2}, \]

and obtains

\[ \tau_2(\beta)=\pm \frac{(\sqrt{Q}-1)}{\sqrt{P} a} \beta,\quad  \text{and} \quad \tau_2(K)\subset K. \]

It follows that for every $\tau\in \Gal(\Qbar/\Q)$, $\tau(K)\subset K$. Therefore, $K/\Q$ is a Galois extension. Let $\tau_1,\tau_2\in \Gal(K/\Q)$ be defined as in (<ref>).
One computes

\[ \tau_1^2(\beta)=-\beta, \quad  \tau_2^2(\beta)=-\beta, \]

\[ \tau_2 \tau_1(\beta)=\tau_2(\frac{(\sqrt{P}-1)}{a} \beta)= \frac{(\sqrt{P}-1)}{a} \frac{(\sqrt{Q}-1)}{\sqrt{P} a} \beta, \]

\[ \tau_1 \tau_2(\beta)=\tau_1(\frac{(\sqrt{Q}-1)}{\sqrt{P} a} \beta) = - \frac{(\sqrt{Q}-1)}{\sqrt{P} a} \frac{(\sqrt{P}-1)}{a}  \beta =\iota \tau_2\tau_1(\beta). \]

Thus, we obtain the relations

\[ \tau_1^4=\tau_2^4=1, \quad \tau_1^2=\tau_2^2, \quad \tau_1 \tau_2 \tau_1^{-1}=\tau_2^{-1},  \]

and hence $K$ is a  Galois CM field with group $Q_8$.

Denote by $D_\ell^+$ the decomposition group at $\ell$ in $G^+=\Gal(K^+/\Q)$. Then $D_\ell=Q_8$ if and only if $D_\ell^+=G^+$. If $\ell\nmid PQ$, then $\ell$ is unramified in $K^+$ and $D_\ell^+$ is cyclic. Thus, $D_\ell^+$ cannot be $G^+$, $D_\ell\neq Q_8$ and $D_\ell$ is cyclic.
If $\ell|P$, then $ ( \frac{Q}{\ell} ) = ( \frac{1}{\ell} )=1$. So $\ell$ is ramified in $\Q(\sqrt{P})$ and splits in $\Q(\sqrt{Q})$.
Thus, $D_\ell^+ \neq G^+$ and hence $D_\ell$ is cyclic. It remains to treat the case $\ell|Q$. If $(\frac{P}{\ell})=1$, then $D_\ell^+\simeq \Z/2\Z$ and hence $D_\ell\simeq \Z/4\Z$. Otherwise, $\ell$ is ramified in $\Q(\sqrt{Q})$ and inert in $\Q(\sqrt{P})$. One has $D_\ell^+=G^+$ and $D_\ell=Q_8$.

Let $P$ and $Q$ be two odd positive integers, and $K=\Q(\sqrt{\alpha})$ with $\alpha=-(P+\sqrt{P})(Q+\sqrt{Q})$ be the Galois CM with group $Q_8$ as in Proposition <ref>. Then
\begin{equation}
\label{eq:tauQ8}
\tau(T^{K,\Q})=\begin{cases}
1/2, & \text{if $\left ( \frac{P}{q} \right) =1$ for all primes  $q|Q$;} \\
2, & \text{otherwise}.
\end{cases}
\end{equation}

Put $G=\Gal(K/\Q)\simeq Q_8$ and $N=\<\iota\>$. We shall show that

\begin{equation}\label{eq:H1LQ8}
\end{equation}

\begin{equation}\label{eq:Sha2LQ8}
\Sha^2(\Lambda)=\begin{cases}
\Z/2\Z \oplus \Z/2\Z, & \text{if $\left ( \frac{P}{q} \right) =1$ for all prime  $q|Q$;} \\
0, & \text{otherwise}.
\end{cases}
\end{equation}

Then (<ref>) follows from (<ref>) and (<ref>).

By Proposition <ref> and (<ref>), we have an exact sequence

\begin{equation}
\begin{tikzcd}
0 \arrow{r} & H^1(\Lambda) \arrow{r}  & \
H^1(\Lambda^1)\simeq N^{\rm ab,\vee} \arrow{r}{ \Ver_{G,N}^{\vee}}  & H^2(\Z)=G^{\rm ab,\vee} \arrow{r} & H^2(\Lambda) \arrow{r} & 0,
\end{tikzcd}
\end{equation}

noting that $H^2(G, \Lambda^1)=0$ when $K/\Q$ is a Galois CM field. Since the $2$-Sylow subgroup of $Q_8$ is not cyclic, the transfer $\Ver_{G,N}$ is not surjective by Proposition <ref>. Therefore, $\Ver_{G,N}^{\vee}$ is not injective and is zero.
This proves $H^1(\Lambda)\simeq \Z/2\Z$ and $H^2(\Z)'\simeq \Sha^2(\Lambda)$ from (<ref>).

If $\left ( \frac{P}{q} \right )=-1$ for some $q|Q$, then by Proposition <ref> $D_q=Q_8$ and $\Sha^2(\Lambda)=0$. Therefore, $\tau(T^{K,\Q})=2$.
On the other hand, suppose $\left ( \frac{P}{q} \right )=1$ for all $q|Q$. By Proposition <ref>, every decomposition group $D$ of the extension $K/\Q$ is cyclic. One has $G^{\rm ab}=G/N\simeq \Z/2\Z \times\Z/2\Z$. Recall

\[ H^2(\Z)'=\{f\in \Hom(G, \Q/\Z): (*)\ f|_D \in {\rm Im} \Ver_{D,D\cap N}^{\vee}, \ \forall\, D\in \scrD. \} \]

Note that $f|_N=0$, so the restriction to $D$ map $f|_D$ is contained in $\Hom(D/N, \Q/\Z)$ whenever $N\subset D$ (also noting that $N\subset D$ if $D$ is not trivial). So if $D=0$ or $D\simeq \Z/2\Z$,
then $f|_D=0$ and condition $(*)$ above is satisfied automatically. Suppose $D\simeq \Z/4\Z$. Since the $2$-Sylow subgroup of $D$ is cyclic, the map $\Ver_{D,N}^\vee:N^\vee \to D^\vee$ is injective and ${\rm Im} \Ver_{D,N}^\vee$ the unique subgroup of $D^\vee$ of order $2$. Then $f|_D\in \Hom(D/N, \Q/\Z)={\rm Im} \Ver_{D,N}^\vee$ and condition $(*)$ is also satisfied automatically. Therefore, $\Sha^2(\Lambda)\simeq H^2(\Z)'=H^2(\Z)\simeq \Z/2\Z \oplus \Z/2\Z$. This proves the proposition.

§.§ The dihedral case
Let $D_n$ denote the dihedral group of order $2n$.

Suppose the CM-field $K$ is Galois over $\Q$ with Galois group $G$,
and let

\[ S:=\{\sigma\in G : \iota \not\in \<\sigma\>\}. \]

If $|S|\ge |G|/2$, then $n_K\le 2$. If the inequality is strict holds, then $n_K=1$.

This is due to Jiangwei Xue. Let $L/\Q$ be the class field corresponding to the open subgroup $\Q^\times N(T(\A))\subset \A^\times$ by class field theory. Then $n_K=[\A^\times: \Q^\times N_{L/\Q}(\A_L^\times)]=[L:\Q]$. Suppose $p$ is a rational prime unramified in $K/\Q$ such that the Artin symbol $(p, K/\Q)$ lies in $S$. Since $p$ splits completely in the fixed field $E=K^{D_p}$ of the decomposition group $D_p=\<(p,K/\Q)\>$ of $G$ at $p$ and $\<\iota\>\cap D_p=\{1\}$ (by the definition of $S$), one has $K=K^+ E$ and that every prime $v$ of $K^+$ lying above $p$ splits in $K$, and therefore  $\Qp^\times \subset N(T(\A)) \subset Q^\times N_{L/\Q}(\A_L^\times)$.
It follows from class field theory that $p$ splits completely in $L$. Thus, the density of $S$ is less than or equal to that of primes splitting completely in $L$.
By the Chebotarev density theorem, one yields $|S|/|G| \le 1/[L:\Q]$. Therefore, $n_K\le |G|/|S|$ and the assertions then follow immediately.

Let $K$ be a Galois CM field with group $G=D_n$ and $T$ the associated CM torus over $\Q$. Then $n$ is even and $\tau(T)=2$.

Write $D_n=\<t,s: t^n=s^2=1, s t s=t^{-1}\>$. One easily sees that the center $Z(D_n)=\{x\in \<t\>: x^2=1\}$ contains an element of order $2$ if and only if $n=2m$ is even.
Since $\iota$ is central of order $2$ in $D_n$, $n$ is even.
In this case $|S|=2m+1$ and $n_K=1$ by Lemma <ref>. Therefore, $\tau(T)=2$.

The criterion in Lemma <ref> does not help to compute $\tau(T)$ for the case $G=Q_8$ or $G=\Z/2^{n} \Z$. However, these cases have been treated in Propositions <ref> and <ref>, respectively.

§ SOME NON-SIMPLE CM CASES

Keep the notation in Section <ref>.
Write $N_i:=\Gal(K_i/K_i^+)=\<\iota_i\>$ with involution $\iota_i$ on $K_i/K_i^+$ and $\iota_i^*\in N_i^{\vee}=\Hom(N_i, \Q/\Z)$ for the unique non-trivial element, that is $\iota_i^*(\iota)=1/2 \mod \Z$. Fix a Galois splitting field $L$ of $T=T^{K,\Q}$ and put $G=\Gal(L/\Q)$.
For a number field $F$, denote by $\Sigma_F:=\Hom_{\Q}(F,\C)=\Hom_{\Q}(F,\Qbar)$ the set of embeddings of $F$ into $\C$. The Galois group $G_\Q:=\Gal(\Qbar/\Q)$ acts on $\Sigma_F$ by $\sigma\cdot \phi=\sigma\circ \phi$ for $\sigma\in G_\Q$ and $\phi\in \Sigma_F$.
Let $\Phi_i$ be a CM type of $K_i/K_i^+$, that is, $\Phi_i \cap \Phi_i \circ \iota_i=\emptyset$ and $\Phi_i \bigcup \Phi_i \circ \iota_i=\Sigma_{K_i}$,
and for each $g\in G$, set $\Phi_i(g):=\{\phi\in \Phi_i: g \phi\not\in \Phi_i\}.$

Let $K$ be a CM algebra and $T=T^{K,\Q}$ the associated CM torus over $\Q$. Then the map

\[ \bigoplus \Ver^{\vee}_{G,N_i}: \bigoplus N_i^{\vee} \to G^{\rm ab, \vee}=\Hom(G, \Q/\Z) \]

sends each element $\sum_{i} a_i \iota_i^*$, for $a_i\in \Z$, to the element $f=\sum_{i} a_i f_i$, where $f_i\in \Hom(G,\Q/\Z)$ is the function on $G$ given by $f_i(g)=|\Phi_i(g)|/2 \mod \Z$.

We first describe the transfer map $\Ver_{G,N_i}: G\to N_i$.
We may regard the CM fields $K_i$ as subfields of $\Qbar$, and put $\wt X_i:=G/H_i=\Sigma_{K_i}$ and $X_i:=G/\wt N_i=\Sigma_{K_i^+}$. Fix a section $\wt \varphi:X_i \to G$ such that induced section $\varphi: X_i \to \wt X_i$ has image $\Phi_i$. Since $N_i=\wt N_i/H_i$ is abelian, modulo $D(\wt N_i) H_i$ is the same as modulo $H_i$. Following the definition of $\Ver_{G,N_i}$, for each $g\in G$,

\[ \Ver_{G,N_i}(g)=\prod_{x\in X_i} n^{\wt \varphi}_{g,x} \mod H_i=\iota_i^{n_i(g)}\in N_i, \]

where the element $n^{\wt \varphi}_{g,x}\in \wt N_i$ is defined in Section <ref> and $n_i(g):=\#\{x\in X_i: n^{\wt \varphi}_{g,x}\not\in H_i\}$.
Let $f_i:=\Ver_{G,N_i}^\vee(\iota_i^*)$.

\[ f_i(g)=\iota_i^*(\Ver_{G,N_i} (g))=\iota_i^*(\iota_i^{n_i(g)})=n_i(g)/2 \in \Q/\Z. \]

Thus, it remains to show $n_i(g)=|\Phi_i(g)|$. Let
$\wt \Phi_i:=\wt \varphi(\Sigma_{K_i^+})$. Then we have bijections $\wt \Phi_i\isoto \Phi_i \isoto \Sigma_{K_i^+}$. Let $x\in \Sigma_{K_i^+}$, and let $\wt \phi\in\wt \Phi_i$ and $\phi\in \Phi_i$ be the corresponding elements. Put
$\wt \phi':=\wt \varphi (gx)$ and $\phi'=\varphi(gx)$, the image of $\wt \phi'$. Then $g\wt \phi$ and $\wt \phi'$ are elements lying over $gx$ and we have $g \wt \phi=\wt \phi' {n}^{\wt \varphi}_{g,x}$. So $g\phi=\phi'$ if and only if ${n}^{\wt \varphi}_{g,x}\in H_i$. On the other hand, since $\phi'\in \Phi_i$ and two elements $g \phi$ and $\phi'$ are lying over the same element $gx$, we have

\[ g\phi\not \in \Phi_i \iff g \phi\neq \phi' \iff {n}^{\wt \varphi}_{g,x}\not\in H_i. \]

Therefore, the bijection $\Phi_i\isoto \Sigma_{K_i^+}$ gives the bijection of subsets:

\[ \{\phi\in \Phi_i: g\phi \not \in \Phi_i\}\isoto \{x\in \Sigma_{K_i^+}: {n}^{\wt \varphi}_{g,x}\not\in H_i\}. \]

This shows the desired equality $|\Phi_i(g)|=n_i(g)$.

By Proposition <ref> and Lemma <ref>, we re-obtain the following result of Li and Rüd <cit.>.

Let $K$ and $T$ be as in Lemma <ref> and $\Lambda=X(T)$ be the character group of $T$. Then

\begin{equation}\label{eq:A.11}
H^1(\Lambda)\simeq \left \{(a_i)\in \{\pm 1\}^r: \sum_{1\le i \le r, a_i=-1} |\Phi_i(g)| \in 2\Z, \ \forall\, g\in G \right \}.
\end{equation}

Indeed after making the identity $\oplus N_i^\vee\simeq (\Z/2\Z)^r$ with $\{\pm 1\}^r$, by Lemma <ref> the map $\oplus V^\vee_{G,N_i}: \oplus N_i^\vee=\{\pm 1\}^r\to \Hom(G,\Q/\Z)$ sends $(a_i)$ to the function $f$ with $f(g)=\sum_{1\le i\le r, a_i=-1} |\Phi_i(g)|/2 \mod \Z$. Thus, $f=0$ precisely when

$\sum_{1\le i \le r, a_i=-1} |\Phi_i(g)| \in 2\Z, \ \forall\, g\in G$, and (<ref>) follows from Proposition <ref>.

Let $K$ be a CM algebra and $T$ the associated CM torus over $\Q$. Then the global index $n_K:=[\A^\times: N(T(\A))\cdot \Q^\times]$ is equal to $|H^2(\Z)'|$.

Clearly the condition of Proposition <ref> is fulfilled for $K$ and $|N_i|=2$ for all $i$, one obtains $\tau(T)=2^r/|H^2(\Z)'|$ by (<ref>). Then this follows immediately from Proposition <ref>.

It is interesting to know whether the abelian group
$\A^\times/ \Q^\times N(T(\A))$ is actually isomorphic to $H^2(\Z)'^\vee$.
Then this would give a cohomological interpretation of the abelian group $\A^\times/ \Q^\times N(T(\A))$.
Note that $\A^\times/ \Q^\times N(T(\A))\simeq H^2(\Z)'^\vee$ if and only if the abelian group $H^2(\Z)'^\vee$ is $2$-torsion. This follows from that these two groups have the same cardinality and that the group $\A^\times/ \Q^\times N(T(\A))$ is $2$-torsion.

Observe that for a Galois extension $K/k$ with group $G=\Gal(K/k)$, class field theory gives a natural isomorphism

\[ \A_k^\times/N_{K/k}(\A_K^\times) k^\times \simeq G^{\rm ab} \simeq H^2(G,\Z)^\vee. \]

This gives a good sample of cohomological interpretation of the abelian group $\A_k^\times/N_{K/k}(\A_K^\times) k^\times$.

Let $K_1$ be a CM field and $K:=K_1^r$ be the $r$-copies of $K_1$. Then
$\tau(T^{K,\Q})=2^{r-1}\cdot \tau(T^{K_1,\Q})$.

Observe that $N(T^{K,\Q}(\A))=N(\A_K^\times) \cap \A^\times$, where $N=N_{K/K^+}$. To see this, the inclusion $\subseteq$ is clear. If $x\in N(\A_K^\times) \cap \A^\times$, then $x=N(y)$ for some $y\in \A_K^\times$. By definition $y\in T(\A)$, and hence $x\in N(T(\A))$. This verifies the other inclusion. It follows that if $K=\prod_{i=1}^r K_i$ is a product of CM fields $K_i$, then

\[ N(T^{K,\Q}(\A))=\A^\times \bigcap^r_{i=1} N_{K_i/K_i^+}(\A_{K_i}^\times) \]

in the sense that the right hand side consists of all elements $x\in \A^\times$ which are contained in $N_{K_i/K_i^+}(\A_{K_i}^\times)$ via the embeddings $\A^\times \embed \A_{K_i^+}^\times$ for all $i$. Thus, if $K_i=K_1$ for all $i$, then
$N(T^{K,\Q}(\A))=\A^\times \cap  N_{K_1/K_1^+}(\A_{K_1}^\times)=N(T^{K_1,\Q}(\A))$ and hence $n_K=n_{K_1}$. By Proposition <ref>, $\tau(T^{K,\Q})=2^{r}/n_K=2^{r-1}\cdot 2/n_{K_1}= 2^{r-1}\cdot \tau(T^{K_1})$.

For any integer $n\ge 0$, there exists a CM torus $T$ over $\Q$ such that

Take $K=E^r$ with $r=n+1\ge 1$, where $E$ is an imaginary quadratic field. Then $\tau(T^{K,\Q})=2^{r-1} \cdot \tau(T^{E,\Q})=2^{n}$.

Suppose that each CM field $K_i$ is Galois over $\Q$ with group $G_i\simeq Q_8$ and that the Galois group $G=\Gal(L/\Q)$ of the compositum $L=K_1\cdots K_r$ over $\Q$ is isomorphic to $G_1\times \dots \times G_r$. Assume further that $\tau(T^{K_i,\Q})=1/2$ for all $i$. Then $\tau(T^{K,\Q})=(1/2)^r$.

By (<ref>) and $\tau(T^{K_i,\Q})=1/2$, one has $H^2(G_i,\Z)'=H^2(G_i,\Z)\simeq (\Z/2\Z)^2$. In particular the condition $(*)_i$ $f_i|_{D_{i}'}\in {\rm Im} \Ver_{D_{i}', D_{i}'\cap N_i}^\vee$ holds for every $f_i\in \Hom(G_i,\Q/\Z)$ and for every decomposition group $D_{i}'$ of $G_i$. To show $\tau(T^{K,\Q})=(1/2)^r$, by (<ref>) one must show that $H^2(G,\Z)'=H^2(G,\Z)\simeq (\Z/2\Z)^{2r}$, that is, the condition $(*)$ $f|_D\in \sum_i {\rm Im} \Ver_{D, D_{\grP_i}\cap N_i}^\vee$ holds for every $f\in \Hom(D,\Q/\Z)$ and every decomposition group $D=D_\grP$, where $D_{\grP_i}$ is the decomposition group of $G_i$ at $\grP_i$ below the prime $\grP$. One could use Lemma <ref> to conclude this. Below we provide a direct argument.

Note that each $D_{\grP_i}$ is a cyclic subgroup of $G_i$, otherwise $D_{\grP_i}\simeq G_i$, $\Sha^2(\Lambda_i)=0$ and hence $\tau(T^{K_i,\Q})=2$, a contradiction. Since $G\simeq \prod_i G_i$, the map $(\pr_i)_i: D\to \prod_{i} D_{\grP_i}$ is injective and hence $D$ is abelian. Therefore it induces a surjective morphism $\pi: \Hom(\prod_i D_{\grP_i}, \Q/\Z) \twoheadrightarrow \Hom(D, \Q/\Z)$ and then we obtain the following cartesian diagram

\begin{equation}\label{diagram_Hom(Di)}
\begin{tikzcd}
\qquad  \Hom(\prod_i D_{\grP_i}, \Q/\Z) \arrow{r}{\pi} & \Hom(D,\Q/\Z)   \\
\qquad \Hom(\prod_i (D_{\grP_i}/N_i), \Q/\Z) \arrow{r}{\pi}\arrow[hook]{u} & \Hom(D/(\prod_{i} N_i), \Q/\Z) \arrow[hook]{u}{}.
\end{tikzcd}
\end{equation}

Note that $G^{\rm ab}=\prod G_i^{\rm ab}=\prod N_i$ and hence the function $f|_D$ is contained in $\Hom(D/(\prod_{i} N_i), \Q/\Z)$. Similarly, the restriction $f_i|_{D_{\grP_i}}$ is contained in $\Hom(D_{\grP_i}/N_i,\Q/\Z)$ which is equal to ${\rm Im} \Ver^\vee_{D_{\grP_i}, D_{\grP_i}\cap N_i}$ by (the argument of) Proposition <ref>. As the image of $\prod_i {\rm Im} \Ver^\vee_{D_{\grP_i}, D_{\grP_i}\cap N_i}$ is equal to $\sum_i {\rm Im} \Ver_{D, D_{\grP_i}\cap N_i}^\vee$ and $\pi$ is surjective, the condition ($*$) holds and hence we get $\tau(T)=(1/2)^r$.

For any positive integer $m$, the integral quadratic polynomial $f(x)=mx^2+1$ represents infinitely many primes.

This is a folklore conjecture, or called a generalized Landau conjecture. When $m=1$, this is one of famous Landau's problems and is still open. Iwaniec [6] showed that there exist infinitely many integers $n>0$ such that $f(n)$ is either a prime or an almost prime, namely, a product of two primes.
Conjecture <ref> is a special case of the Bunyakovsky Conjecture, which is a special case of Schinzel's Hypothesis [17].

Let $f(x)\in \Z[x]$ be an irreducible polynomial of positive degree with positive leading coefficient such that there is no integer $d>1$ such that $d|f(n)$ for all $n\in \Z$. Then there are infinitely many primes in $\{f(n):n\in \Z\}$.

Let s be a positive integer, and $f_1(x), \dots, f_s(x)\in \Z[x]$
be irreducible polynomials with positive leading
coefficients. If there exists no integer $d > 1$ such that it divides $\prod_{k=1}^s f_k(n)$ for
all integers $n$, then there should exist infinitely many positive integers $n$ such that the numbers $f_1(n), \dots,  f_s(n)$ are simultaneously primes.

We call a pair of odd primes $(p,q)$ a Landau pair if $p=1+a^2$
and $q=1+pb^2$ for some integers $a$ and $b$.
Two Landau pairs $(p,q)$ and $(p',q')$ are said to be disjoint if
$\gcd (pq,p'q')=1$.

Note that the set of primes $p$ of form $1+a^2$ with $a\in \Z$ and the set of primes $q$ of form $1+p'b^2$ for some prime $p'$ and some integer $b\in \Z$ are disjoint, because $p-1$ is a square but $q-1$ is not. Thus two Landau pairs $(p,q)$ and $(p',q')$ are disjoint if and only if $\{p,q\}\cap \{p',q'\}=\emptyset$.
If $(p,q)$ is a Landau pair, then the prime $p$ is uniquely determined by $q$ as it is the square-free part of $q-1$. Therefore, two Landau pairs $(p,q)$ and $(p',q')$ are disjoint if and only if $p\neq p'$.

Let $K=\prod_{i=1}^rK_i$ be a product of Galois CM fields $K_i$. Assume that the Galois group $G$ of the composite $K_1 \cdots K_r$ over $\Q$ is isomorphic to
the product

of the Galois groups $G_i=\Gal(K_i/\Q)$ of $K_i$ over $\Q$. Then $\tau(T^{K,\Q})=\prod_{i=1}^r \tau(T^{K_i,\Q})$.

This is Proposition 7.21 of [16] (now we don't need this proposition).

Assume Conjecture <ref> holds true. Then for any integer $n$, there exists a CM torus $T$ over $\Q$ such that $\tau(T)=2^n$.

The part of the theorem where $n\ge 0$ is proved by Corollary <ref>. Thus, it remains to prove the part where $n=-r<0$. By Conjecture <ref>, there are $r$ odd primes $p_i=1+a_i^2$ for some integers $a_i>0$, and for each $p_i$, there is a prime $q_i=1+p_ib_i^2$ for some integer $b_i>0$. Therefore, we obtain $r$ Landau pairs $(p_i,q_i)$ for $i=1,\dots, r$, which are all disjoint.
Put $K_i:=\Q(\sqrt{\alpha_i})$, where $\alpha_i:=-(p_i+\sqrt{p_i})(q_i+\sqrt{q_i})$. Then each $K_i$ is a $Q_8$-CM field and $\tau(T^{K_i,\Q})=1/2$, by Proposition <ref>. We now show
$G\simeq G_1\times \dots \times G_r$. We prove this by induction on $r$. Note that the maximal abelian subextension of $K_i$ over $\Q$ is $K_i^+=\Q(\sqrt{p_i}, \sqrt{q_i})$ and every proper subextension of $K_i$ over $\Q$ is contained $K_i^+$.
Thus, if $K_i$ intersects properly with a number field $L$
(i.e. $K_i \cap L \supsetneq \Q$), then $K_i^+$ intersect properly with the maximal abelian subextension $L_{\rm ab}$ of $L$ over $\Q$.
So if $K_1\cap K_2\supsetneq \Q$, then $\Q(\sqrt{p_1},\sqrt{q_1})\cap \Q(\sqrt{p_2},\sqrt{q_2}) \supsetneq \Q$, which is absurd. This proves the case $r=2$. By the induction hypothesis, the maximal abelian subextension of
$K_1\cdots K_{r-1}$ over $\Q$ is $\Q(\sqrt{p_i},\sqrt{q_i}\, ; i=1,\dots, r-1)$, while the maximal abelian subextension of $K_r$ is $\Q(\sqrt{p_{r}},\sqrt{q_{r}})$. As their intersection is $\Q$, one has $K_1\cdots K_{r-1}\cap K_{r}=\Q$. This proves
$G\simeq \Gal(K_1\cdots K_{r-1}/\Q) \times \Gal(K_r/\Q)\simeq   G_1\times \dots \times G_r$. Now apply Proposition <ref> and we obtain
$\tau(T^{K,\Q})=2^{-r}=2^n$. This completes the proof of the theorem.

We search Landau pairs $(p,q)=(1+4 a^2, 1+p b^2)$ for $1\le a\le 10^6$ and $1\le b \le 10^2$ using a Python program and obtain $709,617$ disjoint Landau pairs of primes. In particular, we prove the following unconditional result by Theorem <ref>.

For any integer $n\ge -709,617$, there exists a CM torus $T$ over $\Q$ such that $\tau(T)=2^n$.

§ ACKNOWLEDGMENTS

The present project grows up from the 2020 NCTS USRP “Arithmetic of CM tori”
where the authors participated.

The authors thank Jeff Achter, Ming-Lun Hsieh,
Wen-Wei Li, Thomas Rüd, Yasuhiro Terakado,
Jiangwei Xue for helpful discussions and specially to
to Thomas Rüd who kindly answered the third named
author's questions and explained his results.

The authors are partially supported by the MoST grant 109-2115-M-001-002-MY3.

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