Let $v\in M_{F}^{\infty}$. There exists a continuous compactly supported
function
$k^{\mathbf{a}}_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{\geq
0}$, which is $F_{v,1}$-invariant and such that for every $\mathbf{x}\in
F^{n}_{v}-\\{0\\}$ one has that

$\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(\rho_{v}(t)\cdot\mathbf{x})d^{*}t=\frac{1}{\lambda_{v,1}(F_{v,1})},$

where $\rho_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\to
F_{v}^{\times}$ is given by $\rho_{v}\mathrel{\mathop{\ordinarycolon}}r\mapsto
r^{1/n_{v}}$. Such function satisfies furthermore for every $\mathbf{x}\in
F^{n}_{v}-\\{0\\}$ that

$\int_{F_{v}^{\times}}k^{\mathbf{a}}_{v}(t\cdot\mathbf{x})d^{*}t={1}.$

###### Proof.

The isomorphism (1)

$\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\times
F_{v,1}\xrightarrow{\sim}F_{v}^{\times}\hskip
28.45274pt(r,x)\mapsto\rho_{v}(r)x$

satisfies $(\widetilde{\rho}_{v})_{*}(d^{*}r\times\lambda_{v,1})=d^{*}x_{v}$
by 3. It induces a topological action of $\mathbb{R}_{>0}\times F_{v,1}$ on
$F^{n}_{v}-\\{0\\}$. Moreover, as $F_{v,1}=\\{1\\}\times F_{v,1}$ is a closed
subgroup of $\mathbb{R}_{>0}\times F_{v,1},$ by [7, Example 1, n0 1, §4,
Chapter III] the restriction of this action to $F_{v,1}$ is continuous and
proper. Let
$q_{v,1}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to(F^{n}_{v}-\\{0\\})/F_{v,1}$
be the corresponding quotient map, by [7, Proposition 2, n0 1, §4, Chapter
III], the map $q_{v,1}$ is proper. Let $\mathbb{R}_{>0}$ acts on
$F^{n}_{v}-\\{0\\}$ via the identification
$\mathbb{R}_{>0}=\mathbb{R}_{>0}\times\\{1\\}$. Lemma 2 provides an action of
$\mathbb{R}_{>0}$ on $(F^{n}_{v}-\\{0\\})/F_{v,1}$ which satisfies

$r\cdot
q_{v,1}(\mathbf{x})=q_{v,1}(r\cdot\mathbf{x})=q_{v,1}(\rho_{v}(r)\cdot\mathbf{x}),$

for $r\in\mathbb{R}_{>0}$ and $\mathbf{x}\in F^{n}_{v}-\\{0\\}$. Moreover, the
canonical map $(F^{n}_{v}-\\{0\\})/F_{v,1}\to[\mathscr{P}(\mathbf{a})(F_{v})]$
induces an identification
$((F^{n}_{v}-\\{0\\})/F_{v,1})/\mathbb{R}_{>0}\xrightarrow{\sim}[\mathscr{P}(\mathbf{a})(F_{v})].$
As $[\mathscr{P}(\mathbf{a})(F_{v})]$ is compact, hence paracompact,
Proposition 3 gives that there exists a continuous compactly supported
function
$k^{\prime}\mathrel{\mathop{\ordinarycolon}}(F^{n}_{v}-\\{0\\})/F_{v,1}\to\mathbb{R}_{\geq
0}$ such that for every $\mathbf{y}\in(F^{n}_{v}-\\{0\\})/F_{v,1}$, we have
that

$\int_{\mathbb{R}_{>0}}k^{\prime}(r\cdot\mathbf{y})d^{*}r=\frac{1}{\lambda_{v,1}(F_{v,1})}.$

Let us set $k^{\mathbf{a}}_{v}=k^{\prime}\circ q_{v,1}$. As $q_{v,1}$ is
proper and $F_{v,1}$-invariant, the map $k^{\mathbf{a}}_{v}$ is compactly
supported and $F_{v,1}$-invariant. Let $\mathbf{x}\in F^{n}_{v}-\\{0\\}$, we
have that

$\displaystyle\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(\rho_{v}(r)\cdot\mathbf{x})d^{*}r$
$\displaystyle=\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(r\cdot\mathbf{x})d^{*}r$
$\displaystyle=\int_{\mathbb{R}_{>0}}k^{\prime}(q_{v,1}(r\cdot\mathbf{x}))d^{*}r$
$\displaystyle=\int_{\mathbb{R}_{>0}}k^{\prime}(r\cdot
q_{v,1}(\mathbf{x}))d^{*}r$ $\displaystyle=\frac{1}{\lambda_{v,1}(F_{v,1})},$

and that

$\displaystyle\int_{F_{v}^{\times}}k^{\mathbf{a}}_{v}(t\cdot\mathbf{x})d^{*}t$
$\displaystyle=\int_{\mathbb{R}_{>0}\times
F_{v,1}}k^{\mathbf{a}}_{v}((r,z)\cdot\mathbf{x}){d^{*}r}d\lambda_{v,1}(z)$
$\displaystyle=\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(r\cdot(z_{j}^{a_{j}}x_{j})_{j})d^{*}r$
$\displaystyle=\int_{F_{v,1}}\frac{1}{\lambda_{v,1}(F_{v,1})}d\lambda_{v,1}$
$\displaystyle=1.$

The statement is proven. ∎

#### 3

The goal of this paragraph is to give several equivalent conditions that make
$f^{-1}dx_{1}\dots dx_{n}$ a measure on $F^{n}_{v}-\\{0\\}$, where $f$ is an
$\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$.

Let $v\in M_{F}.$ We will use the following conventions $a\cdot\infty=\infty$
for $a\in\mathbb{R}_{>0}$ and $\infty^{-1}=0$. If $t\in\mathbb{G}_{m}(F_{v})$,
we will see it as a function
$t\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to F^{n}_{v}-\\{0\\}$
given by its action on $F^{n}_{v}-\\{0\\}$. We say that a function
$f\mathrel{\mathop{\ordinarycolon}}F_{v}^{n}-\\{0\\}\to\mathbb{C}\cup\\{\infty\\}$
is $\mathbf{a}$-homogenous of weighted degree $|\mathbf{a}|$ if
$f(t\cdot\mathbf{x})=|t|^{|\mathbf{a}|}_{v}f(\mathbf{x})$ for every $t\in
F_{v}^{\times}$ and every $\mathbf{x}\in F^{n}_{v}-\\{0\\}$.

######  Lemma 1.

Let
$f\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}\cup\\{\infty\\}$
be an $\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$ such
that

$dx_{1}\dots dx_{n}(\\{\mathbf{x}\in F^{n}_{v}-\\{0\\}|f(\mathbf{x})=0\\})=0.$

Let $\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$. One has that
$\phi f^{-1}\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ if and only if
for every $t\in F_{v}^{\times}$ one has that $(\phi\circ t^{-1})f^{-1}\in
L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ and if $\phi f^{-1}\in
L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ then

$\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots
dx_{n}=\int_{F^{n}_{v}-\\{0\\}}(\phi\circ t^{-1})f^{-1}dx_{1}\dots dx_{n}$

for every $t\in F_{v}^{\times}$.

###### Proof.

Let $t\in\mathbb{G}_{m}(F_{v})$. The action of $t$ on $F_{v}^{n}-\\{0\\}$ is
given by the multiplication to the left by the diagonal matrix $D_{t}$ which
has the diagonal vector $(t^{a_{1}},\dots,t^{a_{n}})$. By using the formula
for the change of the coordinates, we get that

$\displaystyle\int_{F^{n}_{v}-\\{0\\}}(\phi\circ t^{-1})f^{-1}dx_{1}\dots
dx_{n}\hskip-96.73918pt$
$\displaystyle=\int_{F^{n}_{v}-\\{0\\}}|\det(\operatorname{Jac}(D_{t}))|_{v}(\phi\circ
t^{-1})(t\cdot\mathbf{x})f_{v}(t\cdot\mathbf{x})^{-1}dx_{1}\dots dx_{n}$
$\displaystyle=\int_{F^{n}_{v}-\\{0\\}}|t^{|\mathbf{a}|}|_{v}\phi(\mathbf{x})\cdot|t|_{v}^{-|\mathbf{a}|}f(\mathbf{x})^{-1}dx_{1}\dots
dx_{n}$ $\displaystyle=\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots dx_{n},$

if one, and hence every, integral converges absolutely. The statement follows.
∎

The following lemma will be needed to treat the case $v\in M_{F}$.

######  Lemma 2.

Suppose $v\in M_{F}^{\infty}$. Let
$\rho_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\to F_{v}^{\times}$
be the map $r\mapsto r^{1/n_{v}}$.

  1. 1.

Consider the map
$A\mathrel{\mathop{\ordinarycolon}}(F_{v}^{\times})^{n}\to(F_{v}^{\times})^{n}$
given by

$A\mathrel{\mathop{\ordinarycolon}}\mathbf{x}\mapsto((\rho_{v}(|x_{n}|_{v}^{a_{j}/a_{n}})x_{j})_{j=1}^{n-1},x_{n}).$

One has that

$|x_{n}|^{\frac{|\mathbf{a}|}{a_{n}}-1}A_{*}(dx_{1}\dots dx_{n})=dx_{1}\dots
dx_{n}.$

  2. 2.

Let
$f\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}\cup\\{\infty\\}$
be an $\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$ such
that

$dx_{1}\dots dx_{n}(\\{\mathbf{x}\in F^{n}_{v}-\\{0\\}|f(\mathbf{x})=0\\})=0.$

Let $\phi\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}$ be a
compactly supported continuous function. One has that $\phi f^{-1}\in
L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ if and only if
$f^{-1}|_{(F_{v}^{\times})^{n-1}\times F_{v,1}}\in
L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n}\times\lambda_{v,1})$ and if $\phi f^{-1}\in
L(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ then

$\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots
dx_{n}=a_{n}\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{\mathbb{R}_{>0}}{\phi\big{(}\rho_{v}(t)\cdot((x_{j})_{j=1}^{n-1},z)\big{)}d^{*}t}\times\\\
\times\int_{(F_{v}^{\times})^{n-1}}f(x_{1},\dots,x_{n-1},z)^{-1}dx_{1}\dots
dx_{n-1}.$ (3)

###### Proof.

  1. 1.

The Jacobian of $A$ is given by the diagonal matrix having for the diagonal
vector $((\rho_{v}(|x_{n}|^{a_{j}/a_{n}}_{v}))_{j=1}^{n-1},1).$ The $v$-adic
absolute value of the determinant of the Jacobian is equal to

$\big{|}\prod_{j=1}^{n-1}\rho_{v}(|x_{n}|_{v}^{a_{j}/a_{n}})\big{|}_{v}=\prod_{j=1}^{n-1}\big{|}\rho_{v}(|x_{n}|_{v}^{a_{j}/a_{n}})\big{|}_{v}=\prod_{j=1}^{n-1}|x_{n}|_{v}^{a_{j}/a_{n}}=|x_{n}|_{v}^{\frac{|\mathbf{a}|}{a_{n}}-1}.$

Thus, by the formula for the change of variables we get that

$|x_{n}|_{v}^{\frac{|\mathbf{a}|}{a_{n}}-1}A_{*}(dx_{1}\dots
dx_{n})=dx_{1}\dots dx_{n}.$

  2. 2.

Let us define
$B,C\mathrel{\mathop{\ordinarycolon}}(F_{v}^{\times})^{n}\to(F_{v}^{\times})^{n}$
by

$\displaystyle B\mathrel{\mathop{\ordinarycolon}}\mathbf{x}$
$\displaystyle\mapsto(x_{j}\rho_{v}({|x_{n}|_{v}^{{-a_{j}}/{a_{n}}}}))_{j},$
$\displaystyle C\mathrel{\mathop{\ordinarycolon}}\mathbf{x}$
$\displaystyle\mapsto((x_{j})_{j=1}^{n-1},x_{n}\rho_{v}(|x_{n}|_{v})^{-1}).$

Note that for $\mathbf{x}\in(F_{v}^{\times})^{n}$, one has that

$B(A(\mathbf{x}))=B((x_{j}\rho_{v}(|x_{n}|_{v}^{a_{j}/a_{n}}))_{j=1}^{n-1},x_{n})=((x_{j})_{j=1}^{n-1},x_{n}\rho_{v}(|x_{n}|^{-1}))=C(\mathbf{x})$

and that

$A(\mathbf{x})=((\rho_{v}(|x_{n}|_{v}^{\frac{a_{j}}{a_{n}}})x_{j})_{j=1}^{n-1},x_{n})=\rho_{v}(|x_{n}|_{v}^{\frac{1}{a_{j}}})\cdot((x_{j})_{j=1}^{n-1},x_{n}\rho_{v}(|x_{n}|_{v})^{-1})\\\
=\rho_{v}(|x_{n}|_{v}^{\frac{1}{a_{j}}})\cdot C(\mathbf{x}).$

For every $\mathbf{x}\in F^{n}_{v}-\\{0\\}$, we have that

$f_{v}(\mathbf{x})=|x_{n}|_{v}^{|\mathbf{a}|/a_{n}}f_{v}\big{(}\big{(}{x_{j}}\rho_{v}({|x_{n}|_{v}^{{-a_{j}}/{a_{n}}}})\big{)}_{j}\big{)}=f_{v}(B(\mathbf{x})),$

by the fact that $f_{v}$ is $\mathbf{a}$-homogenous of weighted degree
$|\mathbf{a}|$. Let
$\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$. By using that
$dx_{1}\dots dx_{n}(F^{n}_{v}-\\{0\\}-(F_{v}^{\times})^{n})=0,$ because
$F^{n}_{v}-\\{0\\}-(F_{v}^{\times})^{n}$ is contained in a finite union of
hyperplanes of $(F_{v})^{n}$, and the part (1), we get that:

$\displaystyle\begin{split}\int_{F^{n}_{v}-\\{0\\}}&\phi f_{v}^{-1}dx_{1}\dots
dx_{n}\\\ &=\int_{(F_{v}^{\times})^{n}}\phi f_{v}^{-1}dx_{1}\dots dx_{n}\\\
&=\int_{(F_{v}^{\times})^{n}}|x_{n}|_{v}^{-|\mathbf{a}|/a_{n}}\phi\cdot(f_{v}^{-1}\circ
B)dx_{1}\dots dx_{n}\\\
&=\int_{(F_{v}^{\times})^{n}}|x_{n}|_{v}^{-1}\phi\cdot(f_{v}^{-1}\circ
B)A_{*}(dx_{1}\dots dx_{n})\\\ &=\int_{(F_{v}^{\times})^{n}}|x_{n}\circ
A|_{v}^{-1}(\phi\circ A)(f_{v}^{-1}\circ(B\circ A))dx_{1}\dots dx_{n}\\\
&=\int_{(F_{v}^{\times})^{n}}|x_{n}|_{v}^{-1}(\phi\circ
A)f_{v}((x_{j})_{j=1}^{n-1},x_{n}\rho_{v}(|x_{n}|_{v}^{-1}))^{-1}dx_{1}\dots
dx_{n}\\\ &=\int_{(F_{v}^{\times})^{n}}(\phi\circ
A)f_{v}((x_{j})_{j=1}^{n-1},x_{n}\rho_{v}(|x_{n}|_{v}^{-1}))^{-1}dx_{1}\dots
dx_{n-1}d^{*}x_{n}\\\
&=\int_{(F_{v}^{\times})^{n}}\phi(\rho_{v}(|x_{n}|_{v}^{1/a_{n}})\cdot
C(\mathbf{x}))f_{v}(C(\mathbf{x}))^{-1}dx_{1}\dots
dx_{n-1}d^{*}x_{n}.\end{split}$

if one (and hence every) integral converges absolutely. It follows from 3 that
the map

$\widetilde{\rho}_{v}^{-1}\mathrel{\mathop{\ordinarycolon}}F_{v}^{\times}\to\mathbb{R}_{>0}\times
F_{v,1}\hskip 28.45274ptx\mapsto(|x|_{v},{x}{\rho_{v}(|x|_{v})^{-1}})$

satisfies
that$(\widetilde{\rho}_{v}^{-1})_{*}d^{*}x=d^{*}r\times\lambda_{v,1}$. The
last integral is hence equal to

$\int_{F_{v,1}\times\mathbb{R}_{>0}}\int_{(F_{v}^{\times})^{n-1}}\phi(\rho_{v}(r^{1/a_{j}})\cdot((x_{j})_{j=1}^{n-1},z))\times\\\
\times
f_{v}((x_{j})_{j=1}^{n-1},z)^{-1}dx_{1}..dx_{n-1}d^{*}rd\lambda_{v,1}(z).$

Note that setting $u^{a_{n}}=r$ gives $d^{*}r=a_{n}d^{*}u$ and thus the last
integral becomes

$a_{n}\int_{F_{v,1}\times\mathbb{R}_{>0}}\int_{(F_{v}^{\times})^{n-1}}\frac{\phi(\rho_{v}(u)\cdot((x_{j})_{j=1}^{n-1},z))}{f_{v}((x_{j})_{j=1}^{n-1},z)}dx_{1}\dots
dx_{n-1}d^{*}ud\lambda_{v,1}(z).$ (4)

By Fubini theorem, we get that (4) is equal to

$a_{n}\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{\mathbb{R}_{>0}}\phi(\rho_{v}(u\cdot((x_{j})_{j=1}^{n-1},z)))d^{*}u\times\\\
\times\int_{(F_{v}^{\times})^{n-1}}f_{v}((x_{j})_{j=1}^{n-1},z)^{-1}dx_{1}\dots
dx_{n-1}.$ (5)

For every $((x_{j})_{j},z)\in(F_{v}^{\times})^{n-1}\times F_{v,1}$, the map
$\mathbb{R}_{>0}\to\mathbb{C}$ given by
$u\mapsto\phi(\rho_{v}(u)\cdot((x_{j})_{j},z))$ is compactly supported,
because $\phi$ is compactly supported, the map $y\mapsto
y\cdot((x_{j})_{j},z)$ is proper (the action of $F_{v}^{\times}$ on
$(F_{v}^{\times})^{n}$ is proper by 1, thus by [7, Proposition 4, n0 2, §4,
Chapter III], the map $y\mapsto y\cdot((x_{j})_{j},z)$ is proper) and
$\rho_{v}\mathrel{\mathop{\ordinarycolon}}r\mapsto r^{1/n_{v}}$ is proper. It
follows that

$\int_{\mathbb{R}_{>0}}\phi(\rho_{v}(u)\cdot((x_{j})_{j},z))d^{*}u$

converges absolutely. It follows that the integral (5) converges absolutely if
and only if
$\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{(F_{v}^{\times})^{n}}f_{v}((x_{j})_{j=1}^{n-1},z)^{-1}dx_{1}\dots
dx_{n-1}$ converges absolutely. The statement follows.

∎

In the following proposition, we give equivalent conditions to the condition
that $f^{-1}dx_{1}\dots dx_{n}$ is a measure on $F^{n}_{v}-\\{0\\}$.

######  Proposition 6.

Let
$f\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}\cup\\{\infty\\}$
be an $\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$ such
that

$dx_{1}\dots dx_{n}(\\{\mathbf{x}\in F^{n}_{v}-\\{0\\}|f(\mathbf{x})=0\\})=0.$

The following are equivalent:

  1. 1.

For every $\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$ one has
that $\phi f^{-1}\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ and that

$\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})\to\mathbb{C}\hskip
28.45274pt\phi\mapsto\int_{F^{n}_{v}-\\{0\\}}\phi\cdot f^{-1}dx_{1}\dots
dx_{n}$

is a measure on $F^{n}_{v}-\\{0\\}$.

  2. 2.

For every compactly supported function
$\phi\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}$ one has
that $\phi f^{-1}\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$ and that
there exists a unique measure $\omega_{v}$ on
$[\mathscr{P}(\mathbf{a})(F_{v})]$ such that

$\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots
dx_{n}=\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}d\omega_{v}(\mathbf{y})\int_{F_{v}^{\times}}\phi(t\cdot\widetilde{\mathbf{y}})d^{*}t,$
(7)

where $\widetilde{\mathbf{y}}$ is a lift on an element
$\mathbf{y}\in[\mathscr{P}(\mathbf{a})(F_{v})]$.

  3. 3.

One has that $k^{\mathbf{a}}_{v}\cdot f^{-1}\in
L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n}),$ where the function
$k^{\mathbf{a}}_{v}$ is defined in 2.

  4. 4.

If $v\in M_{F}^{0}$, one has that $f^{-1}|_{\mathcal{D}^{\mathbf{a}}_{v}}\in
L^{1}(\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n})$. If $v\in
M_{F}^{\infty}$, one has that $f^{-1}|_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}\in L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n-1}\times\lambda_{v,1})$.

If any of the conditions is satisfied, one has that

$\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}f^{-1}dx_{1}\dots
dx_{n}$ (8)

Moreover, if $v\in M_{F}^{0}$, both quantities in (8) are equal to

$\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}f^{-1}dx_{1}\dots
dx_{n}$

and if $v\in M_{F}^{\infty}$, and are equal to

$\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}f^{-1}dx_{1}\dots dx_{n-1}\times\lambda_{v,1}.$

###### Proof.

Note that the implication $(1)\implies(2)$ follows from 2. By using the fact
that $[\mathscr{P}(\mathbf{a})(F_{v})]$ is compact, hence paracompact, and
Proposition 5, we get the implication $(2)\implies(3)$ and that if the
condition (2) is satisfied, then

$\omega_{v}^{\mathbf{a}}([\mathscr{P}(\mathbf{a})(F_{v})])=\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}f^{-1}dx_{1}\dots
dx_{n}.$

$(3)\implies(4)$. Suppose $v$ is a finite place. As
$k^{\mathbf{a}}_{v}f^{-1}=\frac{\mathbf{1}_{\mathcal{D}^{\mathbf{a}}_{v}}\cdot
f^{-1}}{1-|\pi_{v}|_{v}}$, we deduce that the restriction of
$\mathbf{x}\mapsto f(\mathbf{x})^{-1}$ to $\mathcal{D}^{\mathbf{a}}_{v}$ is
$dx_{1}\dots dx_{n}$-absolutely integrable and that

$\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}f^{-1}dx_{1}\dots
dx_{n}=\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}f^{-1}dx_{1}\dots
dx_{n}.$

Suppose $v$ is infinite. By Lemma 2 one has that

$f^{-1}|_{(F_{v}^{\times})^{n-1}\times F_{v,1}}\in
L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n-1}\times\lambda_{v,1})$

and that

$\displaystyle\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}f^{-1}dx_{1}\dots
dx_{n}\hskip-56.9055pt$
$\displaystyle=a_{n}\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(\rho_{v}(u)\cdot((x_{j})_{j},z))d^{*}u\times$
$\displaystyle\quad\quad\quad\quad\quad\times\int_{(F_{v}^{\times})^{n-1}}f(x_{1},\dots,x_{n-1},z)^{-1}dx_{1}\dots
dx_{n-1}.$

We have by 2 that

$\int_{\mathbb{R}_{>0}}k^{\mathbf{a}}_{v}(\rho_{v}(u)\cdot((x_{j})_{j},z))d^{*}u=\frac{1}{\lambda_{v,1}(F_{v,1})},$

for every $((x_{j})_{j},z)\in(F_{v}^{\times})^{n-1}\times F_{v,1}$. It follows
that

$\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}f^{-1}dx_{1}\dots
dx_{n}=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}f^{-1}dx_{1}\dots dx_{n-1}\times\lambda_{v,1}.$

The implication is proven.

$(4)\implies(1).$ If
$\phi\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}$ is
compactly supported, we set
$||\phi||_{\sup}\mathrel{\mathop{\ordinarycolon}}=\sup_{\mathbf{x}\in
F^{n}_{v}-\\{0\\}}|\phi(\mathbf{x})|.$ Suppose firstly $v\in M_{F}^{0}$ and
let $\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$. It follows
from 1 that
$\cup_{t\in\mathbb{G}_{m}(F_{v})}(t\cdot\mathcal{D}^{\mathbf{a}}_{v})=F^{n}_{v}-\\{0\\}$.
As $\operatorname{supp}(\phi)$ is compact, there exists a finite set
$\\{t_{1},\dots,t_{m}\\}$ such that
$(\cup_{i=1}^{m}(t_{i}\cdot\mathcal{D}^{\mathbf{a}}_{v}))\supset\operatorname{supp}(\phi)$.
For $\mathbf{x}\in\mathcal{D}^{\mathbf{a}}_{v}$ and
$t\in\mathbb{G}_{m}(F_{v})$ one has that
$f(t\cdot\mathbf{x})^{-1}=|t|_{v}^{-|\mathbf{a}|}f(\mathbf{x})^{-1}$. Thus
from the fact that $f^{-1}|_{\mathcal{D}^{\mathbf{a}}_{v}}\in
L^{1}(\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n})$, it follows that
$f^{-1}|_{t\cdot\mathcal{D}^{\mathbf{a}}_{v}}\in
L^{1}(t\cdot\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n})$ and hence that

$f^{-1}|_{\cup_{i=1}^{m}(t_{i}\cdot\mathcal{D}^{\mathbf{a}}_{v})}\in
L^{1}(\cup_{i=1}^{m}(t_{i}\cdot\mathcal{D}^{\mathbf{a}}_{v}),dx_{1}\dots
dx_{n}).$

We deduce that $\phi f^{-1}\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$
and that

$\displaystyle\bigg{|}\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots
dx_{n}\bigg{|}$
$\displaystyle\leq\int_{\sup(\phi)}||\phi||_{\sup}f^{-1}dx_{1}\dots dx_{n}$
$\displaystyle\leq||\phi||_{\sup}\int_{(\cup_{i=1}^{m}(t_{i}\cdot\mathcal{D}^{\mathbf{a}}_{v}))}f^{-1}dx_{1}\dots
dx_{n}.$

It follows furthermore that $\phi\mapsto\int_{F^{n}_{v}-\\{0\\}}\phi
f^{-1}dx_{1}\dots dx_{n}$ is a bounded, hence a continuous operator, i.e.
$f^{-1}dx_{1}\dots dx_{n}$ is a measure on $F^{n}_{v}-\\{0\\}$.

Suppose now $v\in M_{F}^{\infty}.$ For any
$\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C}),$ one has by Lemma 2
that $\phi f^{-1}\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n}).$ We will use
[10, Proposition 6 in n0 3, §1, Chapter III ] to prove the continuity of the
operator $\phi\mapsto\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots dx_{n}.$
By the mentioned proposition, it suffices to pick a sequence of compacts
$\\{K_{\alpha}\\}_{\alpha}$, the interiors of which cover $F^{n}_{v}-\\{0\\}$
and to establish that for every $\alpha$ there exists $M_{\alpha}>0$ such that

$\bigg{|}\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots dx_{n}\bigg{|}\leq
M_{\alpha}||{\phi}||_{\sup}$

for $\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$ with
$\operatorname{supp}(\phi)\subset K_{\alpha}$. We set

$K_{\alpha}\mathrel{\mathop{\ordinarycolon}}=\\{\mathbf{x}\in
F^{n}_{v}-\\{0\\}|\forall j\mathrel{\mathop{\ordinarycolon}}\hskip
2.84544pt\alpha^{-1}\leq|x_{j}|_{v}\leq\alpha\\}\hskip 28.45274pt\text{for
}\alpha\in{\mathbb{Z}_{\geq 2}}.$

Let $\xi_{\alpha}\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{R}_{\geq
0})$ such that $\xi_{\alpha}(\mathbf{x})\geq 1$ for every $\mathbf{x}\in
K_{\alpha}$. For every $\mathbf{x}\in F^{n}_{v}-\\{0\\}$, the map
$\mathbb{R}_{>0}\to\mathbb{R}_{\geq 0}$ given by
$r\mapsto\xi_{\alpha}(\rho_{v}(r)\cdot\mathbf{x})$ is compactly supported,
because $\xi_{\alpha}$ is compactly supported, the map $y\mapsto
y\cdot\mathbf{x}$ is proper (the action of $F_{v}^{\times}$ on
$(F_{v}^{\times})^{n}$ is proper by 1, thus by [7, Proposition 4, n0 2, §4,
Chapter III], the map $y\mapsto y\cdot\mathbf{x}$ is proper) and
$\rho_{v}\mathrel{\mathop{\ordinarycolon}}r\mapsto r^{1/n_{v}}$ is proper. Let
$\phi\in\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})$ with
$\operatorname{supp}(\phi)\subset K_{\alpha}$. We have that

$\displaystyle\bigg{|}\int_{\mathbb{R}_{>0}}{\phi(\rho_{v}(r)\cdot((u_{j})_{j=1}^{n-1},z))d^{*}r}\bigg{|}\hskip-28.45274pt$
$\displaystyle\leq\int_{\mathbb{R}_{>0}}{||\phi||_{\sup}\cdot\xi_{\alpha}(\rho_{v}(r)\cdot((u_{j})_{j=1}^{n-1},z))d^{*}r}$
$\displaystyle\leq||\phi||_{\sup}\int_{\mathbb{R}_{>0}}\xi_{\alpha}(\rho_{v}(t)\cdot((u_{j})_{j=1}^{n-1},z))d^{*}t$
$\displaystyle\leq||\phi||_{\sup}||\xi||_{\sup}\int_{\operatorname{supp}(r\mapsto\xi_{\alpha}(\rho_{v}(r)\cdot\mathbf{x}))}1d^{*}t$
$\displaystyle=||\phi||_{\sup}||\xi||_{\sup}d^{*}t(\operatorname{supp}(r\mapsto\xi_{\alpha}(\rho_{v}(r)\cdot\mathbf{x}))).$

We deduce from (2) that

$\displaystyle\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots
dx_{n}\hskip-85.35826pt$
$\displaystyle=a_{n}\int_{F_{v,1}}d\lambda_{v,1}(z)\int_{\mathbb{R}_{>0}}{\phi\big{(}\rho_{v}(t)\cdot((x_{j})_{j=1}^{n-1},z)\big{)}d^{*}t}\times$
$\displaystyle\quad\quad\quad\quad\times\int_{(F_{v}^{\times})^{n-1}}f(x_{1},\dots,x_{n-1},z)^{-1}dx_{1}\dots
dx_{n-1}$
$\displaystyle\leq||\psi||_{\sup}||\xi||_{\sup}d^{*}t(\operatorname{supp}(\eta))\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}f^{-1}dx_{1}\dots dx_{n-1}\lambda_{v,1}.$

By [9, Proposition 6 in n0 3, §1, Chapter III] the operator

$\mathscr{C}^{0}_{c}(F^{n}_{v}-\\{0\\},\mathbb{C})\to\mathbb{C}\hskip
28.45274pt\phi\mapsto\int_{F^{n}_{v}-\\{0\\}}\phi f^{-1}dx_{1}\dots dx_{n}$

is continuous. We conclude that $f^{-1}dx_{1}\dots dx_{n}$ is a measure. The
statement is proven. ∎

When $f$ satisfies the equivalent conditions of 6, the measure
$f^{-1}dx_{1}\dots dx_{n}$ is $F_{v}^{\times}$-invariant by 1.

######  Example 9.

Suppose $f$ is continuous and $f(\mathbf{x})\in\mathbb{C}-\\{0\\}$ for every
$\mathbf{x}\in F^{n}_{v}-\\{0\\}$. From the fact that a product of a measure
and a continuous function is a measure [9, n0 4, §1, Chapter III], it follows
that $f^{-1}dx_{1}\dots dx_{n}$ is a measure and hence $f$ satisfies the
equivalent conditions of Proposition 6.

#### 4

Proposition 6 provides a measure $\omega_{v}$ on
$[\mathscr{P}(\mathbf{a})(F_{v})]$.

######  Definition 1.

Let $v\in M_{F}$ and let
$f_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{C}\cup\\{\infty\\}$
be an $\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$ such
that

$dx_{1}\dots dx_{n}(\\{\mathbf{x}|f_{v}(\mathbf{x})=0\\})=0$

and such that $f_{v}^{-1}dx_{1}\dots dx_{n}$ is a measure on
$F^{n}_{v}-\\{0\\}$. We define $\omega_{v}$ to be the quotient measure

$\omega_{v}\mathrel{\mathop{\ordinarycolon}}=(f_{v}^{-1}dx_{1}\dots
dx_{n})/d^{*}x$

on $[\mathscr{P}(\mathbf{a})(F_{v})].$

Recall from 2 that $[\mathscr{T}({\mathbf{a}})(F_{v})]$ is the open subset of
$[\mathscr{P}(\mathbf{a})(F_{v})]$ given by the image of
$q^{\mathbf{a}}_{v}((F_{v}^{\times})^{n})$. We prove that the complement of
the open subset
$[\mathscr{T}({\mathbf{a}})(F_{v})]\subset[\mathscr{P}(\mathbf{a})(F_{v})]$ is
$\omega_{v}$-negligible.

######  Lemma 2.

Let $f_{v}$ be as in 1. One has that

$\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})]-[\mathscr{T}({\mathbf{a}})(F_{v})])=0.$

###### Proof.

The preimage
$(q^{\mathbf{a}}_{v})^{-1}([\mathscr{P}(\mathbf{a})(F_{v})]-[\mathscr{T}({\mathbf{a}})(F_{v})])$
is $dx_{1}\dots dx_{n}$-negligible. Now, [9, Proposition 6, n0 3, §2, Chapter
II] gives that
$\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})]-[\mathscr{T}({\mathbf{a}})(F_{v})])=0$.
∎

We explain how to do the integration against $\omega_{v}$.

######  Lemma 3.

Let $v\in M_{F}$. Let $f_{v}$ be as in 1. Let
$h\mathrel{\mathop{\ordinarycolon}}[\mathscr{P}(\mathbf{a})(F_{v})]\to\mathbb{C}$
be a function. Suppose $v\in M_{F}^{0}$. Then $h\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$ if and only if $(h\circ
q^{\mathbf{a}}_{v})\cdot f_{v}^{-1}|_{\mathcal{D}^{\mathbf{a}}_{v}}\in
L^{1}(\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n})$ and if $h\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$, then

$\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}h\omega_{v}=\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}(h\circ
q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots dx_{n}.$

Suppose $v\in M_{F}^{\infty}$. Then $h\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$ if and only if

$(h\circ q^{\mathbf{a}}_{v})\cdot f_{v}^{-1}|_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}\in L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n-1}\times\lambda_{v,1})$

and if $h\in L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$, then

$\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}h\omega_{v}=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(h\circ q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots
dx_{n-1}\times\lambda_{v,1}.$

###### Proof.

As $[\mathscr{P}(\mathbf{a})(F_{v})]$ is compact and hence paracompact, by 5,
one has that $h\in L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$ if and
only if $k^{\mathbf{a}}_{v}\cdot(h\circ q^{\mathbf{a}}_{v})\in
L^{1}(F^{n}_{v}-\\{0\\},f_{v}^{-1}dx_{1}\dots dx_{n})$, and if $h\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$ then

$\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}h\omega_{v}=\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}\cdot(h\circ
q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots dx_{n}.$

The function $q^{\mathbf{a}}_{v}$ is $F_{v}^{\times}$-invariant, hence is such
$h^{-1}\circ q^{\mathbf{a}}_{v}$. It follows that $(h^{-1}\circ
q^{\mathbf{a}}_{v})\cdot f_{v}$ is an $\mathbf{a}$-homogenous function of
weighted degree $|\mathbf{a}|.$ The set where it vanishes coincides with the
set where $f_{v}$ vanishes, thus this set is $dx_{1}\dots dx_{n}$-negligible.
We apply Proposition 6 for the function $(h^{-1}\circ q^{\mathbf{a}}_{v})\cdot
f_{v}$. It follows that $k^{\mathbf{a}}_{v}\cdot((h\circ
q^{\mathbf{a}}_{v})f_{v}^{-1})\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n})$
if and only if

$(h\circ q^{\mathbf{a}}_{v})\cdot
f_{v}^{-1}|_{\mathcal{D}^{\mathbf{a}}_{v}}\in
L^{1}(\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n})$

if $v$ is finite, and if and only if

$(h\circ q^{\mathbf{a}}_{v})\cdot f_{v}^{-1}|_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}\in L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n-1}\times\lambda_{v,1})$

if $v$ is infinite. Moreover, 6 gives that if

$k^{\mathbf{a}}_{v}\cdot(h\circ q^{\mathbf{a}}_{v})f_{v}^{-1}\in
L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots dx_{n}),$

then

$\int_{F^{n}_{v}-\\{0\\}}k^{\mathbf{a}}_{v}(h\circ
q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots
dx_{n}=\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}(h\circ
q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots dx_{n}.$

if $v$ is finite, and

$\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(h\circ q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots
dx_{n-1}\times\lambda_{v,1}.$

if $v$ is infinite. The statement follows. ∎

###  3 Peyre’s constant

In this section from a quasi-toric family of $\mathbf{a}$-homogenous functions
of weighted degree $|\mathbf{a}|$, we will define a measure on the product
space $\prod_{v\in M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})],$ and Peyre’s
constant of the stacks $\mathscr{P}(\mathbf{a})$ and
$\overline{\mathscr{P}(\mathbf{a})}$.

#### 1

In this paragraph, we calculate the volume $\omega_{v}$ for $v\in M_{F}^{0}$
from 1 when the function $f_{v}$ is toric.

Let $v\in M_{F}^{0}$. As in 3, we set
$r_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0}$
for the function

$r_{v}(\mathbf{x})=\sup_{\begin{subarray}{c}j=1,\dots,n\\\ x_{j}\neq
0\end{subarray}}\bigg{\lceil}-\frac{v(x_{j})}{a_{j}}\bigg{\rceil}.$

Let
$f^{\\#}_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0}$
be the toric $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$. Recall that this means

$f^{\\#}_{v}(\mathbf{x})=|\pi_{v}|_{v}^{-|\mathbf{a}|r_{v}(\mathbf{x})}.$

In 1, we have established that $r_{v}|_{\mathcal{D}^{\mathbf{a}}_{v}}=0$, thus
$f^{\\#}_{v}|_{\mathcal{D}^{\mathbf{a}}_{v}}=1,$ where
$\mathcal{D}^{\mathbf{a}}_{v}=(\mathcal{O}_{v})^{n}-(\pi_{v}^{a_{1}}\mathcal{O}_{v}\times\cdots\times\pi_{v}^{a_{n}}\mathcal{O}_{v})$.
For $v\in M_{F}^{0}$ and $s\in\mathbb{C}$, we denote

$\zeta_{v}(s)=\frac{1}{1-|\pi_{v}|_{v}^{s}}.$

######  Lemma 1.

Let $v\in M_{F}^{0}$ and let
$f_{v}^{\\#}\mathrel{\mathop{\ordinarycolon}}F_{v}^{n}-\\{0\\}\to\mathbb{R}_{>0}$
be the toric $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$. Let $\omega^{\\#}_{v}$ be the measure on
$[\mathscr{P}(\mathbf{a})(F_{v})]$ that is given by

$((f_{v}^{\\#})^{-1}dx_{1}\dots dx_{n})/d^{*}x.$

One has that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=\frac{\zeta_{v}(1)}{\zeta_{v}(|\mathbf{a}|)}.$

###### Proof.

By applying 3 and using the fact that
$f_{v}^{\\#}|_{\mathcal{D}^{\mathbf{a}}_{v}}=1$, we get that

$\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}1\omega^{\\#}_{v}=\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}{(f_{v}^{\\#})^{-1}}dx_{1}\dots
dx_{n}=\zeta_{v}(1)\int_{\mathcal{D}^{\mathbf{a}}_{v}}dx_{1}\dots dx_{n}.$

In turn one has that

$dx_{1}\dots
dx_{n}(\mathcal{D}^{\mathbf{a}}_{v})=1-\prod_{j=1}^{n}dx(\pi_{v}^{a_{j}}\mathcal{O}_{v})=1-|\pi_{v}|_{v}^{|\mathbf{a}|}=\zeta_{v}(|\mathbf{a}|)^{-1},$

and the claim follows. ∎

######  Remark 2.

We will generalize the calculation of 1 in 1,when will be calculating the
Fourier transform of a local toric height at a non-archimedean place.

#### 2

Let us calculate the volume $\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])$
when $v\in M_{F}^{\infty}$ and the function $f_{v}=f_{v}^{\\#}$ is the toric
$\mathbf{a}$-homogenous function of weighted degree $|\mathbf{a}|$.

Recall that the toric $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$ is the function

$f_{v}^{\\#}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0}\hskip
28.45274pt\mathbf{x}\mapsto\max_{j=1,\dots,n}(|x_{j}|_{v}^{1/a_{j}})^{|\mathbf{a}|}.$

######  Lemma 1.

Let $v\in M_{F}^{\infty}$ and let
$f_{v}^{\\#}\mathrel{\mathop{\ordinarycolon}}F_{v}^{n}-\\{0\\}\to\mathbb{R}_{>0}$
be the toric $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$. Let $\omega^{\\#}_{v}$ be the measure on
$[\mathscr{P}(\mathbf{a})(F_{v})]$ that is given by

$((f_{v}^{\\#})^{-1}dx_{1}\dots dx_{n})/d^{*}x.$

One has that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=2^{n-1}|\mathbf{a}|$

if $v$ is real, and that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=(2\pi)^{n-1}|\mathbf{a}|$

if $v$ is complex.

###### Proof.

Lemma 3 gives that:

$\displaystyle\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])\hskip-113.81102pt$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(f_{v}^{\\#})^{-1}dx_{1}\dots dx_{n-1}\times\lambda_{v,1}$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}\max_{j=1,\dots,n}(|x_{j}|_{v}^{{1}/{a_{j}}})^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}\times\lambda_{v,1}$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}\max(\max_{j=1,\dots,n-1}(|x_{j}|_{v}^{1/a_{j}}),1)^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}\times\lambda_{v,1}$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{F_{v,1}}\lambda_{v,1}\int_{(F_{v}^{\times})^{n-1}}\max(\max_{j=1,\dots,n-1}(|x_{j}|_{v}^{1/a_{j}}),1)^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}$
$\displaystyle=a_{n}\int_{(F_{v}^{\times})^{n-1}}\max(\max_{j=1,\dots,n-1}(|x_{j}|_{v}^{1/a_{j}}),1)^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}.$

Lemma 3 gives that the homomorphism

$\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\times
F_{v,1}\to F_{v}^{\times}\hskip 28.45274pt(r,z)\mapsto\rho_{v}(r)z,$

where $\rho_{v}(r)=r^{1/[F_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}]}$,
satisfies that
$(\widetilde{\rho}_{v})_{*}(dr\times\lambda_{v,1})=dx|_{F_{v}^{\times}}$. One
has that $|\widetilde{\rho}_{v}(r,z)|_{v}=r$ and we deduce that

$\displaystyle\omega^{\\#}_{v}(\mathscr{P}(\mathbf{a})(F_{v}))\hskip-28.45274pt$
$\displaystyle=a_{n}\int_{(F_{v}^{\times})^{n-1}}\max(\max_{j=1,\dots,n-1}(|x_{j}|_{v}^{1/a_{j}}),1)^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}$ $\displaystyle=a_{n}\int_{(\mathbb{R}_{>0}\times
F_{v,1})^{n-1}}\max(\max_{j=1,\dots,n-1}(|\widetilde{\rho}_{v}(r_{j}z)|_{v}^{1/a_{j}}),1)^{-|\mathbf{a}|}dx_{1}\dots
dx_{n-1}$
$\displaystyle=a_{n}\int_{F_{v,1}^{n-1}}\lambda_{v,1}^{\otimes(n-1)}\int_{\mathbb{R}_{>0}^{n-1}}\max(\max_{j=1,\dots,n-1}(r_{j}^{1/a_{j}}),1)^{-|\mathbf{a}|}dr_{1}\dots
dr_{n-1}$
$\displaystyle=a_{n}(\lambda_{v,1}(F_{v,1}))^{n-1}\int_{\mathbb{R}_{>0}^{n-1}}\max(\max_{j=1,\dots,n-1}(r_{j}^{1/a_{j}}),1)^{-|\mathbf{a}|}dr_{1}\dots
dr_{n-1}.$

Let us evaluate the last integral. Define
$V_{0}\mathrel{\mathop{\ordinarycolon}}=\\{\mathbf{x}\in\mathbb{R}^{n-1}_{>0}|x_{i}\leq
1\\}.$ For $i=1,\dots,n-1$, define

$V_{i}\mathrel{\mathop{\ordinarycolon}}=\\{\mathbf{x}\in\mathbb{R}^{n-1}_{>0}|x_{i}^{1/a_{i}}=\max_{j}(x^{1/a_{j}}_{j})\\}.$

Clearly, for every $i,j\in\\{0,\dots,n-1\\}$ with $i\neq j$, one has that
$V_{i}\cap V_{j}$ is $dr_{1}\dots dr_{n-1}$-negligible. Thus

$\displaystyle\int_{\mathbb{R}^{n-1}_{>0}}\max(\max_{j}(r_{j}^{1/a_{j}}),1)^{-|\mathbf{a}|}dr_{1}\dots
dr_{n-1}\hskip-85.35826pt$
$\displaystyle=\sum_{i=0}^{n-1}\int_{V_{i}}\max(\max_{j}(r_{j}^{1/a_{j}}),1)^{-|\mathbf{a}|}dr_{1}\dots
dr_{n-1}$ $\displaystyle=\int_{V_{0}}1dr_{1}\dots
dr_{n-1}+\sum_{i=1}^{n-1}\int_{V_{i}}r_{i}^{-|\mathbf{a}|/a_{i}}dr_{1}\dots
dr_{n-1}$
$\displaystyle=1+\sum_{i=1}^{n-1}\int_{1}^{\infty}\bigg{(}\prod^{n-1}_{\begin{subarray}{c}j=1\\\
j\neq i\end{subarray}}\int_{0}^{r_{i}^{a_{j}/a_{i}}}1dr_{j}\bigg{)}\cdot
r_{i}^{-|\mathbf{a}|/a_{i}}dr_{i}$
$\displaystyle=1+\sum_{i=1}^{n-1}\int_{1}^{\infty}\bigg{(}\prod^{n-1}_{\begin{subarray}{c}j=1\\\
j\neq
i\end{subarray}}r_{i}^{a_{j}/a_{i}}\bigg{)}r_{i}^{-|\mathbf{a}|/a_{i}}dr_{i}$
$\displaystyle=1+\int_{1}^{\infty}r_{i}^{-1-a_{n}/a_{i}}dr_{i}$
$\displaystyle=1+\sum_{i=1}^{n-1}\frac{a_{i}}{a_{n}}$
$\displaystyle=\frac{|\mathbf{a}|}{a_{n}},$

where the third equality follows from Fubini theorem. We deduce that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=a_{n}\lambda_{v,1}(F_{v,1})^{n-1}\frac{|\mathbf{a}|}{a_{n}}=\lambda_{v,1}(F_{v,1})^{n-1}|\mathbf{a}|.$

Thus if $v$ is real one has that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=2^{n-1}|\mathbf{a}|$

and if $v$ is complex, one has that

$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})])=(2\pi)^{n-1}|\mathbf{a}|.$

∎

#### 3

We define Peyre’s constant for stack $\mathscr{P}(\mathbf{a})$ for quasi-toric
families $(f_{v})_{v}$.

We will define a measure on the product space

$\prod_{v\in M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})].$

The space $\prod_{v\in M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]$ is compact and
Hausdorff, as for every $v\in M_{F}$ by 5 and by 1, the spaces
$[\mathscr{P}(\mathbf{a})(F_{v})]$ are compact and Hausdorff. Let
$(f_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{\geq
0})_{v\in M_{F}}$ be a quasi-toric family of $|\mathbf{a}|$-homogenous
functions of weighted degree $|\mathbf{a}|$ such that for every $v$, one has
that the set $\\{\mathbf{x}\in F^{n}_{v}-\\{0\\}|f_{v}(\mathbf{x})=0\\}$ is
$dx_{1}\dots dx_{n}$-negligible and that $f_{v}^{-1}dx_{1}\dots dx_{n}$ is a
measure on $F^{n}_{v}-\\{0\\}$. For every $v\in M_{F}$, we set
$\omega_{v}=(f_{v}^{-1}dx_{1}\dots dx_{n})/d^{*}x$. Using measures
$\omega_{v}$, we define a product measure on $\prod_{v\in
M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]$ (by [9, Proposition 9, n0 6, §4,
Chapter III] we indeed get a measure on the product).

######  Definition 1.

We define a measure $\omega=\omega((f_{v})_{v})$ on $\prod_{v\in
M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]$ by

$|\mu_{\gcd(\mathbf{a})}(F)|\Delta(F)^{-\frac{n-1}{2}}\operatorname{Res}(\zeta_{F},1)\bigotimes_{v\in
M_{F}^{0}}\big{(}\zeta_{v}(1)^{-1}\omega_{v}\big{)}\bigotimes_{v\in
M_{F}^{\infty}}\omega_{v}.$

We set

$\tau=\tau((f_{v})_{v})=\omega\big{(}\prod_{v\in
M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]\big{)},$

where $\Delta(F)$ is the absolute discriminant of $F$.

We explain how $\omega$ changes, when the quasi-toric family is changed.

######  Lemma 2.

Let $S$ be a finite set of places and for $v\in S$, let
$h_{v}\mathrel{\mathop{\ordinarycolon}}[\mathscr{P}(\mathbf{a})(F_{v})]\to\mathbb{R}_{>0}$
be a continuous function. For $v\in M_{F}-S$, we set $h_{v}=1$. Let us denote
by $h\mathrel{\mathop{\ordinarycolon}}\prod_{v\in
M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]\to\mathbb{R}_{>0}$ the function
$\otimes_{v\in M_{F}}h_{v}$. One has that

$\omega((h_{v}f_{v})_{v})=h^{-1}\omega((f_{v})_{v}).$

###### Proof.

For $v\in M_{F}$, it follows directly from Lemma 4 that

$(((h_{v}\circ q^{\mathbf{a}}_{v})\cdot f_{v})^{-1}dx_{1}\dots
dx_{n})/d^{*}x=(h^{-1}_{v})((f_{v}^{-1}dx_{1}\dots
dx_{n})/d^{*}x)=h^{-1}_{v}\omega_{v}.$

It follows that the measure $\omega(((h_{v,\eta}\circ q^{\mathbf{a}}_{v})\cdot
f_{v})_{v})$ on $\prod_{v\in M_{F}}[\mathscr{P}(\mathbf{a})(F_{v})]$ defined
by the quasi-toric family $((h_{v}\circ q^{\mathbf{a}}_{v})\cdot f_{v})_{v}$
satisfies that

$\displaystyle\omega(((h_{v}\circ q^{\mathbf{a}}_{v})\cdot
f_{v})_{v})\hskip-28.45274pt$
$\displaystyle=\frac{|\mu_{\gcd(\mathbf{a})}(F)|}{\Delta(F)^{\frac{n-1}{2}}}\operatorname{Res}(\zeta_{F},1)\prod_{v\in
M_{F}^{0}}\bigg{(}\zeta_{v}(1)^{-1}h^{-1}_{v}\omega_{v}\bigg{)}\prod_{v\in
M_{F}^{\infty}}h^{-1}_{v}\omega_{v}$ $\displaystyle=h^{-1}\omega.$

∎

We give another expression for $\tau$.

######  Lemma 3.

Let $S$ be the finite set of places $v$ for which $f_{v}$ is not toric. One
has that

$\tau((f_{v})_{v})\\\
=\frac{\operatorname{Res}(\zeta_{F},1)|\mu_{\gcd(\mathbf{a})}(F)|}{\Delta(F)^{\frac{n-1}{2}}\zeta_{F}(|\mathbf{a}|)}\prod_{v\in
S\cap
M_{F}^{0}}\frac{\zeta_{v}(|\mathbf{a}|)\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])}{\zeta_{v}(1)}\times\\\
\times\prod_{v\in
M_{F}^{\infty}}\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})]).$

###### Proof.

Lemma 1 gives that for every $v\in M_{F}^{0}-S$ one has

$\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])\zeta_{v}(1)^{-1}=\zeta_{v}(|\mathbf{a}|)^{-1}$

and thus:

$\displaystyle\tau((f_{v})_{v})|\mu_{\gcd(\mathbf{a})}(F)|^{-1}\Delta(F)^{\frac{n-1}{2}}\hskip-142.26378pt$
$\displaystyle=\operatorname{Res}(\zeta_{F},1)\prod_{v\in S\cap
M_{F}^{0}}\frac{\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])}{\zeta_{v}(1)}\times$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\times\prod_{v\in
M_{F}^{0}-S}\zeta_{v}(|\mathbf{a}|)^{-1}\times\prod_{v\in
M_{F}^{\infty}}\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])$
$\displaystyle=\frac{\operatorname{Res}(\zeta_{F},1)}{\zeta(|\mathbf{a}|)}\prod_{v\in
S\cap
M_{F}^{0}}\bigg{(}\frac{\zeta_{v}(|\mathbf{a}|)\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})])}{\zeta_{v}(1)}\bigg{)}\times\prod_{v\in
M_{F}^{\infty}}\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})]).$

∎

###  4 Haar measure on $[\mathscr{T}({\mathbf{a}})(F_{v})]$

Let $F_{v}^{\times}$ acts on $(F_{v}^{\times})^{n}$ by
$t\cdot\mathbf{x}=(t^{a_{j}}x_{j})_{j}$. This action is proper by 1. The
quotient for this action is $[\mathscr{T}({\mathbf{a}})(F_{v})]$ by 2 and is
locally compact by 1. By 2, one has that the map
$[\mathscr{T}({\mathbf{a}})(F_{v})]\to[\mathscr{P}(\mathbf{a})(F_{v})],$
induced from $F_{v}^{\times}$-invariant map
$(F_{v}^{\times})^{n}\hookrightarrow
F^{n}_{v}-\\{0\\}\to[\mathscr{P}(\mathbf{a})(F_{v})]$ is an open embedding. By
1, the map

$\epsilon\mathrel{\mathop{\ordinarycolon}}F_{v}^{\times}\to(F_{v}^{\times})^{n}\hskip
28.45274ptt\mapsto(t^{a_{j}})_{j}$

is proper, its image
$(F_{v}^{\times})_{\mathbf{a}}\mathrel{\mathop{\ordinarycolon}}=\epsilon(F_{v}^{\times})=\\{(t^{a_{j}})_{j}|t\in
F_{v}^{\times}\\},$ is a closed subgroup of $(F_{v}^{\times})^{n}$ and one has
an identification
$[\mathscr{T}({\mathbf{a}})(F_{v})]=(F_{v}^{\times})^{n}/(F_{v}^{\times})_{\mathbf{a}}$.
Using this identification, we endow $[\mathscr{T}({\mathbf{a}})(F_{v})]$ with
a structure of a topological group (which is necessary abelian). The goal of
this section is to define a Haar measure on
$[\mathscr{T}({\mathbf{a}})(F_{v})]$ and relate it with the measure
$\omega_{v}$ on $[\mathscr{P}(\mathbf{a})(F_{v})]$.

#### 1

Let $v\in M_{F}$. We are going to define a Haar measure on
$[\mathscr{T}({\mathbf{a}})(F_{v})]$.

By the fact that a product of a continuous function and a measure is a measure
[9, n0 4, §1, Chapter III], one has that

$\mathscr{C}^{0}_{c}((F_{v}^{\times})^{n},\mathbb{C})\to\mathbb{C}\hskip
28.45274pt\phi\mapsto\int_{(F_{v}^{\times})^{n}}\prod_{j=1}^{n}|x_{j}|_{v}^{-1}dx_{1}\dots
dx_{n}$

is a measure on $(F_{v}^{\times})^{n}$. The the function

$F^{n}_{v}-\\{0\\}\to\mathbb{R}_{\geq 0}\hskip
28.45274pt\mathbf{x}\mapsto\prod_{j=1}^{n}|x_{j}|_{v}$

is $\mathbf{a}$-homogenous of weighted degree $|\mathbf{a}|$ and the set where
it vanishes is given by $\\{\mathbf{x}\in F^{n}_{v}-\\{0\\}|\exists
j\mathrel{\mathop{\ordinarycolon}}x_{j}=0\\}$, thus this set is $dx_{1}\dots
dx_{n}$-negligible (because it is contained in a finite union of hyperplanes
in $(F_{v})^{n}$.) It follows from 1 that

$\prod_{j=1}^{n}|x_{j}|_{v}^{-1}dx_{1}\dots dx_{n}=d^{*}x_{1}\dots d^{*}x_{n}$

is $F_{v}^{\times}$-invariant measure on $F_{v}^{\times}$.

######  Definition 1.

We define a measure $\mu_{v}$ on $[\mathscr{T}({\mathbf{a}})(F_{v})]$ by

$\mu_{v}\mathrel{\mathop{\ordinarycolon}}=(d^{*}x_{1}\dots
d^{*}x_{n})/d^{*}x_{v}.$

By 3 the measure $\mu_{v}$ is a Haar measure on
$[\mathscr{T}({\mathbf{a}})(F_{v})]$.

######  Lemma 2.

Let $v\in M_{F}$. Let
$f_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{\geq
0}$ be a continuous $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$ such that

$f_{v}^{-1}dx_{1}\dots dx_{n}$

is a measure on $F^{n}_{v}-\\{0\\}$ (this in particular implies that
$dx_{1}\dots dx_{n}(\\{\mathbf{x}|f_{v}(\mathbf{x})=0\\})=0$). Let
$H_{v}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F_{v})]\to\mathbb{R}_{\geq
0}$ be the function given by the continuous $F_{v}^{\times}$-invariant
function

$\mathbf{x}\mapsto f_{v}(\mathbf{x})\prod_{j=1}^{n}|x_{j}|^{-1}_{v}.$

  1. 1.

One has that $\mu_{v}(\\{\mathbf{y}|H_{v}(\mathbf{y})=0\\})=0$.

  2. 2.

One has an equality of the measures
$\mu_{v}=(H_{v})(\omega_{v}|_{[\mathscr{T}({\mathbf{a}})(F_{v})]}).$

###### Proof.

  1. 1.

By the definition of $H_{v}$ one has that
$\\{\mathbf{y}|H_{v}(\mathbf{y})=0\\}=q^{\mathbf{a}}_{v}(\\{\mathbf{x}\in(F_{v}^{\times})^{n}|f_{v}(\mathbf{x})=0\\}).$
Note that as $dx_{1}\dots dx_{n}(\\{\mathbf{x}|f_{v}(\mathbf{x})=0\\})=0,$ it
follows that the set
$\\{\mathbf{x}\in(F_{v}^{\times})^{n}|f_{v}(\mathbf{x})=0\\}$ is $dx_{1}\dots
dx_{n}$-negligible and thus $d^{*}x_{1}\dots d^{*}x_{n}$-negligible. Hence, by
[10, Proposition 6, n0 3, §2, Chapter VII], one has that
$\mu_{v}(q^{\mathbf{a}}_{v}(\\{\mathbf{x}\in(F_{v}^{\times})^{n}|f(\mathbf{x})=0\\}))=0.$

  2. 2.

Observe that

$\displaystyle d^{*}x_{1}\dots
d^{*}x_{n}=\prod_{j=1}^{n}|x_{j}|_{v}^{-1}dx_{1}\dots dx_{n}$
$\displaystyle=f_{v}(\mathbf{x})\prod_{j=1}^{n}|x_{j}|_{v}^{-1}f_{v}(\mathbf{x})^{-1}dx_{1}\dots
dx_{n}$ $\displaystyle=(H_{v}\circ q^{\mathbf{a}}_{v})f_{v}^{-1}dx_{1}\dots
dx_{n}.$

Now Lemma 4 gives precisely that

$\mu_{v}=(d^{*}x_{1}\dots d^{*}x_{n})/d^{*}x=H_{v}(f_{v}^{-1}dx_{1}\dots
dx_{n})/d^{*}x=H_{v}\omega_{v}.$

∎

######  Lemma 3.

Let
$h\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F_{v})]\to\mathbb{C}$
be a function. Suppose $v\in M_{F}^{0}$. One has that $h\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$ if and only if $(h\circ
q^{\mathbf{a}}_{v})\in
L^{1}((F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v},d^{*}x_{1}\dots
d^{*}x_{n})$, and if $h\in L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$,
one has that:

$\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}h\mu_{v}=\frac{1}{1-|\pi_{v}|_{v}}\int_{(F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v}}(h\circ
q^{\mathbf{a}}_{v})d^{*}x_{1}\dots d^{*}x_{n}.$

Suppose $v\in M_{F}^{\infty}$. One has that $h\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$ if and only if $(h\circ
q^{\mathbf{a}}_{v})\in L^{1}((F_{v}^{\times})^{n-1}\times
F_{v,1},d^{*}x_{1}\dots d^{*}x_{n-1}\times\lambda_{v,1})$, and if $h\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$, one has that:

$\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(h\circ q^{\mathbf{a}}_{v})d^{*}x_{1}\dots
d^{*}x_{n-1}\times\lambda_{v,1}.$

###### Proof.

Let
$f_{v}^{\\#}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0}$
be the toric $\mathbf{a}$-homogenous function of weighted degree
$|\mathbf{a}|$. Let
$H^{\\#}_{v}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F_{v})]\to\mathbb{R}_{>0}$
be the induced function from $F_{v}^{\times}$-invariant function
$(F_{v}^{\times})^{n}\to\mathbb{R}_{>0}$ given by $\mathbf{x}\mapsto
f^{\\#}_{v}(\mathbf{x})\prod_{j=1}^{n}|x_{j}|_{v}$. It follows from Lemma 2,
that one has an equality of the measures
$\mu_{v}=H^{\\#}_{v}\omega^{\\#}_{v}|_{[\mathscr{T}({\mathbf{a}})(F_{v})]}$.
We deduce that $h\in L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$ if and
only if $hH^{\\#}_{v}\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\omega^{\\#}_{v}),$ and as by 2 one
has
$\omega^{\\#}_{v}([\mathscr{P}(\mathbf{a})(F_{v})]-[\mathscr{T}({\mathbf{a}})(F_{v})])=0$,
if and only if $hH^{\\#}_{v}\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega^{\\#}_{v})$. Moreover, it
follows that if $h\in L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v}),$ then

$\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}h\mu_{v}=\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}(hH^{\\#}_{v})\omega^{\\#}_{v}.$

Suppose $v\in M_{F}^{0}.$ Recall that by 1 one has that
$f^{\\#}_{v}|_{\mathcal{D}^{\mathbf{a}}_{v}}=1$. By 3, one has that
$hH^{\\#}_{v}\in L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega^{\\#}_{v})$ if
and only if

$(((hH^{\\#}_{v})\circ
q^{\mathbf{a}}_{v})(f_{v}^{\\#})^{-1})|_{\mathcal{D}^{\mathbf{a}}_{v}}=(h\circ
q^{\mathbf{a}}_{v})(H_{v}^{\\#}\circ
q^{\mathbf{a}}_{v})|_{\mathcal{D}^{\mathbf{a}}_{v}}=(h\circ
q^{\mathbf{a}}_{v})(|x_{1}|_{v}^{-1}\cdots|x_{n}|_{v}^{-1})|_{\mathcal{D}^{\mathbf{a}}_{v}}$

is an element of $L^{1}(\mathcal{D}^{\mathbf{a}}_{v},dx_{1}\dots dx_{n}).$
Moreover, Lemma 3 gives that if $hH^{\\#}_{v}\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega^{\\#}_{v})$ then

$\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}hH^{\\#}_{v}\omega^{\\#}_{v}=\frac{1}{1-|\pi_{v}|_{v}}\int_{\mathcal{D}^{\mathbf{a}}_{v}}(h\circ
q^{\mathbf{a}}_{v})|x_{1}|_{v}^{-1}\cdots|x_{n}|_{v}^{-1}dx_{1}\dots dx_{n}.$

As $dx_{1}\dots
dx_{n}(\mathcal{D}^{\mathbf{a}}_{v}-(\mathcal{D}^{\mathbf{a}}_{v}\cap(F_{v}^{\times})^{n}))=0,$
(because
$\mathcal{D}^{\mathbf{a}}_{v}-(\mathcal{D}^{\mathbf{a}}_{v}\cap(F_{v}^{\times})^{n})$
is contained in a finite union of hyperplanes of $F_{v}^{n}$,) the last
integral is equal to

$\frac{1}{1-|\pi_{v}|_{v}}\int_{(F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v}}(h\circ
q^{\mathbf{a}}_{v})d^{*}x_{1}\dots d^{*}x_{n}.$

It follows that one has $h\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$ if and only if $(h\circ
q^{\mathbf{a}}_{v})\in
L^{1}((F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v},d^{*}x_{1}\dots
d^{*}x_{n})$ and if $h\in L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v}),$
then

$\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}h\mu_{v}=\frac{1}{1-|\pi_{v}|_{v}}\int_{(F_{v}^{\times})^{n}}(h\circ
q^{\mathbf{a}}_{v})d^{*}x_{1}\dots d^{*}x_{n}.$

Suppose that $v\in M_{F}^{\infty}$. By 3, one has that $hH^{\\#}_{v}\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega^{\\#}_{v})$ if and only if

$((hH^{\\#}_{v})\circ q^{\mathbf{a}}_{v})(f_{v}^{\\#})^{-1}=((h\circ
q^{\mathbf{a}}_{v})\cdot(H^{\\#}\circ
q^{\mathbf{a}}_{v}))(f_{v}^{\\#})^{-1}=(h\circ
q^{\mathbf{a}}_{v})\prod_{j=1}^{n-1}|x_{j}|_{v}^{-1}$

is an element of $L^{1}((F_{v}^{\times})^{n-1}\times F_{v,1},dx_{1}\dots
dx_{n-1}\times\lambda_{v,1})$ i.e. if and only if

$(h\circ q^{\mathbf{a}}_{v})\in L^{1}((F_{v}^{\times})^{n-1}\times
F_{v,1},d^{*}x_{1}\dots d^{*}x_{n-1}\lambda_{v,1}).$

Moreover, Lemma 3 gives that if $hH^{\\#}_{v}\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega^{\\#}_{v})$ then

$\displaystyle\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}hH^{\\#}_{v}\omega^{\\#}_{v}\hskip-56.9055pt$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(h\circ
q^{\mathbf{a}}_{v})\prod_{j=1}^{n-1}|x_{j}|_{v}^{-1}dx_{1}\dots
dx_{n-1}\times\lambda_{v,1}$
$\displaystyle=\frac{a_{n}}{\lambda_{v,1}(F_{v,1})}\int_{(F_{v}^{\times})^{n-1}\times
F_{v,1}}(h\circ q^{\mathbf{a}}_{v})d^{*}x_{1}\dots
d^{*}x_{n-1}\times\lambda_{v,1}.$

∎

#### 2

Recall that for $v\in M_{F}^{0}$ by 4, the group
$[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]$ identifies with the image of
$\mathcal{O}_{v}^{\times n}$ in $[\mathscr{T}({\mathbf{a}})(F_{v})]$ for the
quotient homomorphism
$(F_{v}^{\times})^{n}\to[\mathscr{T}({\mathbf{a}})(F_{v})]$ and that with this
identification, becomes open and compact subgroup of
$[\mathscr{T}({\mathbf{a}})(F_{v})]$. We calculate the volume of
$[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]$ against $\mu_{v}$.

######  Lemma 1.

Let $v\in M_{F}^{0}$. The measure $\mu_{v}$ is normalized by

$\mu_{v}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])=(1-|\pi_{v}|_{v})^{n-1}=\zeta_{v}(1)^{-(n-1)}$

.

###### Proof.

Let us firstly establish that
$(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])\cap\mathcal{D}^{\mathbf{a}}_{v}=(\mathcal{O}_{v}^{\times})^{n}.$
One has that

$(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])^{-1}=\bigcup_{t\in\mathbb{G}_{m}(F_{v})}t\cdot(\mathcal{O}_{v}^{\times})^{n}.$

Note that if $\mathbf{u}\in(\mathcal{O}_{v}^{\times})^{n},$ then for every
index $j$ one has $v(t^{a_{j}}u_{j})=a_{j}v(t)+v(u_{j})=a_{j}v(t)$. Now if
$v(t)>0$ it follows that
$t\cdot\mathbf{u}\in\prod_{j=1}^{n}(\pi_{v}^{a_{j}}\mathcal{O}_{v}),$ thus
$t\cdot\mathbf{u}\not\in\mathcal{D}^{\mathbf{a}}_{v}$ and if $v(t)<0$ then
$t\cdot\mathbf{u}\not\in\mathcal{O}_{v}^{n}$, thus
$t\cdot\mathbf{u}\not\in\mathcal{D}^{\mathbf{a}}_{v}$. We have obtained if
$v(t)\neq 0$, then $(t\cdot\mathcal{O}_{v}^{\times
n})\cap\mathcal{D}^{\mathbf{a}}_{v}=\emptyset$ and hence,

$\mathcal{D}^{\mathbf{a}}_{v}\cap(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])=\mathcal{D}^{\mathbf{a}}_{v}\cap\bigcup_{t\in\mathbb{G}_{m}(F_{v})}t\cdot\mathcal{O}_{v}^{\times
n}=\mathcal{D}^{\mathbf{a}}_{v}\cap\mathcal{O}_{v}^{\times
n}=\mathcal{O}_{v}^{\times n}.$

We deduce that one has equality of functions

$(\mathbf{1}_{(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])})|_{(F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v}}={\mathbf{1}_{(\mathcal{O}_{v}^{\times})^{n}}}.$

Now, we can calculate $\mu_{v}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]).$
One has by 3 that

$\displaystyle\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\mathbf{1}_{[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]}\mu_{v}$
$\displaystyle=\frac{1}{1-|\pi_{v}|_{v}}\int_{(F_{v}^{\times})^{n}\cap\mathcal{D}^{\mathbf{a}}_{v}}\mathbf{1}_{(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])}d^{*}x_{1}\dots
d^{*}x_{n}$
$\displaystyle=\frac{1}{1-|\pi_{v}|_{v}}\int_{(F_{v}^{\times})^{n}}\mathbf{1}_{(\mathcal{O}_{v}^{\times})^{n}}d^{*}x_{1}\dots
d^{*}x_{n}$
$\displaystyle=\frac{1}{1-|\pi_{v}|_{v}}\bigg{(}\int_{F_{v}^{\times}}\mathbf{1}_{\mathcal{O}_{v}^{\times}}{d^{*}x}\bigg{)}^{n}$
$\displaystyle=(1-|\pi_{v}|_{v})^{n-1}.$

The statement follows. ∎

#### 3

Let us dedicate this paragraph to the definition of a Haar measure on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. Let $n\in\mathbb{Z}_{>0}$ and
let $\mathbf{a}\in\mathbb{Z}^{n}_{>0}$. We set
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ to be the “adelic space” of
$\mathscr{T}({\mathbf{a}})$, i.e.

$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]=\sideset{}{{}^{\prime}}{\prod}_{v\in
M_{F}}[\mathscr{T}({\mathbf{a}})(F_{v})],$

where the restricted product is taken with respect to the sequence of the open
and compact subgroups for
$[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]\subset[\mathscr{T}({\mathbf{a}})(F_{v})]$
for $v\in M_{F}^{0}$.

######  Proposition 1.

Let $I$ be a set and let $I^{\prime}$ be a subset such that $I-I^{\prime}$ is
finite. For $i\in I$, let $G_{i}$ be a locally compact abelian group endowed
with a Haar measure $dg_{i}$. For $i\in I^{\prime}$, let $H_{i}$ be an open
and compact subgroup of $G_{i}$ such that $dg_{i}(H_{i})=1$. Set $G$ to be the
restricted product $\sideset{}{{}^{\prime}}{\prod}_{i\in I}G_{i}$ with respect
to the subgroups $H_{i}\subset G_{i}$ for $i\in I^{\prime}$.

  1. 1.

[10, Proposition 5, n0 5, §1, Chapter VII]Let $S$ be a finite subset of $I$
containing $I-I^{\prime}$. Set

$G_{S}\mathrel{\mathop{\ordinarycolon}}=\prod_{i\in S}G_{i}\times\prod_{i\in
I-S}H_{i}.$

The measure

$\bigotimes_{i\in S}dg_{i}\bigotimes_{i\in I-S}(dg_{i}|_{H_{i}})$

is a Haar measure on $G_{S}$.

  2. 2.

[49, Proposition 5.5] There exists a unique Haar measure $dg$ on $G$ such that
for every finite subset $S\subset I$ which contains $I-I^{\prime}$ one has
that

$dg|_{G_{S}}=\bigotimes_{i\in S}dg_{i}\bigotimes_{i\in I-S}(dg_{i}|_{H_{i}}).$

The measure $dg$ will be called the restricted product Haar measure and, by
the abuse of the notation, may be denoted as $dg=\otimes dg_{i}$. We present a
way how to calculate the integral of a function.

######  Proposition 2 ([49, Proposition 5-6]).

In the situation of 1, let $f_{i}\in L^{1}(G_{i},dg_{i})$ be a continuous
complex valued function such that there exists a finite subset
$I^{\prime\prime}\subset I^{\prime}$ with $I^{\prime}-I^{\prime\prime}$ is
finite such that $f_{i}|H_{i}=1$ for every $i\in I^{\prime\prime}$. The
function

$f\mathrel{\mathop{\ordinarycolon}}(x_{i})_{i}\mapsto\prod_{i}f_{i}(x_{i})$

is continuous. Suppose that

$\prod_{i}\int_{G_{i}}f_{i}dg_{i}$

converges. Then $f\in L^{1}(G)$ and

$\int_{G}fdg=\prod_{i}\int_{G_{i}}f_{i}dg_{i}.$

For $v\in M_{F}^{0}$, we have established in 1 that
$\mu_{v}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])=\zeta_{v}(1)^{-(n-1)}.$
We will apply 1 to define a Haar measure on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$.

######  Definition 3.

Let $\mu_{\mathbb{A}_{F}}$ be the restricted product measure

$\mu_{\mathbb{A}_{F}}=\bigotimes_{v\in
M_{F}^{0}}\zeta_{v}(1)^{(n-1)}\mu_{v}\otimes\bigotimes_{v\in
M_{F}^{\infty}}\mu_{v}.$

The group $\mathbb{A}_{F}^{\times}$ acts on $(\mathbb{A}_{F}^{\times})^{n}$
via the proper homomorphism $(x_{v})_{v}\mapsto((x_{v}^{a_{j}})_{j})_{v}$
(Lemma 2) and one has an identification
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]=(\mathbb{A}_{F}^{\times})^{n}/(\mathbb{A}_{F}^{\times})$
(Lemma 4 together with 3). Endow $\mathbb{A}_{F}^{\times}$ with the Haar
measure

$d^{*}x_{\mathbb{A}_{F}}\mathrel{\mathop{\ordinarycolon}}=\bigotimes_{v\in
M_{F}^{0}}\zeta_{v}(1)d^{*}x_{v}\otimes\bigotimes_{v\in
M_{F}^{\infty}}d^{*}x_{v}.$

Let
$d^{*}\mathbf{x}_{\mathbb{A}_{F}}\mathrel{\mathop{\ordinarycolon}}=d^{*}x_{\mathbb{A}_{F}}^{\otimes
n}$ be the product Haar measure on $(\mathbb{A}_{F}^{\times})^{n}$.

######  Lemma 4.

One has the following equality of the measures on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]=(\mathbb{A}_{F}^{\times})^{n}/\mathbb{A}_{F}^{\times}\mathrel{\mathop{\ordinarycolon}}$

$\mu_{\mathbb{A}_{F}}=d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x_{\mathbb{A}_{F}}.$

###### Proof.

The quotient measure
$d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x_{\mathbb{A}_{F}}$ is a Haar measure
on $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ by 3. Therefore, it suffices
to verify the equality on a single non-trivial compactly supported function on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ which takes non-negative values.
For $v\in M_{F}^{0}$ we set
$\phi_{v}=\mathbf{1}_{(\mathcal{O}_{v}^{\times})^{n}}\mathrel{\mathop{\ordinarycolon}}(F_{v}^{\times})^{n}\to\mathbb{C}$
and for $v\in M_{F}^{\infty}$, we let
$\phi_{v}\mathrel{\mathop{\ordinarycolon}}(F_{v}^{\times})^{n}\to\mathbb{R}_{\geq
0}$ be a non-trivial continuous function with compact support. The function
$\phi=\bigotimes_{v}\phi_{v}$ is continuous by 2 and compactly supported (its
support is the set $\prod_{v\in
M_{F}^{0}}(\mathcal{O}_{v}^{\times})^{n}\times\prod_{v\in
M_{F}^{\infty}}\operatorname{supp}(\phi_{v})$). For $v\in M_{F}$ and
$\mathbf{y}\in[\mathscr{T}({\mathbf{a}})(F_{v})],$ let
$\widetilde{\mathbf{y}}\in(F_{v}^{\times})^{n}$ be its lift and for
$\mathbf{y}\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})],$ let
$\widetilde{\mathbf{y}}\in(\mathbb{A}_{F}^{\times})^{n}$ be its lift. For
$v\in M_{F}^{0}$ we define

$\phi_{v}^{*}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F_{v})]\to\mathbb{R}_{\geq
0}\hskip
28.45274pt\mathbf{y}\mapsto\zeta_{v}(1)\int_{F_{v}^{\times}}\phi_{v}(x\cdot\widetilde{\mathbf{y}})d^{*}x$

and for $v\in M_{F}^{\infty}$ we define

$\phi^{*}_{v}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F_{v})]\to\mathbb{R}_{\geq
0}\hskip
28.45274pt\mathbf{y}\mapsto\int_{F_{v}^{\times}}\phi_{v}(x\cdot\widetilde{\mathbf{y}})d^{*}x,$

the functions are well defined, continuous compactly supported and of non-
negative values by 1. By the definition we have that $(d^{*}x)^{\otimes
n}/d^{*}x=\mu_{v}$ and thus $(d^{*}x)^{\otimes
n}/(\zeta_{v}(1)d^{*}x)=\zeta_{v}(1)^{-1}\mu_{v}$. Now by 2, one has that

$\displaystyle\int_{(F_{v}^{\times})^{n}}(\phi)(d^{*}x)^{\otimes n}$
$\displaystyle=\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}(\mu_{v}/\zeta_{v}(1))\bigg{(}\mathbf{y}\mapsto\int_{F_{v}^{\times}}\phi(x\cdot\widetilde{\mathbf{y}})\zeta_{v}(1)d^{*}x\bigg{)}$
$\displaystyle=\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}(\phi^{*})(\mu_{v}/\zeta_{v}(1))$
$\displaystyle=\zeta_{v}(1)^{-1}\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\phi^{*}\mu_{v}.$

If $v\in M_{F}^{0}$, let us prove that

$\phi^{*}_{v}=\mathbf{1}_{[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]}.$

Indeed, if
$\mathbf{z}\not\in(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]),$
then $\phi_{v}(x\cdot\mathbf{z})=0$ for every $x\in F_{v}^{\times}$, thus
$\phi^{*}_{v}(q^{\mathbf{a}}_{v}(\mathbf{z}))=0$. If
$\mathbf{z}\in(q^{\mathbf{a}}_{v})^{-1}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})],$
then $x\cdot\mathbf{z}\in(\mathcal{O}_{v}^{\times})^{n}$ if and only if
$v(x)=v(z_{1})/a_{1}$, and thus $d^{*}x(\\{x\in
F_{v}^{\times}|x\cdot\mathbf{z}\in(\mathcal{O}_{v}^{\times})^{n}\\})=\zeta_{v}(1)^{-1},$
and hence
$\zeta_{v}(1)\int_{F_{v}^{\times}}\phi_{v}(x\cdot\mathbf{z})d^{*}x=1$. It
follows that
$\phi^{*}_{v}=\mathbf{1}_{[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]}.$

We also define

$\phi^{*}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]\to\mathbb{C}\hskip
28.45274pt\mathbf{y}\mapsto\int_{\mathbb{A}_{F}^{\times}}\phi(x_{\mathbb{A}_{F}}\cdot\widetilde{\mathbf{y}})d^{*}x_{\mathbb{A}_{F}},$

by 1 it is well defined and continuous function. We are going to verify that
the two measures coincide on $\phi^{*}$. For every
$\mathbf{y}\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$, one has that

$\displaystyle\phi^{*}(\mathbf{y})$
$\displaystyle=\int_{\mathbb{A}_{F}^{\times}}\phi(x_{\mathbb{A}_{F}}\cdot\widetilde{\mathbf{y}})d^{*}x_{\mathbb{A}_{F}}$
$\displaystyle=\prod_{v\in
M_{F}^{0}}\zeta_{v}(1)\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\phi_{v}(x\cdot\widetilde{\mathbf{y}}_{v})d^{*}x\times\prod_{v\in
M_{F}^{\infty}}\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\phi_{v}(x\cdot\widetilde{\mathbf{y}}_{v})d^{*}x$
$\displaystyle=\prod_{v\in M_{F}}\phi_{v}^{*}(\mathbf{y}_{v}),$

i.e. $\phi^{*}=\otimes_{v\in M_{F}}\phi^{*}_{v}$. By 2, we have that

$\displaystyle\int_{[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]}(\phi^{*})(d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x_{\mathbb{A}_{F}})=\int_{(\mathbb{A}_{F}^{\times})^{n}}\phi
d^{*}\mathbf{x}_{\mathbb{A}_{F}}.$

On the other side, one has that

$\displaystyle\int_{[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]}\phi^{*}\mu_{\mathbb{A}_{F}}\hskip-56.9055pt$
$\displaystyle=\int_{[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]}\big{(}\otimes_{v}\phi^{*}_{v}\big{)}\mu_{\mathbb{A}_{F}}$
$\displaystyle=\prod_{v\in
M_{F}^{0}}\zeta_{v}(1)^{n-1}\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\phi^{*}_{v}\mu_{v}\times\prod_{v\in
M_{F}^{\infty}}\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}\phi^{*}_{v}\mu_{v}$
$\displaystyle=\prod_{v\in
M_{F}^{0}}\zeta_{v}(1)^{n}\int_{(F_{v}^{\times})^{n}}\phi_{v}d^{*}x_{1}\dots
d^{*}x_{n}\times\prod_{v\in
M_{F}^{\infty}}\int_{(F_{v}^{\times})^{n}}\phi_{v}d^{*}x_{1}\dots d^{*}x_{n}$
$\displaystyle=\int_{(\mathbb{A}_{F}^{\times})^{n}}(\otimes_{v}\phi_{v})d^{*}\mathbf{x}_{\mathbb{A}_{F}}$
$\displaystyle=\int_{(\mathbb{A}_{F}^{\times})^{n}}\phi
d^{*}\mathbf{x}_{\mathbb{A}_{F}}.$

We have verified that the Haar measures
$d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x$ and $\mu_{\mathbb{A}_{F}}$ satisfy
that

$(d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x)(\phi^{*})=\mu_{\mathbb{A}_{F}}(\phi^{*})>0,$

thus $d^{*}\mathbf{x}_{\mathbb{A}_{F}}/d^{*}x=\mu_{\mathbb{A}_{F}}.$ The
statement is proven. ∎

#### 4

In this paragraph we define and calculate the Tamagawa number of the “stacky
torus” $\mathscr{T}({\mathbf{a}})$.

Denote by $\mu_{\mathbb{R}}$ the quotient measure $(d^{*}x)^{\otimes
n}/d^{*}x$ on the quotient $\mathbb{R}^{n}_{>0}/\mathbb{R}_{>0}$ for the
action of $\mathbb{R}_{>0}$ via the proper map $t\mapsto(t^{a_{j}})_{j}$. By 3
the quotient identifies with the quotient group
$\mathbb{R}^{n}_{>0}/(\mathbb{R}_{>0})_{\mathbf{a}}$ and $\mu_{\mathbb{R}}$ is
a Haar measure on it.

######  Definition 1.

We define $\mu_{1}$ to be the Haar measure on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})_{1}]$ normalized by the condition
$\mu_{1}\otimes\mu_{\mathbb{R}}=\mu_{\mathbb{A}_{F}}$ for the identification

$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}\times(\mathbb{R}^{n}_{>0}/(\mathbb{R}_{>0})_{\mathbf{a}})=[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$

given by the isomorphism (2).

We can write $\mu_{1}$ as a quotient measure:

######  Lemma 2.

The measure $\mu_{1}$ identifies with the measure
$d^{*}\mathbf{x}^{1}_{\mathbb{A}_{F}}/d^{*}x^{1}_{\mathbb{A}_{F}}$ on
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}=(\mathbb{A}_{F}^{1})^{n}/\mathbb{A}_{F}^{1},$
where the action is given by the proper morphism

$\mathbb{A}_{F}^{1}\to(\mathbb{A}_{F}^{1})^{n}\hskip
28.45274pt(x_{v})\mapsto((x^{a_{j}}_{v})_{j})_{v}.$

###### Proof.

We have the following bicomplex (we have written the corresponding measures
next to the groups)

${{K_{1}}}$${{K_{2}}}$${{K_{3}}}$${(\mathbb{A}_{F}^{1},dx^{1}_{\mathbb{A}_{F}})}$${(\mathbb{A}_{F}^{\times},d^{*}x_{\mathbb{A}_{F}})}$${{(\mathbb{R}_{>0},d^{*}x)}}$${{((\mathbb{A}_{F}^{1})^{n},(dx^{1}_{\mathbb{A}_{F}})^{\otimes
n})}}$${{((\mathbb{A}_{F}^{\times})^{n},d^{*}\mathbf{x}_{\mathbb{A}_{F}})}}$${{(\mathbb{R}^{n}_{>0},(d^{*}x)^{\otimes
n})}}$${{([\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1},\mu_{1})}}$${{([\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})],\mu_{\mathbb{A}_{F}})}}$${{((\mathbb{R}^{n}/(\mathbb{R}_{>0})_{\mathbf{a}},(d^{*}x)^{\otimes
n}/d^{*}x)},}$

where $K_{1}$, $K_{2}$ are $K_{3}$ are the corresponding kernels, which are
compact by 2, endowed with the probability Haar measures, all terms that are
not drawn and corresponding measures are assumed to be the trivial groups
endowed with the probability Haar measures. All horizontal sequences and the
last vertical sequence are of the trivial measure Euler-Poincaré
characteristic. By 4, the second vertical sequence is of the trivial measure
Euler-Poincaré characteristic. By 2, it follows that the first vertical
sequence is of the trivial measure Euler-Poincaré characteristic. The
statement follows. ∎

For every space $X$, we denote by $\operatorname{count}_{X}$ the counting
measure on $X$. Proposition 2 gives that the image of
$[\mathscr{T}({\mathbf{a}})(F)]$ under the map
$[\mathscr{T}({\mathbf{a}})(i)]$ (which is induced map from the canonical
inclusion $(F^{\times})^{n}\to(\mathbb{A}_{F}^{\times})^{n}$) is contained in
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}$ and that, moreover, it is
discrete, closed and cocompact subgroup of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}$.

The kernel of the map of discrete groups

$F^{\times}\to(F^{\times})^{n}\hskip 28.45274ptt\mapsto(t^{a_{j}})_{j}$

is the finite group $\mu_{\gcd(\mathbf{a})}(F).$ We endow
$[\mathscr{T}({\mathbf{a}})(F)]=(F^{\times})^{n}/F^{\times}$ with the unique
Haar measure which makes the complex

$1\to\mu_{\gcd(\mathbf{a})}(F)\to
F^{\times}\to(F^{\times})^{n}\to[\mathscr{T}({\mathbf{a}})(F)]\to 1$

to have the trivial measure Euler-Poincaré characteristics. This measure is
precisely
$\frac{1}{|\mu_{\gcd(\mathbf{a})}(F)|}\operatorname{count}_{[\mathscr{T}({\mathbf{a}})(F)]}.$
The kernel of the homomorphism
$[\mathscr{T}({\mathbf{a}})(i)]\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F)]\to[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])$
is the finite group $|{}^{1}(F,\mu_{\gcd(\mathbf{a})})|$. We endow the
discrete group
$[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])$ with the
pushforward measure of the measure
$\frac{1}{|\mu_{\gcd(\mathbf{a})}(F)|}\operatorname{count}_{[\mathscr{T}({\mathbf{a}})(F)]}$
on $[\mathscr{T}({\mathbf{a}})(F)]$. This measure is precisely the measure

$\frac{|{}^{1}(F,\mu_{\gcd(\mathbf{a})})|}{|\mu_{\gcd(\mathbf{a})}(F)|}\operatorname{count}_{[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])}.$

######  Definition 3.

We define the Haar measure on the discrete set
$[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])$ by

$\overline{\mu_{1}}\mathrel{\mathop{\ordinarycolon}}=\mu_{1}/\bigg{(}\frac{|{}^{1}(F,\mu_{\gcd(\mathbf{a})})|}{|\mu_{\gcd(\mathbf{a})}(F)|}\operatorname{count}_{[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])}\bigg{)}.$

We define

$\operatorname{Tam}(\mathscr{T}({\mathbf{a}}))\\\
\mathrel{\mathop{\ordinarycolon}}=\operatorname{Res}(\zeta_{F},1)^{-(n-1)}\Delta(F)^{-\frac{n-1}{2}}\overline{\mu_{1}}([\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}/[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])),$

where $\Delta(F)$ is the absolute discriminant of $F$.

######  Proposition 4.

One has that

$\operatorname{Tam}(\mathscr{T}({\mathbf{a}}))=1.$

###### Proof.

Consider the following bicomplex (where $\lambda_{\mathbf{a}}$ stands for the
map $t\mapsto(t^{a_{j}})_{j},$ whatever the domain is; $K_{1}$, $K_{2}$ and
$K_{3}$ are the corresponding kernels and $E$ is the corresponding quotient;
and every term that is not drawn is assumed to be the trivial group):

${1}$${{\mu_{\gcd(\mathbf{a})}(F)}}$${{K_{2}}}$${{K_{3}}}$${1}$${{F^{\times}}}$${\mathbb{A}_{F}^{1}}$${{(\mathbb{A}_{F}^{1}/F^{\times})}}$${1}$${{(F^{\times})^{n}}}$${{(\mathbb{A}_{F}^{1})^{n}}}$${{(\mathbb{A}_{F}^{1})^{n}/(F^{\times})^{n}}}$${{{}^{1}(F,\mu_{\gcd(\mathbf{a})})}}$${{[\mathscr{T}({\mathbf{a}})(F)]}}$${{[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}}}$${E,}$$\scriptstyle{\lambda_{\mathbf{a}}}$$\scriptstyle{\lambda_{\mathbf{a}}}$$\scriptstyle{\lambda_{\mathbf{a}}}$

Endow every finite group in the bicomplex with the probability Haar measure.
The group $K_{2}$ is compact by 2, we endow it with the probability Haar
measure. The group $K_{3}$ is compact, as it is a closed subgroup of the
compact group $\mathbb{A}_{F}^{1}/F^{\times}$ and we endow it with the
probability Haar measure. Endow the discrete groups $F^{\times}$ and
$(F^{\times})^{n}$ with the counting measures. Endow $\mathbb{A}_{F}^{1},$
$(\mathbb{A}_{F}^{1})^{n}$ and
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}$ with the measure
$d^{*}x^{1}_{\mathbb{A}_{F}},$ $(d^{*}x^{1}_{\mathbb{A}_{F}})^{\otimes n}$ and
$\mu_{1}$, respectively. Finally, endow $(\mathbb{A}_{F}^{1}/F^{\times}),$
$(\mathbb{A}_{F}^{1})^{n}/(F^{\times})^{n}$ and $E$ with the unique Haar
measures so that the corresponding rows are of the trivial measure Euler-
Poincaré characteristics, that is $(\mathbb{A}_{F}^{1}/F^{\times})$ and
$(\mathbb{A}_{F}^{1})^{n}/(F^{\times})^{n}$ are endowed with the corresponding
quotient measures, while $E$ is endowed with the measure $\overline{\mu_{1}}.$
We apply 2. The measure Euler-Poincaré characteristics of every row is $1$.
The measure Euler-Poincaré characteristics of the first three columns is $1$.
Proposition 2 gives that the measure Euler-Poincaré characteristics of the
fourth column is $1$. It follows from Part (2) of 1 that

$\displaystyle\overline{\mu_{1}}(E)$
$\displaystyle=\overline{\mu_{1}}([\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}/[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)]))$
$\displaystyle=\frac{d^{*}x^{1}_{\mathbb{A}_{F}}(\mathbb{A}_{F}^{1}/F^{\times})^{n}}{d^{*}x^{1}_{\mathbb{A}_{F}}(\mathbb{A}_{F}^{1}/F^{\times})}$
$\displaystyle=d^{*}x^{1}_{\mathbb{A}_{F}}(\mathbb{A}_{F}^{1}/F^{\times})^{n-1}$
$\displaystyle=(\operatorname{Res}(\zeta_{F},1)\Delta(F)^{\frac{1}{2}})^{n-1},$

where we have used that
$d^{*}x^{1}_{\mathbb{A}_{F}}(\mathbb{A}_{F}^{1}/F^{\times})=\operatorname{Res}(\zeta_{F},1)\Delta(F)^{\frac{1}{2}}$
(see e.g. [59, Page 116]). We obtain that

$\operatorname{Tam}(\mathscr{T}({\mathbf{a}}))=1.$

∎

######  Remark 5.

When $\mathbf{a}=\mathbf{1}$, the result is a classical result that the
Tamagawa number of a split torus is $1$ ([46, Theorem 3.5.1]).

######  Remark 6.

When $n=1$, Oesterlé has calculated in [43, Proposition 2] the volume of the
fundamental domain for the action of the subgroup
$[\mathscr{T}(a)(i)([\mathscr{T}(a)(F)])]$ on
$[\mathscr{T}(a)(\mathbb{A}_{F})]$. The volume of the fundamental domain is
not $1$, because Oesterlé has used a different normalization of the Haar
measure.

##  Chapter 4 Analysis of characters of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$

In this chapter, we will study the characters of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. Later in the chapter, we recall
several facts on the estimates of $L$-functions.

###  1 Characters of $\mathbb{A}_{F}^{\times}$

We are going to define two “norms” for Hecke characters and we are going to
compare them for the characters vanishing on certain compact subgroups. Later
we establish that there are only finitely many characters vanishing on such
subgroup of bounded “norm”. The analogues for the characters of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ are stated and proven in 2.

#### 1

In this paragraph we recall several facts about characters of locally compact
topological groups.

If $G$ is a locally compact topological group, by a character of $G$ we mean a
continuous homomorphism $G\to S^{1}$. Let $G^{*}$ be the group of characters
of $G$ ([5, Definition 2, n0 1, §1, Chapter II]). The group $G^{*}$ is locally
compact by [5, Corollary 2, n0 1, §1, Chapter II]. A morphism of topological
groups $\phi\mathrel{\mathop{\ordinarycolon}}G\to G^{\prime}$ induces a
continuous homomorphism
$\phi^{*}\mathrel{\mathop{\ordinarycolon}}(G^{\prime})^{*}\to
G^{*},\phi^{*}(\chi)=\chi\circ\phi$ (see [5, Paragraph n0 7, §1, Chapter II]).
If $A$ is a subset of $G$, by $A^{\perp}$ we denote the subgroup of $G^{*}$
given by the characters vanishing on $A$.

######  Proposition 1 ([5, Theorem 4, n0 7, §1, Chapter II]).

Let $G$ be a commutative Hausdorff locally compact group. Let
$i\mathrel{\mathop{\ordinarycolon}}G_{1}\to G$ be the inclusion of a closed
subgroup $G_{1}$ into $G.$ Let $G_{2}=G/G_{1}$ and let
$p\mathrel{\mathop{\ordinarycolon}}G\to G_{2}$ be the quotient map. In the
sequence

$G_{2}^{*}\xrightarrow{p^{*}}G^{*}\xrightarrow{i^{*}}G_{1}^{*},$

the homomorphism $p^{*}$ is an isomorphism of $G_{2}^{*}$ onto $G_{1}^{\perp}$
and $i^{*}$ is a strict homomorphism $G^{*}\to G_{1}^{*}$ of kernel
$G_{1}^{\perp}$ (it follows that $G_{1}^{\perp}$ is closed).

Occasionally, we may identify $G_{2}^{*}$ with its image $G_{1}^{\perp}$ under
the homomorphism $p^{*}$. Recall also that if $G=H_{1}\times\cdots H_{n}$, the
canonical homomorphism $G^{*}\to(H^{1})^{*}\times\cdots\times(H^{n})^{*}$ is
an isomorphism of topological groups ([5, Corollary 5, n0 7, §1, Chapter II])
and we may identify $G^{*}$ with $(H^{1})^{*}\times\cdots\times(H^{n})^{*}$
using this isomorphism.

#### 2

We define two “norms” of Hecke characters and we compare them. For $v\in
M_{F}^{\infty}$, in 1 we have defined

$F_{v,1}=\\{x|x\in
F_{v}^{\times}\mathrel{\mathop{\ordinarycolon}}|x|_{v}=1\\}.$

We have furthermore established that

$\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\times
F_{v,1}\xrightarrow{\sim}F_{v}^{\times}\hskip
28.45274pt(r,z)\mapsto\rho_{v}(r)z,$

where $\rho_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\to
F_{v}^{\times}$ is defined by $\rho_{v}(r)=r^{1/n_{v}}$, is an isomorphism of
abelian topological groups. For a character $\chi_{v}\in(F_{v}^{\times})^{*}$,
we set $m(\chi_{v})$ to be the unique real number $m$ such that the character
$\chi_{v}\circ\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{R}_{>0}\to
S^{1}$ is given by $r\mapsto r^{im}$. If $v$ is a real place, we set
$\ell(\chi_{v})$ to be $0$ if the character
$\chi_{v}\circ\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}F_{v,1}\to
S^{1}$ is the trivial character, otherwise we set $\ell(\chi_{v})=1$. If $v$
is a complex place, we set $\ell(\chi_{v})$ to be the unique integer $\ell$
such that
$\chi_{v}\circ\widetilde{\rho}_{v}\mathrel{\mathop{\ordinarycolon}}F_{v,1}\to
S^{1}$ is given by $z\mapsto z^{\ell}$.

Let $\mathbb{A}_{F}^{\times}$ be the group of ideles of $F$ and let
$\mathbb{A}_{F}^{1}$ be the subgroup of $\mathbb{A}_{F}^{\times}$ given by
$(x_{v})_{v}$ which satisfy that $\prod_{v}|x_{v}|_{v}=1$. For a character
$\chi\in(\mathbb{A}_{F}^{\times})^{*}$, let us define

$\displaystyle||\chi||_{\operatorname{discrete}}$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\max_{v\in
M_{F}^{\infty}}(||\ell(\chi_{v})||),$ $\displaystyle||\chi||_{\infty}$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\max_{v\in
M_{F}^{\infty}}(||m(\chi_{v})||).$

Let $K^{0}_{\max}$ be the topological group $\prod_{v\in
M_{F}^{0}}\mathcal{O}_{v}^{\times}.$ For an open subgroup $K\subset
K^{0}_{\max}$, we let $\mathfrak{A}_{K}$ be the subgroup of
$(\mathbb{A}_{F}^{1})^{*}$ given by the characters vanishing on $F^{\times}$
(technically we mean $i(F^{\times})$) and on the compact subgroup
$K\times\prod_{v\in M_{F}^{\infty}}\\{1\\}\subset(\mathbb{A}_{F}^{1})^{*}$. By
the abuse of notation, we may write sometimes $K$ for what is technically
$K\times\prod_{v\in M_{F}^{\infty}}\\{1\\}$. The group $K^{0}_{\max}$ is
compact, therefore, the subgroup $K$ is of finite index in $K^{0}_{\max}$.

The following lemma will be used on several occasions:

######  Lemma 1.

Let $G$ be an abelian group and let $A$ and $B$ be two subgroups such that
$A\subset B$.

  1. 1.

Let $H\subset G$ be a subgroup. The homomorphism $B/A\to(B+H)/(A+H)$ induced
from $A$-invariant homomorphism $B\to(B+H)/(A+H)$ is surjective.

  2. 2.

Suppose $G$ is a topological group and that
$(B\mathrel{\mathop{\ordinarycolon}}A)$ is finite. Then
$(A^{\perp}\mathrel{\mathop{\ordinarycolon}}B^{\perp})$ is finite and
$(A^{\perp}\mathrel{\mathop{\ordinarycolon}}B^{\perp})\leq(B\mathrel{\mathop{\ordinarycolon}}A)$.

###### Proof.

  1. 1.

We have the following commutative diagram, with first two horizontal and all
three vertical sequences exact:

${0}$${A}$${{B}}$${{B/A}}$${0}$${0}$${{A+H}}$${{B+H}}$${(B+H)/(A+H)}$${0}$${{(A+H)/A}}$${{(B+H)/B}}$${E}$${0,}$

where $E=\operatorname{coker}((B/A)\to(B+H)/(A+H))$. By Snake Lemma, the third
horizontal sequence is exact. By the second isomorphism theorem, the
homomorphism $(A+H)/A\to(B+H)/B$ identifies with the homomorphism $H/(H\cap
A)\to H/(H\cap B)$ induced from the inclusion $(H\cap A)\subset(H\cap B)$,
hence is surjective and thus $E=0$. It follows that $B/A\to(B+H)/(A+H)$ is
surjective.

  2. 2.

The kernel of the homomorphism

$A^{\perp}\to\operatorname{Hom}_{\mathbb{Z}}(B,S^{1})\hskip
28.45274pt\chi\mapsto\chi|_{B}$

is the subgroup $B^{\perp}$. We deduce an injective homomorphism

$A^{\perp}/B^{\perp}\to\operatorname{Hom}_{\mathbb{Z}}(B,S^{1}).$ (2)

The image of $(\ref{yuvog})$ is contained in the subgroup of
$\operatorname{Hom}_{\mathbb{Z}}(B,S^{1})$ given by the homomorphisms which
vanish on $A$, i.e. in the image of the canonical homomorphism
$\operatorname{Hom}_{\mathbb{Z}}(B/A,S^{1})\to\operatorname{Hom}_{\mathbb{Z}}(B,S^{1})$.
It follows that

$(A^{\perp}\mathrel{\mathop{\ordinarycolon}}B^{\perp})\leq|\operatorname{Hom}_{\mathbb{Z}}(B/A,S^{1})|=(B\mathrel{\mathop{\ordinarycolon}}A).$

∎

######  Corollary 3.

One has that

$(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)\geq((F^{\times}K^{0}_{\max})\mathrel{\mathop{\ordinarycolon}}(F^{\times}K))\geq(\mathfrak{A}_{K}\mathrel{\mathop{\ordinarycolon}}\mathfrak{A}_{K^{0}_{\max}}).$

###### Proof.

To obtain the first inequality, we apply 1 for $G=\mathbb{A}_{F}^{1}$, $A=K$,
$B=K^{0}_{\max}$ and $H=F^{\times}$. It follows that the homomorphism from 1

$K^{0}_{\max}/K\to(F^{\times}K^{0}_{\max}/F^{\times}K)$

is surjective, thus
$(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)\geq((F^{\times}K^{0}_{\max})\mathrel{\mathop{\ordinarycolon}}(F^{\times}K))$.
The second inequality is the case of 1 for $G=\mathbb{A}_{F}^{1}$,
$A=F^{\times}K$ and $B=F^{\times}K^{0}_{\max}$. ∎

######  Lemma 4.

Let $K\subset K^{0}_{\max}$ be an open subgroup. The group $\mathfrak{A}_{K}$
is finitely generated and its rank is at most $r_{2}$, where $r_{2}$ is the
number of complex places of $F$.

###### Proof.

The group $\mathfrak{A}_{K^{0}_{\max}}$ is the group of Hecke characters
$(\mathbb{A}_{F}^{1}/F^{\times})\to S^{1}$ which are unramified at the finite
places of $F$ and we deduce that the kernel of the homomorphism

$\phi\mathrel{\mathop{\ordinarycolon}}\mathfrak{A}_{K^{0}_{\max}}\to\prod_{v\in
M_{F}^{\infty}}(F_{v,1})^{*}\hskip 56.9055pt\chi\mapsto\prod_{v\in
M_{F}^{\infty}}\chi_{v}|_{F_{v,1}}$

is given by the unramified Hecke characters
$(\mathbb{A}_{F}^{1}/F^{\times})\to S^{1}$, hence is finite by [56]. The group
$(\prod_{v\in M_{F}^{\infty}}(F_{v,1})^{*})$ is finitely generated and of rank
$r_{2}$, because $F_{v,1}^{*}$ is of order $2$ if $v$ is real and is an
infinite cyclic group if $v$ is complex. It follows that

$\operatorname{rk}(\mathfrak{A}_{K^{0}_{\max}})=\operatorname{rk}(\operatorname{Im}(\phi))\leq
r_{2}.$

By 3, one has that $\mathfrak{A}_{K^{0}_{\max}}$ is of finite index in
$\mathfrak{A}_{K},$ thus one has that
$\operatorname{rk}(\mathfrak{A}_{K})=\operatorname{rk}(\mathfrak{A}_{K^{0}_{\max}})\leq
r_{2}.$ The statement is proven.∎

The following lemma will be used in the proof of Proposition 6

######  Lemma 5.

Let $K$ be an open subgroup of $K^{0}_{\max}$. Consider the homomorphism

$\ell^{\mathbb{C}}\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{1}/F^{\times})^{*}\to\mathbb{Z}^{M_{F}^{\mathbb{C}}}\hskip
42.67912pt\chi\mapsto(\ell(\chi_{v}))_{v\in M_{F}^{\mathbb{C}}}$

The group $\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K}$ is finite.

###### Proof.

Firstly, let us establish that
$\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K^{0}_{\max}}$ is finite. The group
$\ker(\ell^{\mathbb{C}})$ (respectively, the group
$\mathfrak{A}_{K^{0}_{\max}}$) is the group of Hecke characters
$(\mathbb{A}_{F}^{1}/F^{\times})\to S^{1}$ which are unramified at the complex
(respectively, at the finite) places of $F$. Hence, the kernel of the map

$\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K^{0}_{\max}}\to\prod_{v\in
M_{F}^{\mathbb{R}}}(F_{v,1})^{*}\hskip 56.9055pt\chi\mapsto\prod_{v\in
M_{F}^{\mathbb{R}}}\chi_{v}|_{F_{v,1}}$

is given by the unramified Hecke characters
$(\mathbb{A}_{F}^{1}/F^{\times})\to S^{1}$, therefore is finite. As
$F_{v,1}^{*}$ is a cyclic group of order $2$, the group $\prod_{v\in
M_{F}^{\mathbb{R}}}(F_{v,1})^{*}$ is finite and we conclude that
$\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K^{0}_{\max}}$ is finite. Now let us
establish that $\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K^{0}_{\max}}$ is of
finite index in $\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K}$. By applying
Snake lemma to the “snake” diagram:

${1}$${1}$${\ker(\ell^{\mathbb{C}})\cap{\mathfrak{A}_{K_{\max}^{0}}}}$${\mathfrak{A}_{K_{\max}^{0}}}$${l^{\mathbb{C}}({\mathfrak{A}_{K_{\max}^{0}}})}$${1}$${1}$${\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K}}$${\mathfrak{A}_{K}}$${\ell^{\mathbb{C}}(\mathfrak{A}_{K})}$${1}$${E}$${\mathfrak{A}_{K}/\mathfrak{A}_{K^{0}_{\max}},}$$\scriptstyle{\ell^{\mathbb{C}}}$$\scriptstyle{\ell^{\mathbb{C}}}$

where
$E=(\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K})/(\ker(\ell^{\mathbb{C}})\cap{\mathfrak{A}_{K_{\max}^{0}}})$
we get an exact sequence

$1\to E\to\mathfrak{A}_{K}/\mathfrak{A}_{K^{0}_{\max}}.$

By 3, one has that $\mathfrak{A}_{K}/\mathfrak{A}_{K^{0}_{\max}}$ is finite,
thus
$E=(\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K})/(\ker(\ell^{\mathbb{C}})\cap{\mathfrak{A}_{K_{\max}^{0}}})$
is finite. Using the fact that
$\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K^{0}_{\max}}$ is finite, we deduce
that $\ker(\ell^{\mathbb{C}})\cap\mathfrak{A}_{K}$ is finite. ∎

The main proposition of this paragraph is the following one.

######  Proposition 6.

For every open subgroup $K\subset K^{0}_{\max}$, there exists a constant
$C=C(K)>0$ such that

$||\chi||_{\infty}\leq C||\chi||_{\operatorname{discrete}}$

for all $\chi\in\mathfrak{A}_{K}$.

###### Proof.

Let $K$ be an open subgroup of $K^{0}_{\max}$. Let
$\ell^{\mathbb{C}}\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{1}/F^{\times})^{*}\to\mathbb{Z}^{M_{F}^{\mathbb{C}}}=\mathbb{Z}^{r_{2}}$
be as in Lemma 5. To simplify notation, in the rest of the proof we will write
$\ell^{\mathbb{C}}$ for $\ell^{\mathbb{C}}|_{\mathfrak{A}_{K}}$. The abelian
group $\ell^{\mathbb{C}}(\mathfrak{A}_{K})$ is finitely generated and free,
let us pick $\chi_{1},\dots,\chi_{k}\in\mathfrak{A}_{K}$ such that
$\ell^{\mathbb{C}}(\chi_{1}),\dots,\ell^{\mathbb{C}}(\chi_{k})$ is a basis of
$\ell^{\mathbb{C}}(\mathfrak{A}^{K})$. Obviously,
$k\leq|M_{F}^{\mathbb{C}}|=r_{2}$. One has an isomorphism

$\mathbb{Z}^{k}\to\ell^{\mathbb{C}}(\mathfrak{A}_{K})\hskip
42.67912pt(d_{1},\dots,d_{k})\mapsto
d_{1}\ell^{\mathbb{C}}(\chi_{1})+\cdots+d_{k}\ell^{\mathbb{C}}(\chi_{k}).$

One can pick a section to the surjective homomorphism
$\ell^{\mathbb{C}}\mathrel{\mathop{\ordinarycolon}}\mathfrak{A}_{K}\to\ell^{\mathbb{C}}(\mathfrak{A}_{K})$,
we obtain an induced splitting

$\ell^{\mathbb{C}}(\mathfrak{A}_{K})\oplus\ker(\ell^{\mathbb{C}})\xrightarrow{\sim}\mathfrak{A}_{K}.$

We deduce an identification

$\mathbb{Z}^{k}\oplus\ker(\ell^{\mathbb{C}})\xrightarrow{\sim}\mathfrak{A}_{K}.$
(7)

A character $\chi\in\mathfrak{A}_{K}$ writes as
$\chi=p_{2}(\chi)\chi_{1}^{q_{1}(p_{1}(\chi))}\cdots\chi_{k}^{q_{k}(p_{1}(\chi))}$.
Let us firstly estimate $||\chi||_{\infty}$ for $\chi\in\mathfrak{A}_{K}$. The
group $\ker(\ell^{\mathbb{C}})$ is finite by 5 and let $r$ be its order. For
every $\chi\in\ker(\ell^{\mathbb{C}})$ one has that

$\mathbf{0}=(m(1_{v}))_{v}=(m(\chi_{v}^{r}))_{v}=(rm(\chi_{v}))_{v},$

hence $m(\chi_{v})=0$ for every $v\in M_{F}^{\infty}$. Now, we can make the
following estimate for every character $\chi\in\mathfrak{A}_{K}$:

$\displaystyle||\chi||_{\infty}$ $\displaystyle=\max_{v\in
M_{F}^{\infty}}|m(\chi_{v})|$ $\displaystyle=\max_{v\in
M_{F}^{\infty}}|m((p_{2}(\chi))_{v})+m(\chi_{1v})q_{1}(p_{1}(\chi))+\cdots
m(\chi_{kv}){q_{k}(p_{1}(\chi))}|$ $\displaystyle=\max_{v\in
M_{F}^{\infty}}|m(\chi_{1v})q_{1}(p_{1}(\chi))+\cdots
m(\chi_{kv}){q_{k}(p_{1}(\chi))}|$ $\displaystyle\leq\max_{v\in
M_{F}^{\infty},i=1,\dots,k}|m(\chi_{iv})|\max_{i=1,\dots,k}|q_{i}(p_{1}(\chi))|$
$\displaystyle=C_{0}\max_{i=1,\dots,k}|q_{i}(p_{1}(\chi))|,$

where we have shortened $C_{0}=\max_{v\in
M_{F}^{\infty},i=1,\dots,k}|m(\chi_{iv})|.$ Let us now estimate
$||\chi||_{\infty}$ for $\chi\in\mathfrak{A}_{K}$. For every
$\chi\in\mathfrak{A}_{K}$ we have that:

$\displaystyle||\chi||_{\operatorname{discrete}}$ $\displaystyle=\max_{v\in
M_{F}^{\infty}}|\ell(\chi_{v})|$ $\displaystyle\geq\max_{v\in
M_{F}^{\mathbb{C}}}|\ell(\chi_{v})|$ $\displaystyle=\max_{v\in
M_{F}^{\mathbb{C}}}|\ell\big{(}(p_{2}(\chi))_{v}\chi_{1v}^{q_{1}(p_{1}(\chi))}\cdots\chi_{kv}^{q_{k}(p_{1}(\chi))}\big{)}|$
$\displaystyle=\max_{v\in
M_{F}^{\mathbb{C}}}|\ell((p_{2}(\chi))_{v})+q_{1}(p_{1}(\chi))\ell(\chi_{1v})+\cdots+q_{k}(p_{1}(\chi))\ell(\chi_{kv})|$
$\displaystyle=\max_{v\in
M_{F}^{\mathbb{C}}}|q_{1}(p_{1}(\chi))\ell(\chi_{1v})+\cdots+q_{k}(p_{1}(\chi))\ell(\chi_{kv})|.$

Tensoring the injective map $\ell^{\mathbb{C}}|_{\mathbb{Z}^{k}}$ with the
flat $\mathbb{Z}$-module $\mathbb{R}$ gives an injective map
$\mathbb{R}^{k}\to\mathbb{R}^{r_{2}}$. The pullback of the norm $(x_{v})_{v\in
M_{F}^{\mathbb{C}}}\mapsto\max_{v}|x_{v}|$ on $\mathbb{R}^{r_{2}}$ along this
map is given by

$\mathbf{x}\mapsto\max_{v\in M_{F}^{\mathbb{C}}}|x_{1}\ell(\chi_{1v})+\cdots
x_{k}\ell(\chi_{k,v})|,$

and is a norm on $\mathbb{R}^{k}$. One can find, hence, a constant $C_{1}>0$
such that

$C_{1}\max_{v\in
M_{F}^{\mathbb{C}}}|x_{1}\ell(\chi_{1v})+\cdots+x_{k}\ell(\chi_{k,v})|\geq\max_{i=1,\dots,k}|x_{k}|$

for every $\mathbf{x}\in\mathbb{R}^{k}$. Thus for every
$\chi\in\mathfrak{A}_{K}$ one has that

$C_{1}||\chi||_{\operatorname{discrete}}\geq\max_{i=1,\dots,k}|q_{k}(p_{1}(\chi))|.$

Let us now prove the wanted inequality. We set $C=C_{1}C_{0}$. For every
$\chi\in\mathfrak{A}_{K}$, we have that

$\displaystyle C||\chi||_{\operatorname{discrete}}\geq
C_{0}C_{1}||\chi||_{\operatorname{discrete}}\geq
C_{0}\max_{i=1,\dots,k}|q_{i}(p_{1}(\chi))|\geq||\chi||_{\infty}.$

The statement is proven ∎

#### 3

In this paragraph, we bound the number of characters $\chi$ for which
$||\chi||_{\operatorname{discrete}}$ is bounded.

######  Lemma 1.

Let $h_{F}=|\operatorname{Cl}(F)|$ be the class number of $F$. For every open
subgroup $K\subset K^{0}_{\max}$ and any $C>0$, there are no more than

$h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)2^{r_{1}}(2C+1)^{r_{2}}$

characters $\chi\in\mathfrak{A}_{K}$ for which
$||\chi||_{\operatorname{discrete}}\leq C$ (here $r_{1}$ and $r_{2}$ are
numbers of real and complex places of $F$, respectively).

###### Proof.

Let $K\subset K^{0}_{\max}$ be an open subgroup. For $v\in
M_{F}^{\mathbb{C}}$, we define
$\widetilde{\ell}(\chi_{v})\mathrel{\mathop{\ordinarycolon}}=\ell(\chi_{v})$,
and for $v\in M_{F}^{\mathbb{R}}$ we define
$\widetilde{\ell}(\chi_{v})\in\mathbb{Z}/2\mathbb{Z}$ by

$\displaystyle\quad\widetilde{\ell}(\chi_{v})\mathrel{\mathop{\ordinarycolon}}=$
$\displaystyle\begin{cases}0&\text{if $\chi_{v}|_{F_{v,1}}$ is the trivial
character on $F_{v,1}$}\\\ 1&\text{otherwise.}\end{cases}$ (2)

(The difference between $\ell$ and $\widetilde{\ell}$ is of technical nature,
recall that we have defined $\ell(\chi_{v})\in\mathbb{Z}$ for $v\in
M_{F}^{\mathbb{R}}$.) For a character $\chi\in\mathfrak{A}_{K}$, the
$r_{1}+r_{2}$-tuple given by $(\ell(\chi_{v}))_{v\in
M_{F}^{\infty}}\in(\mathbb{Z}/2\mathbb{Z})^{r_{1}}\times\mathbb{Z}^{r_{2}}$
will be called the signature of $\chi$.

Let us estimate the number of characters with fixed signature. Let
$(\ell_{v})_{v\in
M_{F}^{\infty}}\in(\mathbb{Z}/2\mathbb{Z})^{r_{1}}\times\mathbb{Z}^{r_{2}}$ be
a signature of some character $\delta\in\mathfrak{A}_{K}$. Then the characters
in $\mathfrak{A}_{K}$ having $(\ell_{v})_{v\in M_{F}^{\infty}}$ for the
signature are in a bijection with the characters in $\mathfrak{A}_{K}$ having
$(0)_{v\in M_{F}^{\infty}}$ for the signature. Indeed, a bijection between the
two sets is given by $\chi\mapsto\chi\delta^{-1}$. The group of
$\chi\in\mathfrak{A}_{K}$ having $(0)_{v\in M_{F}^{\infty}}$ for signature is
given by the subgroup $(F^{\times}({K}\times\prod_{v\in
M_{F}^{\infty}}F_{v,1}))^{\perp}$ of $(\mathbb{A}_{F}^{1})^{*}.$ Lemma 1 gives
that

$\displaystyle(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)$
$\displaystyle\geq\bigg{(}\big{(}F^{\times}(K_{\max}^{0}\times\prod_{v\in
M_{F}^{\infty}}F_{v,1})\big{)}\mathrel{\mathop{\ordinarycolon}}\big{(}F^{\times}(K\times\prod_{v\in
M_{F}^{\infty}}F_{v,1})\big{)}\bigg{)}$
$\displaystyle\geq\bigg{(}(F^{\times}({K}\times\prod_{v\in
M_{F}^{\infty}}F_{v,1}))^{\perp}\mathrel{\mathop{\ordinarycolon}}(F^{\times}({K^{0}_{\max}}\times\prod_{v\in
M_{F}^{\infty}}F_{v,1}))^{\perp}\bigg{)}.$

The group $(F^{\times}({K^{0}_{\max}}\times\prod_{v\in
M_{F}^{\infty}}F_{v,1}))^{\perp}$ is the group of the unramified Hecke
characters $(\mathbb{A}_{F}^{1}/F^{\times})\to S^{1}$ and its order is
$h_{F}$. It follows that the number of characters $\chi\in\mathfrak{A}_{K}$
having $(0)_{v\in M_{F}^{\infty}}$ for the signature is

$|(F^{\times}({K}\times\prod_{v\in M_{F}^{\infty}}F_{v,1}))^{\perp}|\leq
h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K).$

Let us estimate the number of signatures when
$|\lvert\cdot\rvert|_{\operatorname{discrete}}$ is bounded. Note that if for
$C>0$ and a character $\chi\in\mathfrak{A}_{K}$ one has
$||\chi||_{\operatorname{discrete}}<C$, then the signature of $\chi$ lies in
$(\mathbb{Z}/2\mathbb{Z})^{r_{1}}\times([-C,C]^{r_{2}}\cap\mathbb{Z}^{r_{2}})$.
Thus the number of signatures can be bounded by $2^{r_{1}}(2C+1)^{r_{2}}$.

It follows that the number of characters $\chi\in\mathfrak{A}_{K}$ such that
$||\chi||_{\operatorname{discrete}}\leq\leavevmode\nobreak\ C$ is bounded by
$h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)2^{r_{1}}(2C+1)^{r_{2}}$.
The statement is proven. ∎

###  2 Results for the characters of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$

Let $n\geq 1$ be an integer and let $\mathbf{a}\in\mathbb{Z}^{n}_{\geq 1}$. We
make analogous estimates to those in 1 for characters of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. The results are simple
consequences of the corresponding results in 1

#### 1

In this paragraph we explain our notation and define “norms” for the
characters of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ and compare them
for characters vanishing on certain compact subgroups of
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. In this paragraph, we discuss
local characters.

Recall that the subgroups
$(F_{v}^{\times})_{\mathbf{a}}\subset(F_{v}^{\times})^{n}$ and
$(\mathbb{A}_{F}^{\times})_{\mathbf{a}}\subset(\mathbb{A}_{F}^{\times})^{n}$
are closed by 1 and 2, respectively. Recall that
$[\mathscr{T}({\mathbf{a}})(F_{v})]$ identifies with the quotient group
$(F_{v}^{\times})^{n}/(F_{v}^{\times})_{\mathbf{a}}$ by 1. Recall that
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ identifies with the quotient
group $(\mathbb{A}_{F}^{\times})^{n}/(\mathbb{A}_{F}^{\times})_{\mathbf{a}}$
by 2.

######  Lemma 1.

The subgroup $(F_{v}^{\times})_{\mathbf{a}}^{\perp}$ is the subgroup

$\\{(\chi_{j})_{j}|\prod_{j}\chi_{j}^{a_{j}}=1\\}$

of $((F_{v}^{\times})^{*})^{n}$. Moreover,
$(F_{v}^{\times})_{\mathbf{a}}^{\perp}$ is closed in
$((F_{v}^{\times})^{*})^{n},$ and it is the image of the pullback homomorphism
$[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}\to((F_{v}^{\times})^{n})^{*}$. The
subgroup $(\mathbb{A}_{F}^{\times})_{\mathbf{a}}^{\perp}$ is the subgroup
$\\{(\chi_{j})_{j}|\prod_{j}\chi_{j}^{a_{j}}=1\\}$ of
$((\mathbb{A}_{F}^{\times})^{*})^{n}$. Moreover,
$(\mathbb{A}_{F}^{\times})_{\mathbf{a}}^{\perp}$ is closed in
$((\mathbb{A}_{F}^{\times})^{*})^{n},$ and it is the image of the pullback
homomorphism
$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]^{*}\to((\mathbb{A}_{F}^{\times})^{*})^{n}$.

###### Proof.

Note that if $\chi=(\chi_{1},\dots,\chi_{n})$ is a character of
$((F_{v}^{\times})^{*})^{n}$ which vanishes on
$(F_{v}^{\times})_{\mathbf{a}}=\\{(t^{a_{j}})_{j}|t\in F_{v}^{\times}\\},$
then for every $t\in F_{v}^{\times}$, one has that

$1=\chi_{1}(t^{a_{1}})\cdots\chi_{n}(t^{a_{n}})=\chi_{1}(t)^{a_{1}}\cdots\chi_{n}(t)^{a_{n}}.$

The first claim follows. The facts that
$(F_{v}^{\times})_{\mathbf{a}}^{\perp}$ is closed and that it is the image of
the pullback homomorphism follow from 1. Analogous argument shows the claims
about $(\mathbb{A}_{F}^{\times})_{\mathbf{a}}^{\perp}$. ∎

Let $v$ be a place of $F.$ For $j\in\\{1,\dots,n\\},$ we let
$\delta^{j}_{v}\mathrel{\mathop{\ordinarycolon}}F_{v}^{\times}\to(F_{v}^{\times})^{n}$
be the inclusion

$x\mapsto((x_{j})_{j},(1)_{i\neq j}).$

If $\chi\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$ is a character we let
$\chi^{(j)}$ be the character $\chi\circ
q^{\mathbf{a}}_{v}\circ\delta^{j}\mathrel{\mathop{\ordinarycolon}}F_{v}^{\times}\to
S^{1}$. The lemma from above gives that that
$\prod_{j=1}^{n}(\chi^{(j)})^{a_{j}}=1$ for every
$\chi\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$. We define
$\mathbf{m}(\chi)\mathrel{\mathop{\ordinarycolon}}=(m(\chi^{(j)}))_{j}$ and
$\bm{\ell}(\chi)=(\ell(\chi^{(j)}))_{j}.$

If $\chi$ is a character of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$, for
$j\in\\{1,\dots,n\\},$ we will denote by $\delta^{j}_{\mathbb{A}_{F}}$ the
inclusion $x\mapsto((x)_{j},(1)_{i\neq j})$ and by $\chi^{(j)}$ the character
$\chi\circ
q^{\mathbf{a}}_{\mathbb{A}_{F}}\circ\delta^{j}_{\mathbb{A}_{F}}\mathrel{\mathop{\ordinarycolon}}\mathbb{A}_{F}^{\times}\to
S^{1}$. The lemma from above gives that
$\prod_{j=1}^{n}(\chi^{(j)})^{a_{j}}=1$ for every
$\chi\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]^{*}$. Moreover, it is
immediate that

$\chi|_{[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])}=1\implies\text{
for $j=1,\dots,n$ one has }\chi^{(j)}|_{i(F^{\times})}=1.$

For a character $\chi\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]^{*}$, we
denote by $\chi_{v}$ the restriction
$\chi|_{[\mathscr{T}({\mathbf{a}})(F_{v})]}$.

######  Remark 2.

Let $\chi$ be a character of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. A
priori the notation $\chi^{(j)}_{v}$ can make confusion as it may design
$(\chi^{(j)})|_{F_{v}^{\times}}$ and
$(\chi|_{[\mathscr{T}({\mathbf{a}})(F_{v})]})^{(j)}$. The commutativity of the
following diagram shows that no confusion arises:

${F_{v}^{\times}}$${\mathbb{A}_{F}^{\times}}$${{(F_{v}^{\times})^{n}}}$${{(\mathbb{A}_{F}^{\times})^{n}}}$${{[\mathscr{T}({\mathbf{a}})(F_{v})]}}$${{[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})].}}$$\scriptstyle{q^{\mathbf{a}}_{v}}$$\scriptstyle{q^{\mathbf{a}}_{\mathbb{A}_{F}}}$$\scriptstyle{\delta^{j}_{\mathbb{A}_{F}}}$$\scriptstyle{\delta^{j}_{v}}$

Namely, for a character $\chi$, the character $(\chi^{(j)})_{v}$ is the
pullback of $\chi$ for the composite by the two vertical homomorphisms on the
right and the most upper horizontal homomorphism, while $(\chi_{v})^{(j)}$ is
the pullback for the lowest horizontal and then two left vertical
homomorphisms.

For a character $\chi\in[\mathscr{T}^{\mathbf{a}}(\mathbb{A}_{F})]^{*}$, let
us define

$\displaystyle||\chi||_{\operatorname{discrete}}$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\max_{v\in
M_{F}^{\infty}}(||\bm{\ell}(\chi_{v})||),$ $\displaystyle||\chi||_{\infty}$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\max_{v\in
M_{F}^{\infty}}(||\mathbf{m}(\chi_{v})||).$

Note that

$\displaystyle||\chi||_{\operatorname{discrete}}=\max_{v\in
M_{F}^{\infty}}(||\bm{\ell}(\chi_{v})||)$ $\displaystyle=\max_{v\in
M_{F}^{\infty}}\left(\max_{j=1,\dots,n}(|\ell(\chi_{v}^{(j)})|)\right)$
$\displaystyle=\max_{j=1,\dots,n}\bigg{(}\max_{v\in
M_{F}^{\infty}}|\ell(\chi_{v}^{(j)})|\bigg{)}$
$\displaystyle=\max_{j=1,\dots,n}||\chi^{(j)}||_{\operatorname{discrete}};$

and analogously

$||\chi||_{\infty}=\max_{j=1,\dots,n}||\chi^{(j)}||_{\infty}.$

######  Lemma 3.

Suppose that $n=1$ and $a_{1}=a\in\mathbb{Z}_{\geq 1}$. For every
$\chi\in[\mathscr{T}(a)(\mathbb{A}_{F})]^{*}$, one has that
$||\chi||_{\operatorname{discrete}}\leq 1$ and that $||\chi||_{\infty}=0$.

###### Proof.

Let $\chi\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]^{*}$. To simplify
notation, in this proof we write $\widetilde{\chi}$ for $\chi^{(1)}$. For
every $\chi\in[\mathscr{T}(a)(\mathbb{A}_{F})]^{*}$, one has that
$\chi^{a}=1$. It follows that

$(0)_{v\in M_{F}^{\infty}}=(m(1))_{v\in
M_{F}^{\infty}}=(m((\widetilde{\chi}_{v})^{a}))_{v\in
M_{F}^{\infty}}=(am((\widetilde{\chi}_{v})))_{v\in M_{F}^{\infty}},$

hence $(m(\widetilde{\chi}_{v}^{a}))_{v\in M_{F}^{\infty}}=(0)_{v\in
M_{F}^{\infty}}$, and thus $||\widetilde{\chi}||_{\infty}=0$. We obtain that

$||\chi||_{\infty}=||\widetilde{\chi}||_{\infty}=0.$

We have that

$||\chi||_{\operatorname{discrete}}=||\widetilde{\chi}||_{\operatorname{discrete}}\leq\max(1,\max_{v\in
M_{F}^{\mathbb{C}}}(\ell(\chi_{v}))).$

For every $v\in M_{F}^{\mathbb{C}}$, one has that
$0=\ell(1)=\ell(\widetilde{\chi}^{a})=a\ell(\widetilde{\chi}_{v}),$ hence
$\ell(\widetilde{\chi}_{v})=0$. We deduce that
$||\chi||_{\operatorname{discrete}}\leq 1.$ ∎

#### 2

In this paragraph we present analogous results to those of 1 for the
characters of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$.

Let $K^{\mathbf{a}}_{\max}$ be the topological group given by

$K^{\mathbf{a}}_{\max}\mathrel{\mathop{\ordinarycolon}}=\prod_{v\in
M_{F}^{0}}[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})].$

For an open subgroup $K$ of $K_{\max}^{\mathbf{a}}$, we let $\mathfrak{A}_{K}$
be the subgroup of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})_{1}]^{*}$ given
by the characters vanishing on
$[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])$ and on
$K\times\prod_{v\in M_{F}^{\infty}}\\{1\\}$ (we may simply write $K$ for what
is technically $K\times\prod_{v\in M_{F}^{\infty}}\\{1\\}$). The group
$K^{\mathbf{a}}_{\max}$ is compact, therefore, the subgroup $K$ is of finite
index in $K^{\mathbf{a}}_{\max}$. We present another corollary of 1.

######  Corollary 1.

Let $K\subset K^{\mathbf{a}}_{\max}$ be an open subgroup. One has that

$\displaystyle(K^{\mathbf{a}}_{\max}\mathrel{\mathop{\ordinarycolon}}K)$
$\displaystyle\geq([\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])K^{\mathbf{a}}_{\max}\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])K)$
$\displaystyle\geq(\mathfrak{A}_{K}\mathrel{\mathop{\ordinarycolon}}\mathfrak{A}_{K^{0}_{\max}}).$

###### Proof.

The left inequality follows from 1 for
$G=[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1},$ $A=K,$
$B=K^{\mathbf{a}}_{\max}$ and
$H=[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])$. The right
inequality follows from 1 for
$G=[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1},$
$A=[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])K$ and
$B=[\mathscr{T}({\mathbf{a}})(i)]([\mathscr{T}({\mathbf{a}})(F)])K^{\mathbf{a}}_{\max}$.
∎

Note that

$q^{\mathbf{a}}_{\mathbb{A}_{F}}\big{(}(K^{0}_{\max})^{n}\big{)}=(q^{\mathbf{a}}_{v}((\mathcal{O}_{v}^{\times})^{n}))_{v}=([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])_{v}=K^{\mathbf{a}}_{\max}.$

If $K\subset K^{\mathbf{a}}_{\max}$ is an open subgroup, let us set

$\widetilde{K}=\bigcap_{j=1}^{n}(\delta^{j}_{\mathbb{A}_{F}})^{-1}(q^{\mathbf{a}}_{\mathbb{A}_{F}}|_{(K^{0}_{\max})^{n}})^{-1}(K),$

it is an open subgroup of $K^{0}_{\max}$. For a character
$\chi\in\mathfrak{A}_{K}$, one has that

$\chi^{(j)}|_{\widetilde{K}}=(\chi\circ
q^{\mathbf{a}}_{\mathbb{A}_{F}}\circ\delta^{j}_{\mathbb{A}_{F}})|_{\widetilde{K}}=1.$

The following statements are simple corollaries of corresponding statements
for characters of $\mathbb{A}_{F}^{1}$.

######  Corollary 2.

The group $\mathfrak{A}_{K}$ is finitely generated and of rank at most
$nr_{2}$.

###### Proof.

Recall that by 4, the abelian group $\mathfrak{A}_{\widetilde{K}}$ is finitely
generated and of rank at most $r_{2}$. The image of $\mathfrak{A}_{K}$ under
the injective homomorphism

$[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]_{1}^{*}\to((\mathbb{A}_{F}^{1})^{n})^{*}=((\mathbb{A}_{F}^{1})^{*})^{n}\hskip
42.67912pt\chi\mapsto\chi\circ
q^{\mathbf{a}}_{\mathbb{A}_{F}}=(\chi^{(j)})_{j}$

lies in $(\mathfrak{A}_{\widetilde{K}})^{n}.$ It follows that
$\mathfrak{A}_{K}$ is finitely generated and of rank at most $nr_{2}$. ∎

######  Corollary 3.

For every compact group $K\subset K^{\mathbf{a}}_{\max}$, there exists a
constant $C=C(K)>0$ such that for all $\chi\in\mathfrak{A}_{K}$ one has that

$||\chi||_{\infty}\leq C||\chi||_{\operatorname{discrete}}.$

###### Proof.

Proposition 6 gives that there exists constant $C=C(\widetilde{K})>0$ such
that for every $\chi\in\mathfrak{A}_{K}$ one has that

$||\chi||_{\infty}\leq C||\chi||_{\operatorname{discrete}}.$

For $\chi\in\mathfrak{A}_{K}$ we deduce that:

$||\chi||_{\infty}=\max_{j=1,\dots,n}||\chi^{(j)}||_{\infty}\leq
C\max_{j=1,\dots,n}||\chi^{(j)}||_{0}=C||\chi||_{\operatorname{discrete}}.$

It follows that $C(K)=C(\widetilde{K})$ is the wanted constant. ∎

######  Corollary 4.

For every open subgroup $K\subset K^{\mathbf{a}}_{\max}$ and every $C>0$,
there are no more than

$(h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}\widetilde{K})2^{r_{1}}(2C+1)^{r_{2}})^{n}$

characters $\chi\in\mathfrak{A}_{K}$ for which
$||\chi||_{\operatorname{discrete}}\leq C$.

###### Proof.

Let $K\subset K^{\mathbf{a}}_{\max}$ be an open subgroup. Lemma 1 gives that
for every $C>0$, there are no more than

$h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}\widetilde{K})2^{r_{1}}(2C+1)^{r_{2}}$

characters in $\mathfrak{A}_{\widetilde{K}}$ having
$||\chi||_{\operatorname{discrete}}\leq C$. A character
$\chi\in\mathfrak{A}_{K}$ is completely determined by the characters
$\chi^{(j)}\in\mathfrak{A}_{\widetilde{K}}$ for $j=1,\dots,n$. It follows that
there are no more than

$(h_{F}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}\widetilde{K})2^{r_{1}}(2C+1)^{r_{2}})^{n}$

characters $\chi\in\mathfrak{A}_{K}$ having
$||\chi||_{\operatorname{discrete}}=\max_{j}(||\chi^{(j)}||_{\operatorname{discrete}})\leq
C$. ∎

By 3, we have that when $n=1$ and $a=a_{1}\in\mathbb{Z}_{\geq 1}$, we have
that $||\chi||_{\operatorname{discrete}}\leq 1.$ We deduce that

######  Corollary 5.

Suppose that $n=1$ and that $a=a_{1}\in\mathbb{Z}_{\geq 1}$. For every open
subgroup $K\subset K^{a}_{\max}$, the group $\mathfrak{A}_{K}$ is finite.

###  3 Estimates of Rademacher

In this section we recall some bounds on the growth of $L$ functions of Hecke
characters in vertical strips. If $v\in M_{F}^{0}$, for $s\in\mathbb{C}$ and
$\chi_{v}\in(F_{v}^{\times})^{*}$, one defines

$L_{v}(s,\chi_{v})\mathrel{\mathop{\ordinarycolon}}=\frac{1}{1-|\pi_{v}|_{v}^{s}\chi_{v}(\pi_{v})},$

and one writes $\zeta_{v}(s)$ for $L_{v}(s,1)$. For a character
$\chi=(\chi_{v})_{v}\mathrel{\mathop{\ordinarycolon}}\mathbb{A}_{F}^{\times}\to
S^{1}$, we set

$L(s,\chi)\mathrel{\mathop{\ordinarycolon}}=\prod_{v\in
M_{F}^{0}}L_{v}(s,\chi_{v}),$

and we denote

$\zeta_{F}(s)=L(s,1).$

#### 1

In this paragraph we restrict ourselves to the characters of
$\mathbb{A}_{F}^{1}$. The corresponding estimates for the characters of
$\mathbb{A}_{F}^{\times}$ will be established in 2.

Rademacher in [48, Theorem 5] establishes that the growth of the $L$-function
of a character in vertical strip is moderate (i.e. bounded by a polynomial).
As the notation there is cumbersome, let us quote the variant given as [39,
Theorem 14.A, Chapter III] as part (1) of the next theorem. Part (2) is [48,
Theorem 3].

######  Theorem 1.

Let $0<\eta\leq\frac{1}{2}$ and let
$\chi=\prod_{v}\chi_{v}\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{1}/F^{\times})\to\leavevmode\nobreak\
S^{1}$ be a Hecke character. Let $\operatorname{cond}(\chi)$ be the conductor
ideal of $\chi$. We set
$d_{F}=N_{F/\mathbb{Q}}(\operatorname{disc}_{F/\mathbb{Q}}).$

  1. 1.

(Moreno, [39, Theorem 14.A, Chapter III]) Suppose $\chi$ is not the trivial
character. One has that

$|L(s,\chi)|\leq\zeta_{F}(1+\eta)\bigg{(}\frac{d_{F}N_{F/\mathbb{Q}}(\operatorname{cond}({\chi}))}{(2\pi)^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}\prod_{v\in
M_{F}^{\infty}}|{1+s+m(\chi_{v})}|^{n_{v}}\bigg{)}^{\frac{1+\eta-\Re(s)}{2}}$

in the strip $-\eta\leq\Re(s)\leq 1+\eta.$

  2. 2.

(Rademacher, [48, Theorem 3]) One has that

$\bigg{|}\frac{\zeta_{F}(s)(1-s)}{1+s}\bigg{|}\leq
3\zeta_{F}(1+\eta)^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\bigg{(}d_{F}\bigg{(}\frac{|1+s|}{2\pi}\bigg{)}^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\bigg{)}^{\frac{1+\eta-\Re(s)}{2}}$

in the strip $-\eta\leq\Re(s)\leq 1+\eta.$

The following proposition is a corollary of 1. A similar version, but only for
unramified Hecke characters, has been invoked in the analysis of height zeta
functions in [2, Theorem 3.2.3], [17, Corollary 4.2.3], etc. As our metrics at
infinite places are not invariant for the maximal compact subgroups, we
present the following version.

######  Proposition 2.

Let $K\subset K_{\max}^{0}=\prod_{v\in M_{F}^{0}}\mathcal{O}_{v}^{\times}$ be
an open subgroup. For every $\epsilon>0$, there exist $C=C(\epsilon)>0$ and
$\delta=\delta(\epsilon)>0$ such that the following conditions are satisfied
if provided that $\Re(s)\geq 1-\delta\mathrel{\mathop{\ordinarycolon}}$

  1. 1.

for every non trivial Hecke character
$\chi\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{1}/F^{\times})\to
S^{1}$ which vanishes on $K\subset K_{\max}^{0}$ one has

$|L(s,\chi)|\leq C\big{(}(1+|\Im(s)|)(1+||\chi||_{\infty})\big{)}^{\epsilon};$
(3)

  2. 2.

one has

$\bigg{|}\frac{(s-1)\zeta_{F}(s)}{s}\bigg{|}\leq C(1+|\Im(s)|)^{\epsilon}.$
(4)

###### Proof.

Let $\frac{1}{6}>\epsilon>0$. We set

$C=27\zeta_{F}\big{(}1+\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\big{)}^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}d_{F}^{1/[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)^{1/[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}.$

We are going to verify that $C$ and
$\delta=\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}$
verify the above conditions.

Let us prove (3) and (4) in the domain $\Re(s)>\frac{4}{3}.$ For a non-trivial
character $\chi$ one can estimate

$|L(s,\chi)|\leq\zeta_{F}(\Re(s))\leq\zeta_{F}\big{(}\frac{4}{3}\big{)}\leq\zeta_{F}\big{(}1+\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\big{)}\leq
C.$

One also has

$\bigg{|}\frac{(s-1)L(s,1)}{s}\bigg{|}\leq\bigg{|}1-\frac{1}{s}\bigg{|}\zeta_{F}\bigg{(}\frac{4}{3}\bigg{)}\leq\frac{7}{4}\zeta_{F}\bigg{(}\frac{4}{3}\bigg{)}\leq\frac{7}{4}\zeta_{F}(1+\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})\leq
C.$

It follows that (2) and (3) are satisfied in the domain $\Re(s)>\frac{4}{3}.$

Now we prove (3) and (4) in the domain
$1-\frac{1}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\epsilon<\Re(s)<\frac{4}{3}$.
Let us set
$\eta(s)=\frac{1}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\epsilon+\Re(s)-1$,
we have that

$0<\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}<\eta(s)<\frac{1}{6[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}+\frac{4}{3}-1\leq\frac{1}{2}$

and that $-\eta(s)\leq 0\leq\Re(s)<1+\eta(s)$. We will apply 1 for $s$ and
$\eta=\eta(s)$. The following estimate will be used: for every $s$ in the
domain
${1}-\frac{1}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}<\Re(s)<\frac{4}{3}$
one has

$1+|s|\leq 1+|\Re(s)|+|\Im(s)|<3(1+|\Im(s)|).$ (5)

Let us firstly prove the estimate for the non-trivial characters. Using the
first part of Theorem 1, we deduce that for every $\chi\neq 1$ in the domain
$1-\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}<\Re(s)<\frac{4}{3}$
one has

$\displaystyle|L(s,\chi)|\hskip-56.9055pt$
$\displaystyle\leq|\zeta_{F}(1+\eta(s))|^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\bigg{(}\frac{d_{F}N_{F/\mathbb{Q}}(\operatorname{cond}(\chi))}{(2\pi)^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}\prod_{v\in
M_{F}^{\infty}}|{1+s+m(\chi_{v})}|^{n_{v}}\bigg{)}^{\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}$
$\displaystyle\leq\zeta_{F}(1+\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\times$
$\displaystyle\quad\quad\quad\quad\quad\times\bigg{(}\frac{d_{F}N_{F/\mathbb{Q}}(\operatorname{cond}(\chi))}{(2\pi)^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}\prod_{v\in
M_{F}^{\infty}}(1+|s|)(1+|m(\chi_{v})|)\bigg{)}^{\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}$
$\displaystyle\leq\zeta_{F}(1+\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\times$
$\displaystyle\quad\quad\quad\quad\quad\times\bigg{(}\frac{d_{F}N_{F/\mathbb{Q}}(\operatorname{cond}(\chi))}{(2\pi)^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}\prod_{v\in
M_{F}^{\infty}}3({1+|\Im(s)|)(1+|m(\chi_{v})|})\bigg{)}^{\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}$
$\displaystyle\leq\zeta_{F}(1+\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})^{{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}\times$
$\displaystyle\quad\quad\quad\quad\quad\times\frac{((d_{F}N_{F/\mathbb{Q}}(\operatorname{cond}(\chi)))^{\frac{1}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}3(({1+|\Im(s)|)(1+||\chi||_{\infty}})))^{\epsilon}}{(2\pi)^{\epsilon}}.$

Moreover, as $\chi_{v}$ vanishes at $K$, we have

$N_{F/\mathbb{Q}}(\operatorname{cond}(\chi))\leq\prod_{v\in
M^{0}_{F}}(\mathcal{O}_{v}^{\times}\mathrel{\mathop{\ordinarycolon}}K_{v})=(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K),$

where $K_{v}$ is the image in $\mathcal{O}_{v}^{\times}$ of the $v$-adic
projection of $K$. The inequality (3) now follows from the observation that

$\zeta_{F}(1+\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\bigg{(}\frac{(d_{F}(K\mathrel{\mathop{\ordinarycolon}}K^{0}_{\max}))^{\frac{1}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}}3}{2\pi}\bigg{)}^{\epsilon}\\\
\leq
27\zeta_{F}\big{(}1+\frac{\epsilon}{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\big{)}^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}d_{F}^{1/[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}(K^{0}_{\max}\mathrel{\mathop{\ordinarycolon}}K)^{1/[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}=C.$

Let us now consider the trivial character. When
$\Re(s)>1-\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}>\frac{1}{2}$,
one has that

$\bigg{|}\frac{3s}{s-1}\bigg{|}\geq\frac{|s|+1}{|s-1|}\geq\bigg{|}\frac{s+1}{s-1}\bigg{|}.$
(6)

Using the second part of 1 and (6), we deduce that

$\displaystyle\bigg{|}\frac{(s-1)\zeta_{F}(s)}{s}\bigg{|}$ $\displaystyle\leq
9\zeta_{F}(1+\frac{\epsilon}{2[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]})^{[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}\frac{d_{F}^{\epsilon/[F\mathrel{\mathop{\ordinarycolon}}\mathbb{Q}]}3^{\epsilon}(1+||\Im(s)||)^{\epsilon}}{(2\pi)^{\epsilon}}$
$\displaystyle\leq C(1+|\Im(s)|)^{\epsilon}.$

The proposition is proven. ∎

#### 2

We will now present a bound on the growth in the vertical strips of the
$L$-function of a general Hecke character
$\chi\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{\times}/F^{\times})\to
S^{1}$. In 1, we have established an identification:

$\mathbb{A}_{F}^{1}\times\mathbb{R}_{>0}\xrightarrow{\sim}\mathbb{A}_{F}^{\times}.$

For a character $\chi\in(\mathbb{R}_{>0})^{*}$ we denote by $m(\chi)$ the
unique real number $m$ such that $\chi(x)=x^{im}$ for every
$x\in\mathbb{R}_{>0}$. For a character $\chi\in(\mathbb{A}_{F}^{\times})^{*}$
we denote by $\chi_{0}$ the restriction $\chi|_{\mathbb{A}_{F}^{1}}$ and we
write $m(\chi)$ for $m(\chi|_{\mathbb{R}_{>0}})$ so that
$\chi=\chi_{0}\lvert\cdot\rvert^{im(\chi)}.$

For a character
$\chi\mathrel{\mathop{\ordinarycolon}}\mathbb{A}_{F}^{\times}\to S^{1}$, one
has that

$\displaystyle L(s,\chi)=L(s,\chi_{0}\lvert\cdot\rvert^{im(\chi)})$
$\displaystyle=\prod_{v\in
M_{F}^{0}}L_{v}(s,(\chi_{0})_{v}\lvert\cdot\rvert_{v}^{im(\chi)})$
$\displaystyle=\prod_{v\in
M_{F}^{0}}\frac{1}{1-|\pi_{v}|_{v}^{s}(\chi_{0})_{v}(\pi_{v})|\pi_{v}|_{v}^{im(\chi)}}$
$\displaystyle=\prod_{v\in
M_{F}^{0}}\frac{1}{1-|\pi_{v}|_{v}^{s+im(\chi)}(\chi_{0})_{v}(\pi_{v})}$
$\displaystyle=L(s+im(\chi),\chi_{0}).$

The following proposition deduces easily from 2.

######  Corollary 1.

Let $K\subset K^{0}_{\max}$ be an open subgroup. For every $\epsilon>0$, there
exist $C=C(\epsilon)>0$ and $\delta=\delta(\epsilon)>0$ such that the
following conditions are satisfied if provided that $\Re(s)\geq
1-\delta\mathrel{\mathop{\ordinarycolon}}$

  1. 1.

for every non trivial Hecke character
$\chi\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{\times}/F^{\times})\to
S^{1}$ with $\chi_{0}\neq 1$ which vanishes on $K\subset K_{\max}^{0}$ one has

$|L(s,\chi)|\leq
C\big{(}(1+|\Im(s)|)(1+||\chi_{0}||_{\infty})(1+|m(\chi)|)\big{)}^{\epsilon};$
(2)

  2. 2.

for every Hecke character
$\chi\mathrel{\mathop{\ordinarycolon}}(\mathbb{A}_{F}^{\times}/F^{\times})\to
S^{1}$ with $\chi_{0}=1$ one has that

$\bigg{|}\frac{(s+im(\chi)-1)L(s,\chi)}{s+im(\chi)}\bigg{|}\leq
C((1+|\Im(s)|)(1+|m(\chi)|)^{\epsilon}.$ (3)

###### Proof.

Let $\epsilon>0$ and let $C=C(\epsilon)$ and $\delta=\delta(\epsilon)>0$ be
given by 2.

  1. 1.

Let $\chi\mathrel{\mathop{\ordinarycolon}}\mathbb{A}_{F}^{\times}\to S^{1}$ be
a Hecke character which vanishes on $K$ such that $\chi_{0}\neq 1$. Then
$\chi_{0}$ vanishes on $K$ and Proposition 2 gives that

$\displaystyle L(s,\chi)=L(s+im(\chi),\chi_{0})$ $\displaystyle\leq
C((1+|\Im(s)+m(\chi)|)(1+||\chi_{0}||_{\infty}))^{\epsilon}$
$\displaystyle\leq
C((1+|\Im(s)|)(1+|m(\chi)|)(1+||\chi_{0}||_{\infty}))^{\epsilon}.$

  2. 2.

Let $\chi\mathrel{\mathop{\ordinarycolon}}\mathbb{A}_{F}^{\times}\to S^{1}$ be
a Hecke character with $\chi_{0}=1$. Proposition 2 gives that

$\displaystyle\bigg{|}\frac{(s+im(\chi)-1)L(s,\chi)}{s+im(\chi)}\bigg{|}$
$\displaystyle=\bigg{|}\frac{(s+im(\chi)-1)L(s+im(\chi),1)}{s+im(\chi)}\bigg{|}$
$\displaystyle=\bigg{|}\frac{(s+im(\chi)-1)\zeta_{F}(s+im(\chi))}{s+im(\chi)}\bigg{|}$
$\displaystyle\leq C((1+|\Im(s)+m(\chi)|))^{\epsilon}$ $\displaystyle\leq
C((1+|\Im(s)|)(1+|m(\chi)|))^{\epsilon}.$

∎

##  Chapter 5 Fourier transform of the height function

In this chapter we analyse the Fourier transform of the height function, when
the functions $f_{v}$ are smooth. Let $n$ be a positive integer and let
$\mathbf{a}\in\mathbb{Z}^{n}_{>0}$ if $n\geq 2$ and
$a=a_{1}\in\mathbb{Z}_{>1}$ if $n=1$. As before, we use notation $f_{v}^{\\#}$
for the toric $\mathbf{a}$-homogenous function
$F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0}$ of weighted degree $|\mathbf{a}|$. For
$v\in M_{F}^{0}$, we have established in 1 that $f_{v}^{\\#}$ is locally
constant i.e. smooth. Let
$(f_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{>0})_{v}$
be a degree $|\mathbf{a}|$ quasi-toric $\mathbf{a}$-homogenous family of
smooth functions. Let $S$ be the union of the set consisting of the finite
places $v$ at which $f_{v}$ is not toric and the set of the infinite places.
Let $H=H((f_{v})_{v})$ be the corresponding height on
$[\mathscr{P}(\mathbf{a})(F)]$. If $v\in M_{F}$, for a character
$\chi_{v}\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$ and $j\in\\{1,\dots,n\\}$,
we denote by $\chi^{(j)}_{v}$ the character $F_{v}^{\times}\to S^{1}$ given by

$x\mapsto\chi_{v}(q_{v}^{\mathbf{a}}((1)_{\begin{subarray}{c}k=1,\dots,n\\\
k\neq j\end{subarray}},(x)_{k})).$

###  1 Local transform

In this section we study local Fourier transform.

#### 1

In this paragraph we define height pairing.

For $v\in M_{F},$ $\mathbf{s}\in\mathbb{C}^{n},$ $t\in F_{v}^{\times}$ and
$\mathbf{x}\in(F_{v}^{\times})^{n}$ one has that

$f_{v}(t\cdot\mathbf{x})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}}}\prod_{j=1}^{n}|t^{a_{j}}x_{j}|_{v}^{-s_{j}}=|t|_{v}^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}f_{v}(\mathbf{x})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|t|_{v}^{-a_{j}s_{j}}|x_{j}|_{v}^{-s_{j}}\\\
=f_{v}(\mathbf{x})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{-s_{j}}$

i.e. for $v\in M_{F}$ and $\mathbf{s}\in\mathbb{C}^{n}$, the continuous
function

$(F_{v}^{\times})^{n}\to\mathbb{C},\hskip 28.45274pt\mathbf{x}\mapsto
f_{v}(\mathbf{x})^{\mathbf{a}\cdot\mathbf{s}}\prod_{j=1}^{n}|x_{j}|_{v}^{-s_{j}}$
(1)

is $(F_{v}^{\times})_{\mathbf{a}}$-invariant. Let
$H_{v}(\mathbf{s},-)\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F)]\to\mathbb{C}$
be the function induced from $(F_{v})_{\mathbf{a}}$-invariant function (1).
For $\mathbf{x}\in[\mathscr{T}({\mathbf{a}})(F)],$ we write
$H_{v}(\mathbf{s},\mathbf{x})$ for what is technically
$H_{v}(\mathbf{s},[\mathscr{T}({\mathbf{a}})(i_{v})](\mathbf{x})),$ where
$[\mathscr{T}({\mathbf{a}})(i_{v})]\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(i)(F)]\to[\mathscr{T}({\mathbf{a}})(i)(F_{v})]$
is the induced homomorphism from $(F^{\times})_{\mathbf{a}}$-invariant
homomorphism

$(F^{\times})^{n}\hookrightarrow(F_{v}^{\times})^{n}\to[\mathscr{T}({\mathbf{a}})(F_{v})].$

Note that if $\mathbf{x}\in[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]$, one
can choose $\widetilde{\mathbf{x}}\in(\mathcal{O}_{v}^{\times})^{n}$. We
deduce that if $f_{v}=f_{v}^{\\#}$ is the toric $\mathbf{a}$-homogenous
function of weighted degree $|\mathbf{a}|$ then for
$\mathbf{s}\in\mathbb{C}^{n}$, one has

$H_{v}^{\\#}(\mathbf{s},\mathbf{x})=f^{\\#}_{v}(\mathbf{x})^{\frac{-\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{-s_{j}}=1.$

######  Lemma 2.

Let $\mathbf{s}\in\mathbb{C}^{n}$.

  1. 1.

Suppose that $\mathbf{x}\in[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]$. One
has that $H^{\\#}_{v}(\mathbf{s},\mathbf{x})=1$.

  2. 2.

Let $(\mathbf{x}_{v})_{v}\in[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$. The
product

$H(\mathbf{s},\mathbf{x})\mathrel{\mathop{\ordinarycolon}}=\prod_{v\in
M_{F}}H_{v}(\mathbf{s},\mathbf{x})$

is a finite product.

  3. 3.

Let $\mathbf{x}\in[\mathscr{T}({\mathbf{a}})(F)].$ One has that

$H(\mathbf{x})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}=H(\mathbf{s},[\mathscr{T}({\mathbf{a}})(i)](\mathbf{x})).$

Here
$[\mathscr{T}({\mathbf{a}})(i)]\mathrel{\mathop{\ordinarycolon}}[\mathscr{T}({\mathbf{a}})(F)]\to[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$
is the map induced from $(F^{\times})_{\mathbf{a}}$-invariant map

$F^{\times
n}\to(\mathbb{A}_{F}^{\times})^{n}\to[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})],$

where the first map is the diagonal inclusion and the second map is the
quotient map.

###### Proof.

  1. 1.

Let $\widetilde{\mathbf{x}}\in(\mathcal{O}_{v}^{\times})^{n}$ be a lift of
$\mathbf{x}$. One has that
$(\mathcal{O}_{v}^{\times})^{n}\subset(\mathcal{O}_{v}^{n}-\prod_{j=1}^{n}\pi_{v}^{a_{j}}\mathcal{O}_{v})$
and thus by 1, one has $f_{v}^{\\#}|_{(\mathcal{O}_{v}^{\times})^{n}}=1$. We
deduce that

$H^{\\#}_{v}(\mathbf{x})=f_{v}^{\\#}(\widetilde{\mathbf{x}})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|\widetilde{x}_{j}|_{v}^{-s_{j}}=1.$

  2. 2.

By definition of $[\mathscr{T}({\mathbf{a}})(\mathbb{A}_{F})]$ for almost
every $v\in M_{F}$ one has that
$\mathbf{x}_{v}\in[\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})]$. For almost
every $v$, hence, one has that
$H_{v}(\mathbf{s},\mathbf{x}_{v})=H^{\\#}_{v}(\mathbf{s},\mathbf{x}_{v})=1.$
Thus the product defining $H(\mathbf{s},\mathbf{x})$ is a finite product.

  3. 3.

Let $\widetilde{\mathbf{x}}\in F^{\times n}$ be a lift of $\mathbf{x}$. Recall
from 3 that $v$-th coordinate of $[\mathscr{T}({\mathbf{a}})(i)](\mathbf{x})$
is $[\mathscr{T}({\mathbf{a}})(i_{v})](\mathbf{x})$. Using the product
formula, we get that

$\displaystyle H(\mathbf{s},[\mathscr{T}({\mathbf{a}})(i)](\mathbf{x}))$
$\displaystyle=\prod_{v}H_{v}(\mathbf{s},[\mathscr{T}({\mathbf{a}})(i_{v})](\mathbf{x}))$
$\displaystyle=\prod_{v}\bigg{(}f_{v}(\widetilde{\mathbf{x}})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|\widetilde{x}_{j}|_{v}^{-s_{j}}\bigg{)}$
$\displaystyle=\bigg{(}\prod_{v}f_{v}(\widetilde{\mathbf{x}})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\bigg{)}\prod_{v}\prod_{j=1}^{n}|\widetilde{x}_{j}|_{v}^{-s_{j}}$
$\displaystyle=\prod_{v}f^{\\#}_{v}(\widetilde{\mathbf{x}})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}$
$\displaystyle=H(\mathbf{x})^{\frac{\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}.$

∎

We may write $H(\mathbf{s},\mathbf{x})$ for what is technically
$H(\mathbf{s},[\mathscr{T}({\mathbf{a}})(i)](\mathbf{x}))$.

#### 2

In this paragraph we establish that the functions $H_{v}^{-1}(\mathbf{s},-)$
are integrable and that their Fourier transforms are holomorphic and bounded
in $\mathbf{s}$.

Let $v\in M_{F}$. In 1, we have defined a Haar measure on
$[\mathscr{T}({\mathbf{a}})(F_{v})]=(F_{v}^{\times})^{n}/(F_{v}^{\times})_{\mathbf{a}}$
by $(d^{*}x_{1}\dots d^{*}x_{n})/d^{*}x$. If $v\in M_{F}^{0}$, we have
established in Lemma 1 that
$\mu_{v}([\mathscr{T}({\mathbf{a}})(\mathcal{O}_{v})])=\zeta_{v}(1)^{-n+1}$.
For $\mathbf{s}\in\mathbb{C}^{n}$ and a character
$\chi_{v}\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$ we define formally

$\displaystyle\quad\widehat{H}_{v}(\mathbf{s},\chi_{v})\mathrel{\mathop{\ordinarycolon}}=$
$\displaystyle\begin{cases}\zeta_{v}(1)^{n-1}\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}\chi_{v}\mu_{v}&\text{
if $v\in M_{F}^{0}$,}\\\
\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}\chi_{v}\mu_{v}&\text{
if $v\in M_{F}^{\infty}$.}\end{cases}$

In this paragraph, we are going to prove that this integral converges
absolutely when $\mathbf{s}\in\Omega_{>0}$ and that it is a holomorphic
function of $\mathbf{s}$ in this domain. The following result, in a bit weaker
form, has been given as Lemma 8.3 in [18]

######  Lemma 1.

Let $B>0$. For every $\epsilon>0$, the integral

$\int_{\\{x\in F_{v}|\hskip 2.84544pt|x|_{v}\leq B\\}}|x|_{v}^{s-1}dx_{v}$ (2)

converges absolutely and uniformly in the domain
$s\in\mathbb{R}_{>\epsilon}+i\mathbb{R}$. The function that associates to $s$
the value of (2) is holomorphic in the domain $\mathbb{R}_{>0}+i\mathbb{R}$.

###### Proof.

Suppose $v\in M_{F}^{0}$. Let $r$ be the largest integer satisfying that
$(|\pi_{v}|_{v}^{-1})^{r}\leq B$. For every $\epsilon>0$ and every
$s\in\mathbb{R}_{>\epsilon}+i\mathbb{R}$, we have that

$\displaystyle\int_{\\{x\in F_{v}|\hskip 2.84544pt|x|_{v}\leq
B\\}}|x|_{v}^{s-1}dx_{v}$
$\displaystyle=\sum_{k=-\infty}^{r}\int_{|x|_{v}=(|\pi_{v}|_{v}^{-1})^{k}}|x|_{v}^{s-1}dx$
$\displaystyle=\sum_{k=-\infty}^{r}\int_{|x|_{v}=(|\pi_{v}|_{v}^{-1})^{k}}|\pi_{v}|_{v}^{-k(s-1)}dx$
$\displaystyle=\sum_{k=-\infty}^{r}|\pi_{v}|_{v}^{-k(s-1)}dx(\pi_{v}^{-k}\mathcal{O}_{v})$
$\displaystyle=\sum_{k=-\infty}^{r}|\pi_{v}|_{v}^{-ks}$
$\displaystyle=\sum_{k=-r}^{\infty}|\pi_{v}|_{v}^{ks}$
$\displaystyle=|\pi_{v}|_{v}^{-rs}\sum_{k=0}^{\infty}|\pi_{v}|_{v}^{ks}.$

The last series converges absolutely and uniformly in the domain
$\mathbb{R}_{>\epsilon}+i\mathbb{R}.$ Moreover,
$s\mapsto|\pi_{v}|_{v}^{-rs}\sum_{k=0}^{\infty}|\pi_{v}|_{v}^{ks}=\frac{|\pi_{v}|_{v}^{-rs}}{1-|\pi_{v}|_{v}^{s}}$
is a holomorphic function in the domain $\mathbb{R}_{>0}+i\mathbb{R}$. Suppose
$v$ is a real place. For every $\epsilon>0$, we have that

$\int_{|x|_{v}\leq B}|x|_{v}^{s-1}dx_{v}=\int_{|x|\leq
B}|x|^{s-1}dx=2\int_{0}^{B}x^{s-1}dx$

converges absolutely and uniformly for
$s\in\mathbb{R}_{>\epsilon}+i\mathbb{R}$. Moreover,

$s\mapsto
2\int_{0}^{B}x^{s-1}dx=2\frac{x^{s}}{s}\bigg{|}_{x=0}^{x=B}=2\frac{B^{s}}{s}$

is a holomorphic function in $s$ in the domain $\mathbb{R}_{>0}+i\mathbb{R}$.
Suppose $v$ is a complex place. For every $\epsilon>0$, we have that

$\displaystyle\int_{|x|_{v}\leq B}|x|_{v}^{s-1}dx_{v}$
$\displaystyle=\int_{x^{2}+y^{2}\leq B}(x^{2}+y^{2})^{s-1}2dxdy$
$\displaystyle=\int_{0}^{2\pi}\int_{r^{2}\leq B}r^{2(s-1)}2rdrd\phi$
$\displaystyle=4\pi\int_{0}^{\sqrt{B}}r^{2s-1}dr$

converges absolutely and uniformly for
$s\in\mathbb{R}_{>\epsilon}+i\mathbb{R}$. Moreover,

$s\mapsto
4\pi\int_{0}^{\sqrt{B}}r^{2s-1}dr=4\pi\frac{r^{2s}}{2s}\bigg{|}_{r=0}^{\sqrt{B}}=\frac{2\pi
B^{s}}{s}$

is a holomorphic function in $s$ in the domain $\mathbb{R}_{>0}+i\mathbb{R}$.
The statement is proven.∎

######  Proposition 3.

For every $\chi_{v}\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$, the integral
defining $\widehat{H}_{v}(\mathbf{s},\chi_{v})$ converges absolutely in the
domain $\mathbf{s}\in\Omega_{>0}$. Moreover, for every compact
$\mathcal{K}\subset\Omega_{>0}$, there exists $C(\mathcal{K})>0$ such that for
every $\mathbf{s}\in\mathcal{K}$ and every
$\chi_{v}\in[\mathscr{T}({\mathbf{a}})(F_{v})]^{*}$, one has that

$|\widehat{H}_{v}(\mathbf{s},\chi_{v})|\leq C(\mathcal{K}).$

###### Proof.

As our characters are assumed unitary (that is with the values in $S^{1}$), by
the triangle inequality, it suffices to prove the statement when $\chi_{v}=1$.
Let $\mathcal{K}\subset\Omega_{>0}$ be a compact. Let $\omega_{v}$ be the
quotient measure $f_{v}^{-1}dx_{1}\dots dx_{n}/d^{*}x$ on
$(F^{n}_{v}-\\{0\\})/F_{v}^{\times}=[\mathscr{P}(\mathbf{a})(F_{v})]$ (see 1).
By 2, one has an inequality of the measures
$H_{v}(\mathbf{1},-)\omega_{v}|_{[\mathscr{T}({\mathbf{a}})(F_{v})]}=\mu_{v}$.
We deduce that $H_{v}(\mathbf{s},-)^{-1}\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v})$ if and only if
$H_{v}(\mathbf{s},-)^{-1}\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],H_{v}(\mathbf{1},-)\omega_{v}),$ i.e.
if and only if

$H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\omega_{v}).$

Moreover, if $H_{v}(\mathbf{s},-)^{-1}\in
L^{1}([\mathscr{T}({\mathbf{a}})(F_{v})],\mu_{v}),$ then

$\displaystyle\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}\mu_{v}$
$\displaystyle=\int_{[\mathscr{T}({\mathbf{a}})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\omega_{v}$
$\displaystyle=\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\omega_{v},$

where the last equality follows from the fact that
$\omega_{v}([\mathscr{P}(\mathbf{a})(F_{v})]-[\mathscr{T}({\mathbf{a}})(F_{v})])=0$,
which we have established in 2. In 2, we have defined a compactly supported
function
$k^{\mathbf{a}}_{v}\mathrel{\mathop{\ordinarycolon}}F^{n}_{v}-\\{0\\}\to\mathbb{R}_{\geq
0}$ which satisfies that for every $\mathbf{x}\in F^{n}_{v}-\\{0\\}$ one has
that $\int_{F_{v}^{\times}}k^{\mathbf{a}}_{v}(t\cdot\mathbf{x})d^{*}t=1$.
Proposition 5 gives that $H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v})$ if and only if

$\displaystyle((H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-))\circ
q^{\mathbf{a}}_{v})\cdot k^{\mathbf{a}}_{v}\hskip-28.45274pt$
$\displaystyle=\big{(}f_{v}(\mathbf{x})^{\frac{-\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{s_{j}}\cdot
f_{v}(\mathbf{x})\prod_{j=1}^{n}|x_{j}|^{-1}\big{)}\cdot k^{\mathbf{a}}_{v}$
$\displaystyle=f_{v}(\mathbf{x})^{\frac{-\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{s_{j}-1}\cdot
k^{\mathbf{a}}_{v}$ $\displaystyle\in L^{1}(F^{n}_{v}-\\{0\\},dx_{1}\dots
dx_{n}),$

and that if $H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\in
L^{1}([\mathscr{P}(\mathbf{a})(F_{v})],\omega_{v}),$ then

$\displaystyle\int_{[\mathscr{P}(\mathbf{a})(F_{v})]}H_{v}(\mathbf{s},-)^{-1}H_{v}(\mathbf{1},-)\omega_{v}\hskip-28.45274pt$
$\displaystyle=\int_{F^{n}_{v}-\\{0\\}}f_{v}(\mathbf{x})^{\frac{-\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{s_{j}-1}\cdot
k^{\mathbf{a}}_{v}dx_{1}\dots dx_{n}$
$\displaystyle=\int_{\operatorname{supp}(k^{\mathbf{a}}_{v})}f_{v}(\mathbf{x})^{\frac{-\mathbf{a}\cdot\mathbf{s}}{|\mathbf{a}|}}\prod_{j=1}^{n}|x_{j}|_{v}^{s_{j}-1}\cdot
k^{\mathbf{a}}_{v}dx_{1}\dots dx_{n}.$

For every $\mathbf{s}\in\mathcal{K}$, the function $\mathbf{x}\mapsto
f_{v}(\mathbf{x})^{-\frac{\mathbf{a}\cdot\Re(\mathbf{s})}{|\mathbf{a}|}}k^{\mathbf{a}}_{v}(\mathbf{x})$
is non vanishing and continuous, moreover it can be uniformly bounded for
$\mathbf{s}\in\mathcal{K}$ and