Spectral action on noncommutative torus

TL;DR

This paper computes the spectral action on the noncommutative torus using a Chamseddine--Connes formula, emphasizing the role of Diophantine conditions and establishing holomorphic continuation results for related series.

Contribution

It introduces a method to evaluate the spectral action on noncommutative tori through zeta function computations, highlighting the significance of Diophantine conditions.

Findings

Spectral action explicitly computed for noncommutative torus

Holomorphic continuation of series established

Diophantine condition's importance demonstrated

Abstract

The spectral action on noncommutative torus is obtained, using a Chamseddine--Connes formula via computations of zeta functions. The importance of a Diophantine condition is outlined. Several results on holomorphic continuation of series of holomorphic functions are obtained in this context.

Equations

$${\mathcal{D}}_A:=\mathcal{D}+A+ϵJA{J}^{-1}$$

$$A=\sum _i{a}_i[\mathcal{D},b_i],$$

$$J\mathcal{D}=ϵ\mathcal{D}J.$$

$$\SS(\mathcal{D}_{A},\Phi,\Lambda):=\operatorname{Tr}\big{(}\Phi(\mathcal{D}_{A}/\Lambda)\big{)}$$

$$\text{SS}({\mathcal{D}}_A,\Phi ,\Lambda )=\sum _{0<k\in S{d}^{+}}{\Phi }_k{\Lambda }^{k}to0.0ptto5.83331pt\int \text{hfil}\text{hss}-|D_A{|}^{-k}+\Phi (0){\zeta }_{D_A}(0)+\mathcal{O}({\Lambda }^{-1})$$

$${\zeta }_{\mathcal{D}+A}(0)-{\zeta }_{\mathcal{D}}(0)=\sum _{q=1}^n\frac{(-1{)}^q}{q}to0.0ptto5.83331pt\int \text{hfil}\text{hss}-(A{\mathcal{D}}^{-1}{)}^q,$$

$${\zeta }^P(s):={\sum }_{k\in {\mathbb{Z}}^n}^{\prime }\frac{P(k)}{|k{|}^s},s\in \mathbb{C}$$

$$\underset{s=j+n}{\mathrm{Res}}{\zeta }^P(s)={\int }_{u\in S^{n-1}}{P}_j(u)dS(u).$$

$${\partial }^{\alpha }\left(P(x)|x{|}^{-s}\right){\ll }_{P,\alpha ,n}(1+|s|{)}^{|\alpha {|}_1}|x{|}^{-\sigma -|\alpha {|}_1+\delta (P)}$$

$${\partial }^{\alpha }\left(|x{|}^{-s}\right)=\alpha !\sum _{\genfrac{}{}{0ex}{}{\beta ,\mu \in {\mathbb{N}}_0^n}{\beta +2\mu =\alpha }}\left(\genfrac{}{}{0ex}{}{-s/2}{|\beta {|}_1+|\mu {|}_1}\right)\frac{(|\beta {|}_1+|\mu {|}_1)!}{\beta !\mu !}\frac{x^{\beta }}{|x{|}^{\sigma +2(|\beta {|}_1+|\mu {|}_1)}}.$$

$${\partial }^{\alpha }\left(|x{|}^{-s}\right){\ll }_{\alpha ,n}(1+|s|{)}^{|\alpha {|}_1}|x{|}^{-\sigma -|\alpha {|}_1}.$$

$${\partial }^{\alpha }\left(x^{\gamma }|x{|}^{-s}\right)$$

$${\ll }_{\gamma ,\alpha ,n}\sum _{\beta \le \alpha ;\beta \le \gamma }{x}^{\gamma -\beta }(1+|s|{)}^{|\alpha {|}_1-|\beta {|}_1}|x{|}^{-\sigma -|\alpha {|}_1+|\beta {|}_1}$$

$${\ll }_{\gamma ,\alpha ,n}(1+|s|{)}^{|\alpha {|}_1}|x{|}^{-\sigma -|\alpha {|}_1+|\gamma {|}_1}.$$

$${\Delta }_P(s):={\sum }_{k\in {\mathbb{Z}}^n}^{\prime }\frac{P(k)}{|k{|}^s}-{\int }_{{\mathbb{R}}^n\setminus B^n}\frac{P(x)}{|x{|}^s}dx$$

$$0\le \psi (x)\le 1,\psi (x)=1\text{ if }|x|\ge 1\text{and}\psi (x)=0\text{ if }|x|\le 1/2.$$

$$\sum _{k\in \mathbb{Z}}h(k)={\int }_{\mathbb{R}}h(t)+\frac{(-1{)}^N}{(N+1)!}{\int }_{\mathbb{R}}{B}_{N+1}(t)h^{(N+1)}(t)dt$$

$$\sum _{m_n\in \mathbb{Z}}\psi (m^{\prime },m_n)P(m^{\prime },m_n)|(m^{\prime },m_n){|}^{-s}={\int }_{\mathbb{R}}\psi (m^{\prime },t)P(m^{\prime },t)|(m^{\prime },t){|}^{-s}dt+{\mathcal{R}}_N(m^{\prime };s)$$

$$\int_{\mathbb{R}}{\Big{|}}B_{N+1}(t)~{}\tfrac{\partial^{N+1}}{{\partial x_{n}}^{N+1}}\left(\psi(m^{\prime},t)~{}P(m^{\prime},t)~{}|(m^{\prime},t)|^{-s}\right){\Big{|}}~{}dt\ll_{P,n,N}(1+|s|)^{N+1}~{}(|m^{\prime}|+1)^{-\sigma-N+\delta(P)}.$$

$$s\mapsto \sum _{(m^{\prime },m_n)\in {\mathbb{Z}}^{n-1}\times \mathbb{Z}}\psi (m^{\prime },m_n)P(m^{\prime },m_n)|(m^{\prime },m_n){|}^{-s}-\sum _{m^{\prime }\in {\mathbb{Z}}^{n-1}}{\int }_{\mathbb{R}}\psi (m^{\prime },t)P(m^{\prime },t)|(m^{\prime },t){|}^{-s}dt$$

$$s\mapsto {\sum }_{m\in {\mathbb{Z}}^n}\psi (m)P(m)|m{|}^{-s}-{\int }_{{\mathbb{R}}^n}\psi (x)P(x)|x{|}^{-s}dx$$

$${\int }_{{\mathbb{R}}^n\setminus B^n}\frac{P_j(x)}{|x{|}^s}dx={\int }_1^{\infty }\frac{{\rho }^{j+n-1}}{{\rho }^s}{\int }_{S^{n-1}}{P}_j(u)dS(u)=\frac{1}{j+n-s}{\int }_{S^{n-1}}{P}_j(u)dS(u).$$

$$g(s):={\sum }_{l\in ({\mathbb{Z}}^n{)}^q}b(l)f_{\Theta {\sum }_i{\epsilon }_i{l}_i}(s)$$

$$\underset{s=d+n}{\mathrm{Res}}g(s)=V{\int }_{u\in S^{n-1}}P(u)dS(u).$$

$$g_0(s):={\sum }_{l\in ({\mathbb{Z}}^n{)}^q\setminus \mathcal{Z}}b(l)f_{\Theta {\sum }_{i=1}^q{\epsilon }_i{l}_i}(s)$$

$$\mathcal{F}:=\{\frac{P(X)}{(X_1^2+\cdots +X_n^2+1{)}^{r/2}}:P(X)\in \mathbb{C}[X_1,\dots ,X_n]\text{mbox}andr\in {\mathbb{N}}_0\}.$$

$$F_0(G;a;s):={\sum }_{k\in {\mathbb{Z}}^n}^{\prime }\frac{G(k)}{|k{|}^s}{e}^{2\pi ik.a}\text{and}{F}_1(G;a;s):={\sum }_{k\in {\mathbb{Z}}^n}\frac{G(k)}{(|k{|}^2+1{)}^{s/2}}{e}^{2\pi ik.a}.$$

$$F_{i}(G;a;s)\leq C_{i}(1+|s|)^{B_{i}}\,\big{(}d\left(a.u,\mathbb{Z}\right)\big{)}^{-A_{i}}.$$

$$\frac{|G(k)e^{2\pi ik.a}|}{(|k{|}^2+1{)}^{\sigma /2}}=$$

$$|F_1(G;a;s)|{\ll }_G\sum _{k\in {\mathbb{Z}}^n}\frac{1}{(|k{|}^2+1{)}^{(-N+g_0)/2}}{\ll }_G\sum _{k\in {\mathbb{Z}}^n}\frac{1}{(|k{|}^2+1{)}^{(n+1)/2}}{\ll }_{G}1.$$

$$F_1(G;a;s)$$

Full text

CENTRE DE PHYSIQUE THÉORIQUE$\,{}^{1}$

CNRS–Luminy, Case 907

13288 Marseille Cedex 9

FRANCE

Spectral action on noncommutative torus

D. Essouabri2, B. Iochum1,3, C. Levy1,3 and A. Sitarz4,5

Dedicated to Alain Connes on the occasion of his 60th birthday

Abstract

The spectral action on noncommutative torus is obtained, using a Chamseddine–Connes formula via computations of zeta functions. The importance of a Diophantine condition is outlined. Several results on holomorphic continuation of series of holomorphic functions are obtained in this context.

February 2007

PACS numbers: 11.10.Nx, 02.30.Sa, 11.15.Kc

MSC–2000 classes: 46H35, 46L52, 58B34

CPT-P06-2007

1 UMR 6207

– Unité Mixte de Recherche du CNRS et des Universités Aix-Marseille I, Aix-Marseille II et de l’Université du Sud Toulon-Var

– Laboratoire affilié à la FRUMAM – FR 2291

2 Université de Caen (Campus II), Laboratoire de Math. Nicolas Oresme (CNRS UMR 6139), B.P. 5186, 14032 Caen, France, essoua@math.unicaen.fr

3 Also at Université de Provence, iochum@cpt.univ-mrs.fr, levy@cpt.univ-mrs.fr

4 Institute of Physics, Jagiellonian University, Reymonta 4, 30-059 Kraków, Poland

sitarz@if.uj.edu.pl

5 Partially supported by MNII Grant 115/E-343/SPB/6.PR UE/DIE 50/2005–2008

1 Introduction

The spectral action introduced by Chamseddine–Connes plays an important role [3] in noncommutative geometry. More precisely, given a spectral triple $(\mathcal{A},\mathcal{H},\mathcal{D})$ where $\mathcal{A}$ is an algebra acting on the Hilbert space $\mathcal{H}$ and $\mathcal{D}$ is a Dirac-like operator (see [8, 23]), they proposed a physical action depending only on the spectrum of the covariant Dirac operator

[TABLE]

where $A$ is a one-form represented on $\mathcal{H}$, so has the decomposition

[TABLE]

with $a_{i}$, $b_{i}\in\mathcal{A}$, $J$ is a real structure on the triple corresponding to charge conjugation and $\epsilon\in\{\,1,-1\,\}$ depending on the dimension of this triple and comes from the commutation relation

[TABLE]

This action is defined by

[TABLE]

where $\Phi$ is any even positive cut-off function which could be replaced by a step function up to some mathematical difficulties investigated in [16]. This means that $\SS$ counts the spectral values of $|\mathcal{D}_{A}|$ less than the mass scale $\Lambda$ (note that the resolvent of $\mathcal{D}_{A}$ is compact since, by assumption, the same is true for $\mathcal{D}$, see Lemma 3.1 below).

In [18], the spectral action on NC-tori has been computed only for operators of the form $\mathcal{D}+A$ and computed for $\mathcal{D}_{A}$ in [20]. It appears that the implementation of the real structure via $J$, does change the spectral action, up to a coefficient when the torus has dimension 4. Here we prove that this can be also directly obtained from the Chamseddine–Connes analysis of [4] that we follow quite closely. Actually,

[TABLE]

where $D_{A}=\mathcal{D}_{A}+P_{A}$, $P_{A}$ the projection on $\operatorname{Ker}\mathcal{D}_{A}$, $\Phi_{k}={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}\int_{0}^{\infty}\Phi(t)\,t^{k/2-1}\,dt$ and $Sd^{+}$ is the strictly positive part of the dimension spectrum of $(\mathcal{A},\mathcal{H},\mathcal{D})$. As we will see, $Sd^{+}=\{\,1,2,\cdots,n\,\}$ and \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits|D_{A}|^{-n}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits|D|^{-n}. Moreover, the coefficient $\zeta_{D_{A}}(0)$ related to the constant term in (5) can be computed from the unperturbed spectral action since it has been proved in [4] (with an invertible Dirac operator and a 1-form $A$ such that $\mathcal{D}+A$ is also invertible) that

[TABLE]

using $\zeta_{X}(s)=\operatorname{Tr}(|X|^{-s})$. We will see how this formula can be extended to the case a noninvertible Dirac operator and noninvertible perturbation of the form $\mathcal{D}+\widetilde{A}$ where $\widetilde{A}:=A+\varepsilon JAJ^{-1}$.

All this results on spectral action are quite important in physics, especially in quantum field theory and particle physics, where one adds to the effective action some counterterms explicitly given by (6), see for instance [2, 3, 4, 5, 17, 22, 18, 20, 28, 35, 36, 37, 38].

Since the computation of zeta functions is crucial here, we investigate in section 2 residues of series and integrals. This section contains independent interesting results on the holomorphy of series of holomorphic functions. In particular, the necessity of a Diophantine constraint is naturally emphasized.

In section 3, we revisit the notions of pseudodifferential operators and their associated zeta functions and of dimension spectrum. The reality operator $J$ is incorporated and we pay a particular attention to kernels of operators which can play a role in the constant term of (5). This section concerns general spectral triple with simple dimension spectrum.

Section 4 is devoted to the example of the noncommutative torus. It is shown that it has a vanishing tadpole.

In section 5, all previous technical points are then widely used for the computation of terms in (5) or (6).

Finally, the spectral action (6) is obtained in section 6 and we conjecture that the noncommutative spectral action of $\mathcal{D}_{A}$ has terms proportional to the spectral action of $\mathcal{D}+A$ on the commutative torus.

2 Residues of series and integral, holomorphic continuation,

etc

Notations:

In the following, the prime in ${\sum}^{\prime}$ means that we omit terms with division by zero in the summand. $B^{n}$ (resp. $S^{n-1}$) is the closed ball (resp. the sphere) of $\mathbb{R}^{n}$ with center [math] and radius 1 and the Lebesgue measure on $S^{n-1}$ will be noted $dS$.

For any $x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}$ we denote by $|x|=\sqrt{x_{1}^{2}+\dots+x_{n}^{2}}$ the euclidean norm and $|x|_{1}:=|x_{1}|+\dots+|x_{n}|$.

$\mathbb{N}=\{1,2,\dots\}$ is the set of positive integers and $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ the set of non negative integers.

By $f(x,y)\ll_{y}g(x)$ uniformly in $x$, we mean that $|f(x,y)|\leq a(y)\,|g(x)|$ for all $x$ and $y$ for some $a(y)>0$.

2.1 Residues of series and integral

In order to be able to compute later the residues of certain series, we prove here the following

Theorem 2.1**.**

Let $P(X)=\sum_{j=0}^{d}P_{j}(X)\in\mathbb{C}[X_{1},\cdots,X_{n}]$ be a polynomial function where $P_{j}$ is the homogeneous part of $P$ of degree $j$. The function

[TABLE]

has a meromorphic continuation to the whole complex plane $\mathbb{C}$.

Moreover $\zeta^{P}(s)$ is not entire if and only if $\mathcal{P}_{P}:=\{j\ :\ \int_{u\in S^{n-1}}P_{j}(u)\,dS(u)\neq 0\}\neq\varnothing$. In that case, $\zeta^{P}$ has only simple poles at the points $j+n$, $j\in\mathcal{P}_{P}$, with

[TABLE]

The proof of this theorem is based on the following lemmas.

Lemma 2.2**.**

For any polynomial $P\in\mathbb{C}[X_{1},\dots,X_{n}]$ of total degree $\delta(P):=\sum_{i=1}^{n}deg_{X_{i}}P$ and any $\alpha\in\mathbb{N}_{0}^{n}$, we have

[TABLE]

uniformly in $x\in\mathbb{R}^{n}$ verifying $|x|\geq 1$, where $\sigma=\Re(s)$.

Proof.

By linearity, we may assume without loss of generality that $P(X)=X^{\gamma}$ is a monomial. It is easy to prove (for example by induction on $|\alpha|_{1}$) that for all $\alpha\in\mathbb{N}_{0}^{n}$ and $x\in\mathbb{R}^{n}\setminus\{0\}$:

[TABLE]

It follows that for all $\alpha\in\mathbb{N}_{0}^{n}$, we have uniformly in $x\in\mathbb{R}^{n}$ verifying $|x|\geq 1$:

[TABLE]

By Leibniz formula and (7), we have uniformly in $x\in\mathbb{R}^{n}$ verifying $|x|\geq 1$:

[TABLE]

Lemma 2.3**.**

Let $P\in\mathbb{C}[X_{1},\dots,X_{n}]$ be a polynomial of degree $d$. Then, the difference

[TABLE]

which is defined for $\Re(s)>d+n$, extends holomorphically on the whole complex plane $\mathbb{C}$.

Proof.

We fix in the sequel a function $\psi\in C^{\infty}(\mathbb{R}^{n},\mathbb{R})$ verifying for all $x\in\mathbb{R}^{n}$

[TABLE]

The function $f(x,s):=\psi(x)~{}P(x)~{}|x|^{-s}$, $x\in\mathbb{R}^{n}$ and $s\in\mathbb{C}$, is in ${\cal C}^{\infty}(\mathbb{R}^{n}\times\mathbb{C})$ and depends holomorphically on $s$.

Lemma 2.2 above shows that $f$ is a “gauged symbol” in the terminology of [24, p. 4]. Thus [24, Theorem 2.1] implies that $\Delta_{P}(s)$ extends holomorphically on the whole complex plane $\mathbb{C}$. However, to be complete, we will give here a short proof of Lemma 2.3:

It follows from the classical Euler–Maclaurin formula that for any function $h:\mathbb{R}\rightarrow\mathbb{C}$ of class ${\cal C}^{N+1}$ verifying $\lim_{|t|\rightarrow+\infty}h^{(k)}(t)=0$ and $\int_{\mathbb{R}}|h^{(k)}(t)|~{}dt<+\infty$ for any $k=0\dots,N+1$, that we have

[TABLE]

where $B_{N+1}$ is the Bernoulli function of order $N+1$ (it is a bounded periodic function.)

Fix $m^{\prime}\in\mathbb{Z}^{n-1}$ and $s\in\mathbb{C}$. Applying this to the function $h(t):=\psi(m^{\prime},t)~{}P(m^{\prime},t)~{}|(m^{\prime},t)|^{-s}$ (we use Lemma 2.2 to verify hypothesis), we obtain that for any $N\in\mathbb{N}_{0}$:

[TABLE]

where ${\cal R}_{N}(m^{\prime};s):=\tfrac{(-1)^{N}}{(N+1)!}\int_{\mathbb{R}}B_{N+1}(t)~{}\tfrac{\partial^{N+1}}{{\partial x_{n}}^{N+1}}\left(\psi(m^{\prime},t)~{}P(m^{\prime},t)~{}|(m^{\prime},t)|^{-s}\right)~{}dt$.

By Lemma 2.2,

[TABLE]

Thus $\sum_{m^{\prime}\in\mathbb{Z}^{n-1}}{\cal R}_{N}(m^{\prime};s)$ converges absolutely and define a holomorphic function in the half plane $\{\sigma=\Re(s)>\delta(P)+n-N\}$.

Since $N$ is an arbitrary integer, by letting $N\rightarrow\infty$ and using $(\ref{*1})$ above, we conclude that:

[TABLE]

has a holomorphic continuation to the whole complex plane $\mathbb{C}$.

After $n$ iterations, we obtain that

[TABLE]

has a holomorphic continuation to the whole $\mathbb{C}$.

To finish the proof of Lemma 2.3, it is enough to notice that:

$\bullet$ $\psi(0)=0$ and $\psi(m)=1$, $\forall m\in\mathbb{Z}^{n}\setminus\{0\}$;

$\bullet$ $s\mapsto\int_{B^{n}}\psi(x)~{}P(x)~{}|x|^{-s}~{}dx=\int_{\{x\in\mathbb{R}^{n}:1/2\leq|x|\leq 1\}}\psi(x)~{}P(x)~{}|x|^{-s}~{}dx$ is a holomorphic function on $\mathbb{C}$. ∎

Proof of Theorem 2.1.

Using the polar decomposition of the volume form $dx=\rho^{n-1}\,d\rho\,dS$ in $\mathbb{R}^{n}$, we get for $\Re(s)>d+n$,

[TABLE]

Lemma 2.3 now gives the result. ∎

2.2 Holomorphy of certain series

Before stating the main result of this section, we give first in the following some preliminaries from Diophantine approximation theory:

Definition 2.4**.**

*(i) Let $\delta>0$. A vector $a\in\mathbb{R}^{n}$ is said to be $\delta-$diophantine if there exists $c>0$ such that $|q.a-m|\geq c\,|q|^{-\delta}$, $\forall q\in\mathbb{Z}^{n}\setminus\{\,0\,\}$ and $\forall m\in\mathbb{Z}$.

We note ${\cal BV}(\delta)$ the set of $\delta-$diophantine vectors and ${\cal BV}:=\cup_{\delta>0}{\cal BV}(\delta)$ the set of diophantine vectors.*

(ii) A matrix $\Theta\in{\cal M}_{n}(\mathbb{R})$ (real $n\times n$ matrices) will be said to be diophantine if there exists $u\in\mathbb{Z}^{n}$ such that ${}^{t}\Theta(u)$ is a diophantine vector of $\mathbb{R}^{n}$.

Remark. A classical result from Diophantine approximation asserts that for all $\delta>n$, the Lebesgue measure of $\mathbb{R}^{n}\setminus{\cal BV}(\delta)$ is zero (i.e almost any element of $\mathbb{R}^{n}$ is $\delta-$diophantine.)

Let $\Theta\in{\cal M}_{n}(\mathbb{R})$. If its row of index $i$ is a diophantine vector of $\mathbb{R}^{n}$ (i.e. if $L_{i}\in{\cal BV}$) then ${}^{t}\Theta(e_{i})\in{\cal BV}$ and thus $\Theta$ is a diophantine matrix. It follows that almost any matrix of ${\cal M}_{n}(\mathbb{R})\approx\mathbb{R}^{n^{2}}$ is diophantine.

The goal of this section is to show the following

Theorem 2.5**.**

Let $P\in\mathbb{C}[X_{1},\cdots,X_{n}]$ be a homogeneous polynomial of degree $d$ and let $b$ be in $\mathcal{S}(\mathbb{Z}^{n}\times\dots\times\mathbb{Z}^{n})$ ($q$ times, $q\in\mathbb{N}$). Then,

(i) Let $a\in\mathbb{R}^{n}$. We define $f_{a}(s):={\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\frac{P(k)}{|k|^{s}}\,e^{2\pi ik.a}$.

*1. If $a\in\mathbb{Z}^{n}$, then $f_{a}$ has a meromorphic continuation to the whole complex plane $\mathbb{C}$.

Moreover if $S$ is the unit sphere and $dS$ its Lebesgue measure, then $f_{a}$ is not entire if and only if $\int_{u\in S^{n-1}}P(u)\,dS(u)\neq 0$. In that case, $f_{a}$ has only a simple pole at the point $d+n$, with $\underset{s=d+n}{\operatorname{Res}}\,f_{a}(s)=\int_{u\in S^{n-1}}P(u)\,dS(u)$.*

2. If $a\in\mathbb{R}^{n}\setminus\mathbb{Z}^{n}$, then $f_{a}(s)$ extends holomorphically to the whole complex plane $\mathbb{C}$.

(ii) Suppose that $\Theta\in{\cal M}_{n}(\mathbb{R})$ is diophantine. For any $(\varepsilon_{i})_{i}\in\{-1,0,1\}^{q}$, the function

[TABLE]

extends meromorphically to the whole complex plane $\mathbb{C}$ with only one possible pole on $s=d+n$.

Moreover, if we set ${\cal Z}:=\{l\in(\mathbb{Z}^{n})^{q}\ :\ \sum_{i=1}^{q}\varepsilon_{i}l_{i}=0\}$ and $V:=\sum_{l\in{\cal Z}}\,b(l)$, then

1. If $V\int_{S^{n-1}}P(u)\,dS(u)\neq 0$, then $s=d+n$ is a simple pole of $g(s)$ and

[TABLE]

2. If $V\int_{S^{n-1}}P(u)\,dS(u)=0$, then $g(s)$ extends holomorphically to the whole complex plane $\mathbb{C}$.

(iii) Suppose that $\Theta\in\mathcal{M}_{n}(\mathbb{R})$ is diophantine. For any $(\varepsilon_{i})_{i}\in\{-1,0,1\}^{q}$, the function

[TABLE]

where ${\cal Z}:=\{l\in(\mathbb{Z}^{n})^{q}\ :\ \sum_{i=1}^{q}\varepsilon_{i}l_{i}=0\}$ extends holomorphically to the whole complex plane $\mathbb{C}$.

Proof of Theorem 2.5: First we remark that

If $a\in\mathbb{Z}^{n}$ then $f_{a}(s)={\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\frac{P(k)}{|k|^{s}}$. So, the point $(i.1)$ follows from Theorem 2.1;

$g(s):=\sum_{l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}}\,b(l)\,f_{\Theta\,\sum_{i}\varepsilon_{i}l_{i}}(s)+\left(\sum_{l\in{\cal Z}}\,b(l)\right){\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\frac{P(k)}{|k|^{s}}$. Thus, the point $(ii)$ rises

easily from $(iii)$ and Theorem 2.1.

So, to complete the proof, it remains to prove the items $(i.2)$ and $(iii)$.

The direct proof of $(i.2)$ is easy but is not sufficient to deduce $(iii)$ of which the proof is more delicate and requires a more precise (i.e. more effective) version of $(i.2)$. The next lemma gives such crucial version, but before, let us give some notations:

[TABLE]

We set $g=$deg$(G)=$deg$(P)-r\in\mathbb{Z}$, the degree of $G=\frac{P(X)}{(X_{1}^{2}+\dots+X_{n}^{2}+1)^{r/2}}\in{\cal F}$.

By convention we set deg$(0)=-\infty$.

Lemma 2.6**.**

Let $a\in\mathbb{R}^{n}$. We assume that $d\left(a.u,\mathbb{Z}\right):=\inf_{m\in\mathbb{Z}}|a.u-m|>0$ for some $u\in\mathbb{Z}^{n}$. For all $G\in{\cal F}$, we define formally,

[TABLE]

Then for all $N\in\mathbb{N}$, all $G\in{\cal F}$ and all $i\in\{0,1\}$, there exist positive constants $C_{i}:=C_{i}(G,N,u)$, $B_{i}:=B_{i}(G,N,u)$ and $A_{i}:=A_{i}(G,N,u)$ such that $s\mapsto F_{i}(G;\alpha;s)$ extends holomorphically to the half-plane $\{\Re(s)>-N\}$ and verifies in it:

[TABLE]

Remark 2.7**.**

The important point here is that we obtain an explicit bound of $F_{i}(G;\alpha;s)$ in $\{\Re(s)>-N\}$ which depends on the vector $a$ only through $d(a.u,\mathbb{Z})$, so depends on $u$ and indirectly on $a$ (in the sequel, $a$ will vary.) In particular the constants $C_{i}:=C_{i}(G,N,u)$, $B_{i}=B_{i}(G,N)$ and $A_{i}:=A_{i}(G,N)$ do not depend on the vector $a$ but only on $u$. This is crucial for the proof of items $(ii)$ and $(iii)$ of Theorem 2.5!

2.2.1 Proof of Lemma 2.6 for $i=1$:

Let $N\in\mathbb{N}_{0}$ be a fixed integer, and set $g_{0}:=n+N+1$.

We will prove Lemma 2.6 by induction on $g=$deg$(G)\in\mathbb{Z}$. More precisely, in order to prove case $i=1$, it suffices to prove that:

Lemma 2.6 is true for all $G\in{\cal F}$ verifying deg$(G)\leq-g_{0}$.

Let $g\in\mathbb{Z}$ with $g\geq-g_{0}+1$. If Lemma 2.6 is true for all $G\in{\cal F}$ such that deg$(G)\leq g-1$,

then it is also true for all $G\in{\cal F}$ satisfying deg$(G)=g$.

$\bullet$ Step 1: Checking Lemma 2.6 for deg$(G)\leq-g_{0}:=-(n+N+1)$.

Let $G(X)=\frac{P(X)}{(X_{1}^{2}+\dots+X_{n}^{2}+1)^{r/2}}\in{\cal F}$ verifying deg$(G)\leq-g_{0}$. It is easy to see that we have uniformly in $s=\sigma+i\tau\in\mathbb{C}$ and in $k\in\mathbb{Z}^{n}$:

[TABLE]

It follows that $F_{1}(G;a;s)=\sum_{k\in\mathbb{Z}^{n}}\frac{G(k)}{(|k|^{2}+1)^{s/2}}\,e^{2\pi i\,k.a}$ converges absolutely and defines a holomorphic function in the half plane $\{\sigma>-N\}$. Therefore, we have for any $s\in\{\Re(s)>-N\}$:

[TABLE]

Thus, Lemma 2.6 is true when deg$(G)\leq-g_{0}$.

$\bullet$ Step 2: Induction.

Now let $g\in\mathbb{Z}$ satisfying $g\geq-g_{0}+1$ and suppose that Lemma 2.6 is valid for all $G\in{\cal F}$ verifying deg$(G)\leq g-1$. Let $G\in{\cal F}$ with deg$(G)=g$. We will prove that $G$ also verifies conclusions of Lemma 2.6:

There exist $P\in\mathbb{C}[X_{1},\dots,X_{n}]$ of degree $d\geq 0$ and $r\in\mathbb{N}_{0}$ such that $G(X)=\frac{P(X)}{(X_{1}^{2}+\dots+X_{n}^{2}+1)^{r/2}}$ and $g=$deg$(G)=d-r$.

Since $G(k)\ll(|k|^{2}+1)^{g/2}$ uniformly in $k\in\mathbb{Z}^{n}$, we deduce that $F_{1}(G;a;s)$ converges absolutely in $\{\sigma=\Re(s)>n+g\}$.

Since $k\mapsto k+u$ is a bijection from $\mathbb{Z}^{n}$ into $\mathbb{Z}^{n}$, it follows that we also have for $\Re(s)>n+g$

[TABLE]

Let $M:=\sup(N+n+g,0)\in\mathbb{N}_{0}$. We have uniformly in $k\in\mathbb{Z}^{n}$

[TABLE]

Thus, for $\sigma=\Re(s)>n+d$,

[TABLE]

Set $I:=\left\{(\alpha,j)\in\mathbb{N}_{0}^{n}\times\{0,\dots,M\}\mid|\alpha|_{1}\leq d\right\}$ and $I^{*}:=I\setminus\{\,(0,0)\,\}$.

Set also $G_{(\alpha,j);u}(X):=\tfrac{\partial^{\alpha}P(X)\left(2X.u+|u|^{2}\right)^{j}}{(|X|^{2}+1)^{(r+2j)/2}}\in{\cal F}$ for all $(\alpha,j)\in I^{*}$.

Since $M\geq N+n+g$, it follows from (9) that

[TABLE]

where $s\mapsto R_{N}(G;a;u;s)$ is a holomorphic function in the half plane $\{\sigma=\Re(s)>-N\}$, in which it satisfies the bound $R_{N}(G;a;u;s)\ll_{G,N,u}1$.

Moreover it is easy to see that, for any $(\alpha,j)\in I^{*}$,

[TABLE]

Relation (10) and the induction hypothesis imply then that

[TABLE]

Since $|1-e^{2\pi i\,u.a}|=2|\sin(\pi u.a)|\geq d\left(u.a,\mathbb{Z}\right)$, then (11) implies that $F_{1}(G;a;s)$ satisfies conclusions of Lemma 2.6. This completes the induction and the proof for $i=1$.

2.2.2 Proof of Lemma 2.6 for $i=0$:

Let $N\in\mathbb{N}$ be a fixed integer. Let $G(X)=\frac{P(X)}{(X_{1}^{2}+\dots+X_{n}^{2}+1)^{r/2}}\in{\cal F}$ and $g=$ deg$(G)=d-r$ where $d\geq 0$ is the degree of the polynomial $P$. Set also $M:=\sup(N+g+n,0)\in\mathbb{N}_{0}$.

Since $P(k)\ll|k|^{d}$ for $k\in\mathbb{Z}^{n}\setminus\{\,0\,\}$, it follows that $F_{0}(G;a;s)$ and $F_{1}(G;a;s)$ converge absolutely in the half plane $\{\sigma=\Re(s)>n+g\}$.

Moreover, we have for $s=\sigma+i\tau\in\mathbb{C}$ verifying $\sigma>n+g$:

[TABLE]

In addition we have uniformly in $s=\sigma+i\tau\in\mathbb{C}$ verifying $\sigma>-N$,

[TABLE]

So (2.2.2) and Lemma 2.6 for $i=1$ imply that Lemma 2.6 is also true for $i=0$. This completes the proof of Lemma 2.6. ∎

2.2.3 Proof of item $(i.2)$ of Theorem 2.5:

Since $a\in\mathbb{R}^{n}\setminus\mathbb{Z}^{n}$, there exists $i_{0}\in\{1,\dots,n\}$ such that $a_{i_{0}}\not\in\mathbb{Z}$. In particular $d(a.e_{i_{0}},\mathbb{Z})=d(a_{i_{0}},\mathbb{Z})>0$. Therefore, $a$ satisfies the assumption of Lemma 2.6 with $u=e_{i_{0}}$. Thus, for all $N\in\mathbb{N}$, $s\mapsto f_{a}(s)=F_{0}(P;a;s)$ has a holomorphic continuation to the half-plane $\{\Re(s)>-N\}$. It follows, by letting $N\rightarrow\infty$, that $s\mapsto f_{a}(s)$ has a holomorphic continuation to the whole complex plane $\mathbb{C}$.

2.2.4 Proof of item $(iii)$ of Theorem 2.5:

Let $\Theta\in{\cal M}_{n}(\mathbb{R})$, $(\varepsilon_{i})_{i}\in\{-1,0,1\}^{q}$ and $b\in{\cal S}(\mathbb{Z}^{n}\times\mathbb{Z}^{n})$. We assume that $\Theta$ is a diophantine matrix. Set ${\cal Z}:=\{\,l=(l_{1},\dots,l_{q})\in(\mathbb{Z}^{n})^{q}\,:\,\sum_{i}\varepsilon_{i}l_{i}=0\,\}$ and $P\in\mathbb{C}[X_{1},\dots,X_{n}]$ of degree $d\geq 0$.

It is easy to see that for $\sigma>n+d$:

[TABLE]

So

[TABLE]

converges absolutely in the half plane $\{\Re(s)>n+d\}$.

Moreover with the notations of Lemma 2.6, we have for all $s=\sigma+i\tau\in\mathbb{C}$ verifying $\sigma>n+d$:

[TABLE]

But $\Theta$ is diophantine, so there exists $u\in\mathbb{Z}^{n}$ and $\delta,c>0$ such

[TABLE]

We deduce that $\forall l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z},$

[TABLE]

It follows that there exists $u\in\mathbb{Z}^{n}$, $\delta>0$ and $c>0$ such that

[TABLE]

Therefore, for any $l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}$, the vector $a=\Theta\,\sum_{i}\varepsilon_{i}l_{i}$ verifies the assumption of Lemma 2.6 with the same $u$. Moreover $\delta$ and $c$ in (14) are also independent on $l$.

We fix now $N\in\mathbb{N}$. Lemma 2.6 implies that there exist positive constants $C_{0}:=C_{0}(P,N,u)$, $B_{0}:=B_{i}(P,N,u)$ and $A_{0}:=A_{0}(P,N,u)$ such that for all $l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}$, $s\mapsto F_{0}(P;\Theta\,\sum_{i}\varepsilon_{i}l_{i};s)$ extends holomorphically to the half plane $\{\Re(s)>-N\}$ and verifies in it the bound

[TABLE]

This and (14) imply that for any compact set $K$ included in the half plane $\{\Re(s)>-N\}$, there exist two constants $C:=C(P,N,c,\delta,u,K)$ and $D:=D(P,N,c,\delta,u)$ (independent on $l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}$) such that

[TABLE]

It follows that $s\mapsto\sum_{l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}}b(l)F_{0}(P;\Theta\,{\sum}_{i}\varepsilon_{i}l_{i};s)$ has a holomorphic continuation to the half plane $\{\Re(s)>-N\}$.

This and ( 13) imply that $s\mapsto g_{0}(s)=\sum_{l\in(\mathbb{Z}^{n})^{q}\setminus{\cal Z}}b(l)f_{\Theta\,\sum_{i}\varepsilon_{i}l_{i}}(s)$ has a holomorphic continuation to $\{\Re(s)>-N\}$. Since $N$ is an arbitrary integer, by letting $N\rightarrow\infty$, it follows that $s\mapsto g_{0}(s)$ has a holomorphic continuation to the whole complex plane $\mathbb{C}$ which completes the proof of the theorem. ∎

Remark 2.8**.**

*By equation (11), we see that a Diophantine condition is sufficient to get Lemma 2.6. Our Diophantine condition appears also (in equivalent form) in Connes [7, Prop. 49] (see Remark 4.2 below). The following heuristic argument shows that our condition seems to be necessary in order to get the result of Theorem 2.5:

For simplicity we assume $n=1$ (but the argument extends easily to any $n$).

Let $\theta\in\mathbb{R}\setminus\mathbb{Q}$. We know (see this reflection formula in [15, p. 6]) that for any $l\in\mathbb{Z}\setminus\{0\}$,*

[TABLE]

So, for any $(a_{l})\in{\cal S}(\mathbb{Z})$, the existence of meromorphic continuation of $g_{0}(s):=\sum_{l\in\mathbb{Z}}^{\prime}a_{l}\,g_{\theta l}(s)$ is equivalent to the existence of meromorphic continuation of

[TABLE]

So, for at least one $\sigma_{0}\in\mathbb{R}$, we must have $\tfrac{|a_{l}|}{|\theta l+k|^{\sigma_{0}}}=O(1){\mbox{ uniformly in }}k,l\in\mathbb{Z}^{*}.$

It follows that for any $(a_{l})\in{\cal S}(\mathbb{Z})$, $|\theta l+k|\gg|a_{l}|^{1/\sigma_{0}}$ uniformly in $k,l\in\mathbb{Z}^{*}$. Therefore, our Diophantine condition seems to be necessary.

2.2.5 Commutation between sum and residue

Let $p\in\mathbb{N}$. Recall that $\mathcal{S}((\mathbb{Z}^{n})^{p})$ is the set of the Schwartz sequences on $(\mathbb{Z}^{n})^{p}$. In other words, $b\in\mathcal{S}((\mathbb{Z}^{n})^{p})$ if and only if for all $r\in\mathbb{N}_{0}$, $(1+|l_{1}|^{2}+\cdots|l_{p}|^{2})^{r}\,|b(l_{1},\cdots,l_{p})|^{2}$ is bounded on $(\mathbb{Z}^{n})^{p}$. We note that if $Q\in\mathbb{R}[X_{1},\cdots,X_{np}]$ is a polynomial, $(a_{j})\in\mathcal{S}(\mathbb{Z}^{n})^{p}$, $b\in\mathcal{S}(\mathbb{Z}^{n})$ and $\phi$ a real-valued function, then $l:=(l_{1},\cdots,l_{p})\mapsto\widetilde{a}(l)\,b(-\widehat{l}_{p})\,Q(l)\,e^{i\phi(l)}$ is a Schwartz sequence on $(\mathbb{Z}^{n})^{p}$, where

[TABLE]

In the following, we will use several times the fact that for any $(k,l)\in(\mathbb{Z}^{n})^{2}$ such that $k\neq 0$ and $k\neq-l$, we have

[TABLE]

Lemma 2.9**.**

There exists a polynomial $P\in\mathbb{R}[X_{1},\cdots,X_{p}]$ of degree $4p$ and with positive coefficients such that for any $k\in\mathbb{Z}^{n}$, and $l:=(l_{1},\cdots,l_{p})\in(\mathbb{Z}^{n})^{p}$ such that $k\neq 0$ and $k\neq-\widehat{l}_{i}$ for all $1\leq i\leq p$, the following holds:

[TABLE]

Proof.

Let’s fix $i$ such that $1\leq i\leq p$. Using two times (16), Cauchy–Schwarz inequality and the fact that $|k+\widehat{l}_{i}|^{2}\geq 1$, we get

[TABLE]

Since $|k|\geq 1$, and $|\widehat{l}_{i}|^{j}\leq|\widehat{l}_{i}|^{4}$ if $1\leq j\leq 4$, we find

[TABLE]

Taking P(X_{1},\cdots,X_{p}):=5^{p}\big{(}1+4({\sum}_{j=1}^{p}X_{j})^{4}\big{)}^{p} now gives the result. ∎

Lemma 2.10**.**

Let $b\in\mathcal{S}((\mathbb{Z}^{n})^{p})$, $p\in\mathbb{N}$, $P_{j}\in\mathbb{R}[X_{1},\cdots,X_{n}]$ be a homogeneous polynomial function of degree $j$, $k\in\mathbb{Z}^{n}$, $l:=(l_{1},\cdots,l_{p})\in(\mathbb{Z}^{n})^{p}$, $r\in\mathbb{N}_{0}$, $\phi$ be a real-valued function on $\mathbb{Z}^{n}\times(\mathbb{Z}^{n})^{p}$ and

[TABLE]

with $h(s,k,l):=0$ if, for $k\neq 0$, one of the denominators is zero.

For all $s\in\mathbb{C}$ such that $\Re(s)>n+j-r-2p$, the series

[TABLE]

is absolutely summable. In particular,

[TABLE]

Proof.

Let $s=\sigma+i\tau\in\mathbb{C}$ such that $\sigma>n+j-r-2p$. By Lemma 2.9 we get, for $k\neq 0$,

[TABLE]

where $P(l):=P(|l_{1}|,\cdots,|l_{p}|)$ and $P$ is a polynomial of degree $4p$ with positive coefficients. Thus, $|h(s,k,l)|\leq F(l)\,G(k)$ where $F(l):=|b(l)|\,P(l)$ and $G(k):=|P_{j}(k)||k|^{-r-\sigma-2p}$. The summability of $\sum_{l\in(\mathbb{Z}^{n})^{p}}F(l)$ is implied by the fact that $b\in\mathcal{S}((\mathbb{Z}^{n})^{p})$. The summability of ${\sum}^{\prime}_{k\in\mathbb{Z}^{n}}G(k)$ is a consequence of the fact that $\sigma>n+j-r-2p$. Finally, as a product of two summable series, ${\sum_{k,l}}F(l)G(k)$ is a summable series, which proves that ${\sum_{k,l}}h(s,k,l)$ is also absolutely summable. ∎

Definition 2.11**.**

Let $f$ be a function on $D\times(\mathbb{Z}^{n})^{p}$ where $D$ is an open neighborhood of [math] in $\mathbb{C}$.

We say that $f$ satisfies (H1) if and only if there exists $\rho>0$ such that

(i) for any $l$, $s\mapsto f(s,l)$ extends as a holomorphic function on $U_{\rho}$, where $U_{\rho}$ is the open disk of center 0 and radius $\rho$,

*(ii) the series $\sum_{l\in(\mathbb{Z}^{n})^{p}}\left\lVert H(\cdot,l)\right\rVert_{\infty,\rho}$ is summable,where $\left\lVert H(\cdot,l)\right\rVert_{\infty,\rho}:=\sup_{s\in U_{\rho}}|H(s,l)|$.

We say that $f$ satisfies (H2) if and only if there exists $\rho>0$ such that*

(i) for any $l$, $s\mapsto f(s,l)$ extends as a holomorphic function on $U_{\rho}-\{0\}$,

(ii) for any $\delta$ such that $0<\delta<\rho$, the series $\sum_{l\in(\mathbb{Z}^{n})^{p}}\left\lVert H(\cdot,l)\right\rVert_{\infty,\delta,\rho}$ is summable, where $\left\lVert H(\cdot,l)\right\rVert_{\infty,\delta,\rho}:=\sup_{\delta<|s|<\rho}|H(s,l)|$.

Remark 2.12**.**

Note that (H1) implies (H2). Moreover, if $f$ satisfies (H1) (resp. (H2) for $\rho>0$, then it is straightforward to check that $f:s\mapsto\sum_{l\in(\mathbb{Z}^{n})^{p}}f(s,l)$ extends as an holomorphic function on $U_{\rho}$ (resp. on $U_{\rho}\setminus\{\,0\,\}$).

Corollary 2.13**.**

With the same notations of Lemma 2.10, suppose that $r+2p-j>n$, then, the function $H(s,l):={\sum}^{\prime}_{k\in\mathbb{Z}^{n}}h(s,k,l)$ satisfies (H1).

Proof.

$(i)$ Let’s fix $\rho>0$ such that $\rho<r+2p-j-n$. Since $r+2p-j>n$, $U_{\rho}$ is inside the half-plane of absolute convergence of the series defined by $H(s,l)$. Thus, $s\mapsto H(s,l)$ is holomorphic on $U_{\rho}$.

$(ii)$ Since \big{|}|k|^{-s}\big{|}\leq|k|^{\rho} for all $s\in U_{\rho}$ and $k\in\mathbb{Z}^{n}\setminus\{\,0\,\}$, we get as in the above proof

[TABLE]

Since $\rho<r+2p-j-n$, the series ${\sum}^{\prime}_{k\in\mathbb{Z}^{n}}|P_{j}(k)||k|^{-r+\rho-2p}$ is summable.

Thus, $\left\lVert H(\cdot,l)\right\rVert_{\infty,\rho}\leq K\,F(l)$ where $K:={{\sum_{k}}}^{\prime}|P_{j}(k)||k|^{-r+\rho-2p}<\infty$. We have already seen that the series $\sum_{l}F(l)$ is summable, so we get the result. ∎

We note that if $f$ and $g$ both satisfy (H1) (or (H2)), then so does $f+g$. In the following, we will use the equivalence relation

[TABLE]

Lemma 2.14**.**

Let $f$ and $g$ be two functions on $D\times(\mathbb{Z}^{n})^{p}$ where $D$ is an open neighborhood of [math] in $\mathbb{C}$, such that $f\sim g$ and such that $g$ satisfies (H2). Then

[TABLE]

Proof.

Since $f\sim g$, $f$ satisfies (H2) for a certain $\rho>0$. Let’s fix $\eta$ such that $0<\eta<\rho$ and define $C_{\eta}$ as the circle of center 0 and radius $\eta$. We have

[TABLE]

where $I=[0,2\pi]$ and $u(t,l):=\tfrac{1}{2\pi}\eta e^{it}f(\eta\,e^{it},l)$. The fact that $f$ satisfies (H2) entails that the series $\sum_{l\in(\mathbb{Z}^{n})^{p}}\left\lVert f(\cdot,l)\right\rVert_{\infty,C_{\eta}}$ is summable. Thus, since $\left\lVert u(\cdot,l)\right\rVert_{\infty}=\tfrac{1}{2\pi}\eta\left\lVert f(\cdot,l)\right\rVert_{\infty,C_{\eta}}$, the series $\sum_{l\in(\mathbb{Z}^{n})^{p}}\left\lVert u(\cdot,l)\right\rVert_{\infty}$ is summable, so, as a consequence, $\int_{I}\sum_{l\in(\mathbb{Z}^{n})^{p}}u(t,l)dt=\sum_{l\in(\mathbb{Z}^{n})^{p}}\int_{I}u(t,l)dt$ which gives the result. ∎

2.3 Computation of residues of zeta functions

Since, we will have to compute residues of series, let us introduce the following

Definition 2.15**.**

[TABLE]

where $\zeta(s)$ is the Riemann zeta function (see [25] or [14]).

By the symmetry $k\rightarrow-k$, it is clear that these functions $\zeta_{p_{1},\dots,p_{n}}$ all vanish for odd values of $p_{i}$.

Let us now compute $\zeta_{0,\cdots,0,1_{i},0\cdots,0,1_{j},0\cdots,0}(s)$ in terms of $Z_{n}(s)$:

Since $\zeta_{0,\cdots,0,1_{i},0\cdots,0,1_{j},0\cdots,0}(s)=A_{i}(s)\,\delta_{ij}$, exchanging the components $k_{i}$ and $k_{j}$, we get

[TABLE]

Similarly,

[TABLE]

but it is difficult to write explicitly $\zeta_{p_{1},\dots,p_{n}}(s)$ in terms of $Z_{n}(s-4)$ and other $Z_{n}(s-m)$ when at least four indices $p_{i}$ are non zero.

When all $p_{i}$ are even, $\zeta_{p_{1},\dots,p_{n}}(s)$ is a nonzero series of fractions $\tfrac{P(k)}{|k|^{s}}$ where $P$ is a homogeneous polynomial of degree $p_{1}+\cdots+p_{n}$. Theorem 2.1 now gives us the following

Proposition 2.16**.**

$\zeta_{p_{1},\dots,p_{n}}$* has a meromorphic extension to the whole plane with a unique pole at $n+p_{1}+\cdots+p_{n}$. This pole is simple and the residue at this pole is*

[TABLE]

*when all $p_{i}$ are even or this residue is zero otherwise.

In particular, for $n=2$,*

[TABLE]

and for $n=4$,

[TABLE]

Proof.

Equation (17) follows from Theorem (2.1)

[TABLE]

and standard formulae (see for instance [32, VIII,1;22]). Equation (18) is a straightforward consequence of Equation (17). Equation (19) can be checked for the cases $i=j\neq l=m$ and $i=j=l=m$. ∎

Note that $Z_{n}(s)$ is an Epstein zeta function associated to the quadratic form $q(x):=x_{1}^{2}+...+x_{n}^{2}$, so $Z_{n}$ satisfies the following functional equation

[TABLE]

Since $\pi^{s-n/2}\Gamma(n/2-s/2)\,\Gamma(s/2)^{-1}=0$ for any negative even integer $n$ and $Z_{n}(s)$ is meromorphic on $\mathbb{C}$ with only one pole at $s=n$ with residue $2\pi^{n/2}\Gamma(n/2)^{-1}$ according to previous proposition, so we get $Z_{n}(0)=-1$. We have proved that

[TABLE]

2.4 Meromorphic continuation of a class of zeta functions

Let $n,q\in\mathbb{N}$, $q\geq 2$, and $p=(p_{1},\dots,p_{q-1})\in\mathbb{N}_{0}^{q-1}$.

Set $I:=\{i\mid p_{i}\neq 0\}$ and assume that $I\neq\emptyset$ and

[TABLE]

We will use in the sequel also the following notations:

• for $x=(x_{1},\dots,x_{t})\in\mathbb{R}^{t}$ recall that $|x|_{1}=|x_{1}|+\dots+|x_{t}|$ and $|x|=\sqrt{x_{1}^{2}+\dots+x_{t}^{2}}$;

• for all $\alpha=(\alpha_{i})_{i\in I}\in{\cal I}=\prod_{i\in I}\mathbb{N}_{0}^{p_{i}}$,

[TABLE]

2.4.1 A family of polynomials

In this paragraph we define a family of polynomials which plays an important role later.

Consider first the variables:

• for $X_{1},\dots,X_{n}$ we set $X=(X_{1},\dots,X_{n})$;

• for any $i=1,\dots,2q$, we consider the variables $Y_{i,1},\dots,Y_{i,n}$ and set $Y_{i}:=(Y_{i,1},\dots,Y_{i,n})$ and $Y:=(Y_{1},\dots,Y_{2q})$;

• for $Y=(Y_{1},\dots,Y_{2q})$, we set for any $1\leq j\leq q$, $\widetilde{Y}_{j}:=Y_{1}+\cdots+Y_{j}+Y_{q+1}+\cdots+Y_{q+j}$.

We define for all $\alpha=(\alpha_{i})_{i\in I}\in{\cal I}=\prod_{i\in I}\mathbb{N}_{0}^{p_{i}}$ the polynomial

[TABLE]

It is clear that $P_{\alpha}(X,Y)\in\mathbb{Z}[X,Y]$, deg${}_{X}P_{\alpha}\leq|\alpha|_{1}$ and deg${}_{Y}P_{\alpha}\leq 2|\alpha|_{1}$.

Let us fix a polynomial $Q\in\mathbb{R}[X_{1},\cdots,X_{n}]$ and note $d:=\deg Q$. For $\alpha\in{\cal I}$, we want to expand $P_{\alpha}(X,Y)\,Q(X)$ in homogeneous polynomials in $X$ and $Y$ so defining

[TABLE]

where $d_{\beta}:=\sum_{1}^{n}\beta_{i}$, we set

[TABLE]

where $c_{\alpha,\beta}\in\mathbb{R}$, $X^{\beta}:=X_{1}^{\beta_{1}}\cdots X_{n}^{\beta_{n}}$ and $Y^{\beta}:=Y_{1,1}^{\beta_{n+1}}\cdots Y_{2q,n}^{\beta_{(2q+1)n}}$. By definition, $X^{\beta}$ is a homogeneous polynomial of degree in $X$ equals to $d_{\beta}$. We note

[TABLE]

2.4.2 Residues of a class of zeta functions

In this section we will prove the following result, used in Proposition 5.4 for the computation of the spectrum dimension of the noncommutative torus:

Theorem 2.17**.**

(i) Let $\tfrac{1}{2\pi}\Theta$ be a diophantine matrix, and \widetilde{a}\in\mathcal{S}\big{(}(\mathbb{Z}^{n})^{2q}\big{)}. Then

[TABLE]

has a meromorphic continuation to the whole complex plane $\mathbb{C}$ with at most simple possible poles at the points $s=n+d+|p|_{1}-m$ where $m\in\mathbb{N}_{0}$.

(ii) Let $m\in\mathbb{N}_{0}$ and set $I(m):=\{\,(\alpha,\beta)\in\mathcal{I}\times\mathbb{N}_{0}^{(2q+1)n}\,:\,\beta\in L(\alpha)\text{ and }m=2|\alpha|_{1}-d_{\beta}+d\,\}$. Then $I(m)$ is a finite set and $s=n+d+|p|_{1}-m$ is a pole of $f$ if and only if

[TABLE]

with $Z:=\{l\,:\,\sum_{1}^{q}l_{j}=0\}$ and the convention $\sum_{\emptyset}=0$. In that case $s=n+d+|p|_{1}-m$ is a simple pole of residue $\underset{s=n+d+|p|_{1}-m}{\operatorname{Res}}\,f(s)=C(f,m)$.

In order to prove the theorem above we need the following

Lemma 2.18**.**

For all $N\in\mathbb{N}$ we have

[TABLE]

uniformly in $k\in\mathbb{Z}^{n}$ and $l\in(\mathbb{Z}^{n})^{2q}$ verifying $|k|>U(l):=36\,(\sum_{i=1,\,i\neq q}^{2q-1}|l_{i}|)^{4}$.

Proof.

For $i=1,\dots,q-1$, we have uniformly in $k\in\mathbb{Z}^{n}$ and $l\in(\mathbb{Z}^{n})^{2q}$ verifying $|k|>U(l)$,

[TABLE]

In that case,

[TABLE]

where for all $i=1,\dots,q-1$ and for all $u\in\mathbb{N}_{0}$,

[TABLE]

with the convention $P^{i}_{0}(k,l):=1$.

In particular $P^{i}_{u}(k,l)\in\mathbb{Z}[k,l]$, $\deg_{k}P^{i}_{u}\leq u$ and $\deg_{l}P^{i}_{u}\leq 2u$. Inequality (23) implies that for all $i=1,\dots,q-1$ and for all $u\in\mathbb{N}$,

[TABLE]

uniformly in $k\in\mathbb{Z}^{n}$ and $l\in(\mathbb{Z}^{n})^{2q}$ verifying $|k|>U(l)$.

Let $N\in\mathbb{N}$. We deduce from the previous that for any $k\in\mathbb{Z}^{n}$ and $l\in(\mathbb{Z}^{n})^{2q}$ verifying $|k|>U(l)$ and for all $i=1,\dots,q-1$, we have

[TABLE]

It follows that for any $N\in\mathbb{N}$, we have uniformly in $k\in\mathbb{Z}^{n}$ and $l\in(\mathbb{Z}^{n})^{2q}$ verifying $|k|>U(l)$ and for all $i\in I$,

[TABLE]

where $P^{i}_{\alpha_{i}}(k,l)=\prod_{j=1}^{p_{i}}P^{i}_{\alpha_{i,j}}(k,l)$ for all $\alpha_{i}=(\alpha_{i,1},\dots,\alpha_{i,p_{i}})\in\{0,\dots,N\}^{p_{i}}$ and

[TABLE]

where $P_{\alpha}(k,l)=\prod_{i\in I}P^{i}_{\alpha_{i}}(k,l)=\prod_{i\in I}\prod_{j=1}^{p_{i}}P^{i}_{\alpha_{i,j}}(k,l)$. ∎

Proof of Theorem 2.17.

$(i)$ All $n$, $q$, $p=(p_{1},\dots,p_{q-1})$ and $\widetilde{a}\in{\cal S}\left((\mathbb{Z}^{n})^{2q}\right)$ are fixed as above and we define formally for any $l\in(\mathbb{Z}^{n})^{2q}$

[TABLE]

Thus, still formally,

[TABLE]

It is clear that $F(l,s)$ converges absolutely in the half plane $\{\sigma=\Re(s)>n+d+|p|_{1}\}$ where $d=\deg Q$.

Let $N\in\mathbb{N}$. Lemma 2.18 implies that for any $l\in(\mathbb{Z}^{n})^{2q}$ and for $s\in\mathbb{C}$ such that $\sigma>n+|p|_{1}+d$,

[TABLE]

where $s\mapsto G_{N}(l,s)$ is a holomorphic function in the half-plane $D_{N}:=\{\sigma>n+d+|p|_{1}-\tfrac{N+1}{2}\}$ and verifies in it the bound $G_{N}(l,s)\ll_{N,\sigma}1$ uniformly in $l$.

It follows that

[TABLE]

where

[TABLE]

In particular there exists $A(N)>0$ such that $s\mapsto R_{N}(l,s)$ extends holomorphically to the half-plane $D_{N}$ and verifies in it the bound $R_{N}(l,s)\ll_{N,\sigma}1+|l|^{A(N)}$ uniformly in $l$.

Let us note formally

[TABLE]

Equation (26) and $R_{N}(l,s)\ll_{N,\sigma}1+|l|^{A(N)}$ imply that

[TABLE]

where $\sim_{N}$ means modulo a holomorphic function in $D_{N}$.

Recall the decomposition $\genfrac{(}{)}{0.0pt}{1}{1/2}{\alpha}\,P_{\alpha}(k,l)\,Q(k)=\sum_{\beta\in L(\alpha)}M_{\alpha,\beta}(l)\,k^{\beta}$ and we decompose similarly $h_{\alpha}(s)=\sum_{\beta\in L(\alpha)}h_{\alpha,\beta}(s).$ Theorem 2.5 now implies that for all $\alpha=(\alpha_{i})_{i\in I}\in\prod_{i\in I}\{0,\dots,N\}^{p_{i}}$ and $\beta\in L(\alpha)$,

• the map $s\mapsto h_{\alpha,\beta}(s)$ has a meromorphic continuation to the whole complex plane $\mathbb{C}$ with only one simple possible pole at $s=n+|p|_{1}-2|\alpha|_{1}+d_{\beta}$,

• the residue at this point is equal to

[TABLE]

where $\mathcal{Z}:=\{l\in(\mathbb{Z})^{n})^{2q}\,:\,\sum_{1}^{q}l_{j}=0\}$. If the right hand side is zero, $h_{\alpha,\beta}(s)$ is holomorphic on $\mathbb{C}$.

By (27), we deduce therefore that $f(s)$ has a meromorphic continuation on the halfplane $D_{N}$, with only simple possible poles in the set $\{\,n+|p|_{1}+k\,:\,-2N|p|_{1}\leq k\leq d\,\}.$ Taking now $N\to\infty$ yields the result.

$(ii)$ Let $m\in\mathbb{N}_{0}$ and set $I(m):=\{\,(\alpha,\beta)\in\mathcal{I}\times\mathbb{N}_{0}^{(2q+1)n}\,:\,\beta\in L(\alpha)\text{ and }m=2|\alpha|_{1}-d_{\beta}+d\,\}$. If $(\alpha,\beta)\in I(m)$, then $|\alpha|_{1}\leq m$ and $|\beta|_{1}\leq 3m+d$, so $I(m)$ is finite.

With a chosen $N$ such that $2N|p|_{1}+d>m$, we get by (27) and (28)

[TABLE]

with the convention $\sum_{\emptyset}=0$. Thus, $n+d+|p|_{1}-m$ is a pole of $f$ if and only if $C(f,m)\neq 0$. ∎

3 Noncommutative integration on a simple spectral triple

In this section, we revisit the notion of noncommutative integral pioneered by Alain Connes, paying particular attention to the reality (Tomita–Takesaki) operator $J$ and to kernels of perturbed Dirac operators by symmetrized one-forms.

3.1 Kernel dimension

We will have to compare here the kernels of $\mathcal{D}$ and $\mathcal{D}_{A}$ which are both finite dimensional:

Lemma 3.1**.**

Let $(\mathcal{A},\mathcal{H},\mathcal{D})$ be a spectral triple with a reality operator $J$ and chirality $\chi$. If $A\in\Omega^{1}_{\mathcal{D}}$ is a one-form, the fluctuated Dirac operator

[TABLE]

(where $\mathcal{D}J=\epsilon\,J\mathcal{D}$, $\epsilon=\pm 1$) is an operator with compact resolvent, and in particular its kernel $\operatorname{Ker}\mathcal{D}_{A}$ is a finite dimensional space. This space is invariant by $J$ and $\chi$.

Proof.

Let $T$ be a bounded operator and let $z$ be in the resolvent of $\mathcal{D}+T$ and $z^{\prime}$ be in the resolvent of $\mathcal{D}$. Then

[TABLE]

Since $(\mathcal{D}-z^{\prime})^{-1}$ is compact by hypothesis and since the term in bracket is bounded, $\mathcal{D}+T$ has a compact resolvent. Applying this to $T=A+\epsilon JAJ^{-1}$, $\mathcal{D}_{A}$ has a finite dimensional kernel (see for instance [27, Theorem 6.29]).

Since according to the dimension, $J^{2}=\pm 1$, $J$ commutes or anticommutes with $\chi$, $\chi$ commutes with the elements in the algebra $\mathcal{A}$ and $\mathcal{D}\chi=-\chi\mathcal{D}$ (see [10] or [23, p. 405]), we get $\mathcal{D}_{A}\chi=-\chi\mathcal{D}_{A}$ and $\mathcal{D}_{A}J=\pm J\mathcal{D}_{A}$ which gives the result. ∎

3.2 Pseudodifferential operators

Let $(A,\mathcal{D},\mathcal{H})$ be a given real regular spectral triple of dimension $n$.

We note

[TABLE]

$P_{0}$ and $P_{A}$ are thus finite-rank selfadjoint bounded operators. We remark that $D$ and $D_{A}$ are selfadjoint invertible operators with compact inverses.

Remark 3.2**.**

Since we only need to compute the residues and the value at 0 of the $\zeta_{D}$, $\zeta_{D_{A}}$ functions, it is not necessary to define the operators $\mathcal{D}^{-1}$ or $\mathcal{D}_{A}^{-1}$ and the associated zeta functions. However, we can remark that all the work presented here could be done using the process of Higson in [26] which proves that we can add any smoothing operator to $\mathcal{D}$ or $\mathcal{D}_{A}$ such that the result is invertible without changing anything to the computation of residues.

Define for any $\alpha\in\mathbb{R}$

[TABLE]

where $F_{t}(T):=e^{it|D|}\,T\,e^{-it|D|}=e^{it|\mathcal{D}|}\,T\,e^{-it|\mathcal{D}|}$ since $|D|=|\mathcal{D}|+P_{0}$. Define

[TABLE]

It has been shown in [13] that $OP^{0}=\bigcap_{p\geq 0}\operatorname{Dom}(\delta^{p})$. In particular, $OP^{0}$ is a subalgebra of $\mathcal{B}(\mathcal{H})$ (while elements of $OP^{\alpha}$ are not necessarily bounded for $\alpha>0$) and $\mathcal{A}\subseteq OP^{0}$, $J\mathcal{A}J^{-1}\subseteq OP^{0}$, $[\mathcal{D},\mathcal{A}]\subseteq OP^{0}$. Note that $P_{0}\in OP^{-\infty}$ and $\delta(OP^{0})\subseteq OP^{0}$.

For any $t>0$, $\mathcal{D}^{t}$ and and $|\mathcal{D}|^{t}$ are in $OP^{t}$ and for any $\alpha\in\mathbb{R},D^{\alpha}$ and $|D|^{\alpha}$ are in $OP^{\alpha}$. By hypothesis, $|D|^{-n}\in\mathcal{L}^{(1,\infty)}(\mathcal{H})$ so for any $\alpha>n$, $OP^{-\alpha}\subseteq\mathcal{L}^{1}(\mathcal{H})$.

Lemma 3.3**.**

[13]**

(i) For any $T\in OP^{0}$ and $s\in\mathbb{C}$, $\sigma_{s}(T)\in OP^{0}$.

(ii) For any $\alpha,\beta\in\mathbb{R}$, $OP^{\alpha}OP^{\beta}\subseteq OP^{\alpha+\beta}$.

(iii) If $\alpha\leq\beta$, $OP^{\alpha}\subseteq OP^{\beta}$.

(iv) For any $\alpha$, $\delta(OP^{\alpha})\subseteq OP^{\alpha}$.

(v) For any $\alpha$ and $T\in OP^{\alpha}$, $\nabla(T)\in OP^{\alpha+1}$.

Proof.

See the appendix. ∎

Remark 3.4**.**

Any operator in $OP^{\alpha}$, where $\alpha\in\mathbb{R}$, extends as a continuous linear operator from $\operatorname{Dom}|D|^{\alpha+1}$ to $\operatorname{Dom}|D|$ where the $\operatorname{Dom}|D|^{\alpha}$ spaces have their natural norms (see [13, 26]).

We now introduce a definition of pseudodifferential operators in a slightly different way than in [13, 9, 26] which in particular pays attention to the reality operator $J$ and the kernel of $\mathcal{D}$ and allows $\mathcal{D}$ and $|D|^{-1}$ to be a pseudodifferential operators. It is more in the spirit of [4].

Definition 3.5**.**

Let us define $\mathcal{D}(\mathcal{A})$ as the polynomial algebra generated by $\mathcal{A}$, $J\mathcal{A}J^{-1}$, $\mathcal{D}$ and $|\mathcal{D}|$.

A pseudodifferential operator is an operator $T$ such that there exists $d\in\mathbb{Z}$ such that for any $N\in\mathbb{N}$, there exist $p\in\mathbb{N}_{0}$, $P\in\mathcal{D}(\mathcal{A})$ and $R\in OP^{-N}$ ($p$, $P$ and $R$ may depend on $N$) such that $P\,D^{-2p}\in OP^{d}$ and

[TABLE]

Define $\Psi(\mathcal{A})$ as the set of pseudodifferential operators and $\Psi(\mathcal{A})^{k}:=\Psi(\mathcal{A})\cap OP^{k}$.

Note that if $A$ is a 1-form, $A$ and $JAJ^{-1}$ are in $\mathcal{D}(\mathcal{A})$ and moreover $\mathcal{D}(\mathcal{A})\subseteq\cup_{p\in\mathbb{N}_{0}}OP^{p}$. Since $|\mathcal{D}|\in\mathcal{D}(\mathcal{A})$ by construction and $P_{0}$ is a pseudodifferential operator, for any $p\in\mathbb{Z}$, $|D|^{p}$ is a pseudodifferential operator (in $OP^{p}$.) Let us remark also that $\mathcal{D}(\mathcal{A})\subseteq\Psi(\mathcal{A})\subseteq\cup_{k\in\mathbb{Z}}OP^{k}$.

Lemma 3.6**.**

[13, 9*]**

The set of all pseudodifferential operators $\Psi(\mathcal{A})$ is an algebra. Moreover, if $T\in\Psi(\mathcal{A})^{d}$ and $T\in\Psi(\mathcal{A})^{d^{\prime}}$, then $TT^{\prime}\in\Psi(\mathcal{A})^{d+d^{\prime}}$.*

Proof.

See the appendix. ∎

Due to the little difference of behavior between scalar and nonscalar pseudodifferential operators (i.e. when coefficients like $[\mathcal{D},a]$, $a\in\mathcal{A}$ appears in $P$ of Definition 3.5), it is convenient to also introduce

Definition 3.7**.**

Let $\mathcal{D}_{1}(\mathcal{A})$ be the algebra generated by $\mathcal{A}$, $J\mathcal{A}J^{-1}$ and $\mathcal{D}$, and $\Psi_{1}(\mathcal{A})$ be the set of pseudodifferential operators constructed as before with $\mathcal{D}_{1}(\mathcal{A})$ instead of $\mathcal{D}(\mathcal{A})$. Note that $\Psi_{1}(\mathcal{A})$ is subalgebra of $\Psi(\mathcal{A})$.

Remark that $\Psi_{1}(\mathcal{A})$ does not necessarily contain operators such as $|D|^{k}$ where $k\in\mathbb{Z}$ is odd. This algebra is similar to the one defined in [4].

3.3 Zeta functions and dimension spectrum

For any operator $B$ and if $X$ is either $D$ or $D_{A}$, we define

[TABLE]

The dimension spectrum $Sd(\mathcal{A},\mathcal{H},\mathcal{D})$ of a spectral triple has been defined in [9, 13]. It is extended here to pay attention to the operator $J$ and to our definition of pseudodifferential operator.

Definition 3.8**.**

The spectrum dimension of the spectral triple is the subset $Sd(\mathcal{A},\mathcal{H},\mathcal{D})$ of all poles of the functions \zeta_{D}^{P}:=s\mapsto\operatorname{Tr}\big{(}P|D|^{-s}\big{)} where $P$ is any pseudodifferential operator in $OP^{0}$. The spectral triple $(\mathcal{A},\mathcal{H},\mathcal{D})$ is simple when these poles are all simple.

Remark 3.9**.**

If $Sp(\mathcal{A},\mathcal{H},\mathcal{D})$ denotes the set of all poles of the functions s\mapsto\operatorname{Tr}\big{(}P|D|^{-s}\big{)} where $P$ is any pseudodifferential operator, then, $Sd(\mathcal{A},\mathcal{H},\mathcal{D})\subseteq Sp(\mathcal{A},\mathcal{H},\mathcal{D})$.

When $Sp(\mathcal{A},\mathcal{H},\mathcal{D})=\mathbb{Z}$, $Sd(\mathcal{A},\mathcal{H},\mathcal{D})=\{\,n-k\ :\ k\in\mathbb{N}_{0}\,\}$: indeed, if $P$ is a pseudodifferential operator in $OP^{0}$, and $q\in\mathbb{N}$ is such that $q>n$, $P|D|^{-s}$ is in $OP^{-\Re(s)}$ so is trace-class for $s$ in a neighborhood of $q$; as a consequence, $q$ cannot be a pole of s\mapsto\operatorname{Tr}\big{(}P|D|^{-s}\big{)}.

Remark 3.10**.**

$Sp(\mathcal{A},\mathcal{H},\mathcal{D})$* is also the set of poles of functions s\mapsto\operatorname{Tr}\big{(}B|D|^{-s-2p}\big{)} where $p\in\mathbb{N}_{0}$ and $B\in\mathcal{D}(\mathcal{A})$.*

3.4 The noncommutative integral \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits

We already defined the one parameter group $\sigma_{z}(T):=|D|^{z}T|D|^{-z},\,z\in\mathbb{C}$.

Introducing the notation (recall that $\nabla(T)=[\mathcal{D}^{2},T]$) for an operator $T$,

[TABLE]

we get from [4, (2.44)] the following expansion for $T\in OP^{q}$

[TABLE]

where $g(z,r):=\tfrac{1}{r!}(\tfrac{z}{2})\cdots(\tfrac{z}{2}-(r-1))=\genfrac{(}{)}{0.0pt}{1}{z/2}{r}$ with the convention $g(z,0):=1$.

We define the noncommutative integral by

[TABLE]

Proposition 3.11**.**

[13]** If the spectral triple is simple, \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits is a trace on $\Psi(\mathcal{A})$.

Proof.

See the appendix. ∎

4 Residues of $\zeta_{D_{A}}$ for a spectral triple with

simple dimension spectrum

We fix a regular spectral triple $(\mathcal{A},\mathcal{H},\mathcal{D})$ of dimension $n$ and a self-adjoint 1-form $A$.

Recall that

[TABLE]

where $P_{A}$ is the projection on $\operatorname{Ker}\mathcal{D}_{A}$. Remark that $\widetilde{A}\in\mathcal{D}(\mathcal{A})\cap OP^{0}$ and $\mathcal{D}_{A}\in\mathcal{D}(\mathcal{A})\cap OP^{1}$.

We note

[TABLE]

As the following lemma shows, $V_{A}$ is a smoothing operator:

Lemma 4.1**.**

(i) $\bigcap_{k\geq 1}\operatorname{Dom}(\mathcal{D}_{A})^{k}\subseteq\bigcap_{k\geq 1}\operatorname{Dom}|D|^{k}$.

(ii) $\operatorname{Ker}\mathcal{D}_{A}\subseteq\bigcap_{k\geq 1}\operatorname{Dom}|D|^{k}$.

(iii) For any $\alpha,\beta\in\mathbb{R}$, $|D|^{\beta}P_{A}|D|^{\alpha}$ is bounded.

(iv) $P_{A}\in OP^{-\infty}$.

Proof.

$(i)$ Let us define for any $p\in\mathbb{N}$, $R_{p}:=(\mathcal{D}_{A})^{p}-\mathcal{D}^{p}$, so $R_{p}\in OP^{p-1}$ and R_{p}\big{(}\operatorname{Dom}|D|^{p}\big{)}\subseteq\operatorname{Dom}|D| (see Remark 3.4).

Let us fix $k\in\mathbb{N}$, $k\geq 2$. Since $\operatorname{Dom}\mathcal{D}_{A}=\operatorname{Dom}\mathcal{D}=\operatorname{Dom}|D|$, we have

[TABLE]

Let $\phi\in\operatorname{Dom}(\mathcal{D}_{A})^{k}$. We prove by recurrence that for any $j\in\{\,1,\cdots,k-1\,\}$, $\phi\in\operatorname{Dom}|D|^{j+1}$:

We have $\phi\in\operatorname{Dom}|D|$ and $(\mathcal{D}+R_{1})\,\phi\in\operatorname{Dom}|D|$. Thus, since $R_{1}\,\phi\in\operatorname{Dom}|D|$, $\mathcal{D}\phi\in\operatorname{Dom}|D|$, which proves that $\phi\in\operatorname{Dom}|D|^{2}$. Hence, case $j=1$ is done.

Suppose now that $\phi\in\operatorname{Dom}|D|^{j+1}$ for a $j\in\{\,1,\cdots,k-2\,\}$. Since $(\mathcal{D}^{j+1}+R_{j+1})\,\phi\in\operatorname{Dom}|D|$, and $R_{j+1}\,\phi\in\operatorname{Dom}|D|$, we get $\mathcal{D}^{j+1}\,\phi\in\operatorname{Dom}|D|$, which proves that $\phi\in\operatorname{Dom}|D|^{j+2}$.

Finally, if we set $j=k-1$, we get $\phi\in\operatorname{Dom}|D|^{k}$, so $\operatorname{Dom}(\mathcal{D}_{A})^{k}\subseteq\operatorname{Dom}|D|^{k}$.

$(ii)$ follows from $\operatorname{Ker}\mathcal{D}_{A}\subseteq\bigcap_{k\geq 1}\operatorname{Dom}(\mathcal{D}_{A})^{k}$ and $(i)$.

$(iii)$ Let us first check that $|D|^{\alpha}P_{A}$ is bounded. We define $D_{0}$ as the operator with domain $\operatorname{Dom}D_{0}=\operatorname{Im}P_{A}\cap\operatorname{Dom}|D|^{\alpha}$ and such that $D_{0}\,\phi=|D|^{\alpha}\,\phi.$ Since $\operatorname{Dom}D_{0}$ is finite dimensional, $D_{0}$ extends as a bounded operator on $\mathcal{H}$ with finite rank. We have

[TABLE]

so $|D|^{\alpha}P_{A}$ is bounded. We can remark that by $(ii)$, $\operatorname{Dom}D_{0}=\operatorname{Im}P_{A}$ and $\operatorname{Dom}|D|^{\alpha}P_{A}=\mathcal{H}$.

Let us prove now that $P_{A}|D|^{\alpha}$ is bounded: Let $\phi\in\operatorname{Dom}P_{A}|D|^{\alpha}=\operatorname{Dom}|D|^{\alpha}$. By $(ii)$, we have $\operatorname{Im}P_{A}\subseteq\operatorname{Dom}|D|^{\alpha}$ so we get

[TABLE]

$(iv)$ For any $k\in\mathbb{N}_{0}$ and $t\in\mathbb{R}$, $\delta^{k}(P_{A})|D|^{t}$ is a linear combination of terms of the form $|D|^{\beta}P_{A}|D|^{\alpha}$, so the result follows from $(iii)$. ∎

Remark 4.2**.**

We will see later on the noncommutative torus example how important is the difference between $\mathcal{D}_{A}$ and $\mathcal{D}+A$. In particular, the inclusion $\operatorname{Ker}\mathcal{D}\subseteq\operatorname{Ker}\mathcal{D}+A$ is not satisfied since $A$ does not preserve $\operatorname{Ker}\mathcal{D}$ contrarily to $\widetilde{A}$.

The coefficient of the nonconstant term $\Lambda^{k}$ ($k>0$) in the expansion (5) of the spectral action $S(\mathcal{D}_{A},\Phi,\Lambda)$ is equal to the residue of $\zeta_{D_{A}}(s)$ at $k$. We will see in this section how we can compute these residues in term of noncommutative integral of certain operators.

Define for any operator $T$, $p\in\mathbb{N}$, $s\in\mathbb{C}$,

[TABLE]

with $dt:=dt_{1}\cdots dt_{p}$.

Remark that if $T\in OP^{\alpha}$, then $\sigma_{z}(T)\in OP^{\alpha}$ for $z\in\mathbb{C}$ and $K_{p}(T,s)\in OP^{\alpha p}$.

Let us define

[TABLE]

thus $X\in\mathcal{D}_{1}(\mathcal{A})\cap OP^{1}$ and by Lemma 4.1,

[TABLE]

We will use

[TABLE]

which makes sense since $D_{A}^{2}=\mathcal{D}_{A}^{2}+P_{A}$ is invertible for any $A$.

By definition of $X_{V}$, we get

[TABLE]

Lemma 4.3**.**

[4]**

(i) $Y$ is a pseudodifferential operator in $OP^{-1}$ with the following expansion for any $N\in\mathbb{N}$

[TABLE]

(ii) For any $N\in\mathbb{N}$ and $s\in\mathbb{C}$,

[TABLE]

Proof.

$(i)$ We follow [4, Lemma 2.2]. By functional calculus, $Y=\int_{0}^{\infty}I(\lambda)\,d\lambda$, where

[TABLE]

By (30), \big{(}(D^{2}+\lambda)^{-1}X_{V}\big{)}^{p}\sim\big{(}(D^{2}+\lambda)^{-1}X\big{)}^{p}\mod OP^{-\infty} and we get

[TABLE]

We set A_{p}(X):=\big{(}(D^{2}+\lambda)^{-1}X\big{)}^{p}(D^{2}+\lambda)^{-1} and $L:=(D^{2}+\lambda)^{-1}\in OP^{-2}$ for a fixed $\lambda$. Since $[D^{2}+\lambda,X]\sim\nabla(X)\mod OP^{-\infty}$, a recurrence proves that if $T$ is an operator in $OP^{r}$, then, for $q\in\mathbb{N}_{0}$,

[TABLE]

With $A_{p}(X)=LXA_{p-1}(X)$, another recurrence gives, for any $q\in\mathbb{N}_{0}$,

[TABLE]

which entails that

[TABLE]

With $\int_{0}^{\infty}(D^{2}+\lambda)^{-(|k|_{1}+p+1)}d\lambda=\tfrac{1}{|k|_{1}+p}D^{-2(|k|_{1}+p)}$, we get the result provided we control the remainders. Such a control is given in [4, (2.27)].

$(ii)$ We have $|D_{A}|^{-s}=e^{B-(s/2)Y}e^{-B}\,|D|^{-s}$ where $B:=(-s/2)\log(D^{2})$. Following [4, Theorem 2.4], we get

[TABLE]

and each $K_{p}(Y,s)$ is in $OP^{-p}$. ∎

Corollary 4.4**.**

For any $p\in\mathbb{N}$ and $r_{1},\cdots,r_{p}\in\mathbb{N}_{0}$, $\varepsilon^{r_{1}}(Y)\cdots\varepsilon^{r_{p}}(Y)\in\Psi_{1}(\mathcal{A})$.

Proof.

If for any $q\in\mathbb{N}$ and $k=(k_{1},\cdots,k_{q})\in\mathbb{N}_{0}^{q}$,

[TABLE]

then, $\Gamma_{q}^{k}(X)\in OP^{|k|_{1}+q}$. For any $N\in\mathbb{N}$,

[TABLE]

Note that the $\Gamma_{q}^{k}(X)$ are in $\mathcal{D}_{1}(\mathcal{A})$, which, with (33) proves that $Y$ and thus $\varepsilon^{r}(Y)=\nabla^{r}(Y)D^{-2r}$, are also in $\Psi_{1}(\mathcal{A})$. ∎

We remark, as in [11], that the fluctuations leave invariant the first term of the spectral action (5). This is a generalization of the fact that in the commutative case, the noncommutative integral depends only on the principal symbol of the Dirac operator $\mathcal{D}$ and this symbol is stable by adding a gauge potential like in $\mathcal{D}+A$. Note however that the symmetrized gauge potential $A+\epsilon JAJ^{-1}$ is always zero in this case for any selfadjoint one-form $A$.

Lemma 4.5**.**

If the spectral triple is simple, formula (6) can be extended as

[TABLE]

Proof.

Since the spectral triple is simple, equation (32) entails that

[TABLE]

Thus, with (29), we get \zeta_{D_{A}}(0)-\zeta_{D}(0)=-{\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits Y. Replacing $A$ by $\widetilde{A}$, the same proof as in [4] gives

[TABLE]

Lemma 4.6**.**

For any $k\in\mathbb{N}_{0}$,

[TABLE]

where

[TABLE]

Proof.

By Lemma 4.3 $(ii)$, $|D_{A}|^{-s}\sim|D|^{-s}+\sum_{p=1}^{k}K_{p}(Y,s)|D|^{-s}\mod OP^{-(k+1)-\Re(s)}$, where the convention $\sum_{\emptyset}=0$ is used. Thus, we get for $s$ in a neighborhood of $n-k$,

[TABLE]

which gives

[TABLE]

Let us fix $1\leq p\leq k$ and $N\in\mathbb{N}$. By (29) we get

[TABLE]

If we now take $N=k-p$, we get for $s$ in a neighborhood of $n-k$

[TABLE]

so (35) gives the result. ∎

Our operators $|D_{A}|^{k}$ are pseudodifferential operators:

Lemma 4.7**.**

For any $k\in\mathbb{Z}$, $|D_{A}|^{k}\in\Psi^{k}(\mathcal{A})$.

Proof.

Using (4), we see that $K_{p}(Y,s)$ is a pseudodifferential operator in $OP^{-p}$, so (31) proves that $|D_{A}|^{k}$ is a pseudodifferential operator in $OP^{k}$. ∎

The following result is quite important since it shows that one can use \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits for $D$ or $D_{A}$:

Proposition 4.8**.**

If the spectral triple is simple, \underset{s=0}{\operatorname{Res}}\,\operatorname{Tr}\big{(}P|D_{A}|^{-s}\big{)}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits P for any pseudodifferential operator $P$. In particular, for any $k\in\mathbb{N}_{0}$

[TABLE]

Proof.

Suppose $P\in OP^{k}$ with $k\in\mathbb{Z}$ and let us fix $p\geq 1$. With (4), we see that for any $N\in\mathbb{N}$,

[TABLE]

Thus if we take $N=n-p+k$, we get

[TABLE]

Since $s=0$ is a zero of the analytic function $s\mapsto h(s,r,p)$ and $s\mapsto\operatorname{Tr}P\varepsilon^{r_{1}}(Y)\cdots\varepsilon^{r_{p}}(Y)|D|^{-s}$ has only simple poles by hypothesis, we see that \underset{s=0}{\operatorname{Res}}\,h(s,r,p)\,\operatorname{Tr}\big{(}P\varepsilon^{r_{1}}(Y)\cdots\varepsilon^{r_{p}}(Y)|D|^{-s}\big{)}=0 and

[TABLE]

Using (31), $P|D_{A}|^{-s}\sim P|D|^{-s}+\sum_{p=1}^{k+n}PK_{p}(Y,s)|D|^{-s}\mod OP^{-n-1-\Re(s)}$ and thus,

[TABLE]

The result now follows from (37) and (38). To get the last equality, one uses the pseudodifferential operator $|D_{A}|^{-(n-k)}$. ∎

Proposition 4.9**.**

If the spectral triple is simple, then

[TABLE]

Proof.

Lemma 4.6 and previous proposition for $k=0$. ∎

Lemma 4.10**.**

If the spectral triple is simple,

[TABLE]

Proof.

$(i)$ By (31),

[TABLE]

where for the last equality we use the simple dimension spectrum hypothesis. Lemma 4.3 $(i)$ yields $Y\sim XD^{-2}\mod OP^{-2}$ and $Y|D|^{-(n-1)}\sim X|D|^{-n-1}\mod OP^{-n-1}\subseteq\mathcal{L}^{1}(\mathcal{H})$. Thus,

[TABLE]

$(ii)$ Lemma 4.6 $(ii)$ gives

[TABLE]

We have $h(s,0,1)=-\tfrac{s}{2}$, $h(s,1,1)={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\tfrac{s}{2})^{2}$ and $h(s,0,2)={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\tfrac{s}{2})^{2}$. Using again Lemma 4.3 $(i)$,

[TABLE]

Thus,

[TABLE]

Moreover, using \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\nabla(X)|D|^{-k}=0 for any $k\geq 0$ since \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits is a trace,

[TABLE]

Similarly, since $Y\sim XD^{-2}$ mod $OP^{-2}$ and $Y^{2}\sim X^{2}D^{-4}\mod OP^{-3}$, we get

[TABLE]

Thus,

[TABLE]

Finally,

[TABLE]

and the result follows from Proposition 4.8. ∎

Corollary 4.11**.**

If the spectral triple is simple and satisfies \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits|D|^{-(n-2)}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\widetilde{A}\mathcal{D}|D|^{-n}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathcal{D}\widetilde{A}|D|^{-n}=0, then

[TABLE]

Proof.

By previous lemma,

[TABLE]

Since $\nabla(\widetilde{A})\in OP^{1}$, the trace property of \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits yields the result. ∎

5 The noncommutative torus

5.1 Notations

Let $C^{\infty}(\mathbb{T}^{n}_{\Theta})$ be the smooth noncommutative $n$-torus associated to a non-zero skew-symmetric deformation matrix $\Theta\in M_{n}(\mathbb{R})$ (see [6], [30]). This means that $C^{\infty}(\mathbb{T}^{n}_{\Theta})$ is the algebra generated by $n$ unitaries $u_{i}$, $i=1,\dots,n$ subject to the relations

[TABLE]

and with Schwartz coefficients: an element $a\in C^{\infty}(\mathbb{T}_{\Theta}^{n})$ can be written as $a=\sum_{k\in\mathbb{Z}^{n}}a_{k}\,U_{k}$, where $\{a_{k}\}\in\SS(\mathbb{Z}^{n})$ with the Weyl elements defined by $U_{k}:=e^{-\frac{i}{2}k.\chi k}\,u_{1}^{k_{1}}\cdots u_{n}^{k_{n}}$, $k\in\mathbb{Z}^{n}$, relation (40) reads

[TABLE]

where $\chi$ is the matrix restriction of $\Theta$ to its upper triangular part. Thus unitary operators $U_{k}$ satisfy $U_{k}^{*}=U_{-k}$ and $[U_{k},U_{l}]=-2i\sin(\frac{1}{2}k.\Theta l)\,U_{k+l}$.

Let $\tau$ be the trace on $C^{\infty}(\mathbb{T}^{n}_{\Theta})$ defined by \tau\big{(}\sum_{k\in\mathbb{Z}^{n}}a_{k}\,U_{k}\big{)}:=a_{0} and $\mathcal{H}_{\tau}$ be the GNS Hilbert space obtained by completion of $C^{\infty}(\mathbb{T}_{\Theta}^{n})$ with respect of the norm induced by the scalar product $\langle a,b\rangle:=\tau(a^{*}b)$. On $\mathcal{H}_{\tau}=\{\,\sum_{k\in\mathbb{Z}^{n}}a_{k}\,U_{k}\,:\,\{a_{k}\}_{k}\in l^{2}(\mathbb{Z}^{n})\,\}$, we consider the left and right regular representations of $C^{\infty}(\mathbb{T}_{\Theta}^{n})$ by bounded operators, that we denote respectively by $L(.)$ and $R(.)$.

Let also $\delta_{\mu}$, $\mu\in\{\,1,\dots,n\,\}$, be the $n$ (pairwise commuting) canonical derivations, defined by

[TABLE]

We need to fix notations: let $\mathcal{A}_{\Theta}:=C^{\infty}(\mathbb{T}_{\Theta}^{n})$ acting on $\mathcal{H}:=\mathcal{H}_{\tau}\otimes\mathbb{C}^{2^{m}}$ with $n=2m$ or $n=2m+1$ (i.e., $m=\lfloor\tfrac{n}{2}\rfloor$ is the integer part of $\tfrac{n}{2}$), the square integrable sections of the trivial spin bundle over $\mathbb{T}^{n}$.

Each element of $\mathcal{A}_{\Theta}$ is represented on $\mathcal{H}$ as $L(a)\otimes 1_{2^{m}}$ where $L$ (resp. $R$) is the left (resp. right) multiplication. The Tomita conjugation $J_{0}(a):=a^{*}$ satisfies $[J_{0},\delta_{\mu}]=0$ and we define $J:=J_{0}\otimes C_{0}$ where $C_{0}$ is an operator on $\mathbb{C}^{2^{m}}$. The Dirac operator is given by

[TABLE]

where we use hermitian Dirac matrices $\gamma$. It is defined and symmetric on the dense subset of $\mathcal{H}$ given by $C^{\infty}(\mathbb{T}_{\Theta}^{n})\otimes\mathbb{C}^{2^{m}}$. We still note $\mathcal{D}$ its selfadjoint extension. This implies

[TABLE]

and

[TABLE]

where $(e_{i})$ is the canonical basis of $\mathbb{C}^{2^{m}}$. Moreover, $C_{0}^{2}=\pm 1_{2^{m}}$ depending on the parity of $m$. Finally, one introduces the chirality (which in the even case is $\chi:=id\otimes(-i)^{m}\gamma^{1}\cdots\gamma^{n}$) and this yields that $(\mathcal{A}_{\Theta},\mathcal{H},\mathcal{D},J,\chi)$ satisfies all axioms of a spectral triple, see [8, 23].

The perturbed Dirac operator $V_{u}\mathcal{D}V_{u}^{*}$ by the unitary

[TABLE]

defined for every unitary $u\in\mathcal{A}$, $uu^{*}=u^{*}u=U_{0}$, must satisfy condition (3) (which is equivalent to $\mathcal{H}$ being endowed with a structure of $\mathcal{A}_{\Theta}$-bimodule). This yields the necessity of a symmetrized covariant Dirac operator:

[TABLE]

since $V_{u}\mathcal{D}V_{u}^{*}=\mathcal{D}_{L(u)\otimes 1_{2^{m}}[\mathcal{D},L(u^{*})\otimes 1_{2^{m}}]}$: in fact, for $a\in\mathcal{A}_{\Theta}$, using $J_{0}L(a)J_{0}^{-1}=R(a^{*})$, we get

[TABLE]

and that the representation $L$ and the anti-representation $R$ are $\mathbb{C}$-linear, commute and satisfy

[TABLE]

This induces some covariance property for the Dirac operator: one checks that for all $k\in\mathbb{Z}^{n}$,

[TABLE]

so with (44), we get $U_{k}[\mathcal{D},U_{k}^{*}]+\epsilon JU_{k}[\mathcal{D},U_{k}^{*}]J^{-1}=0$ and

[TABLE]

Moreover, we get the gauge transformation:

[TABLE]

where the gauged transform one-form of $A$ is

[TABLE]

with the shorthand $L(u)\otimes 1_{2^{m}}\longrightarrow u$.

As a consequence, the spectral action is gauge invariant:

[TABLE]

An arbitrary selfadjoint one-form $A$, can be written as

[TABLE]

thus

[TABLE]

Defining

[TABLE]

we get $\mathcal{D}_{A}^{2}=-g^{{\alpha}_{1}{\alpha}_{2}}(\delta_{{\alpha}_{1}}+\tilde{A}_{{\alpha}_{1}})(\delta_{{\alpha}_{2}}+\tilde{A}_{{\alpha}_{2}})\otimes 1_{2^{m}}-\tfrac{1}{2}\Omega_{{\alpha}_{1}{\alpha}_{2}}\otimes\gamma^{{\alpha}_{1}{\alpha}_{2}}$ where

[TABLE]

with

[TABLE]

In summary,

[TABLE]

5.2 Kernels and dimension spectrum

We now compute the kernel of the perturbed Dirac operator:

Proposition 5.1**.**

(i) $\operatorname{Ker}\mathcal{D}=U_{0}\otimes\mathbb{C}^{2^{m}}$, so $\dim\operatorname{Ker}\mathcal{D}=2^{m}$.

(ii) For any selfadjoint one-form $A$, $\operatorname{Ker}\mathcal{D}\subseteq\operatorname{Ker}\mathcal{D}_{A}$.

(iii) For any unitary $u\in\mathcal{A}$, $\operatorname{Ker}\mathcal{D}_{\gamma_{u}(A)}=V_{u}\,\operatorname{Ker}\mathcal{D}_{A}$.

Proof.

$(i)$ Let $\psi=\sum_{k,j}c_{k,j}\,U_{k}\otimes e_{j}\in\operatorname{Ker}\mathcal{D}$. Thus, $0=\mathcal{D}^{2}\psi=\sum_{k,i}c_{k,j}|k|^{2}\,U_{k}\otimes e_{j}$ which entails that $c_{k,j}|k|^{2}=0$ for any $k\in\mathbb{Z}^{n}$ and $1\leq j\leq 2^{m}$. The result follows.

$(ii)$ Let $\psi\in\operatorname{Ker}\mathcal{D}$. So, $\psi=U_{0}\otimes v$ with $v\in\mathbb{C}^{2^{m}}$ and from (50), we get

[TABLE]

since $U_{0}$ is the unit of the algebra, which proves that $\psi\in\operatorname{Ker}\mathcal{D}_{A}$.

$(iii)$ This is a direct consequence of (47). ∎

Corollary 5.2**.**

Let $A$ be a selfadjoint one-form. Then $\operatorname{Ker}\mathcal{D}_{A}=\operatorname{Ker}\mathcal{D}$ in the following cases:

(i) $A_{u}:=L(u)\otimes 1_{2^{m}}[\mathcal{D},L(u^{*})\otimes 1_{2^{m}}]$ when $u$ is a unitary in $\mathcal{A}$.

(ii) $||A||<\tfrac{1}{2}$.

(iii) The matrix $\tfrac{1}{2\pi}\Theta$ has only integral coefficients.

Proof.

$(i)$ This follows from previous result because $V_{u}(U_{0}\otimes v)=U_{0}\otimes v$ for any $v\in\mathbb{C}^{2^{m}}$.

$(ii)$ Let $\psi=\sum_{k,j}c_{k,j}\,U_{k}\otimes e_{j}$ be in $\operatorname{Ker}\mathcal{D}_{A}$ (so $\sum_{k,j}|c_{k,j}|^{2}<\infty$) and $\phi:=\sum_{j}c_{0,j}\,U_{0}\otimes e_{j}$. Thus $\psi^{\prime}:=\psi-\phi\in\text{Ker }\mathcal{D}_{A}$ since $\phi\in\operatorname{Ker}\mathcal{D}\subseteq\operatorname{Ker}\mathcal{D}_{A}$ and

[TABLE]

Defining $X_{k}:=\sum_{\alpha}k_{\alpha}\gamma_{\alpha}$, $X_{k}^{2}=\sum_{\alpha}|k_{\alpha}|^{2}\,1_{2^{m}}$ is invertible and the vectors $\{\,U_{k}\otimes X_{k}e_{j}\,\}_{0\neq k\in\mathbb{Z}^{n},\,j}$ are orthogonal in $\mathcal{H}$, so

[TABLE]

which is possible only if $c_{k,j}=0,\,\forall k,\,j$ that is $\psi^{\prime}=0$ et $\psi=\phi\in\text{Ker }\mathcal{D}$.

$(iii)$ This is a consequence of the fact that the algebra is commutative, thus $A+\epsilon JAJ^{-1}=0$. ∎

Note that if $\widetilde{A}_{u}:=A_{u}+\epsilon JA_{u}J^{-1}$, then by (45), $\widetilde{A}_{U_{k}}=0$ for all $k\in\mathbb{Z}^{n}$ and $\left\lVert A_{U_{k}}\right\rVert=|k|$, but for an arbitrary unitary $u\in\mathcal{A}$, $\widetilde{A}_{u}\neq 0$ so $\mathcal{D}_{A_{u}}\neq\mathcal{D}$.

Naturally the above result is also a direct consequence of the fact that the eigenspace of an isolated eigenvalue of an operator is not modified by small perturbations. However, it is interesting to compute the last result directly to emphasize the difficulty of the general case:

Let $\psi=\sum_{l\in\mathbb{Z}^{n},1\leq j\leq 2^{m}}c_{l,j}\,U_{l}\otimes e_{j}\in\operatorname{Ker}\mathcal{D}_{A}$, so $\sum_{l\in\mathbb{Z}^{n},1\leq j\leq 2^{m}}|c_{l,j}|^{2}<\infty$. We have to show that $\psi\in$ Ker $\mathcal{D}$ that is $c_{l,j}=0$ when $l\neq 0$.

Taking the scalar product of $\langle U_{k}\otimes e_{i}|$ with

[TABLE]

we obtain

[TABLE]

If $A_{\alpha}=\sum_{\alpha,l}a_{\alpha,l}\,U_{l}\otimes\gamma^{\alpha}$ with $\{\,a_{\alpha,l}\,\}_{l}\in\SS(\mathbb{Z}^{n})$, note that $[U_{l},U_{m}]=-2i\sin(\tfrac{1}{2}l.\Theta m)\,U_{l+m}$ and

[TABLE]

Thus

[TABLE]

We conjecture that $\operatorname{Ker}\mathcal{D}=\operatorname{Ker}\mathcal{D}_{A}$ at least for generic $\Theta$’s:

the constraints (53) should imply $c_{l,j}=0$ for all $j$ and all $l\neq 0$ meaning $\psi\in\operatorname{Ker}\mathcal{D}$. When $\tfrac{1}{2\pi}\Theta$ has only integer coefficients, the sin part of these constraints disappears giving the result.

Lemma 5.3**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, Sp\big{(}C^{\infty}(\mathbb{T}^{n}_{\Theta}),\mathcal{H},\mathcal{D}\big{)}=\mathbb{Z} and all these poles are simple.

Proof.

Let $B\in\mathcal{D}(\mathcal{A})$ and $p\in\mathbb{N}_{0}$. Suppose that $B$ is of the form

[TABLE]

where $r\in\mathbb{N}$, $a_{i}\in\mathcal{A}$, $b_{i}\in J\mathcal{A}J^{-1}$, $q_{i},p_{i}\in\mathbb{N}_{0}$. We note $a_{i}=:\sum_{l}a_{i,l}\,U_{l}$ and $b_{i}=:\sum_{i}b_{i,l}\,U_{l}$. With the shorthand $k_{\mu_{1},\mu_{q_{i}}}:=k_{\mu_{1}}\cdots k_{\mu_{q_{i}}}$ and $\gamma^{\mu_{1},\mu_{q_{i}}}=\gamma^{\mu_{1}}\cdots\gamma^{\mu_{q_{i}}}$, we get

[TABLE]

which gives, after $r$ iterations,

[TABLE]

where $\widetilde{a}_{l}:=a_{1,l_{1}}\cdots a_{r,l_{r}}$ and $\widetilde{b}_{l^{\prime}}:=b_{1,l^{\prime}_{1}}\cdots b_{r,l^{\prime}_{r}}$.

Let us note $F_{\mu}(k,l,l^{\prime}):=\prod_{i=1}^{r-1}|k+\widehat{l}_{i}+\widehat{l}^{\prime}_{i}|^{p_{i}}(k+\widehat{l}_{i}+\widehat{l}^{\prime}_{i})_{\mu^{i}_{1},\mu^{i}_{q_{i}}}$ and $\gamma^{\mu}:=\gamma^{\mu^{r-1}_{1},\mu^{r-1}_{q_{r-1}}}\cdots\gamma^{\mu^{1}_{1},\mu^{1}_{q_{1}}}$. Thus, with the shortcut $\sim_{c}$ meaning modulo a constant function towards the variable $s$,

[TABLE]

Since $U_{l_{r}}\cdots U_{l_{1}}U_{k}=U_{k}U_{l_{r}}\cdots U_{l_{1}}e^{-i\sum_{1}^{r}l_{i}.\Theta k}$ we get

[TABLE]

where $\phi$ is a real valued function. Thus,

[TABLE]

The function $f_{\mu}(s)$ can be decomposed has a linear combination of zeta function of type described in Theorem 2.17 (or, if $r=1$ or all the $p_{i}$ are zero, in Theorem 2.5). Thus, s\mapsto\operatorname{Tr}\big{(}B|D|^{-2p-s}\big{)} has only poles in $\mathbb{Z}$ and each pole is simple. Finally, by linearity, we get the result. ∎

The dimension spectrum of the noncommutative torus is simple:

Proposition 5.4**.**

(i) If $\tfrac{1}{2\pi}\Theta$ is diophantine, the spectrum dimension of \big{(}C^{\infty}(\mathbb{T}^{n}_{\Theta}),\mathcal{H},\mathcal{D}\big{)} is equal to the set $\{\,n-k\,:\,k\in\mathbb{N}_{0}\,\}$ and all these poles are simple.

(ii) $\zeta_{D}(0)=0.$

Proof.

$(i)$ Lemma 5.3 and Remark 3.9.

$(ii)$ $\zeta_{D}(s)={\sum}_{k\in\mathbb{Z}^{n}}\sum_{1\leq j\leq 2^{m}}\langle U_{k}\otimes e_{j}\,,\,|D|^{-s}U_{k}\otimes e_{j}\rangle=2^{m}({\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\frac{1}{|k|^{s}}+1)=2^{m}(\,Z_{n}(s)+1).$ By (21), we get the result. ∎

We have computed $\zeta_{D}(0)$ relatively easily but the main difficulty of the present work is precisely to calculate $\zeta_{D_{A}}(0)$.

5.3 Noncommutative integral computations

We fix a self-adjoint 1-form $A$ on the noncommutative torus of dimension $n$.

Proposition 5.5**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, then the first elements of the expansion (5) are given by

[TABLE]

We need few technical lemmas:

Lemma 5.6**.**

On the noncommutative torus, for any $t\in\mathbb{R}$,

[TABLE]

Proof.

Using notations of (49), we have

[TABLE]

since $\langle U_{k},[A_{\alpha},U_{k}]\rangle=0$. Similarly

[TABLE]

Any element $h$ in the algebra generated by $\mathcal{A}$ and $[\mathcal{D},\mathcal{A}]$ can be written as a linear combination of terms of the form ${a_{1}}^{p_{1}}\cdots{a_{n}}^{p_{r}}$ where $a_{i}$ are elements of $\mathcal{A}$ or $[\mathcal{D},\mathcal{A}]$. Such a term can be written as a series $b:=\sum a_{1,\alpha_{1},l_{1}}\cdots a_{q,\alpha_{q},l_{q}}U_{l_{1}}\cdots U_{l_{q}}\otimes\gamma^{\alpha_{1}}\cdots\gamma^{\alpha_{q}}$ where $a_{i,\alpha_{i}}$ are Schwartz sequences and when $a_{i}=:\sum_{l}a_{l}U_{l}\in\mathcal{A}$, we set $a_{i,\alpha,l}=a_{i,l}$ with $\gamma^{\alpha}=1$. We define

[TABLE]

By linearity, $L$ is defined as a linear form on the whole algebra generated by $\mathcal{A}$ and $[\mathcal{D},\mathcal{A}]$.

Lemma 5.7**.**

If $h$ is an element of the algebra generated by $\mathcal{A}$ and $[\mathcal{D},\mathcal{A}]$,

[TABLE]

In particular, \operatorname{Tr}\big{(}h|D|^{-s}\big{)} has at most one pole at $s=n$.

Proof.

We get with $b$ of the form $\sum a_{1,\alpha_{1},l_{1}}\cdots a_{q,\alpha_{q},l_{q}}U_{l_{1}}\cdots U_{l_{q}}\otimes\gamma^{\alpha_{1}}\cdots\gamma^{\alpha_{q}}$,

[TABLE]

The results follows now from linearity of the trace. ∎

Lemma 5.8**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, the function s\mapsto\operatorname{Tr}\big{(}\varepsilon JAJ^{-1}A|D|^{-s}\big{)} extends meromorphically on the whole plane with only one possible pole at $s=n$. Moreover, this pole is simple and

[TABLE]

Proof.

With $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}$, we get $\epsilon JAJ^{-1}=R(iA_{\alpha})\otimes\gamma^{\alpha}$, and by multiplication $\varepsilon JAJ^{-1}A=R(A_{\beta})L(A_{\alpha})\otimes\gamma^{\beta}\gamma^{\alpha}$. Thus,

[TABLE]

Theorem 2.5 $(ii)$ entails that ${\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\,\sum_{l}a_{\alpha,l}\,a^{\alpha}_{-l}\,e^{ik.\Theta l}\,|k|^{-s}$ extends meromorphically on the whole plane $\mathbb{C}$ with only one possible pole at $s=n$. Moreover, this pole is simple and we have

[TABLE]

Equation (20) now gives the result. ∎

Lemma 5.9**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, then for any $t\in\mathbb{R}$,

[TABLE]

where $X=\widetilde{A}\mathcal{D}+\mathcal{D}\widetilde{A}+{\widetilde{A}}^{2}$ and $A=:-i\sum_{l}a_{\alpha,l}\,U_{l}\otimes\gamma^{\alpha}$.

Proof.

By Lemma 5.6, we get \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits X|D|^{-t}=\operatorname{Res}_{s=0}\operatorname{Tr}({\widetilde{A}}^{2}|D|^{-s-t}). Since $A$ and $\varepsilon JAJ^{-1}$ commute, we have $\widetilde{A}^{2}=A^{2}+JA^{2}J^{-1}+2\varepsilon JAJ^{-1}A$. Thus,

[TABLE]

Since $|D|$ and $J$ commute, we have with Lemma 5.7,

[TABLE]

Thus Lemma 5.8 entails that $\operatorname{Tr}({\widetilde{A}}^{2}|D|^{-s-t})$ is holomorphic at 0 if $t\neq n$. When $t=n$,

[TABLE]

which gives the result. ∎

Lemma 5.10**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, then

[TABLE]

Proof.

With $\mathcal{D}J=\varepsilon J\mathcal{D}$, we get

[TABLE]

Let us first compute \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits A\mathcal{D}A\mathcal{D}|D|^{-2-n}. We have, with $A=:-iL(A_{\alpha})\otimes\gamma^{\alpha}=:-i\sum_{l}a_{\alpha,l}U_{l}\otimes\gamma^{\alpha}$,

[TABLE]

where $\gamma^{\alpha,\mu}:=\gamma^{\alpha_{2}}\gamma^{\mu_{2}}\gamma^{\alpha_{1}}\gamma^{\mu_{1}}$. Thus,

[TABLE]

We have also, with $\varepsilon JAJ^{-1}=iR(A_{\alpha})\otimes\gamma^{a}$,

[TABLE]

which gives

[TABLE]

Thus,

[TABLE]

With ${\sum}^{\prime}_{k}\tfrac{k_{\mu_{1}}k_{\mu_{2}}}{|k|^{s+2+n}}=\tfrac{\delta_{\mu_{1}\mu_{2}}}{n}Z_{n}(s+n)$ and $C_{n}:=\operatorname{Res}_{s=0}Z_{n}(s+n)=2\pi^{n/2}\Gamma(n/2)^{-1}$ we obtain

[TABLE]

Since $\operatorname{Tr}(\gamma^{\alpha_{2}}\gamma^{\mu}\gamma^{\alpha_{1}}\gamma_{\mu})=2^{m}(2-n)\delta^{\alpha_{2},\alpha_{1}}$, we get

[TABLE]

Equation (55) now proves the lemma. ∎

Lemma 5.11**.**

If $\tfrac{1}{2\pi}\Theta$ is diophantine, then for any $P\in\Psi_{1}(\mathcal{A})$ and $q\in\mathbb{N}$, $q$ odd,

[TABLE]

Proof.

There exist $B\in\mathcal{D}_{1}(\mathcal{A})$ and $p\in\mathbb{N}_{0}$ such that $P=BD^{-2p}+R$ where $R$ is in $OP^{-q-1}$. As a consequence, \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits P|D|^{-(n-q)}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits B|D|^{-n-2p+q}. Assume $B=a_{r}b_{r}\mathcal{D}^{q_{r-1}}a_{r-1}b_{r-1}\cdots\mathcal{D}^{q_{1}}a_{1}b_{1}$ where $r\in\mathbb{N}$, $a_{i}\in\mathcal{A}$, $b_{i}\in J\mathcal{A}J^{-1}$, $q_{i}\in\mathbb{N}$. If we prove that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits B|D|^{-n-2p+q}=0, then the general case will follow by linearity. We note $a_{i}=:\sum_{l}a_{i,l}\,U_{l}$ and $b_{i}=:\sum_{l}b_{i,l}\,U_{l}$. With the shorthand $k_{\mu_{1},\mu_{q_{i}}}:=k_{\mu_{1}}\cdots k_{\mu_{q_{i}}}$ and $\gamma^{\mu_{1},\mu_{q_{i}}}=\gamma^{\mu_{1}}\cdots\gamma^{\mu_{q_{i}}}$, we get

[TABLE]

which gives, after iteration,

[TABLE]

where $\widetilde{a}_{l}:=a_{1,l_{1}}\cdots a_{r,l_{r}}$ and $\widetilde{b}_{l^{\prime}}:=b_{1,l^{\prime}_{1}}\cdots b_{r,l^{\prime}_{r}}$. Let’s note $Q_{\mu}(k,l,l^{\prime}):=\prod_{i=1}^{r-1}(k+\widehat{l}_{i}+\widehat{l}^{\prime}_{i})_{\mu^{i}_{1},\mu^{i}_{q_{i}}}$ and $\gamma^{\mu}:=\gamma^{\mu^{r-1}_{1},\mu^{r-1}_{q_{r-1}}}\cdots\gamma^{\mu^{1}_{1},\mu^{1}_{q_{1}}}$. Thus,

[TABLE]

Since $U_{l_{r}}\cdots U_{l_{1}}U_{k}=U_{k}U_{l_{r}}\cdots U_{l_{1}}e^{-i\sum_{1}^{r}l_{i}.\Theta k}$, we get

[TABLE]

where $\phi$ is a real valued function. Thus,

[TABLE]

We decompose $Q_{\mu}(k,l,l^{\prime})$ as a sum $\sum_{h=0}^{r}M_{h,\mu}(l,l^{\prime})\,Q_{h,\mu}(k)$ where $Q_{h,\mu}$ is a homogeneous polynomial in $(k_{1},\cdots,k_{n})$ and $M_{h,\mu}(l,l^{\prime})$ is a polynomial in \big{(}(l_{1})_{1},\cdots,(l_{r})_{n},(l^{\prime}_{1})_{1},\cdots,(l^{\prime}_{r})_{n}\big{)}.

Similarly, we decompose $f_{\mu}(s)$ as $\sum_{h=0}^{r}f_{h,\mu}(s)$. Theorem 2.5 $(ii)$ entails that $f_{h,\mu}(s)$ extends meromorphically to the whole complex plane $\mathbb{C}$ with only one possible pole for $s+2p+n-q=n+d$ where $d:=\text{deg }Q_{h,\mu}$. In other words, if $d+q-2p\neq 0$, $f_{h,\mu}(s)$ is holomorphic at $s=0$. Suppose now $d+q-2p=0$ (note that this implies that $d$ is odd, since $q$ is odd by hypothesis), then, by Theorem 2.5 $(ii)$

[TABLE]

where $V:=\sum_{l,l^{\prime}\in Z}M_{h,\mu}(l,l^{\prime})\,e^{i\phi(l,l^{\prime})}\,\delta_{\sum_{1}^{r}l_{i}+l^{\prime}_{i},0}\,\widetilde{a}_{l}\,\widetilde{b}_{l^{\prime}}$ and $Z:=\{\,l,l^{\prime}\,:\,\sum_{i=1}^{r}l_{i}=0\,\}$. Since $d$ is odd, $Q_{h,\mu}(-u)=-Q_{h,\mu}(u)$ and $\int_{u\in S^{n-1}}Q_{h,\mu}(u)\,dS(u)=0$. Thus, $\underset{s=0}{\operatorname{Res}}\ f_{h,\mu}(s)=0$ in any case, which gives the result. ∎

As we have seen, the crucial point of the preceding lemma is the decomposition of the numerator of the series $f_{\mu}(s)$ as polynomials in $k$. This has been possible because we restricted our pseudodifferential operators to $\Psi_{1}(\mathcal{A})$.

Proof of Proposition 5.5. The top element follows from Proposition 4.9 and according to (20),

[TABLE]

For the second equality, we get from Lemmas 5.7 and 4.6

[TABLE]

Corollary 4.4 and Lemma 5.11 imply that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\varepsilon^{r_{1}}(Y)\cdots\varepsilon^{r_{p}}(Y)|D|^{-(n-k)}=0, which gives the result.

Last equality follows from Lemma 5.10 and Corollary 4.11. ∎

6 The spectral action

Here is the main result of this section.

Theorem 6.1**.**

Consider the $n$-NC-torus \big{(}C^{\infty}(\mathbb{T}^{n}_{\Theta}),\mathcal{H},\mathcal{D}\big{)} where $n\in\mathbb{N}$ and $\tfrac{1}{2\pi}\Theta$ is a real $n\times n$ skew-symmetric real diophantine matrix, and a selfadjoint one-form $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}$. Then, the full spectral action of $\mathcal{D}_{A}=\mathcal{D}+A+\epsilon JAJ^{-1}$ is

(i) for $n=2$,

[TABLE]

(ii) for $n=4$,

[TABLE]

(iii) More generally, in

[TABLE]

$c_{n-2}(A)=0$, $c_{n-k}(A)=0$ for $k$ odd. In particular, $c_{0}(A)=0$ when $n$ is odd.

This result (for $n=4$) has also been obtained in [20] using the heat kernel method. It is however interesting to get the result via direct computations of (5) since it shows how this formula is efficient. As we will see, the computation of all the noncommutative integrals require a lot of technical steps. One of the main points, namely to isolate where the Diophantine condition on $\Theta$ is assumed, is outlined here.

Remark 6.2**.**

Note that all terms must be gauge invariants, namely, according to (48), invariant by $A_{\alpha}\longrightarrow\gamma_{u}(A_{\alpha})=uA_{\alpha}u^{*}+u\delta_{\alpha}(u^{*})$. A particular case is $u=U_{k}$ where $U_{k}\delta_{\alpha}(U_{k}^{*})=-ik_{\alpha}U_{0}$.

In the same way, note that there is no contradiction with the commutative case where, for any selfadjoint one-form $A$, $\mathcal{D}_{A}=\mathcal{D}$ (so $A$ is equivalent to 0!), since we assume in Theorem 6.1 that $\Theta$ is diophantine, so $\mathcal{A}$ cannot be commutative.

Conjecture 6.3**.**

The constant term of the spectral action of $\mathcal{D}_{A}$ on the noncommutative n-torus is proportional to the constant term of the spectral action of $\mathcal{D}+A$ on the commutative n-torus.

Remark 6.4**.**

The appearance of a Diophantine condition for $\Theta$ has been characterized in dimension 2 by Connes [7, Prop. 49] where in this case, $\Theta=\theta\genfrac{(}{)}{0.0pt}{1}{\,0\quad 1}{-1\,\,\,\,0}$ with $\theta\in\mathbb{R}$. In fact, the Hochschild cohomology $H(\mathcal{A}_{\Theta},{\mathcal{A}_{\Theta}}^{*})$ satisfies dim $H^{j}(\mathcal{A}_{\Theta},{\mathcal{A}_{\Theta}}^{*})=2$ (or $1$) for $j=1$ (or $j=2$) if and only if the irrational number $\theta$ satisfies a Diophantine condition like $|1-e^{i2\pi n\theta}|^{-1}=\mathcal{O}(n^{k})$ for some $k$.

Recall that when the matrix $\Theta$ is quite irrational (see [23, Cor. 2.12]), then the C∗-algebra generated by $\mathcal{A}_{\Theta}$ is simple.

Remark 6.5**.**

It is possible to generalize above theorem to the case $\mathcal{D}=-i\,{g^{\mu}}_{\nu}\,\delta_{\mu}\otimes\gamma^{\nu}$ instead of (43) when $g$ is a positive definite constant matrix. The formulae in Theorem 6.1 are still valid.

6.1 Computations of \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits

In order to get this theorem, let us prove a few technical lemmas.

We suppose from now on that $\Theta$ is a skew-symmetric matrix in $\mathcal{M}_{n}(\mathbb{R})$. No other hypothesis is assumed for $\Theta$, except when it is explicitly stated.

When $A$ is a selfadjoint one-form, we define for $n\in N$, $q\in\mathbb{N}$, $2\leq q\leq n$ and $\sigma\in\{-,+\}^{q}$

[TABLE]

Lemma 6.6**.**

We have for any $q\in\mathbb{N}$,

[TABLE]

Proof.

Since $P_{0}\in OP^{-\infty}$, $D^{-1}=\mathcal{D}D^{-2}\mod OP^{-\infty}$ and \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\widetilde{A}D^{-1})^{q}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\widetilde{A}\mathcal{D}D^{-2})^{q}. ∎

Lemma 6.7**.**

Let $A$ be a selfadjoint one-form, $n\in\mathbb{N}$ and $q\in\mathbb{N}$ with $2\leq q\leq n$ and $\sigma\in\{-,+\}^{q}$. Then

[TABLE]

Proof.

Let us first check that $JP_{0}=P_{0}J$. Since $\mathcal{D}J=\varepsilon J\mathcal{D}$, we get $\mathcal{D}JP_{0}=0$ so $JP_{0}=P_{0}JP_{0}$. Since $J$ is an antiunitary operator, we get $P_{0}J=P_{0}JP_{0}$ and finally, $P_{0}J=JP_{0}$. As a consequence, we get $JD^{2}=D^{2}J$, $J\mathcal{D}D^{-2}=\varepsilon\mathcal{D}D^{-2}J$, $J\mathbb{A}^{+}J^{-1}=\mathbb{A}^{-}$ and $J\mathbb{A}^{-}J^{-1}=\mathbb{A}^{+}.$

In summary, $J\mathbb{A}^{\sigma_{i}}J^{-1}=\mathbb{A}^{-\sigma_{i}}$.

The trace property of \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits now gives

[TABLE]

Definition 6.8**.**

In [4] has been introduced the vanishing tadpole hypothesis:

[TABLE]

By the following lemma, this condition is satisfied for the noncommutative torus, a fact more or less already known within the noncommutative community [34].

Lemma 6.9**.**

*Let $n\in\mathbb{N}$, $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}=-i\sum_{l\in\mathbb{Z}^{n}}a_{\alpha,l}\,U_{l}\otimes\gamma^{\alpha}$, $A_{\alpha}\in\mathcal{A}_{\Theta}$, $\{\,a_{\alpha,l}\,\}_{l}\in\SS(\mathbb{Z}^{n})$, be a hermitian one-form. Then,

(i) \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits A^{p}D^{-q}=\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\epsilon JAJ^{-1})^{p}D^{-q}=0 for $p\geq 0$ and $1\leq q<n$ (case $p=q=1$ is tadpole hypothesis.)

(ii) If $\tfrac{1}{2\pi}\Theta$ is diophantine, then \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits BD^{-q}=0 for $1\leq q<n$ and any $B$ in the algebra generated by $\mathcal{A}$, $[\mathcal{D},\mathcal{A}]$, $J\mathcal{A}J^{-1}$ and $J[\mathcal{D},\mathcal{A}]J^{-1}$.*

Proof.

$(i)$ Let us compute

[TABLE]

With $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}$ and $\epsilon JAJ^{-1}=R(iA_{\alpha})\otimes\gamma^{\alpha}$, we get

[TABLE]

and

[TABLE]

We note $\widetilde{a}_{\alpha,l}:=a_{\alpha_{1},l_{1}}\cdots a_{\alpha_{p},l_{p}}$. Since

[TABLE]

and

[TABLE]

we get, with

[TABLE]

[TABLE]

Thus, \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits A^{p}(\epsilon JAJ^{-1})^{p^{\prime}}D^{-q}=\underset{s=0}{\operatorname{Res}}\ f(s) where

[TABLE]

It is straightforward to check that the series {\sum}^{\prime}_{k,l,l^{\prime}}g_{\mu,\alpha,\alpha^{\prime}}(s,k,l,l^{\prime})\,\tau\big{(}U_{l,l^{\prime}}\big{)} is absolutely summable if $\Re(s)>R$ for a $R>0$. Thus, we can exchange the summation on $k$ and $l,l^{\prime}$, which gives

[TABLE]

If we suppose now that $p^{\prime}=0$, we see that,

[TABLE]

which is, by Proposition 2.16, analytic at 0. In particular, for $p=q=1$, we see that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits AD^{-1}=0, i.e. the vanishing tadpole hypothesis is satisfied. Similarly, if we suppose $p=0$, we get

[TABLE]

which is holomorphic at 0.

$(ii)$ Adapting the proof of Lemma 5.11 to our setting (taking $q_{i}=0$, and adding gamma matrices components), we see that

[TABLE]

where $\gamma^{(\mu,\alpha,\beta)}$ is a complicated product of gamma matrices. By Theorem 2.5 $(ii)$, since we suppose here that $\tfrac{1}{2\pi}\Theta$ is diophantine, this residue is 0. ∎

6.1.1 Even dimensional case

Corollary 6.10**.**

Same hypothesis as in Lemma 6.9.

(i) Case $n=2$:

[TABLE]

(ii) Case $n=4$: with the shorthand $\delta_{\mu_{1},\ldots,\mu_{4}}:=\delta_{\mu_{1}\mu_{2}}\delta_{\mu_{3}\mu_{4}}+\delta_{\mu_{1}\mu_{3}}\delta_{\mu_{2}\mu_{4}}+\delta_{\mu_{1}\mu_{4}}\delta_{\mu_{2}\mu_{3}}$,

[TABLE]

Proof.

$(i,ii)$ The same computation as in Lemma 6.9 $(i)$ (with $p^{\prime}=0$, $p=q=n$) gives

[TABLE]

and the result follows from Proposition 2.16. ∎

We will use few notations:

If $n\in\mathbb{N}$, $q\geq 2$, $l:=(l_{1},\cdots,l_{q-1})\in(\mathbb{Z}^{n})^{q-1}$, $\alpha:=(\alpha_{1},\cdots,\alpha_{q})\in\{1,\cdots,n\}^{q}$, $k\in\mathbb{Z}^{n}\backslash\{0\}$, $\sigma\in\{-,+\}^{q}$, $(a_{i})_{1\leq i\leq n}\in(\mathcal{S}(\mathbb{Z}^{n}))^{n}$,

[TABLE]

with the convention $\sum_{2\leq j\leq q-1}=0$ when $q=2$, and $g_{\mu}(s,k,l)=0$ whenever $\widehat{l}_{i}=-k$ for a $1\leq i\leq q-1$.

Lemma 6.11**.**

Let $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}=-i\sum_{l\in\mathbb{Z}^{n}}a_{\alpha,l}\,U_{l}\otimes\gamma^{\alpha}$ where $A_{\alpha}=-A_{\alpha}^{*}\in\mathcal{A}_{\Theta}$ and $\{\,a_{\alpha,l}\,\}_{l}\in\SS(\mathbb{Z}^{n})$, with $n\in\mathbb{N}$, be a hermitian one-form, and let $2\leq q\leq n$, $\sigma\in\{-,+\}^{q}$.

Then, \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{\sigma}=\underset{s=0}{\operatorname{Res}}\ f(s) where

[TABLE]

Proof.

By definition, \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{\sigma}=\underset{s=0}{\operatorname{Res}}\ f(s) where

[TABLE]

Let $r\in\mathbb{Z}^{n}$ and $v\in\mathbb{C}^{2^{m}}$. Since $A=L(-iA_{\alpha})\otimes\gamma^{\alpha}$, and $\epsilon JAJ^{-1}=R(iA_{\alpha})\otimes\gamma^{\alpha}$, we get

[TABLE]

With $U_{l}U_{r}=e^{\tfrac{i}{2}r.\Theta l}U_{r+l}$ and $U_{r}U_{l}=e^{-\tfrac{i}{2}r.\Theta l}U_{r+l}$, we obtain, for any $1\leq j\leq q$,

[TABLE]

We now apply $q$ times this formula to get

[TABLE]

with

[TABLE]

Thus,

[TABLE]

where in the last sum $l_{q}$ is fixed to $-\sum_{1\leq j\leq q-1}l_{j}$ and thus,

[TABLE]

By Lemma 2.10, there exists a $R>0$ such that for any $s\in\mathbb{C}$ with $\Re(s)>R$, the family

[TABLE]

is absolutely summable as a linear combination of families of the type considered in that lemma. As a consequence, we can exchange the summations on $k$ and $l$, which gives the result. ∎

In the following, we will use the shorthand

[TABLE]

Lemma 6.12**.**

Suppose $n=4$. Then, with the same hypothesis of Lemma 6.11,

[TABLE]

(iv) Suppose $\tfrac{1}{2\pi}\Theta$ diophantine. Then the crossed terms in \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\mathbb{A}^{+}+\mathbb{A}^{-})^{q} vanish: if $C$ is the set of all $\sigma\in\{-,+\}^{q}$ with $2\leq q\leq 4$, such that there exist $i,j$ satisfying $\sigma_{i}\neq\sigma_{j}$, we have \sum_{\sigma\in C}\,\mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{\sigma}=0.

Proof.

$(i)$ Lemma 6.11 entails that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{++}=\underset{s=0}{\operatorname{Res}}\sum_{l\in\mathbb{Z}^{n}}-f(s,l) where

[TABLE]

We will now reduce the computation of the residue of an expression involving terms like $|k+l|^{2}$ in the denominator to the computation of residues of zeta functions. To proceed, we use (16) into an expression like the one appearing in $f(s,l)$. We see that the last term on the righthandside yields a $Z_{n}(s)$ while the first one is less divergent by one power of $k$. If this is not enough, we repeat this operation for the new factor of $|k+l|^{2}$ in the denominator. For $f(s,l)$, which is quadratically divergent at $s=0$, we have to repeat this operation three times before ending with a convergent result. All the remaining terms are expressible in terms of $Z_{n}$ functions. We get, using three times (16),

[TABLE]

Let us define

[TABLE]

so that $f(s,l)=f_{\alpha,\mu}(s,l)\operatorname{Tr}(\gamma^{\alpha_{2}}\gamma^{\mu_{2}}\gamma^{\alpha_{1}}\gamma^{\mu_{1}})$. Equation (57) gives

[TABLE]

with obvious identifications. Note that the function

[TABLE]

is a linear combination of functions of the type $H(s,l)$ satisfying the hypothesis of Corollary 2.13. Thus, $r(s,l)$ satisfies (H1) and with the previously seen equivalence relation modulo functions satisfying this hypothesis we get $f_{\alpha,\mu}(s,l)\sim f_{1}(s,l)-f_{2}(s,l)+f_{3}(s,l)$.

Let’s now compute $f_{1}(s,l)$.

[TABLE]

Proposition 2.1 entails that $s\mapsto{\sum_{k\in\mathbb{Z}^{n}}}^{\prime}\tfrac{k_{\mu_{1}}k_{\mu_{2}}}{|k|^{s+4}}$ is holomorphic at 0. Thus, $f_{1}(s,l)$ satisfies (H1), and $f_{\alpha,\mu}(s,l)\sim-f_{2}(s,l)+f_{3}(s,l)$.

Let’s now compute $f_{2}(s,l)$ modulo (H1). We get, using several times Proposition 2.1,

[TABLE]

Recall that ${\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\tfrac{k_{i}k_{j}}{|k|^{s+6}}=\tfrac{\delta_{ij}}{n}Z_{n}(s+4)$. Thus,

[TABLE]

Finally, let us compute $f_{3}(s,l)$ modulo (H1) following the same principles:

[TABLE]

In conclusion,

[TABLE]

Proposition (2.1) entails that $Z_{n}(s+4)$ and $s\mapsto{\sum_{k\in\mathbb{Z}^{n}}}^{\prime}\tfrac{k_{i}k_{j}k_{\mu_{1}}k_{\mu_{2}}}{|k|^{s+8}}$ extend holomorphically in a punctured open disk centered at 0. Thus, $g_{\alpha,\mu}(s,l)$ satisfies (H2) and we can apply Lemma 2.14 to get

[TABLE]

The problem is now reduced to the computation of $\underset{s=0}{\operatorname{Res}}\ g(s,l)$. Recall that Res${}_{s=0}\,Z_{4}(s+4)=2\pi^{2}$ by (20) or (17), and

[TABLE]

Thus,

[TABLE]

We will use

[TABLE]

where $s(P)$ is the signature of the permutation $P$ when $P_{2m-1}<P_{2m}$ for $1\leq m\leq n$. This gives

[TABLE]

Thus,

[TABLE]

Finally,

[TABLE]

$(ii)$ Lemma 6.11 entails that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{+++}=\underset{s=0}{\operatorname{Res}}\sum_{(l_{1},l_{2})\in(\mathbb{Z}^{n})^{2}}f(s,l) where

[TABLE]

and $\widetilde{a}_{\alpha,l}:=a_{\alpha_{1},l_{1}}\,a_{\alpha_{2},l_{2}}\,a_{\alpha_{3},-\widehat{l}_{2}}$ with $\widehat{l}_{2}:=l_{1}+l_{2}$.

We use the same technique as in $(i)$:

[TABLE]

and thus,

[TABLE]

where the remain $R(k,l)$ is a term of order at most $-6$ in $k$. Equation (60) gives

[TABLE]

where $r(s,l)$ corresponds to $R(k,l)$. Note that the function

[TABLE]

is a linear combination of functions of the type $H(s,l)$ satisfying the hypothesis of Corollary (2.13). Thus, $r(s,l)$ satisfies (H1) and $f_{\alpha,\mu}(s,l)\sim f_{1}(s,l)$.

Let us compute $f_{1}(s,l)$ modulo (H1)

[TABLE]

Since $g_{\alpha,\mu}(s,l)$ satisfies (H2), we can apply Lemma 2.14 to get

[TABLE]

Recall that $l_{3}:=-l_{1}-l_{2}=-\widehat{l}_{2}$. By (17) and (19),

[TABLE]

We decompose $X_{l}$ in five terms: $X_{l}=2^{m}\ \tfrac{\pi^{2}}{2}\ i\,e^{\tfrac{i}{2}l_{1}\Theta l_{2}}\,\widetilde{a}_{\alpha,l}\,(T_{1}+T_{2}+T_{3}+T_{4}+T_{5})$ where

[TABLE]

With the shorthand $p:=-l_{1}-2l_{3}$, $q:=2l_{1}+l_{3}$, $r:=-p-q=-l_{1}+l_{3}$, we compute each $T_{i}$, and find

[TABLE]

Thus,

[TABLE]

and

[TABLE]

where $S_{1}$, $S_{2}$ and $S_{3}$ correspond to respectively $q^{\alpha_{3}}\delta^{\alpha_{1}\alpha_{2}}$, $r^{\alpha_{2}}\delta^{\alpha_{1}\alpha_{3}}$ and $p^{\alpha_{1}}\delta^{\alpha_{2}\alpha_{3}}$. In $S_{1}$, we permute the $l_{i}$ variables the following way: $l_{1}\mapsto l_{3}$, $l_{2}\mapsto l_{1}$, $l_{3}\mapsto l_{2}$. Therefore, $l_{3}.\Theta\,l_{1}\mapsto l_{3}.\Theta\,l_{1}$ and $q\mapsto r$. With a similar permutation of the $\alpha_{i}$, we see that $S_{1}=S_{2}$. We apply the same principles to prove that $S_{1}=S_{3}$ (using permutation $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{3}$, $l_{3}\mapsto l_{1}$). Thus,

[TABLE]

where $S_{4}$ correspond to $l_{1}$ and $S_{5}$ to $l_{2}$. We permute the $l_{i}$ variables in $S_{5}$ the following way: $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{1}$, $l_{3}\mapsto l_{3}$, with a similar permutation on the $\alpha_{i}$. Since $l_{1}.\Theta\,l_{2}\mapsto-l_{1}.\Theta\,l_{2}$, we finally get

[TABLE]

$(iii)$ Lemma 6.11 entails that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{++++}=\underset{s=0}{\operatorname{Res}}\sum_{(l_{1},l_{2},l_{3})\in(\mathbb{Z}^{n})^{3}}f_{\mu,\alpha}(s,l)\operatorname{Tr}\gamma^{\mu,\alpha} where

[TABLE]

Using (16) and Corollary 2.13 successively, we find

[TABLE]

Since the function ${\sum_{k\in\mathbb{Z}^{n}}}^{\prime}e^{\tfrac{i}{2}\theta}\,\tfrac{k_{\mu_{1}}k_{\mu_{2}}k_{\mu_{3}}k_{\mu_{4}}}{|k|^{s+8}}\,\widetilde{a}_{\alpha,l}$ satisfies (H2), Lemma 2.14 entails that

[TABLE]

Therefore, with (19), we get $X_{l}=\tfrac{\pi^{2}}{12}\,\widetilde{a}_{\alpha,l}\,e^{\tfrac{i}{2}\theta}\,(A+B+C)$, where

[TABLE]

Using successively $\{\gamma^{\mu},\gamma^{\nu}\}=2\delta^{\mu\nu}$ and $\gamma^{\mu}\gamma_{\mu}=2^{m}\ 1_{2^{m}}$, we see that

[TABLE]

Thus, A+B+C=8\ 2^{m}\big{(}\delta^{\alpha_{4}\alpha_{3}}\delta^{\alpha_{2}\alpha_{1}}+\delta^{\alpha_{4}\alpha_{1}}\delta^{\alpha_{3}\alpha_{2}}-2\delta^{\alpha_{4}\alpha_{2}}\delta^{\alpha_{3}\alpha_{1}}\big{)}, and

[TABLE]

By (62), we get

[TABLE]

where

[TABLE]

We now proceed to the following permutations of the $l_{i}$ variables in the $T_{1}$ term : $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{1}$, $l_{3}\mapsto l_{4}$, $l_{4}\mapsto l_{3}$. While $\sum_{i}l_{i}$ is invariant, $\theta$ is modified : $\theta\mapsto l_{2}.\Theta l_{1}+l_{2}.\Theta l_{4}+l_{1}.\Theta l_{4}$. With $\delta_{0,\sum_{i}l_{i}}$ in factor, we can let $l_{4}$ be $-l_{1}-l_{2}-l_{3}$, so that $\theta\mapsto-\theta$. We also permute the $\alpha_{i}$ in the same way. Thus,

[TABLE]

Therefore,

[TABLE]

The same principles are applied to $T_{2}$ and $T_{3}$. Namely, the permutation $l_{1}\mapsto l_{1}$, $l_{2}\mapsto l_{3}$, $l_{3}\mapsto l_{2}$, $l_{4}\mapsto l_{4}$ in $T_{2}$ and the permutation $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{3}$, $l_{3}\mapsto l_{1}$, $l_{4}\mapsto l_{4}$ in $T_{3}$ (the $\alpha_{i}$ variables are permuted the same way) give

[TABLE]

where $\phi:=l_{1}.\Theta\ l_{2}+l_{1}.\Theta\ l_{3}-l_{2}.\Theta\ l_{3}$. Finally, we get

[TABLE]

$(iv)$ Suppose $q=2$. By Lemma 6.11, we get

[TABLE]

where

[TABLE]

and $\eta:={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\sigma_{1}-\sigma_{2})\in\{-1,1\}$. As in the proof of $(i)$, since the presence of the phase does not change the fact that $r(s,l)$ satisfies (H1), we get

[TABLE]

where

[TABLE]

Suppose that $l=0$. Then $f_{2}(s,0)=f_{3}(s,0)=0$ and Proposition 2.1 entails that

[TABLE]

is holomorphic at 0 and so is $f_{\alpha,\mu}(s,0)$.

Since $\tfrac{1}{2\pi}\Theta$ is diophantine, Theorem 2.5 $3$ gives us the result.

Suppose $q=3$. Then Lemma 6.11 implies that

[TABLE]

where

[TABLE]

and $\varepsilon_{i}:={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\sigma_{i}-\sigma_{3})\in\{-1,0,1\}$. By hypothesis $(\varepsilon_{1},\varepsilon_{2})\neq(0,0)$. There are six possibilities for the values of $(\varepsilon_{1},\varepsilon_{2})$, corresponding to the six possibilities for the values of $\sigma$: $(-,-,+)$, $(-,+,+)$, $(+,-,+)$, $(+,+,-)$, $(-,+,-)$, and $(+,-,-)$. As in $(ii)$, we see that

[TABLE]

With $Z:=\{(l_{1},l_{2})\ :\ \varepsilon_{1}l_{1}+\varepsilon_{2}l_{2}=0\}$, Theorem 2.5 $(iii)$ entails that $\sum_{l\in(\mathbb{Z}^{n})^{2}\setminus Z}f_{\mu,\alpha}(s,l)$ is holomorphic at 0. To conclude we need to prove that

[TABLE]

is holomorphic at 0. By definition, $\lambda_{\sigma}=i\sigma_{1}\sigma_{2}\sigma_{3}$ and as a consequence, we check that

[TABLE]

which implies that $\sum_{\sigma}g(\sigma)=0$. The result follows.

Suppose finally that $q=4$. Again, Lemma 6.11 implies that

[TABLE]

where

[TABLE]

and $\varepsilon_{i}:={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\sigma_{i}-\sigma_{4})\in\{-1,0,1\}$. By hypothesis $(\varepsilon_{1},\varepsilon_{2},\varepsilon_{3})\neq(0,0,0)$. There are fourteen possibilities for the values of $(\varepsilon_{1},\varepsilon_{2},\varepsilon_{3})$, corresponding to the fourteen possibilities for the values of $\sigma$: $(-,-,-,+)$, $(-,-,+,+)$, $(-,+,-,+)$, $(+,-,-,+)$, $(-,+,+,+)$, $(+,-,+,+)$, $(+,+,-,+)$, $(+,+,+,-)$, $(-,-,+,-)$, $(-,+,-,-)$, $(+,-,-,-)$, $(-,+,+,-)$, $(+,-,+,-)$ and $(+,+,-,-)$. As in $(ii)$, we see that, with the shorthand $\theta_{\sigma}:=\sigma_{2}l_{1}.\Theta l_{2}+\sigma_{3}(l_{1}+l_{2}).\Theta l_{3}$,

[TABLE]

With $Z_{\sigma}:=\{(l_{1},l_{2},l_{3})\ :\ \sum_{i=1}^{3}\varepsilon_{i}l_{i}=0\}$, Theorem 2.5 $(iii)$, the series $\sum_{l\in(\mathbb{Z}^{n})^{3}\setminus Z_{\sigma}}f_{\mu,\alpha}(s,l)$ is holomorphic at 0. To conclude, we need to prove that

[TABLE]

Let $C$ be the set of the fourteen values of $\sigma$ and $C_{7}$ be the set of the seven first values of $\sigma$ given above. Lemma 6.7 implies

[TABLE]

Thus, in the following, we restrict to these seven values. Let us note $F_{\mu}(s):={\sum}^{\prime}_{k\in\mathbb{Z}^{n}}\tfrac{k_{\mu_{1}}k_{\mu_{2}}k_{\mu_{3}}k_{\mu_{4}}}{|k|^{s+8}}$ so that

[TABLE]

Recall from (62) that

[TABLE]

As a consequence, we get, with $\widetilde{a}_{\alpha,l}:=a_{\alpha_{1},l_{1}}\cdots a_{\alpha_{4},l_{4}}$,

[TABLE]

We proceed to the following change of variable in $T_{1}$: $l_{1}\mapsto l_{1}$, $l_{2}\mapsto l_{3}$, $l_{3}\mapsto l_{2}$, $l_{4}\mapsto l_{4}$. Thus, we get $\theta_{\sigma}\mapsto\psi_{\sigma}:=\sigma_{2}l_{1}.\Theta l_{3}+\sigma_{3}(l_{1}+l_{3}).\Theta l_{2}$, and $\sum_{i=1}^{3}\varepsilon_{i}l_{i}\mapsto\varepsilon_{1}l_{1}+\varepsilon_{3}l_{2}+\varepsilon_{2}l_{3}=:u_{\sigma}(l)$. With a similar permutation on the $\alpha_{i}$, we get

[TABLE]

We proceed to the following change of variable in $T_{2}$: $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{3}$, $l_{3}\mapsto l_{1}$, $l_{4}\mapsto l_{4}$. Thus, we get $\theta_{\sigma}\mapsto\phi_{\sigma}:=\sigma_{2}l_{2}.\Theta l_{3}+\sigma_{3}(l_{2}+l_{3}).\Theta l_{1}$, and $\sum_{i=1}^{3}\varepsilon_{i}l_{i}\mapsto\varepsilon_{3}l_{1}+\varepsilon_{1}l_{2}+\varepsilon_{2}l_{3}=:v_{\sigma}(l)$. After a similar permutation on the $\alpha_{i}$, we get

[TABLE]

Finally, we proceed to the following change of variable in $T_{3}$: $l_{1}\mapsto l_{2}$, $l_{2}\mapsto l_{1}$, $l_{3}\mapsto l_{4}$, $l_{4}\mapsto l_{3}$. Thus, we get $\theta_{\sigma}\mapsto-\theta_{\sigma}$, and $\sum_{i=1}^{3}\varepsilon_{i}l_{i}\mapsto(\varepsilon_{2}-\varepsilon_{3})l_{1}+(\varepsilon_{1}-\varepsilon_{3})l_{2}-\varepsilon_{3}l_{3}=:w_{\sigma}(l)$. With a similar permutation on the $\alpha_{i}$, we get

[TABLE]

As a consequence, we get

[TABLE]

where K_{\sigma}(l_{1},l_{2},l_{3})=\lambda_{\sigma}\big{(}e^{\tfrac{i}{2}\psi_{\sigma}}\,\delta_{u_{\sigma}(l),0}+e^{\tfrac{i}{2}\phi_{\sigma}}\,\delta_{v_{\sigma}(l),0}-e^{\tfrac{i}{2}\theta_{\sigma}}\,\delta_{\sum_{i=1}^{3}\varepsilon_{i}l_{i},0}-e^{-\tfrac{i}{2}\theta_{\sigma}}\,\delta_{w_{\sigma}(l),0}\big{)}.

The computation of $K_{\sigma}(l_{1},l_{2},l_{3})$ for the seven values of $\sigma$ yields

[TABLE]

Thus,

[TABLE]

and

[TABLE]

The following change of variables: $l_{1}\mapsto l_{2}$, $l_{1}\mapsto l_{2}$, $l_{3}\mapsto l_{4}$, $l_{4}\mapsto l_{3}$ gives

[TABLE]

so

[TABLE]

Finally, the change of variables: $l_{2}\mapsto l_{4},l_{4}\mapsto l_{2}$ gives

[TABLE]

which entails that $\sum_{\sigma\in C_{7}}g(\sigma)=0$. ∎

Lemma 6.13**.**

Suppose $n=4$ and $\tfrac{1}{2\pi}\Theta$ diophantine. For any self-adjoint one-form $A$,

[TABLE]

Proof.

By (34) and Lemma 6.6 we get

[TABLE]

By Lemma 6.12 $(iv)$, we see that the crossed terms all vanish. Thus, with Lemma 6.7, we get

[TABLE]

By definition,

[TABLE]

Thus

[TABLE]

One checks that the term in $a^{q}$ of $\tau(F_{\alpha_{1}\alpha_{2}}F^{\alpha_{1}\alpha_{2}})$ corresponds to the term \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\mathbb{A}^{+})^{q} given by Lemma 6.12. For $q=2$, this is

[TABLE]

For $q=3$, we compute the crossed terms:

[TABLE]

which gives the following $a^{3}$-term in $\tau(F_{\alpha_{1}\alpha_{2}}F^{\alpha_{1}\alpha_{2}})$

[TABLE]

For $q=4$, this is

[TABLE]

which corresponds to the term \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits(\mathbb{A}^{+})^{4}. We get finally,

[TABLE]

Equations (65) and (66) yield the result. ∎

Lemma 6.14**.**

Suppose $n=2$. Then, with the same hypothesis as in Lemma 6.11,

[TABLE]

(ii) Suppose $\tfrac{1}{2\pi}\Theta$ diophantine. Then

[TABLE]

Proof.

$(i)$ Lemma 6.11 entails that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{++}=\underset{s=0}{\operatorname{Res}}\sum_{l\in\mathbb{Z}^{2}}-f(s,l) where

[TABLE]

and $\widetilde{a}_{\alpha,l}:=a_{\alpha_{1},l}\,a_{\alpha_{2},-l}$. This time, since $n=2$, it is enough to apply just once (16) to obtain an absolutely convergent series. Indeed, we get with (16)

[TABLE]

and the function $r(s,l):={\sum}^{\prime}_{k\in\mathbb{Z}^{2}}\tfrac{k_{\mu_{1}}(k+l)_{\mu_{2}}(2k.l+|l|^{2})}{|k|^{s+4}|k+l|^{2}}\ \widetilde{a}_{\alpha,l}$ is a linear combination of functions of the type $H(s,l)$ satisfying the hypothesis of Corollary 2.13. As a consequence, $r(s,l)$ satisfies (H1) and

[TABLE]

Note that the function $(s,l)\mapsto h_{\mu,\alpha}(s,l):={\sum}^{\prime}_{k\in\mathbb{Z}^{2}}\tfrac{k_{\mu_{1}}k_{\mu_{2}}}{|k|^{s+4}}\ \widetilde{a}_{\alpha,l}$ satisfies (H2). Thus, Lemma 2.14 yields

[TABLE]

By Proposition 2.16, we get $\underset{s=0}{\operatorname{Res}}\ h_{\mu,\alpha}(s,l)=\delta_{\mu_{1}\mu_{2}}\,\pi\,\widetilde{a}_{\alpha,l}$. Therefore,

[TABLE]

according to (59).

$(ii)$ By Lemma 6.11, we obtain that \mathop{\mathchoice{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.00006pt\hbox{\displaystyle\int}\hfil}\hss}\hbox{-}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}{\hbox{\hbox to0.0pt{\hbox to5.83331pt{\kern-1.4pt\hbox{\textstyle\int}\hfil}\hss}\hbox{\vbox{\hrule width=4.0pt}}\kern 1.00006pt}}}\nolimits\mathbb{A}^{-+}=\underset{s=0}{\operatorname{Res}}\sum_{l\in\mathbb{Z}^{2}}\lambda_{\sigma}f_{\alpha,\mu}(s,l)\operatorname{Tr}(\gamma^{\alpha_{2}}\gamma^{\mu_{2}}\gamma^{\alpha_{1}}\gamma^{\mu_{1}}) where $\lambda_{\sigma}=-(-i)^{2}=1$ and

[TABLE]

and $\eta:={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}(\sigma_{1}-\sigma_{2})=-1$. As in the proof of $(i)$, since the presence of the phase does not change the fact that $r(s,l)$ satisfies (H1), we get

[TABLE]

Since $\tfrac{1}{2\pi}\Theta$ is diophantine, the functions $s\mapsto\sum_{l\in\mathbb{Z}^{2}\backslash\{0\}}g_{\alpha,\mu}(s,l)$ are holomorphic at $s=0$ by Theorem 2.5 $3$. As a consequence,

[TABLE]

Recall from Proposition 2.1 that $\operatorname{Res}_{s=0}\,{\sum}^{\prime}_{k\in\mathbb{Z}^{2}}\,\tfrac{k_{i}k_{j}}{|k|^{s+4}}=\delta_{ij}\,\pi$. Thus, again with (59),

[TABLE]

Lemma 6.15**.**

Suppose $n=2$ and $\tfrac{1}{2\pi}\Theta$ diophantine. For any self-adjoint one-form $A$,

[TABLE]

Proof.

As in Lemma 6.13, we use (34) and Lemma 6.6 so the result follows from Lemma 6.14. ∎

6.1.2 Odd dimensional case

Lemma 6.16**.**

Suppose $n$ odd and $\tfrac{1}{2\pi}\Theta$ diophantine. Then for any self-adjoint 1-form $A$ and $\sigma\in\{-,+\}^{q}$ with $2\leq q\leq n$,

[TABLE]

Proof.

Since $\mathbb{A}^{\sigma}\in\Psi_{1}(\mathcal{A})$, Lemma 5.11 with $k=n$ gives the result. ∎

Corollary 6.17**.**

With the same hypothesis of Lemma 6.16, for any self-adjoint one-form $A$, $\zeta_{D_{A}}(0)-\zeta_{D}(0)=0.$

Proof.

As in Lemma 6.13, we use (34) and Lemma 6.6 so the result follows from Lemma 6.16. ∎

6.2 Proof of the main result

Proof of Theorem 6.1..

$(i)$ By (5) and Proposition 5.5, we get

[TABLE]

where $\Phi_{2}={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}\int_{0}^{\infty}\Phi(t)\,dt$. By Lemma 6.15, $\zeta_{D_{A}}(0)-\zeta_{D}(0)=0$ and from Proposition 5.4, $\zeta_{D}(0)=0$, so we get the result.

$(ii)$ Similarly, $\SS(\mathcal{D}_{A},\Phi,\Lambda)\,=8\pi^{2}\,\Phi_{4}\,\Lambda^{4}+\Phi(0)\,\zeta_{D_{A}}(0)+\mathcal{O}(\Lambda^{-2})$ with $\Phi_{4}={\mathchoice{\tfrac{1}{2}}{\tfrac{1}{2}}{{\scriptstyle\frac{1}{2}}}{{\scriptstyle\frac{1}{2}}}}\int_{0}^{\infty}\Phi(t)\,t\,dt$. Lemma 6.13 implies that $\zeta_{D_{A}}(0)-\zeta_{D}(0)=-c\,\tau(F_{\mu\nu}F^{\mu\nu})$ and by Proposition 5.4, $\zeta_{D_{A}}(0)=-c\,\tau(F_{\mu\nu}F^{\mu\nu})$ leading to the result.

$(iii)$ is a direct consequence of (5), Propositions 5.4, 5.5, and Corollary 6.17. ∎

Appendix A Appendix

A.1 Proof of Lemma 3.3

$(i)$ We have $|D|T|D|^{-1}=T+\delta(T)|D|^{-1}$ and $|D|^{-1}T|D|=T-|D|^{-1}\delta(T)$. A recurrence proves that for any $k\in\mathbb{N}$, $|D|^{k}T|D|^{-k}=\sum_{q=0}^{k}\genfrac{(}{)}{0.0pt}{1}{q}{k}\,\delta^{q}(T)|D|^{-q}$ and we get $|D|^{-k}T|D|^{k}=\sum_{q=0}^{k}(-1)^{q}\genfrac{(}{)}{0.0pt}{1}{q}{k}\,|D|^{-q}\delta^{q}(T)$.

As a consequence, since $T$, $|D|^{-q}$ and $\delta^{q}(T)$ are in $OP^{0}$ for any $q\in\mathbb{N}$, for any $k\in\mathbb{Z}$, $|D|^{k}T|D|^{-k}\in OP^{0}$. Let us fix $p\in\mathbb{N}_{0}$ and define $F_{p}(s):=\delta^{p}(|D|^{s}T|D|^{-s})$ for $s\in\mathbb{C}$. Since for $k\in\mathbb{Z}$, $F_{p}(k)$ is bounded, a complex interpolation proves that $F_{p}(s)$ is bounded, which gives $|D|^{s}T|D|^{-s}\in OP^{0}$.

$(ii)$ Let $T\in OP^{\alpha}$ and $T^{\prime}\in OP^{\beta}$. Thus, $T|D|^{-\alpha}$, $T^{\prime}|D|^{-\beta}$ are in $OP^{0}$. By $(i)$ we get $|D|^{\beta}T|D|^{-\alpha}|D|^{-\beta}\in OP^{0}$, so $T^{\prime}|D|^{-\beta}|D|^{\beta}T|D|^{-\beta-\alpha}\in OP^{0}$. Thus, $T^{\prime}T|D|^{-(\alpha+\beta)}\in OP^{0}$.

$(iii)$ For $T\in OP^{\alpha}$, $|D|^{\alpha-\beta}$ and $T|D|^{-\alpha}$ are in $OP^{0}$, thus $T|D|^{-\beta}=T|D|^{-\alpha}|D|^{\alpha-\beta}\in OP^{0}$.

$(iv)$ follows from $\delta(OP^{0})\subseteq OP^{0}$.

$(v)$ Since $\nabla(T)=\delta(T)|D|+|D|\delta(T)-[P_{0},T]$, the result follows from $(ii)$, $(iv)$ and the fact that $P_{0}$ is in $OP^{-\infty}$.

A.2 Proof of Lemma 3.6

The non-trivial part of the proof is the stability under the product of operators. Let $T,T^{\prime}\in\Psi(\mathcal{A})$. There exist $d,d^{\prime}\in\mathbb{Z}$ such that for any $N\in\mathbb{N}$, $N>|d|+|d^{\prime}|$, there exist $P,P^{\prime}$ in $\mathcal{D}(\mathcal{A})$, $p,p^{\prime}\in\mathbb{N}_{0}$, $R\in OP^{-N-d^{\prime}}$, $R^{\prime}\in OP^{-N-d}$ such that $T=PD^{-2p}+R$, $T^{\prime}=P^{\prime}D^{-2p^{\prime}}+R^{\prime}$, $PD^{-2p}\in OP^{d}$ and $P^{\prime}D^{-2p^{\prime}}\in OP^{d^{\prime}}$.

Thus, $TT^{\prime}=PD^{-2p}P^{\prime}D^{-2p^{\prime}}+RP^{\prime}D^{-2p^{\prime}}+PD^{-2p}R^{\prime}+RR^{\prime}$.

We also have $RP^{\prime}D^{-2p^{\prime}}\in OP^{-N-d^{\prime}+d^{\prime}}=OP^{-N}$ and similarly, $PD^{-2p}R^{\prime}\in OP^{-N}$. Since $RR^{\prime}\in OP^{-2N}$, we get

[TABLE]

If $p=0$, then $TT^{\prime}\sim QD^{-2p^{\prime}}\mod OP^{-N}$ where $Q=PP^{\prime}\in\mathcal{D}(\mathcal{A})$ and $QD^{-2p^{\prime}}\in OP^{d+d^{\prime}}$. Suppose $p\neq 0$. A recurrence proves that for any $q\in\mathbb{N}_{0}$,

[TABLE]

By Lemma 3.3 $(v)$, the remainder is in $OP^{d^{\prime}+2p^{\prime}-q-3}$, since $P^{\prime}\in OP^{d^{\prime}+2p^{\prime}}$. Another recurrence gives for any $q\in\mathbb{N}_{0}$,

[TABLE]

Thus, with $q_{N}=N+d+d^{\prime}-1$,

[TABLE]

The last sum can be written $Q_{N}D^{-2r_{N}}$ where $r_{N}:=p\,q_{N}+(p+p^{\prime})$. Since $Q_{N}\in\mathcal{D}(\mathcal{A})$ and $Q_{N}D^{-2r_{N}}\in OP^{d+d^{\prime}}$, the result follows.

A.3 Proof of Proposition 3.11

Let $P\in OP^{k_{1}},\,Q\in OP^{k_{2}}\in\Psi(\mathcal{A})$. With [Q,|D|^{-s}]=\big{(}Q-\sigma_{-s}(Q)\big{)}\,|D|^{-s} and the equivalence $Q-\sigma_{-s}(Q)\sim-\sum_{r=1}^{N}g(-s,r)\,\varepsilon^{r}(Q)\mod OP^{-N-1+k_{2}}$, we get

[TABLE]

which gives, if we choose $N=n+k_{1}+k_{2}$,

[TABLE]

By hypothesis s\mapsto\operatorname{Tr}\big{(}P\varepsilon^{r}(Q)|D|^{-s}\big{)} has only simple poles. Thus, since $s=0$ is a zero of the analytic function $s\mapsto g(-s,r)$ for any $r\geq 1$, we have \underset{s=0}{\operatorname{Res}}\ g(-s,r)\,\operatorname{Tr}\big{(}P\varepsilon^{r}(Q)|D|^{-s}\big{)}=0, which entails that \underset{s=0}{\operatorname{Res}}\ \operatorname{Tr}\big{(}P[Q,|D|^{-s}]\big{)}=0 and thus

[TABLE]

When $s\in\mathbb{C}$ with $\Re(s)>2\max(k_{1}+n+1,k_{2})$, the operator $P|D|^{-s/2}$ is trace-class while $|D|^{-s/2}Q$ is bounded, so \operatorname{Tr}\big{(}P|D|^{-s}Q\big{)}=\operatorname{Tr}\big{(}|D|^{-s/2}QP|D|^{-s/2}\big{)}=\operatorname{Tr}\big{(}\sigma_{-s/2}(QP)|D|^{-s}\big{)}. Thus, using (29) again,

[TABLE]

As before, for any $r\geq 1$, \underset{s=0}{\operatorname{Res}}\ g(-s/2,r)\operatorname{Tr}\big{(}\varepsilon^{r}(QP)|D|^{-s}\big{)}=0 since $g(0,r)=0$ and the spectral triple is simple. Finally,

[TABLE]

Acknowledgments

We thank Pierre Duclos, Emilio Elizalde, Victor Gayral, Thomas Krajewski, Sylvie Paycha, Joe Varilly, Dmitri Vassilevich and Antony Wassermann for helpful discussions and Stéphane Louboutin for his help with Proposition 2.16.

A. Sitarz would like to thank the CPT-Marseilles for its hospitality and the Université de Provence for its financial support and acknowledge the support of Alexander von Humboldt Foundation through the Humboldt Fellowship.