Spectral action on noncommutative torus
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 Diophan-
tine 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
http://arxiv.org/abs/0704.0564v2
1 Introduction
The spectral action introduced by Chamseddine–Connes plays an important role [3] in noncom-
mutative geometry. More precisely, given a spectral triple (A,H,D) where A is an algebra acting
on the Hilbert space H and D is a Dirac-like operator (see [8, 23]), they proposed a physical
action depending only on the spectrum of the covariant Dirac operator
DA := D +A+ ǫ JAJ−1 (1)
where A is a one-form represented on H, so has the decomposition
ai[D, bi], (2)
with ai, bi ∈ A, J is a real structure on the triple corresponding to charge conjugation and
ǫ ∈ { 1,−1 } depending on the dimension of this triple and comes from the commutation relation
JD = ǫDJ. (3)
This action is defined by
S(DA,Φ,Λ) := Tr
Φ(DA/Λ)
where Φ 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 S counts the spectral
values of |DA| less than the mass scale Λ (note that the resolvent of DA is compact since, by
assumption, the same is true for D, see Lemma 3.1 below).
In [18], the spectral action on NC-tori has been computed only for operators of the form D+A
and computed for DA 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,
S(DA,Φ,Λ) =
0<k∈Sd+
− |DA|−k +Φ(0) ζDA(0) +O(Λ−1) (5)
where DA = DA + PA, PA the projection on KerDA, Φk = 12
Φ(t) tk/2−1 dt and Sd+ is the
strictly positive part of the dimension spectrum of (A,H,D). As we will see, Sd+ = { 1, 2, · · · , n }
|DA|−n =
|D|−n. Moreover, the coefficient ζDA(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 D +A is also invertible) that
ζD+A(0) − ζD(0) =
(−1)q
−(AD−1)q, (6)
using ζX(s) = Tr(|X|−s). We will see how this formula can be extended to the case a noninvert-
ible Dirac operator and noninvertible perturbation of the form D+ Ã where Ã := A+ εJAJ−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 [2–5,17,18,20,22,28,35–38].
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 noncom-
mutative spectral action of DA has terms proportional to the spectral action of D + A on the
commutative torus.
2 Residues of series and integral, holomorphic continuation, etc
Notations:
In the following, the prime in
means that we omit terms with division by zero in the summand.
Bn (resp. Sn−1) is the closed ball (resp. the sphere) of Rn with center 0 and radius 1 and the
Lebesgue measure on Sn−1 will be noted dS.
For any x = (x1, . . . , xn) ∈ Rn we denote by |x| =
x21 + · · ·+ x2n the euclidean norm and
|x|1 := |x1|+ · · · + |xn|.
N = {1, 2, . . . } is the set of positive integers and N0 = N ∪ {0} the set of non negative integers.
By f(x, y) ≪y g(x) uniformly in x, we mean that |f(x, y)| ≤ 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) =
j=0 Pj(X) ∈ C[X1, · · · ,Xn] be a polynomial function where Pj is
the homogeneous part of P of degree j. The function
ζP (s) :=
P (k)
, s ∈ C
has a meromorphic continuation to the whole complex plane C.
Moreover ζP (s) is not entire if and only if PP := {j :
u∈Sn−1
Pj(u) dS(u) 6= 0} 6= ∅. In that
case, ζP has only simple poles at the points j + n, j ∈ PP , with
s=j+n
ζP (s) =
u∈Sn−1
Pj(u) dS(u).
The proof of this theorem is based on the following lemmas.
Lemma 2.2. For any polynomial P ∈ C[X1, . . . ,Xn] of total degree δ(P ) :=
i=1 degXiP and
any α ∈ Nn0 , we have
P (x)|x|−s
≪P,α,n (1 + |s|)|α|1 |x|−σ−|α|1+δ(P )
uniformly in x ∈ Rn verifying |x| ≥ 1, where σ = ℜ(s).
Proof. By linearity, we may assume without loss of generality that P (X) = Xγ is a monomial.
It is easy to prove (for example by induction on |α|1) that for all α ∈ Nn0 and x ∈ Rn \ {0}:
|x|−s
β,µ∈Nn0
β+2µ=α
|β|1+|µ|1
) (|β|1+|µ|1)!
β! µ!
|x|σ+2(|β|1+|µ|1)
It follows that for all α ∈ Nn0 , we have uniformly in x ∈ Rn verifying |x| ≥ 1:
|x|−s
≪α,n (1 + |s|)|α|1 |x|−σ−|α|1 . (7)
By Leibniz formula and (7), we have uniformly in x ∈ Rn verifying |x| ≥ 1:
xγ |x|−s
∂β(xγ) ∂α−β
|x|−s
≪γ,α,n
β≤α;β≤γ
xγ−β (1 + |s|)|α|1−|β|1 |x|−σ−|α|1+|β|1
≪γ,α,n (1 + |s|)|α|1 |x|−σ−|α|1+|γ|1 .
Lemma 2.3. Let P ∈ C[X1, . . . ,Xn] be a polynomial of degree d. Then, the difference
∆P (s) :=
P (k)
Rn\Bn
P (x)
which is defined for ℜ(s) > d+ n, extends holomorphically on the whole complex plane C.
Proof. We fix in the sequel a function ψ ∈ C∞(Rn,R) verifying for all x ∈ Rn
0 ≤ ψ(x) ≤ 1, ψ(x) = 1 if |x| ≥ 1 and ψ(x) = 0 if |x| ≤ 1/2.
The function f(x, s) := ψ(x) P (x) |x|−s, x ∈ Rn and s ∈ C, is in C∞(Rn × 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 ∆P (s) extends holomorphically on the whole complex plane 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 : R → C of class
CN+1 verifying lim|t|→+∞ h(k)(t) = 0 and
|h(k)(t)| dt < +∞ for any k = 0 . . . , N + 1, that we
have ∑
h(k) =
h(t) +
(−1)N
(N+1)!
BN+1(t) h
(N+1)(t) dt
where BN+1 is the Bernoulli function of order N + 1 (it is a bounded periodic function.)
Fix m′ ∈ Zn−1 and s ∈ C. Applying this to the function h(t) := ψ(m′, t) P (m′, t) |(m′, t)|−s (we
use Lemma 2.2 to verify hypothesis), we obtain that for any N ∈ N0:
ψ(m′,mn) P (m
′,mn) |(m′,mn)|−s =
ψ(m′, t) P (m′, t) |(m′, t)|−s dt+RN (m′; s) (8)
where RN (m′; s) := (−1)
(N+1)!
BN+1(t)
N+1 (ψ(m
′, t) P (m′, t) |(m′, t)|−s) dt.
By Lemma 2.2,
∣∣∣BN+1(t) ∂
ψ(m′, t) P (m′, t) |(m′, t)|−s
) ∣∣∣ dt ≪P,n,N (1 + |s|)N+1 (|m′|+ 1)−σ−N+δ(P ).
m′∈Zn−1 RN (m′; s) converges absolutely and define a holomorphic function in the half
plane {σ = ℜ(s) > δ(P ) + n−N}.
Since N is an arbitrary integer, by letting N → ∞ and using (8) above, we conclude that:
(m′,mn)∈Zn−1×Z
ψ(m′,mn) P (m
′,mn) |(m′,mn)|−s−
m′∈Zn−1
ψ(m′, t) P (m′, t) |(m′, t)|−s dt
has a holomorphic continuation to the whole complex plane C.
After n iterations, we obtain that
ψ(m) P (m) |m|−s −
ψ(x) P (x) |x|−s dx
has a holomorphic continuation to the whole C.
To finish the proof of Lemma 2.3, it is enough to notice that:
• ψ(0) = 0 and ψ(m) = 1, ∀m ∈ Zn \ {0};
• s 7→
ψ(x) P (x) |x|−s dx =
{x∈Rn:1/2≤|x|≤1}
ψ(x) P (x) |x|−s dx is a holomorphic
function on C.
Proof of Theorem 2.1. Using the polar decomposition of the volume form dx = ρn−1 dρ dS in
Rn, we get for ℜ(s) > d+ n,
Rn\Bn
Pj(x)
ρj+n−1
Pj(u) dS(u) =
j+n−s
Pj(u) dS(u).
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 δ > 0. A vector a ∈ Rn is said to be δ−diophantine if there exists c > 0
such that |q.a−m| ≥ c |q|−δ, ∀q ∈ Zn \ { 0 } and ∀m ∈ Z.
We note BV(δ) the set of δ−diophantine vectors and BV := ∪δ>0BV(δ) the set of diophantine
vectors.
(ii) A matrix Θ ∈ Mn(R) (real n × n matrices) will be said to be diophantine if there exists
u ∈ Zn such that tΘ(u) is a diophantine vector of Rn.
Remark. A classical result from Diophantine approximation asserts that for all δ > n, the
Lebesgue measure of Rn \ BV(δ) is zero (i.e almost any element of Rn is δ−diophantine.)
Let Θ ∈ Mn(R). If its row of index i is a diophantine vector of Rn (i.e. if Li ∈ BV) then
tΘ(ei) ∈ BV and thus Θ is a diophantine matrix. It follows that almost any matrix of Mn(R) ≈
is diophantine.
The goal of this section is to show the following
Theorem 2.5. Let P ∈ C[X1, · · · ,Xn] be a homogeneous polynomial of degree d and let b be in
S(Zn × · · · × Zn) (q times, q ∈ N). Then,
(i) Let a ∈ Rn. We define fa(s) :=
P (k)
e2πik.a.
1. If a ∈ Zn, then fa has a meromorphic continuation to the whole complex plane C.
Moreover if S is the unit sphere and dS its Lebesgue measure, then fa is not entire if and only
u∈Sn−1
P (u) dS(u) 6= 0. In that case, fa has only a simple pole at the point d + n, with
s=d+n
fa(s) =
u∈Sn−1
P (u) dS(u).
2. If a ∈ Rn \ Zn, then fa(s) extends holomorphically to the whole complex plane C.
(ii) Suppose that Θ ∈ Mn(R) is diophantine. For any (εi)i ∈ {−1, 0, 1}q , the function
g(s) :=
l∈(Zn)q
b(l) fΘ
i εili
extends meromorphically to the whole complex plane C with only one possible pole on s = d+n.
Moreover, if we set Z := {l ∈ (Zn)q :
i=1 εili = 0} and V :=
l∈Z b(l), then
1. If V
P (u) dS(u) 6= 0, then s = d+ n is a simple pole of g(s) and
s=d+n
g(s) = V
u∈Sn−1
P (u) dS(u).
2. If V
P (u) dS(u) = 0, then g(s) extends holomorphically to the whole complex plane C.
(iii) Suppose that Θ ∈ Mn(R) is diophantine. For any (εi)i ∈ {−1, 0, 1}q , the function
g0(s) :=
l∈(Zn)q\Z
b(l) fΘ
i=1 εili
where Z := {l ∈ (Zn)q :
i=1 εili = 0} extends holomorphically to the whole complex plane C.
Proof of Theorem 2.5: First we remark that
If a ∈ Zn then fa(s) =
P (k)
. So, the point (i.1) follows from Theorem 2.1;
g(s) :=
l∈(Zn)q\Z b(l) fΘ
i εili
(s) +
l∈Z b(l)
P (k)
. 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:
F := { P (X)
(X21+···+X
r/2 : P (X) ∈ C[X1, . . . ,Xn] and r ∈ N0}.
We set g =deg(G) =deg(P )− r ∈ Z, the degree of G = P (X)
(X21+···+X
r/2 ∈ F .
By convention we set deg(0) = −∞.
Lemma 2.6. Let a ∈ Rn. We assume that d (a.u,Z) := infm∈Z |a.u−m| > 0 for some u ∈ Zn.
For all G ∈ F , we define formally,
F0(G; a; s) :=
e2πi k.a and F1(G; a; s) :=
(|k|2+1)s/2
e2πi k.a.
Then for all N ∈ N, all G ∈ F and all i ∈ {0, 1}, there exist positive constants Ci := Ci(G,N, u),
Bi := Bi(G,N, u) and Ai := Ai(G,N, u) such that s 7→ Fi(G;α; s) extends holomorphically to
the half-plane {ℜ(s) > −N} and verifies in it:
Fi(G; a; s) ≤ Ci(1 + |s|)Bi
d (a.u,Z)
Remark 2.7. The important point here is that we obtain an explicit bound of Fi(G;α; s) in
{ℜ(s) > −N} which depends on the vector a only through d(a.u,Z), so depends on u and
indirectly on a (in the sequel, a will vary.) In particular the constants Ci := Ci(G,N, u),
Bi = Bi(G,N) and Ai := Ai(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 ∈ N0 be a fixed integer, and set g0 := n+N + 1.
We will prove Lemma 2.6 by induction on g =deg(G) ∈ Z. More precisely, in order to prove
case i = 1, it suffices to prove that:
Lemma 2.6 is true for all G ∈ F verifying deg(G) ≤ −g0.
Let g ∈ Z with g ≥ −g0+1. If Lemma 2.6 is true for all G ∈ F such that deg(G) ≤ g−1,
then it is also true for all G ∈ F satisfying deg(G) = g.
• Step 1: Checking Lemma 2.6 for deg(G) ≤ −g0 := −(n+N + 1).
Let G(X) =
P (X)
(X21+···+X
r/2 ∈ F verifying deg(G) ≤ −g0. It is easy to see that we have
uniformly in s = σ + iτ ∈ C and in k ∈ Zn:
|G(k) e2πi k.a|
(|k|2+1)σ/2
|P (k)|
(|k|2+1)(r+σ)/2
≪G 1(|k|2+1)(r+σ−deg(P ))/2 ≪G
(|k|2+1)(σ−deg(G))/2
≪G 1(|k|2+1)(σ+g0)/2 .
It follows that F1(G; a; s) =
(|k|2+1)s/2
e2πi k.a converges absolutely and defines a holo-
morphic function in the half plane {σ > −N}. Therefore, we have for any s ∈ {ℜ(s) > −N}:
|F1(G; a; s)| ≪G
(|k|2+1)(−N+g0)/2
(|k|2+1)(n+1)/2
≪G 1.
Thus, Lemma 2.6 is true when deg(G) ≤ −g0.
• Step 2: Induction.
Now let g ∈ Z satisfying g ≥ −g0+1 and suppose that Lemma 2.6 is valid for all G ∈ F verifying
deg(G) ≤ g − 1. Let G ∈ F with deg(G) = g. We will prove that G also verifies conclusions of
Lemma 2.6:
There exist P ∈ C[X1, . . . ,Xn] of degree d ≥ 0 and r ∈ N0 such that G(X) = P (X)(X21+···+X2n+1)r/2
and g =deg(G) = d− r.
Since G(k) ≪ (|k|2 +1)g/2 uniformly in k ∈ Zn, we deduce that F1(G; a; s) converges absolutely
in {σ = ℜ(s) > n+ g}.
Since k 7→ k + u is a bijection from Zn into Zn, it follows that we also have for ℜ(s) > n+ g
F1(G; a; s) =
P (k)
(|k|2+1)(s+r)/2
e2πi k.a =
P (k+u)
(|k+u|2+1)(s+r)/2
e2πi (k+u).a
= e2πi u.a
P (k+u)
(|k|2+2k.u+|u|2+1)(s+r)/2
e2πi k.a
= e2πi u.a
α∈Nn0 ;|α|1=α1+···+αn≤d
∂αP (k)
(|k|2+2k.u+|u|2+1)(s+r)/2
e2πi k.a
= e2πi u.a
|α|1≤d
∂αP (k)
(|k|2+1)(s+r)/2
2k.u+|u|2
(|k|2+1)
)−(s+r)/2
e2πi k.a.
Let M := sup(N + n+ g, 0) ∈ N0. We have uniformly in k ∈ Zn
2k.u+|u|2
(|k|2+1)
)−(s+r)/2
−(s+r)/2
)(2k.u+|u|2)j
(|k|2+1)j
+OM,u
( (1+|s|)M+1
(|k|2+1)(M+1)/2
Thus, for σ = ℜ(s) > n+ d,
F1(G; a; s) = e
2πi u.a
|α|1≤d
∂αP (k)
(|k|2+1)(s+r)/2
2k.u+|u|2
(|k|2+1)
)−(s+r)/2
e2πi k.a
= e2πi u.a
|α|1≤d
−(s+r)/2
∂αP (k)(2k.u+|u|2)
(|k|2+1)(s+r+2j)/2
e2πi k.a
+OG,M,u
(1 + |s|)M+1
(|k|2+1)(σ+M+1−g)/2
. (9)
Set I := {(α, j) ∈ Nn0 × {0, . . . ,M} | |α|1 ≤ d} and I∗ := I \ { (0, 0) }.
Set also G(α,j);u(X) :=
∂αP (X)(2X.u+|u|2)
(|X|2+1)(r+2j)/2
∈ F for all (α, j) ∈ I∗.
Since M ≥ N + n+ g, it follows from (9) that
(1 − e2πi u.a) F1(G; a; s) = e2πi u.a
(α,j)∈I∗
−(s+r)/2
G(α,j);u;α; s
+RN (G; a;u; s) (10)
where s 7→ RN (G; a;u; s) is a holomorphic function in the half plane {σ = ℜ(s) > −N}, in
which it satisfies the bound RN (G; a;u; s) ≪G,N,u 1.
Moreover it is easy to see that, for any (α, j) ∈ I∗,
G(α,j);u
= deg(∂αP ) + j − (r + 2j) ≤ d− |α|1 + j − (r + 2j) = g − |α|1 − j ≤ g − 1.
Relation (10) and the induction hypothesis imply then that
(1− e2πi u.a) F1(G; a; s) verifies the conclusions of Lemma 2.6. (11)
Since |1− e2πi u.a| = 2| sin(πu.a)| ≥ d (u.a,Z), then (11) implies that F1(G; a; s) satisfies conclu-
sions 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 ∈ N be a fixed integer. Let G(X) = P (X)
(X21+···+X
r/2 ∈ F and g = deg(G) = d− r where
d ≥ 0 is the degree of the polynomial P . Set also M := sup(N + g + n, 0) ∈ N0.
Since P (k) ≪ |k|d for k ∈ Zn\{ 0 }, it follows that F0(G; a; s) and F1(G; a; s) converge absolutely
in the half plane {σ = ℜ(s) > n+ g}.
Moreover, we have for s = σ + iτ ∈ C verifying σ > n+ g:
F0(G; a; s) =
k∈Zn\{ 0 }
(|k|2+1−1)s/2
e2πi k.a =
′ G(k)
(|k|2+1)s/2
|k|2+1
)−s/2
e2πi k.a
(−1)j G(k)
(|k|2+1)(s+2j)/2
e2πi k.a
(1 + |s|)M+1
′ |G(k)|
(|k|2+1)(σ+2M+2)/2
(−1)jF1(G; a; s + 2j)
(1 + |s|)M+1
′ |G(k)|
(|k|2+1)(σ+2M+2)/2
. (12)
In addition we have uniformly in s = σ + iτ ∈ C verifying σ > −N ,
′ |G(k)|
(|k|2+1)(σ+2M+2)/2
′ |k|g
(|k|2+1)(−N+2M+2)/2
|k|n+1
< +∞.
So (12) 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 ∈ Rn \ Zn, there exists i0 ∈ {1, . . . , n} such that ai0 6∈ Z. In particular d(a.ei0 ,Z) =
d(ai0 ,Z) > 0. Therefore, a satisfies the assumption of Lemma 2.6 with u = ei0 . Thus, for all
N ∈ N, s 7→ fa(s) = F0(P ; a; s) has a holomorphic continuation to the half-plane {ℜ(s) > −N}.
It follows, by letting N → ∞, that s 7→ fa(s) has a holomorphic continuation to the whole
complex plane C.
2.2.4 Proof of item (iii) of Theorem 2.5:
Let Θ ∈ Mn(R), (εi)i ∈ {−1, 0, 1}q and b ∈ S(Zn × Zn). We assume that Θ is a diophantine
matrix. Set Z := { l = (l1, . . . , lq) ∈ (Zn)q :
i εili = 0 } and P ∈ C[X1, . . . ,Xn] of degree
d ≥ 0.
It is easy to see that for σ > n+ d:
l∈(Zn)q\Z
|b(l)|
′ |P (k)|
|e2πi k.Θ
i εili | ≪P
l∈(Zn)q\Z
|b(l)|
|k|σ−d
l∈(Zn)q\Z
|b(l)|
< +∞.
g0(s) :=
l∈(Zn)q\Z
b(l) fΘ
i εili
(s) =
l∈(Zn)q\Z
′ P (k)
e2πi k.Θ
i εili
converges absolutely in the half plane {ℜ(s) > n+ d}.
Moreover with the notations of Lemma 2.6, we have for all s = σ + iτ ∈ C verifying σ > n+ d:
g0(s) =
l∈(Zn)q\Z
b(l)fΘ
i εili
(s) =
l∈(Zn)q\Z
b(l)F0(P ; Θ
εili; s) (13)
But Θ is diophantine, so there exists u ∈ Zn and δ, c > 0 such
|q. tΘu−m| ≥ c (1 + |q|)−δ , ∀q ∈ Zn \ { 0 }, ∀m ∈ Z.
We deduce that ∀l ∈ (Zn)q \ Z,
.u−m| = |
.tΘu−m| ≥ c
1 + |
εili|
)−δ ≥ c (1 + |l|)−δ.
It follows that there exists u ∈ Zn, δ > 0 and c > 0 such that
∀l ∈ (Zn)q \ Z, d
εili).u;Z
≥ c (1 + |l|)−δ . (14)
Therefore, for any l ∈ (Zn)q \Z, the vector a = Θ
i εili verifies the assumption of Lemma 2.6
with the same u. Moreover δ and c in (14) are also independent on l.
We fix now N ∈ N. Lemma 2.6 implies that there exist positive constants C0 := C0(P,N, u),
B0 := Bi(P,N, u) and A0 := A0(P,N, u) such that for all l ∈ (Zn)q \ Z, s 7→ F0(P ; Θ
i εili; s)
extends holomorphically to the half plane {ℜ(s) > −N} and verifies in it the bound
F0(P ; Θ
εili; s) ≤ C0 (1 + |s|)B0 d
εili).u;Z
This and (14) imply that for any compact set K included in the half plane {ℜ(s) > −N},
there exist two constants C := C(P,N, c, δ, u,K) and D := D(P,N, c, δ, u) (independent on
l ∈ (Zn)q \ Z) such that
∀s ∈ K and ∀l ∈ (Zn)q \ Z, F0(P ; Θ
εili; s) ≤ C (1 + |l|)D . (15)
It follows that s 7→
l∈(Zn)q\Z b(l)F0(P ; Θ
iεili; s) has a holomorphic continuation to the half
plane {ℜ(s) > −N}.
This and ( 13) imply that s 7→ g0(s) =
l∈(Zn)q\Z b(l)fΘ
i εili
(s) has a holomorphic contin-
uation to {ℜ(s) > −N}. Since N is an arbitrary integer, by letting N → ∞, it follows that
s 7→ g0(s) has a holomorphic continuation to the whole complex plane 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 θ ∈ R \Q. We know (see this reflection formula in [15, p. 6]) that for any l ∈ Z \ {0},
gθl(s) :=
e2πiθlk
s−1/2
) hθl(1− s) where hθl(s) :=
|θl+k|s
So, for any (al) ∈ S(Z), the existence of meromorphic continuation of g0(s) :=
l∈Z al gθl(s) is
equivalent to the existence of meromorphic continuation of
h0(s) :=
al hθl(s) =
|θl+k|s
So, for at least one σ0 ∈ R, we must have |al||θl+k|σ0 = O(1) uniformly in k, l ∈ Z
It follows that for any (al) ∈ S(Z), |θl + k| ≫ |al|1/σ0 uniformly in k, l ∈ Z∗. Therefore, our
Diophantine condition seems to be necessary.
2.2.5 Commutation between sum and residue
Let p ∈ N. Recall that S((Zn)p) is the set of the Schwartz sequences on (Zn)p. In other words,
b ∈ S((Zn)p) if and only if for all r ∈ N0, (1 + |l1|2 + · · · |lp|2)r |b(l1, · · · , lp)|2 is bounded on
(Zn)p. We note that if Q ∈ R[X1, · · · ,Xnp] is a polynomial, (aj) ∈ S(Zn)p, b ∈ S(Zn) and φ
a real-valued function, then l := (l1, · · · , lp) 7→ ã(l) b(−l̂p)Q(l) eiφ(l) is a Schwartz sequence on
(Zn)p, where
ã(l) := a1(l1) · · · ap(lp),
l̂i := l1 + . . .+ li.
In the following, we will use several times the fact that for any (k, l) ∈ (Zn)2 such that k 6= 0
and k 6= −l, we have
|k + l|2 =
|k|2 −
2k.l + |l|2
|k|2|k + l|2 . (16)
Lemma 2.9. There exists a polynomial P ∈ R[X1, · · · ,Xp] of degree 4p and with positive
coefficients such that for any k ∈ Zn, and l := (l1, · · · , lp) ∈ (Zn)p such that k 6= 0 and k 6= −l̂i
for all 1 ≤ i ≤ p, the following holds:
|k + l̂1|2 . . . |k + l̂p|2
≤ 1|k|2p P (|l1|, · · · , |lp|).
Proof. Let’s fix i such that 1 ≤ i ≤ p. Using two times (16), Cauchy–Schwarz inequality and
the fact that |k + l̂i|2 ≥ 1, we get
|k+bli|2
2|k||bli|+|bli|
(2|k||bli|+|bli|
|k|4|k+bli|2
|l̂i|+
|l̂i|2 + 4|k|3 |l̂i|
3 + 1
|l̂i|4.
Since |k| ≥ 1, and |l̂i|j ≤ |l̂i|4 if 1 ≤ j ≤ 4, we find
|k+bli|2
|l̂i|j ≤ 5|k|2
1 + 4|l̂i|4
1 + 4(
|lj |)4
|k+bl1|2...|k+blp|2
|k|2p
1 + 4(
|lj |)4
Taking P (X1, · · · ,Xp) := 5p
1 + 4(
j=1Xj)
now gives the result.
Lemma 2.10. Let b ∈ S((Zn)p), p ∈ N, Pj ∈ R[X1, · · · ,Xn] be a homogeneous polynomial
function of degree j, k ∈ Zn, l := (l1, · · · , lp) ∈ (Zn)p, r ∈ N0, φ be a real-valued function on
Zn × (Zn)p and
h(s, k, l) :=
b(l)Pj(k) e
iφ(k,l)
|k|s+r|k + l̂1|2 · · · |k + l̂p|2
with h(s, k, l) := 0 if, for k 6= 0, one of the denominators is zero.
For all s ∈ C such that ℜ(s) > n+ j − r − 2p, the series
H(s) :=
(k,l)∈(Zn)p+1
h(s, k, l)
is absolutely summable. In particular,
l∈(Zn)p
h(s, k, l) =
l∈(Zn)p
h(s, k, l) .
Proof. Let s = σ + iτ ∈ C such that σ > n+ j − r − 2p. By Lemma 2.9 we get, for k 6= 0,
|h(s, k, l)| ≤ |b(l)Pj(k)| |k|−r−σ−2p P (l),
where P (l) := P (|l1|, · · · , |lp|) and P is a polynomial of degree 4p with positive coefficients.
Thus, |h(s, k, l)| ≤ F (l)G(k) where F (l) := |b(l)|P (l) and G(k) := |Pj(k)||k|−r−σ−2p. The
summability of
l∈(Zn)p F (l) is implied by the fact that b ∈ S((Zn)p). The summability of∑′
k∈ZnG(k) is a consequence of the fact that σ > n + j − r − 2p. Finally, as a product of two
summable series,
k,lF (l)G(k) is a summable series, which proves that
k,lh(s, k, l) is also
absolutely summable.
Definition 2.11. Let f be a function on D× (Zn)p where D is an open neighborhood of 0 in C.
We say that f satisfies (H1) if and only if there exists ρ > 0 such that
(i) for any l, s 7→ f(s, l) extends as a holomorphic function on Uρ, where Uρ is the open
disk of center 0 and radius ρ,
(ii) the series
l∈(Zn)p ‖H(·, l)‖∞,ρ is summable,where ‖H(·, l)‖∞,ρ := sups∈Uρ |H(s, l)|.
We say that f satisfies (H2) if and only if there exists ρ > 0 such that
(i) for any l, s 7→ f(s, l) extends as a holomorphic function on Uρ − {0},
(ii) for any δ such that 0 < δ < ρ, the series
l∈(Zn)p ‖H(·, l)‖∞,δ,ρ is summable, where
‖H(·, l)‖∞,δ,ρ := supδ<|s|<ρ |H(s, l)|.
Remark 2.12. Note that (H1) implies (H2). Moreover, if f satisfies (H1) (resp. (H2) for
ρ > 0, then it is straightforward to check that f : s 7→
l∈(Zn)p f(s, l) extends as an holomorphic
function on Uρ (resp. on Uρ \ { 0 }).
Corollary 2.13. With the same notations of Lemma 2.10, suppose that r + 2p − j > n, then,
the function H(s, l) :=
k∈Znh(s, k, l) satisfies (H1).
Proof. (i) Let’s fix ρ > 0 such that ρ < r + 2p − j − n. Since r + 2p − j > n, Uρ is inside
the half-plane of absolute convergence of the series defined by H(s, l). Thus, s 7→ H(s, l) is
holomorphic on Uρ.
(ii) Since
∣∣|k|−s
∣∣ ≤ |k|ρ for all s ∈ Uρ and k ∈ Zn \ { 0 }, we get as in the above proof
|h(s, k, l)| ≤ |b(l)Pj(k)| |k|−r+ρ−2p P (|l1|, · · · , |lp|).
Since ρ < r + 2p − j − n, the series
k∈Zn |Pj(k)||k|−r+ρ−2p is summable.
Thus, ‖H(·, l)‖∞,ρ ≤ K F (l) where K :=
′|Pj(k)||k|−r+ρ−2p <∞. We have already seen that
the series
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
f ∼ g ⇐⇒ f − g satisfies (H1).
Lemma 2.14. Let f and g be two functions on D × (Zn)p where D is an open neighborhood of
0 in C, such that f ∼ g and such that g satisfies (H2). Then
l∈(Zn)p
f(s, l) =
l∈(Zn)p
g(s, l) .
Proof. Since f ∼ g, f satisfies (H2) for a certain ρ > 0. Let’s fix η such that 0 < η < ρ and
define Cη as the circle of center 0 and radius η. We have
g(s, l) = Res
f(s, l) = 1
f(s, l) ds =
u(t, l)dt .
where I = [0, 2π] and u(t, l) := 1
ηeitf(η eit, l). The fact that f satisfies (H2) entails that the
series
l∈(Zn)p ‖f(·, l)‖∞,Cη is summable. Thus, since ‖u(·, l)‖∞ =
η ‖f(·, l)‖∞,Cη , the series∑
l∈(Zn)p ‖u(·, l)‖∞ is summable, so, as a consequence,
l∈(Zn)p u(t, l)dt =
l∈(Zn)p
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.
ζ(s) :=
Zn(s) :=
|k|−s,
ζp1,...,pn(s) :=
1 · · · k
|k|s , for pi ∈ N,
where ζ(s) is the Riemann zeta function (see [25] or [14]).
By the symmetry k → −k, it is clear that these functions ζp1,...,pn all vanish for odd values of pi.
Let us now compute ζ0,··· ,0,1i,0··· ,0,1j ,0··· ,0(s) in terms of Zn(s):
Since ζ0,··· ,0,1i,0··· ,0,1j ,0··· ,0(s) = Ai(s) δij , exchanging the components ki and kj , we get
ζ0,··· ,0,1i,0··· ,0,1j ,0··· ,0(s) =
Zn(s− 2).
Similarly,
|k|s+8
n(n−1)
Zn(s+ 4)− 1n−1
|k|s+8
but it is difficult to write explicitly ζp1,...,pn(s) in terms of Zn(s− 4) and other Zn(s−m) when
at least four indices pi are non zero.
When all pi are even, ζp1,...,pn(s) is a nonzero series of fractions
P (k)
where P is a homogeneous
polynomial of degree p1 + · · ·+ pn. Theorem 2.1 now gives us the following
Proposition 2.16. ζp1,...,pn has a meromorphic extension to the whole plane with a unique pole
at n+ p1 + · · ·+ pn. This pole is simple and the residue at this pole is
s=n+p1+···+pn
ζp1,...,pn(s) = 2
)···Γ(
n+p1+···+pn
when all pi are even or this residue is zero otherwise.
In particular, for n = 2,
′ kikj
|k|s+4
= δij π , (18)
and for n = 4,
′ kikj
|k|s+6
= δij
′ kikjklkm
|k|s+8
= (δijδlm + δilδjm + δimδjl)
. (19)
Proof. Equation (17) follows from Theorem (2.1)
s=n+p1+···+pn
ζp1,...,pn(s) =
k∈Sn−1
1 · · · kpnn dS(k)
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 6= l = m and
i = j = l = m.
Note that Zn(s) is an Epstein zeta function associated to the quadratic form q(x) := x
1+...+x
so Zn satisfies the following functional equation
Zn(s) = π
s−n/2Γ(n/2− s/2)Γ(s/2)−1 Zn(n− s).
Since πs−n/2Γ(n/2−s/2) Γ(s/2)−1 = 0 for any negative even integer n and Zn(s) is meromorphic
on C with only one pole at s = n with residue 2πn/2Γ(n/2)−1 according to previous proposition,
so we get Zn(0) = −1. We have proved that
Zn(s+ n) = 2π
n/2 Γ(n/2)−1, (20)
Zn(0) = −1. (21)
2.4 Meromorphic continuation of a class of zeta functions
Let n, q ∈ N, q ≥ 2, and p = (p1, . . . , pq−1) ∈ Nq−10 .
Set I := {i | pi 6= 0} and assume that I 6= ∅ and
I := {α = (αi)i∈I | ∀i ∈ I αi = (αi,1, . . . , αi,pi) ∈ N
0 } =
We will use in the sequel also the following notations:
- for x = (x1, . . . , xt) ∈ Rt recall that |x|1 = |x1|+ · · ·+ |xt| and |x| =
x21 + · · ·+ x2t ;
- for all α = (αi)i∈I ∈ I =
i∈I N
|α|1 =
|αi|1 =
|αi,j| and
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 X1, . . . ,Xn we set X = (X1, . . . ,Xn);
- for any i = 1, . . . , 2q, we consider the variables Yi,1, . . . , Yi,n and set Yi := (Yi,1, . . . , Yi,n) and
Y := (Y1, . . . , Y2q);
- for Y = (Y1, . . . , Y2q), we set for any 1 ≤ j ≤ q, Ỹj := Y1 + · · ·+ Yj + Yq+1 + · · ·+ Yq+j.
We define for all α = (αi)i∈I ∈ I =
i∈I N
0 the polynomial
Pα(X,Y ) :=
(2〈X, Ỹi〉+ |Ỹi|2)αi,j . (22)
It is clear that Pα(X,Y ) ∈ Z[X,Y ], degXPα ≤ |α|1 and degY Pα ≤ 2|α|1.
Let us fix a polynomial Q ∈ R[X1, · · · ,Xn] and note d := degQ. For α ∈ I, we want to expand
Pα(X,Y )Q(X) in homogeneous polynomials in X and Y so defining
L(α) := {β ∈ N(2q+1)n0 : |β|1 − dβ ≤ 2|α|1 and dβ ≤ |α|1 + d }
where dβ :=
1 βi, we set
Pα(X,Y )Q(X) =:
β∈L(α)
cα,βX
where cα,β ∈ R, Xβ := Xβ11 · · ·X
n and Y
β := Y
1,1 · · ·Y
β(2q+1)n
2q,n . By definition, X
β is a
homogeneous polynomial of degree in X equals to dβ . We note
Mα,β(Y ) := cα,β Y
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 1
Θ be a diophantine matrix, and ã ∈ S
(Zn)2q
. Then
s 7→ f(s) :=
l∈[(Zn)q]2
|k + l̃i|pi |k|−sQ(k) eik.Θ
has a meromorphic continuation to the whole complex plane C with at most simple possible poles
at the points s = n+ d+ |p|1 −m where m ∈ N0.
(ii) Let m ∈ N0 and set I(m) := { (α, β) ∈ I × N(2q+1)n0 : β ∈ L(α) and m = 2|α|1 − dβ + d }.
Then I(m) is a finite set and s = n+ d+ |p|1 −m is a pole of f if and only if
C(f,m) :=
(α,β)∈I(m)
Mα,β(l)
u∈Sn−1
uβ dS(u) 6= 0,
with Z := {l :
1 lj = 0} and the convention
∅ = 0. In that case s = n + d + |p|1 −m is a
simple pole of residue Res
s=n+d+|p|1−m
f(s) = C(f,m).
In order to prove the theorem above we need the following
Lemma 2.18. For all N ∈ N we have
|k + l̃i|pi =
α=(αi)i∈I∈
i∈I{0,...,N}
) Pα(k,l)
|k|2|α|1−|p|1
+ON (|k||p|1−(N+1)/2)
uniformly in k ∈ Zn and l ∈ (Zn)2q verifying |k| > U(l) := 36 (
∑2q−1
i=1, i 6=q |li|)4.
Proof. For i = 1, . . . , q − 1, we have uniformly in k ∈ Zn and l ∈ (Zn)2q verifying |k| > U(l),
∣∣2〈k,eli〉+|eli|2
. (23)
In that case,
|k + l̃i| =
|k|2 + 2〈k, l̃i〉+ |l̃i|2
= |k|
2〈k,eli〉+|eli|
|k|2u−1
P iu(k, l)
where for all i = 1, . . . , q − 1 and for all u ∈ N0,
P iu(k, l) :=
2〈k, l̃i〉+ |l̃i|2
with the convention P i0(k, l) := 1.
In particular P iu(k, l) ∈ Z[k, l], degk P iu ≤ u and degl P iu ≤ 2u. Inequality (23) implies that for
all i = 1, . . . , q − 1 and for all u ∈ N,
|k|2u
|P iu(k, l)| ≤
uniformly in k ∈ Zn and l ∈ (Zn)2q verifying |k| > U(l).
Let N ∈ N. We deduce from the previous that for any k ∈ Zn and l ∈ (Zn)2q verifying |k| > U(l)
and for all i = 1, . . . , q − 1, we have
|k + l̃i| =
|k|2u−1
P iu(k, l) +O
|k| |
|k|)−u
|k|2u−1
P iu(k, l) +ON
|k|(N−1)/2
It follows that for any N ∈ N, we have uniformly in k ∈ Zn and l ∈ (Zn)2q verifying |k| > U(l)
and for all i ∈ I,
|k + l̃i|pi =
αi∈{0,...,N}
|k|2|αi|1−pi
P iαi(k, l) +ON
|k|(N+1)/2−pi
where P iαi(k, l) =
j=1 P
(k, l) for all αi = (αi,1, . . . , αi,pi) ∈ {0, . . . , N}pi and
|k + l̃i|pi =
α=(αi)∈
i∈I{0,...,N}
|k|2|α|1−|p|1
Pα(k, l) +ON
|k|(N+1)/2−|p|1
where Pα(k, l) =
i∈I P
(k, l) =
j=1 P
(k, l).
Proof of Theorem 2.17.
(i) All n, q, p = (p1, . . . , pq−1) and ã ∈ S
(Zn)2q
are fixed as above and we define formally for
any l ∈ (Zn)2q
F (l, s) :=
|k + l̃i|pi Q(k) eik.Θ
1 lj |k|−s. (24)
Thus, still formally,
f(s) :=
l∈(Zn)2q
ãl F (l, s). (25)
It is clear that F (l, s) converges absolutely in the half plane {σ = ℜ(s) > n + d + |p|1} where
d = degQ.
Let N ∈ N. Lemma 2.18 implies that for any l ∈ (Zn)2q and for s ∈ C such that σ > n+ |p|1+d,
F (l, s) =
|k|≤U(l)
|k + l̃i|pi Q(k) eik.Θ
1 lj |k|−s
α=(αi)i∈I∈
i∈I{0,...,N}
|k|>U(l)
|k|s+2|α|1−|p|1
Pα(k, l)Q(k) e
1 lj +GN (l, s).
where s 7→ GN (l, s) is a holomorphic function in the half-plane DN := {σ > n+ d+ |p|1 − N+12 }
and verifies in it the bound GN (l, s) ≪N,σ 1 uniformly in l.
It follows that
F (l, s) =
α=(αi)i∈I∈
i∈I{0,...,N}
Hα(l, s) +RN (l, s), (26)
where
Hα(l, s) :=
′ (1/2
|k|s+2|α|1−|p|1
Pα(k, l)Q(k) e
1 lj ,
RN (l, s) :=
|k|≤U(l)
|k + l̃i|pi Q(k) eik.Θ
1 lj |k|−s
|k|≤U(l)
α=(αi)i∈I∈
i∈I{0,...,N}
) Pα(k,l)
|k|s+2|α|1−|p|1
Q(k) eik.Θ
1 lj +GN (l, s).
In particular there exists A(N) > 0 such that s 7→ RN (l, s) extends holomorphically to the
half-plane DN and verifies in it the bound RN (l, s) ≪N,σ 1 + |l|A(N) uniformly in l.
Let us note formally
hα(s) :=
ãlHα(l, s).
Equation (26) and RN (l, s) ≪N,σ 1 + |l|A(N) imply that
f(s) ∼N
α=(αi)i∈I∈
i∈I{0,...,N}
hα(s), (27)
where ∼N means modulo a holomorphic function in DN .
Recall the decomposition
Pα(k, l)Q(k) =
β∈L(α)Mα,β(l) k
β and we decompose similarly
hα(s) =
β∈L(α) hα,β(s). Theorem 2.5 now implies that for all α = (αi)i∈I ∈
i∈I{0, . . . , N}pi
and β ∈ L(α),
- the map s 7→ hα,β(s) has a meromorphic continuation to the whole complex plane C with
only one simple possible pole at s = n+ |p|1 − 2|α|1 + dβ ,
- the residue at this point is equal to
s=n+|p|1−2|α|1+dβ
hα,β(s) =
ãlMα,β(l)
u∈Sn−1
uβdS(u) (28)
where Z := {l ∈ (Z)n)2q :
1 lj = 0}. If the right hand side is zero, hα,β(s) is holomorphic on
By (27), we deduce therefore that f(s) has a meromorphic continuation on the halfplane DN ,
with only simple possible poles in the set {n+ |p|1+k : −2N |p|1 ≤ k ≤ d }. Taking now N → ∞
yields the result.
(ii) Let m ∈ N0 and set I(m) := { (α, β) ∈ I × N(2q+1)n0 : β ∈ L(α) and m = 2|α|1 − dβ + d }.
If (α, β) ∈ I(m), then |α|1 ≤ m and |β|1 ≤ 3m+ d, so I(m) is finite.
With a chosen N such that 2N |p|1 + d > m, we get by (27) and (28)
s=n+d+|p|1−m
f(s) =
(α,β)∈I(m)
Mα,β(l)
u∈Sn−1
uβ dS(u) = C(f,m)
with the convention
∅ = 0. Thus, n+d+ |p|1−m is a pole of f if and only if C(f,m) 6= 0.
3 Noncommutative integration on a simple spectral triple
In this section, we revisit the notion of noncommutative integral pioneered by Alain Connes, pay-
ing 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 D and DA which are both finite dimensional:
Lemma 3.1. Let (A,H,D) be a spectral triple with a reality operator J and chirality χ. If
A ∈ Ω1D is a one-form, the fluctuated Dirac operator
DA := D +A+ ǫJAJ−1
(where DJ = ǫ JD, ǫ = ±1) is an operator with compact resolvent, and in particular its kernel
KerDA is a finite dimensional space. This space is invariant by J and χ.
Proof. Let T be a bounded operator and let z be in the resolvent of D + T and z′ be in the
resolvent of D. Then
(D + T − z)−1 = (D − z′)−1 [1− (T + z′ − z)(D + T − z)−1].
Since (D− z′)−1 is compact by hypothesis and since the term in bracket is bounded, D+ T has
a compact resolvent. Applying this to T = A+ ǫJAJ−1, DA has a finite dimensional kernel (see
for instance [27, Theorem 6.29]).
Since according to the dimension, J2 = ±1, J commutes or anticommutes with χ, χ commutes
with the elements in the algebraA andDχ = −χD (see [10] or [23, p. 405]), we get DAχ = −χDA
and DAJ = ±JDA which gives the result.
3.2 Pseudodifferential operators
Let (A,D,H) be a given real regular spectral triple of dimension n.
We note
P0 the projection on KerD , PA the projection on KerDA ,
D := D + P0 ,DA := DA + PA .
P0 and PA are thus finite-rank selfadjoint bounded operators. We remark that D and DA 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 ζD, ζDA
functions, it is not necessary to define the operators D−1 or D−1A and the associated zeta func-
tions. 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 D or DA such that
the result is invertible without changing anything to the computation of residues.
Define for any α ∈ R
OP 0 := {T : t 7→ Ft(T ) ∈ C∞
R,B(H)
OPα := {T : T |D|−α ∈ OP 0 }.
where Ft(T ) := e
it|D| T e−it|D| = eit|D| T e−it|D| since |D| = |D|+ P0. Define
δ(T ) := [|D|, T ],
∇(T ) := [D2, T ],
σs(T ) := |D|sT |D|−s, s ∈ C.
It has been shown in [13] that OP 0 =
p≥0Dom(δ
p). In particular, OP 0 is a subalgebra of B(H)
(while elements of OPα are not necessarily bounded for α > 0) and A ⊆ OP 0, JAJ−1 ⊆ OP 0,
[D,A] ⊆ OP 0. Note that P0 ∈ OP−∞ and δ(OP 0) ⊆ OP 0.
For any t > 0, Dt and and |D|t are in OP t and for any α ∈ R,Dα and |D|α are in OPα. By
hypothesis, |D|−n ∈ L(1,∞)(H) so for any α > n, OP−α ⊆ L1(H).
Lemma 3.3. [13]
(i) For any T ∈ OP 0 and s ∈ C, σs(T ) ∈ OP 0.
(ii) For any α, β ∈ R, OPαOP β ⊆ OPα+β .
(iii) If α ≤ β, OPα ⊆ OP β.
(iv) For any α, δ(OPα) ⊆ OPα.
(v) For any α and T ∈ OPα, ∇(T ) ∈ OPα+1.
Proof. See the appendix.
Remark 3.4. Any operator in OPα, where α ∈ R, extends as a continuous linear operator from
Dom |D|α+1 to Dom |D| where the Dom |D|α spaces have their natural norms (see [13,26]).
We now introduce a definition of pseudodifferential operators in a slightly different way than
in [9,13,26] which in particular pays attention to the reality operator J and the kernel of D and
allows D and |D|−1 to be a pseudodifferential operators. It is more in the spirit of [4].
Definition 3.5. Let us define D(A) as the polynomial algebra generated by A, JAJ−1, D and
A pseudodifferential operator is an operator T such that there exists d ∈ Z such that for any
N ∈ N, there exist p ∈ N0, P ∈ D(A) and R ∈ OP−N (p, P and R may depend on N) such
that P D−2p ∈ OP d and
T = P D−2p +R .
Define Ψ(A) as the set of pseudodifferential operators and Ψ(A)k := Ψ(A) ∩OP k.
Note that if A is a 1-form, A and JAJ−1 are in D(A) and moreover D(A) ⊆ ∪p∈N0OP p. Since
|D| ∈ D(A) by construction and P0 is a pseudodifferential operator, for any p ∈ Z, |D|p is a
pseudodifferential operator (in OP p.) Let us remark also that D(A) ⊆ Ψ(A) ⊆ ∪k∈ZOP k.
Lemma 3.6. [9, 13] The set of all pseudodifferential operators Ψ(A) is an algebra. Moreover,
if T ∈ Ψ(A)d and T ∈ Ψ(A)d′, then TT ′ ∈ Ψ(A)d+d′ .
Proof. See the appendix.
Due to the little difference of behavior between scalar and nonscalar pseudodifferential operators
(i.e. when coefficients like [D, a], a ∈ A appears in P of Definition 3.5), it is convenient to also
introduce
Definition 3.7. Let D1(A) be the algebra generated by A, JAJ−1 and D, and Ψ1(A) be the
set of pseudodifferential operators constructed as before with D1(A) instead of D(A). Note that
Ψ1(A) is subalgebra of Ψ(A).
Remark that Ψ1(A) does not necessarily contain operators such as |D|k where k ∈ 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 DA, we define
ζBX(s) := Tr
B|X|−s
ζX(s) := Tr
|X|−s
The dimension spectrum Sd(A,H,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(A,H,D) of all
poles of the functions ζPD := s 7→ Tr
P |D|−s
where P is any pseudodifferential operator in
OP 0. The spectral triple (A,H,D) is simple when these poles are all simple.
Remark 3.9. If Sp(A,H,D) denotes the set of all poles of the functions s 7→ Tr
P |D|−s
where
P is any pseudodifferential operator, then, Sd(A,H,D) ⊆ Sp(A,H,D).
When Sp(A,H,D) = Z, Sd(A,H,D) = {n − k : k ∈ N0 }: indeed, if P is a pseudodifferential
operator in OP 0, and q ∈ N is such that q > n, P |D|−s is in OP−ℜ(s) so is trace-class for s in
a neighborhood of q; as a consequence, q cannot be a pole of s 7→ Tr
P |D|−s
Remark 3.10. Sp(A,H,D) is also the set of poles of functions s 7→ Tr
B|D|−s−2p
where
p ∈ N0 and B ∈ D(A).
3.4 The noncommutative integral
We already defined the one parameter group σz(T ) := |D|zT |D|−z, z ∈ C.
Introducing the notation (recall that ∇(T ) = [D2, T ]) for an operator T ,
ε(T ) := ∇(T )D−2,
we get from [4, (2.44)] the following expansion for T ∈ OP q
σz(T ) ∼
g(z, r) εr(T ) mod OP−N−1+q (29)
where g(z, r) := 1
) · · · (z
− (r − 1)) =
with the convention g(z, 0) := 1.
We define the noncommutative integral by
− T := Res
ζTD(s) = Res
T |D|−s
Proposition 3.11. [13] If the spectral triple is simple,
is a trace on Ψ(A).
Proof. See the appendix.
4 Residues of ζDA for a spectral triple with simple dimension
spectrum
We fix a regular spectral triple (A,H,D) of dimension n and a self-adjoint 1-form A.
Recall that
DA := D + Ã where Ã := A+ εJAJ−1,
DA := DA + PA
where PA is the projection on KerDA. Remark that Ã ∈ D(A) ∩OP 0 and DA ∈ D(A) ∩OP 1.
We note
VA := PA − P0.
As the following lemma shows, VA is a smoothing operator:
Lemma 4.1. (i)
k≥1Dom(DA)k ⊆
k≥1Dom |D|k.
(ii) KerDA ⊆
k≥1Dom |D|k.
(iii) For any α, β ∈ R, |D|βPA|D|α is bounded.
(iv) PA ∈ OP−∞.
Proof. (i) Let us define for any p ∈ N, Rp := (DA)p−Dp, so Rp ∈ OP p−1 and Rp
Dom |D|p
Dom |D| (see Remark 3.4).
Let us fix k ∈ N, k ≥ 2. Since DomDA = DomD = Dom |D|, we have
Dom(DA)k = {φ ∈ Dom |D| : (Dj +Rj)φ ∈ Dom |D| , ∀j 1 ≤ j ≤ k − 1 }.
Let φ ∈ Dom(DA)k. We prove by recurrence that for any j ∈ { 1, · · · , k − 1 }, φ ∈ Dom |D|j+1:
We have φ ∈ Dom |D| and (D +R1)φ ∈ Dom |D|. Thus, since R1 φ ∈ Dom |D|, Dφ ∈ Dom |D|,
which proves that φ ∈ Dom |D|2. Hence, case j = 1 is done.
Suppose now that φ ∈ Dom |D|j+1 for a j ∈ { 1, · · · , k − 2 }. Since (Dj+1 +Rj+1)φ ∈ Dom |D|,
and Rj+1 φ ∈ Dom |D|, we get Dj+1 φ ∈ Dom |D|, which proves that φ ∈ Dom |D|j+2.
Finally, if we set j = k − 1, we get φ ∈ Dom |D|k, so Dom(DA)k ⊆ Dom |D|k.
(ii) follows from KerDA ⊆
k≥1Dom(DA)k and (i).
(iii) Let us first check that |D|αPA is bounded. We define D0 as the operator with domain
DomD0 = ImPA ∩Dom |D|α and such that D0 φ = |D|α φ. Since DomD0 is finite dimensional,
D0 extends as a bounded operator on H with finite rank. We have
φ∈Dom |D|αPA, ‖φ‖≤1
‖|D|αPA φ‖ ≤ sup
φ∈DomD0, ‖φ‖≤1
‖|D|α φ‖ = ‖D0‖ <∞
so |D|αPA is bounded. We can remark that by (ii), DomD0 = ImPA and Dom |D|αPA = H.
Let us prove now that PA|D|α is bounded: Let φ ∈ DomPA|D|α = Dom |D|α. By (ii), we have
ImPA ⊆ Dom |D|α so we get
‖PA|D|α φ‖ ≤ sup
ψ∈ImPA, ‖ψ‖≤1
| < ψ, |D|α φ > | ≤ sup
ψ∈ImPA, ‖ψ‖≤1
| < |D|αψ, φ > |
≤ sup
ψ∈ImPA, ‖ψ‖≤1
‖|D|αψ‖ ‖φ‖ = ‖D0‖ ‖φ‖ .
(iv) For any k ∈ N0 and t ∈ R, δk(PA)|D|t is a linear combination of terms of the form
|D|βPA|D|α, so the result follows from (iii).
Remark 4.2. We will see later on the noncommutative torus example how important is the
difference between DA and D + A. In particular, the inclusion KerD ⊆ KerD + A is not
satisfied since A does not preserve KerD contrarily to Ã.
The coefficient of the nonconstant term Λk (k > 0) in the expansion (5) of the spectral action
S(DA,Φ,Λ) is equal to the residue of ζDA(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 ∈ N, s ∈ C,
Kp(T, s) := (− s2)
0≤t1≤···≤tp≤1
σ−st1(T ) · · · σ−stp(T ) dt
with dt := dt1 · · · dtp.
Remark that if T ∈ OPα, then σz(T ) ∈ OPα for z ∈ C and Kp(T, s) ∈ OPαp.
Let us define
X := D2A −D2 = ÃD +DÃ+ Ã2,
XV := X + VA,
thus X ∈ D1(A) ∩OP 1 and by Lemma 4.1,
XV ∼ X mod OP−∞. (30)
We will use
Y := log(D2A)− log(D2)
which makes sense since D2A = D2A + PA is invertible for any A.
By definition of XV , we get
Y = log(D2 +XV )− log(D2).
Lemma 4.3. [4]
(i) Y is a pseudodifferential operator in OP−1 with the following expansion for any N ∈ N
k1,··· ,kp=0
(−1)|k|1+p+1
|k|1+p
∇kp(X∇kp−1(· · ·X∇k1(X) · · · ))D−2(|k|1+p) mod OP−N−1.
(ii) For any N ∈ N and s ∈ C,
|DA|−s ∼ |D|−s +
Kp(Y, s)|D|−s mod OP−N−1−ℜ(s). (31)
Proof. (i) We follow [4, Lemma 2.2]. By functional calculus, Y =
I(λ) dλ, where
I(λ) ∼
(−1)p+1
(D2 + λ)−1XV
(D2 + λ)−1 mod OP−N−3.
By (30),
(D2 + λ)−1XV
(D2 + λ)−1X
mod OP−∞ and we get
I(λ) ∼
(−1)p+1
(D2 + λ)−1X
(D2 + λ)−1 mod OP−N−3.
We set Ap(X) :=
(D2 + λ)−1X
(D2 + λ)−1 and L := (D2 + λ)−1 ∈ OP−2 for a fixed λ. Since
[D2 + λ,X] ∼ ∇(X) mod OP−∞, a recurrence proves that if T is an operator in OP r, then,
for q ∈ N0,
A1(T ) = LTL ∼
(−1)k∇k(T )Lk+2 mod OP r−q−5.
With Ap(X) = LXAp−1(X), another recurrence gives, for any q ∈ N0,
Ap(X) ∼
k1,··· ,kp=0
(−1)|k|1∇kp(X∇kp−1(· · ·X∇k1(X) · · · ))L|k|1+p+1 mod OP−q−p−3,
which entails that
I(λ) ∼
(−1)p+1
k1,··· ,kp=0
(−1)|k|1∇kp(X∇kp−1(· · ·X∇k1(X) · · · ))L|k|1+p+1 mod OP−N−3.
(D2 + λ)−(|k|1+p+1)dλ = 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 |DA|−s = eB−(s/2)Y e−B |D|−s where B := (−s/2) log(D2). Following [4, Theorem
2.4], we get
|DA|−s = |D|−s +
Kp(Y, s)|D|−s . (32)
and each Kp(Y, s) is in OP
Corollary 4.4. For any p ∈ N and r1, · · · , rp ∈ N0, εr1(Y ) · · · εrp(Y ) ∈ Ψ1(A).
Proof. If for any q ∈ N and k = (k1, · · · , kq) ∈ Nq0,
Γkq(X) :=
(−1)|k|1+q+1
|k|1+q
∇kq(X∇kq−1(· · ·X∇k1(X) · · · )),
then, Γkq (X) ∈ OP |k|1+q. For any N ∈ N,
k1,··· ,kq=0
Γkq(X)D
−2(|k|1+q) mod OP−N−1. (33)
Note that the Γkq(X) are in D1(A), which, with (33) proves that Y and thus εr(Y ) = ∇r(Y )D−2r,
are also in Ψ1(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 D and this symbol is stable
by adding a gauge potential like in D+A. Note however that the symmetrized gauge potential
A+ ǫJAJ−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
ζDA(0)− ζD(0) =
(−1)q
−(ÃD−1)q. (34)
Proof. Since the spectral triple is simple, equation (32) entails that
ζDA(0)− ζD(0) = Tr(K1(Y, s)|D|−s)|s=0 .
Thus, with (29), we get ζDA(0) − ζD(0) = −12
Y . Replacing A by Ã, the same proof as in [4]
gives
− Y =
(−1)q
−(ÃD−1)q.
Lemma 4.6. For any k ∈ N0,
s=n−k
ζDA(s) = Res
s=n−k
ζD(s) +
r1,··· ,rp=0
s=n−k
h(s, r, p) Tr
εr1(Y ) · · · εrp(Y )|D|−s
where
h(s, r, p) := (−s/2)p
0≤t1≤···≤tp≤1
g(−st1, r1) · · · g(−stp, rp) dt .
Proof. By Lemma 4.3 (ii), |DA|−s ∼ |D|−s +
p=1Kp(Y, s)|D|−s mod OP−(k+1)−ℜ(s), where
the convention
∅ = 0 is used. Thus, we get for s in a neighborhood of n− k,
|DA|−s − |D|−s −
Kp(Y, s)|D|−s ∈ OP−(k+1)−ℜ(s) ⊆ L1(H)
which gives
s=n−k
ζDA(s) = Res
s=n−k
ζD(s) +
s=n−k
Kp(Y, s)|D|−s
. (35)
Let us fix 1 ≤ p ≤ k and N ∈ N. By (29) we get
Kp(Y, s) ∼ (− s2)
0≤t1≤···tp≤1
r1,··· ,rp=0
g(−st1, r1) · · · g(−stp, rp)
εr1(Y ) · · · εrp(Y ) dt mod OP−N−p−1. (36)
If we now take N = k − p, we get for s in a neighborhood of n− k
Kp(Y, s)|D|−s −
r1,··· ,rp=0
h(s, r, p) εr1(Y ) · · · εrp(Y )|D|−s ∈ OP−k−1−ℜ(s) ⊆ L1(H)
so (35) gives the result.
Our operators |DA|k are pseudodifferential operators:
Lemma 4.7. For any k ∈ Z, |DA|k ∈ Ψk(A).
Proof. Using (36), we see that Kp(Y, s) is a pseudodifferential operator in OP
−p, so (31) proves
that |DA|k is a pseudodifferential operator in OP k.
The following result is quite important since it shows that one can use
for D or DA:
Proposition 4.8. If the spectral triple is simple, Res
P |DA|−s
P for any pseudodiffer-
ential operator P . In particular, for any k ∈ N0
− |DA|−(n−k) = Res
s=n−k
ζDA(s).
Proof. Suppose P ∈ OP k with k ∈ Z and let us fix p ≥ 1. With (36), we see that for any N ∈ N,
PKp(Y, s)|D|−s ∼
r1,··· ,rp=0
h(s, r, p)Pεr1(Y ) · · · εrp(Y )|D|−s mod OP−N−p−1+k−ℜ(s).
Thus if we take N = n− p+ k, we get
PKp(Y, s)|D|−s
n−p+k∑
r1,··· ,rp=0
h(s, r, p) Tr
Pεr1(Y ) · · · εrp(Y )|D|−s
Since s = 0 is a zero of the analytic function s 7→ h(s, r, p) and s 7→ TrPεr1(Y ) · · · εrp(Y )|D|−s
has only simple poles by hypothesis, we see that Res
h(s, r, p) Tr
Pεr1(Y ) · · · εrp(Y )|D|−s
PKp(Y, s)|D|−s
= 0. (37)
Using (31), P |DA|−s ∼ P |D|−s +
p=1 PKp(Y, s)|D|−s mod OP−n−1−ℜ(s) and thus,
Tr(P |DA|−s) =
− P +
PKp(Y, s)|D|−s
. (38)
The result now follows from (37) and (38). To get the last equality, one uses the pseudodiffer-
ential operator |DA|−(n−k).
Proposition 4.9. If the spectral triple is simple, then
− |DA|−n =
− |D|−n. (39)
Proof. Lemma 4.6 and previous proposition for k = 0.
Lemma 4.10. If the spectral triple is simple,
− |DA|−(n−1) =
− |D|−(n−1) − (n−1
− X|D|−n−1.
− |DA|−(n−2) =
− |D|−(n−2) + n−2
− X|D|−n + n
− X2|D|−2−n
Proof. (i) By (31),
s=n−1
ζDA(s)− ζD(s) = Res
s=n−1
(−s/2)Tr
Y |D|−s
= −n−1
Y |D|−(n−1)|D|−s
where for the last equality we use the simple dimension spectrum hypothesis. Lemma 4.3 (i)
yields Y ∼ XD−2 mod OP−2 and Y |D|−(n−1) ∼ X|D|−n−1 mod OP−n−1 ⊆ L1(H). Thus,
Y |D|−(n−1)|D|−s
= Res
X|D|−n−1|D|−s
− X|D|−n−1.
(ii) Lemma 4.6 (ii) gives
s=n−2
ζDA(s) = Res
s=n−2
ζD(s) + Res
s=n−2
h(s, r, 1) Tr
εr(Y )|D|−s
+ h(s, 0, 2) Tr
Y 2|D|−s
We have h(s, 0, 1) = − s
, h(s, 1, 1) = 1
)2 and h(s, 0, 2) = 1
)2. Using again Lemma 4.3 (i),
Y ∼ XD−2 − 1
∇(X)D−4 − 1
X2D−4 mod OP−3.
Thus,
s=n−2
Y |D|−s
− X|D|−n − 1
−(∇(X) +X2)|D|−2−n.
Moreover, using
∇(X)|D|−k = 0 for any k ≥ 0 since
is a trace,
s=n−2
ε(Y )|D|−s
= Res
s=n−2
∇(X)D−4|D|−s
− ∇(X)|D|−2−n = 0.
Similarly, since Y ∼ XD−2 mod OP−2 and Y 2 ∼ X2D−4 mod OP−3, we get
s=n−2
Y 2|D|−s
= Res
s=n−2
X2D−4|D|−s
− X2|D|−2−n.
Thus,
s=n−2
ζDA(s) = Res
s=n−2
ζD(s)+(−n−22 )(
− X|D|−n − 1
−(∇(X) +X2)|D|−2−n)
− ∇(X)|D|−2−n + 1
− X2|D|−2−n.
Finally,
s=n−2
ζDA(s) = Res
s=n−2
ζD(s) + (−n−22 )
− X|D|−n − 1
− X2|D|−2−n
− X2|D|−2−n
and the result follows from Proposition 4.8.
Corollary 4.11. If the spectral triple is simple and satisfies
|D|−(n−2) =
ÃD|D|−n =∫
DÃ|D|−n = 0, then
− |DA|−(n−2) = n(n−2)4
− ÃDÃD|D|−n−2 + n−2
− Ã2|D|−n
Proof. By previous lemma,
s=n−2
ζDA(s) =
− Ã2|D|−n + n
−( ÃDÃD +DÃDÃ+ ÃD2Ã+DÃ2D )|D|−n−2
Since ∇(Ã) ∈ OP 1, the trace property of
yields the result.
5 The noncommutative torus
5.1 Notations
Let C∞(TnΘ) be the smooth noncommutative n-torus associated to a non-zero skew-symmetric
deformation matrix Θ ∈Mn(R) (see [6], [30]). This means that C∞(TnΘ) is the algebra generated
by n unitaries ui, i = 1, . . . , n subject to the relations
ui uj = e
iΘij uj ui, (40)
and with Schwartz coefficients: an element a ∈ C∞(TnΘ) can be written as a =
k∈Zn ak Uk,
where {ak} ∈ S(Zn) with the Weyl elements defined by Uk := e−
k.χk u
1 · · · uknn , k ∈ Zn,
relation (40) reads
UkUq = e
k.Θq Uk+q, and UkUq = e
−ik.Θq UqUk (41)
where χ is the matrix restriction of Θ to its upper triangular part. Thus unitary operators Uk
satisfy U∗k = U−k and [Uk, Ul] = −2i sin(
k.Θl)Uk+l.
Let τ be the trace on C∞(TnΘ) defined by τ
k∈Zn ak Uk
:= a0 and Hτ be the GNS Hilbert
space obtained by completion of C∞(TnΘ) with respect of the norm induced by the scalar product
〈a, b〉 := τ(a∗b). On Hτ = {
k∈Zn ak Uk : {ak}k ∈ l2(Zn) }, we consider the left and right
regular representations of C∞(TnΘ) by bounded operators, that we denote respectively by L(.)
and R(.).
Let also δµ, µ ∈ { 1, . . . , n }, be the n (pairwise commuting) canonical derivations, defined by
δµ(Uk) := ikµUk. (42)
We need to fix notations: let AΘ := C∞(TnΘ) acting on H := Hτ ⊗ C2
with n = 2m or
n = 2m+ 1 (i.e., m = ⌊n
⌋ is the integer part of n
), the square integrable sections of the trivial
spin bundle over Tn.
Each element of AΘ is represented on H as L(a) ⊗ 12m where L (resp. R) is the left (resp.
right) multiplication. The Tomita conjugation J0(a) := a
∗ satisfies [J0, δµ] = 0 and we define
J := J0 ⊗ C0 where C0 is an operator on C2
. The Dirac operator is given by
D := −i δµ ⊗ γµ, (43)
where we use hermitian Dirac matrices γ. It is defined and symmetric on the dense subset of H
given by C∞(TnΘ)⊗ C2
. We still note D its selfadjoint extension. This implies
α = −εγαC0, (44)
D Uk ⊗ ei = kµUk ⊗ γµei,
where (ei) is the canonical basis of C
2m . Moreover, C20 = ±12m depending on the parity of m.
Finally, one introduces the chirality (which in the even case is χ := id⊗ (−i)mγ1 · · · γn) and this
yields that (AΘ,H,D, J, χ) satisfies all axioms of a spectral triple, see [8, 23].
The perturbed Dirac operator VuDV ∗u by the unitary
Vu :=
L(u)⊗ 12m
L(u)⊗ 12m
defined for every unitary u ∈ A, uu∗ = u∗u = U0, must satisfy condition (3) (which is equivalent
toH being endowed with a structure ofAΘ-bimodule). This yields the necessity of a symmetrized
covariant Dirac operator:
DA := D +A+ ǫJ AJ−1
since VuDV ∗u = DL(u)⊗12m [D,L(u∗)⊗12m ]: in fact, for a ∈ AΘ, using J0L(a)J
0 = R(a
∗), we get
L(a)⊗ γα
J−1 = −R(a∗)⊗ γα
and that the representation L and the anti-representation R are C-linear, commute and satisfy
[δα, L(a)] = L(δαa), [δα, R(a)] = R(δαa).
This induces some covariance property for the Dirac operator: one checks that for all k ∈ Zn,
L(Uk)⊗ 12m [D, L(U∗k )⊗ 12m ] = 1⊗ (−kµγµ), (45)
so with (44), we get Uk[D, U∗k ] + ǫJUk[D, U∗k ]J−1 = 0 and
VUk D V
= D = DL(Uk)⊗12m [D,L(U∗k )⊗12m ]. (46)
Moreover, we get the gauge transformation:
VuDAV ∗u = Dγu(A) (47)
where the gauged transform one-form of A is
γu(A) := u[D, u∗] + uAu∗, (48)
with the shorthand L(u)⊗ 12m −→ u.
As a consequence, the spectral action is gauge invariant:
S(DA,Φ,Λ) = S(Dγu(A),Φ,Λ).
An arbitrary selfadjoint one-form A, can be written as
A = L(−iAα)⊗ γα, Aα = −A∗α ∈ AΘ, (49)
DA = −i
δα + L(Aα)−R(Aα)
⊗ γα. (50)
Defining
Ãα := L(Aα)−R(Aα),
we get D2A = −gα1α2(δα1 + Ãα1)(δα2 + Ãα2)⊗ 12m − 12Ωα1α2 ⊗ γ
α1α2 where
γα1α2 := 1
(γα1γα2 − γα2γα1),
Ωα1α2 := [δα1 + Ãα1 , δα2 + Ãα2 ] = L(Fα1α2)−R(Fα1α2)
Fα1α2 := δα1(Aα2)− δα2(Aα1) + [Aα1 , Aα2 ]. (51)
In summary,
D2A = −δα1α2
δα1 + L(Aα1)−R(Aα1)
δα2 + L(Aα2)−R(Aα2)
⊗ 12m
L(Fα1α2)−R(Fα1α2)
⊗ γα1α2 . (52)
5.2 Kernels and dimension spectrum
We now compute the kernel of the perturbed Dirac operator:
Proposition 5.1. (i) KerD = U0 ⊗C2
, so dimKerD = 2m.
(ii) For any selfadjoint one-form A, KerD ⊆ KerDA.
(iii) For any unitary u ∈ A, KerDγu(A) = Vu KerDA.
Proof. (i) Let ψ =
k,j ck,j Uk ⊗ ej ∈ KerD. Thus, 0 = D2ψ =
k,i ck,j|k|2 Uk ⊗ ej which
entails that ck,j|k|2 = 0 for any k ∈ Zn and 1 ≤ j ≤ 2m. The result follows.
(ii) Let ψ ∈ KerD. So, ψ = U0 ⊗ v with v ∈ C2
and from (50), we get
DAψ = Dψ + (A+ ǫJAJ−1)ψ = (A+ ǫJAJ−1)ψ = −i[Aα, U0]⊗ γαv = 0
since U0 is the unit of the algebra, which proves that ψ ∈ KerDA.
(iii) This is a direct consequence of (47).
Corollary 5.2. Let A be a selfadjoint one-form. Then KerDA = KerD in the following cases:
(i) Au := L(u)⊗ 12m [D, L(u∗)⊗ 12m ] when u is a unitary in A.
(ii) ||A|| < 1
(iii) The matrix 1
Θ has only integral coefficients.
Proof. (i) This follows from previous result because Vu(U0 ⊗ v) = U0 ⊗ v for any v ∈ C2
(ii) Let ψ =
k,j ck,j Uk ⊗ ej be in KerDA (so
k,j |ck,j|2 <∞) and φ :=
j c0,j U0 ⊗ ej . Thus
ψ′ := ψ − φ ∈ Ker DA since φ ∈ KerD ⊆ KerDA and
06=k∈Zn, j
ck,j kα Uk ⊗ γαej ||2 = ||Dψ′||2 = || − (A+ ǫJAJ−1)ψ′||2 ≤ 4||A||2||ψ′||2 < ||ψ′||2.
Defining Xk :=
α kαγα, X
α |kα|2 12m is invertible and the vectors {Uk ⊗Xkej }06=k∈Zn, j
are orthogonal in H, so
06=k∈Zn, j
|kα|2
|ck,j|2 <
06=k∈Zn, j
|ck,j|2
which is possible only if ck,j = 0, ∀k, j that is ψ′ = 0 et ψ = φ ∈ Ker D.
(iii) This is a consequence of the fact that the algebra is commutative, thus A+ǫJAJ−1 = 0.
Note that if Ãu := Au + ǫJAuJ
−1, then by (45), ÃUk = 0 for all k ∈ Zn and ‖AUk‖ = |k|, but
for an arbitrary unitary u ∈ A, Ãu 6= 0 so DAu 6= D.
Naturally the above result is also a direct consequence of the fact that the eigenspace of an iso-
lated 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 ψ =
l∈Zn,1≤j≤2m cl,j Ul⊗ ej ∈ KerDA, so
l∈Zn,1≤j≤2m |cl,j |2 <∞. We have to show that
ψ ∈ Ker D that is cl,j = 0 when l 6= 0.
Taking the scalar product of 〈Uk ⊗ ei| with
0 = DAψ =
l, α, j
cl, j(l
αUl − i[Aα, Ul])⊗ γαej ,
we obtain
l, α, j
cl, j
lαδk,l − i〈Uk, [Aα, Ul]〉
〈ei, γαej〉.
If Aα =
α,l aα,l Ul ⊗ γα with { aα,l }l ∈ S(Zn), note that [Ul, Um] = −2i sin(12 l.Θm)Ul+m and
〈Uk, [Aα, Ul]〉 =
l′∈Zn
aα,l′(−2i sin(12 l
′.Θl)〈Uk, Ul′+l〉 = −2i aα,k−l sin(12k.Θl).
cl, j
lαδk,l − 2aα,k−l sin(12k.Θl)
〈ei, γαej〉, ∀k ∈ Zn, ∀i, 1 ≤ i ≤ 2m. (53)
We conjecture that KerD = KerDA at least for generic Θ’s:
the constraints (53) should imply cl,j = 0 for all j and all l 6= 0 meaning ψ ∈ KerD. When 12πΘ
has only integer coefficients, the sin part of these constraints disappears giving the result.
Lemma 5.3. If 1
Θ is diophantine, Sp
C∞(TnΘ),H,D
= Z and all these poles are simple.
Proof. Let B ∈ D(A) and p ∈ N0. Suppose that B is of the form
B = arbrDqr−1|D|pr−1ar−1br−1 · · · Dq1 |D|p1a1b1
where r ∈ N, ai ∈ A, bi ∈ JAJ−1, qi, pi ∈ N0. We note ai =:
l ai,l Ul and bi =:
i bi,l Ul.
With the shorthand kµ1,µqi := kµ1 · · · kµqi and γ
µ1,µqi = γµ1 · · · γµqi , we get
Dq1 |D|p1a1b1 Uk ⊗ ej =
l1, l
a1,l1b1,l′1
Ul1UkUl′1
|k + l1 + l′1|p1 (k + l1 + l′1)µ1,µq1 ⊗ γ
µ1,µq1 ej
which gives, after r iterations,
BUk⊗ej =
ãlb̃lUlr · · ·Ul1UkUl′1 · · ·Ul′r
|k+ l̂i+ l̂′i|pi(k+ l̂i+ l̂′i)µi1,µiqi ⊗γ
qr−1 · · · γµ
q1 ej
where ãl := a1,l1 · · · ar,lr and b̃l′ := b1,l′1 · · · br,l′r .
Let us note Fµ(k, l, l
′) :=
i=1 |k + l̂i + l̂′i|pi(k + l̂i + l̂′i)µi1,µiqi and γ
µ := γ
µr−11 ,µ
qr−1 · · · γµ
Thus, with the shortcut ∼c meaning modulo a constant function towards the variable s,
B|D|−2p−s
ãlb̃l′ τ
U−kUlr · · ·Ul1UkUl′1 · · ·Ul′r
)Fµ(k,l,l′)
|k|s+2p
Tr(γµ) .
Since Ulr · · ·Ul1Uk = UkUlr · · ·Ul1e−i
1 li.Θk we get
U−kUlr · · ·Ul1UkUl′1 · · ·Ul′r
= δPr
1 li+l
eiφ(l,l
′) e−i
1 li.Θk
where φ is a real valued function. Thus,
B|D|−2p−s
eiφ(l,l
′) δPr
1 li+l
ãlb̃l′
Fµ(k,l,l
′) e−i
1 li.Θk
|k|s+2p
Tr(γµ)
∼c fµ(s)Tr(γµ).
The function fµ(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 pi are zero, in Theorem 2.5). Thus, s 7→ Tr
B|D|−2p−s
has only poles in 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 1
Θ is diophantine, the spectrum dimension of
C∞(TnΘ),H,D
equal to the set {n− k : k ∈ N0 } and all these poles are simple.
(ii) ζD(0) = 0.
Proof. (i) Lemma 5.3 and Remark 3.9.
(ii) ζD(s) =
1≤j≤2m〈Uk ⊗ ej , |D|−sUk ⊗ ej〉 = 2m(
+ 1) = 2m(Zn(s) + 1).
By (21), we get the result.
We have computed ζD(0) relatively easily but the main difficulty of the present work is precisely
to calculate ζDA(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 1
Θ is diophantine, then the first elements of the expansion (5) are given
− |DA|−n =
− |D|−n = 2m+1πn/2 Γ(n
)−1. (54)
− |DA|n−k = 0 for k odd.
− |DA|n−2 = 0.
We need few technical lemmas:
Lemma 5.6. On the noncommutative torus, for any t ∈ R,
− ÃD|D|−t =
− DÃ|D|−t = 0.
Proof. Using notations of (49), we have
Tr(ÃD|D|−s) ∼c
〈Uk ⊗ ej,−ikµ|k|−s[Aα, Uk]⊗ γαγµej〉
∼c −iTr(γαγµ)
kµ|k|−s〈Uk, [Aα, Uk]〉 = 0
since 〈Uk, [Aα, Uk]〉 = 0. Similarly
Tr(DÃ|D|−s) ∼c
〈Uk ⊗ ej , |k|−s
aα,l 2 sin
(l + k)µUl+k ⊗ γµγαej〉
∼c 2Tr(γµγα)
aα,l sin
(l + k)µ |k|−s〈Uk, Ul+k〉 = 0.
Any element h in the algebra generated by A and [D,A] can be written as a linear combination of
terms of the form a1
p1 · · · anpr where ai are elements of A or [D,A]. Such a term can be written
as a series b :=
a1,α1,l1 · · · aq,αq,lqUl1 · · ·Ulq ⊗ γα1 · · · γαq where ai,αi are Schwartz sequences
and when ai =:
l alUl ∈ A, we set ai,α,l = ai,l with γα = 1. We define
L(b) := τ
a1,α1,l1 · · · aq,αq,lqUl1 · · ·Ulq
Tr(γα1 · · · γαq ).
By linearity, L is defined as a linear form on the whole algebra generated by A and [D,A].
Lemma 5.7. If h is an element of the algebra generated by A and [D,A],
h|D|−s
∼c L(h)Zn(s).
In particular, Tr
h|D|−s
has at most one pole at s = n.
Proof. We get with b of the form
a1,α1,l1 · · · aq,αq,lqUl1 · · ·Ulq ⊗ γα1 · · · γαq ,
b|D|−s
a1,α1,l1 · · · aq,αq,lqUl1 · · ·UlqUk〉 Tr(γα1 · · · γαq )|k|−s
∼c τ(
a1,α1,l1 · · · aq,αq,lqUl1 · · ·Ulq)Tr(γα1 · · · γαq )Zn(s) = L(b)Zn(s).
The results follows now from linearity of the trace.
Lemma 5.8. If 1
Θ is diophantine, the function s 7→ Tr
εJAJ−1A|D|−s
extends meromor-
phically on the whole plane with only one possible pole at s = n. Moreover, this pole is simple
εJAJ−1A|D|−s
= aα,0 a
m+1πn/2 Γ(n/2)−1.
Proof. With A = L(−iAα) ⊗ γα, we get ǫJAJ−1 = R(iAα) ⊗ γα, and by multiplication
εJAJ−1A = R(Aβ)L(Aα)⊗ γβγα. Thus,
εJAJ−1A|D|−s
〈Uk, AαUkAβ〉 |k|−s Tr(γβγα)
aα,l aβ,−l e
ik.Θl |k|−sTr(γβγα)
∼c 2m
aα,l a
ik.Θl |k|−s.
Theorem 2.5 (ii) entails that
l aα,l a
ik.Θl |k|−s extends meromorphically on the
whole plane C with only one possible pole at s = n. Moreover, this pole is simple and we
aα,l a
ik.Θl |k|−s = aα,0 aα0 Res
Zn(s).
Equation (20) now gives the result.
Lemma 5.9. If 1
Θ is diophantine, then for any t ∈ R,
− X|D|−t = δt,n 2m+1
aα,l a
−l + aα,0 a
2πn/2 Γ(n/2)−1.
where X = ÃD +DÃ+ Ã2 and A =: −i
l aα,l Ul ⊗ γα.
Proof. By Lemma 5.6, we get
X|D|−t = Ress=0Tr(Ã2|D|−s−t). Since A and εJAJ−1 commute,
we have Ã2 = A2 + JA2J−1 + 2εJAJ−1A. Thus,
Tr(Ã2|D|−s−t) = Tr(A2|D|−s−t) + Tr(JA2J−1|D|−s−t) + 2Tr(εJAJ−1A|D|−s−t).
Since |D| and J commute, we have with Lemma 5.7,
Ã2|D|−s−t
∼c 2L(A2)Zn(s+ t) + 2Tr
εJAJ−1A|D|−s−t
Thus Lemma 5.8 entails that Tr(Ã2|D|−s−t) is holomorphic at 0 if t 6= n. When t = n,
Ã2|D|−s−t
= 2m+1
aα,l a
−l + aα,0 a
2πn/2 Γ(n/2)−1, (55)
which gives the result.
Lemma 5.10. If 1
Θ is diophantine, then
− ÃDÃD|D|−2−n = −n−2
− Ã2|D|−n.
Proof. With DJ = εJD, we get
− ÃDÃD|D|−2−n = 2
− ADAD|D|−2−n + 2
− εJAJ−1DAD|D|−2−n.
Let us first compute
ADAD|D|−2−n. We have, with A =: −iL(Aα)⊗ γα =: −i
l aα,lUl ⊗ γα,
ADAD|D|−s−2−n
l1,l2
aα2,l2 aα1,l1 τ(U−kUl2Ul1Uk)
kµ1(k+l1)µ2
|k|s+2+n
Tr(γα,µ)
where γα,µ := γα2γµ2γα1γµ1 . Thus,
− ADAD|D|−2−n = −
aα2,−l aα1,lRes
′ kµ1kµ2
|k|s+2+n
Tr(γα,µ).
We have also, with εJAJ−1 = iR(Aα)⊗ γa,
εJAJ−1DAD|D|−s−2−n
l1,l2
aα2,l2aα1,l1τ(U−kUl1UkUl2)
kµ1 (k+l1)µ2
|k|s+2+n
Tr(γα,µ).
which gives
− εJAJ−1DAD|D|−2−n = aα2,0aα1,0Res
′ kµ1kµ2
|k|s+2+n
Tr(γα,µ).
Thus,
− ÃDÃD|D|−2−n =
aα2,0aα1,0 −
aα2,−laα1,l
Ress=0
′ kµ1kµ2
|k|s+2+n
Tr(γα,µ).
kµ1kµ2
|k|s+2+n
δµ1µ2
Zn(s+ n) and Cn := Ress=0 Zn(s+ n) = 2π
n/2Γ(n/2)−1 we obtain
− ÃDÃD|D|−2−n =
aα2,0aα1,0 −
aα2,−laα1,l
Tr(γα2γµγα1γµ).
Since Tr(γα2γµγα1γµ) = 2
m(2− n)δα2,α1 , we get
− ÃDÃD|D|−2−n = 2m
− aα,0 aα0 +
aα,−l a
)Cn(n−2)
Equation (55) now proves the lemma.
Lemma 5.11. If 1
Θ is diophantine, then for any P ∈ Ψ1(A) and q ∈ N, q odd,
− P |D|−(n−q) = 0.
Proof. There exist B ∈ D1(A) and p ∈ N0 such that P = BD−2p + R where R is in OP−q−1.
As a consequence,
P |D|−(n−q) =
B|D|−n−2p+q. Assume B = arbrDqr−1ar−1br−1 · · · Dq1a1b1
where r ∈ N, ai ∈ A, bi ∈ JAJ−1, qi ∈ N. If we prove that
B|D|−n−2p+q = 0, then the general
case will follow by linearity. We note ai =:
l ai,l Ul and bi =:
l bi,l Ul. With the shorthand
kµ1,µqi := kµ1 · · · kµqi and γ
µ1,µqi = γµ1 · · · γµqi , we get
Dq1a1b1Uk ⊗ ej =
a1,l1 b1,l′1 Ul1UkUl
(k + l1 + l
1)µ1,µq1 ⊗ γ
µ1,µq1ej
which gives, after iteration,
B Uk ⊗ ej =
ãlb̃lUlr · · ·Ul1UkUl′1 · · ·Ul′r
(k + l̂i + l̂
i)µi1,µiqi
qr−1 · · · γµ
where ãl := a1,l1 · · · ar,lr and b̃l′ := b1,l′1 · · · br,l′r . Let’s note Qµ(k, l, l
′) :=
i=1 (k + l̂i + l̂
i)µi1,µiqi
and γµ := γ
qr−1 · · · γµ
q1 . Thus,
− B |D|−n−2p+q = Res
ãl b̃l′ τ
U−kUlr · · ·Ul1UkUl′1 · · ·Ul′r
) Qµ(k,l,l′)
|k|s+2p+n−q
Tr(γµ) .
Since Ulr · · ·Ul1Uk = UkUlr · · ·Ul1e−i
1 li.Θk, we get
U−kUlr · · ·Ul1UkUl′1 · · ·Ul′r
= δPr
1 li+l
eiφ(l,l
′) e−i
1 li.Θk
where φ is a real valued function. Thus,
− B |D|−n−2p+q = Res
eiφ(l,l
′) δPr
1 li+l
ãl b̃l′
Qµ(k,l,l
′)e−i
1 li.Θk
|k|s+2p+n−q
Tr(γµ)
=: Res
fµ(s)Tr(γ
We decompose Qµ(k, l, l
′) as a sum
h=0Mh,µ(l, l
′)Qh,µ(k) where Qh,µ is a homogeneous poly-
nomial in (k1, · · · , kn) and Mh,µ(l, l′) is a polynomial in
(l1)1, · · · , (lr)n, (l′1)1, · · · , (l′r)n
Similarly, we decompose fµ(s) as
h=0 fh,µ(s). Theorem 2.5 (ii) entails that fh,µ(s) extends
meromorphically to the whole complex plane C with only one possible pole for s+2p+n−q = n+d
where d := deg Qh,µ. In other words, if d+ q− 2p 6= 0, fh,µ(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)
fh,µ(s) = V
u∈Sn−1
Qh,µ(u) dS(u)
where V :=
l,l′∈ZMh,µ(l, l
′) eiφ(l,l
′) δPr
1 li+l
,0 ãl b̃l′ and Z := { l, l′ :
i=1 li = 0 }. Since d is
odd, Qh,µ(−u) = −Qh,µ(u) and
u∈Sn−1
Qh,µ(u) dS(u) = 0. Thus, Res
fh,µ(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 numer-
ator of the series fµ(s) as polynomials in k. This has been possible because we restricted our
pseudodifferential operators to Ψ1(A).
Proof of Proposition 5.5. The top element follows from Proposition 4.9 and according to (20),
− |D|−n = Res
|D|−s−n
= 2mRes
Zn(s + n) =
2m+1πn/2
Γ(n/2)
For the second equality, we get from Lemmas 5.7 and 4.6
s=n−k
ζDA(s) =
r1,··· ,rp=0
h(n− k, r, p)
− εr1(Y ) · · · εrp(Y )|D|−(n−k).
Corollary 4.4 and Lemma 5.11 imply that
εr1(Y ) · · · εrp(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
C∞(TnΘ),H,D
where n ∈ N and 1
Θ is a real n×n
skew-symmetric real diophantine matrix, and a selfadjoint one-form A = L(−iAα)⊗ γα. Then,
the full spectral action of DA = D +A+ ǫJAJ−1 is
(i) for n = 2,
S(DA,Φ,Λ) = 4πΦ2Λ2 +O(Λ−2),
(ii) for n = 4,
S(DA,Φ,Λ) = 8π2 Φ4Λ4 − 4π
Φ(0) τ(FµνF
µν) +O(Λ−2),
(iii) More generally, in
S(DA,Φ,Λ) =
Φn−k cn−k(A)Λ
n−k +O(Λ−1),
cn−2(A) = 0, cn−k(A) = 0 for k odd. In particular, c0(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
Θ is assumed, is outlined here.
Remark 6.2. Note that all terms must be gauge invariants, namely, according to (48), invariant
by Aα −→ γu(Aα) = uAαu∗ + uδα(u∗). A particular case is u = Uk where Ukδα(U∗k ) = −ikαU0.
In the same way, note that there is no contradiction with the commutative case where, for any
selfadjoint one-form A, DA = D (so A is equivalent to 0!), since we assume in Theorem 6.1 that
Θ is diophantine, so A cannot be commutative.
Conjecture 6.3. The constant term of the spectral action of DA on the noncommutative n-torus
is proportional to the constant term of the spectral action of D+A on the commutative n-torus.
Remark 6.4. The appearance of a Diophantine condition for Θ has been characterized in di-
mension 2 by Connes [7, Prop. 49] where in this case, Θ = θ
with θ ∈ R. In fact, the
Hochschild cohomology H(AΘ,AΘ∗) satisfies dim Hj(AΘ,AΘ∗) = 2 (or 1) for j = 1 (or j = 2)
if and only if the irrational number θ satisfies a Diophantine condition like |1−ei2πnθ|−1 = O(nk)
for some k.
Recall that when the matrix Θ is quite irrational (see [23, Cor. 2.12]), then the C∗-algebra
generated by AΘ is simple.
Remark 6.5. It is possible to generalize above theorem to the case D = −i gµν δµ ⊗ γν instead
of (43) when g is a positive definite constant matrix. The formulae in Theorem 6.1 are still
valid.
6.1 Computations of
In order to get this theorem, let us prove a few technical lemmas.
We suppose from now on that Θ is a skew-symmetric matrix in Mn(R). No other hypothesis is
assumed for Θ, except when it is explicitly stated.
When A is a selfadjoint one-form, we define for n ∈ N , q ∈ N, 2 ≤ q ≤ n and σ ∈ {−,+}q
+ := ADD−2,
− := ǫJAJ−1DD−2,
σ := Aσq · · ·Aσ1 .
Lemma 6.6. We have for any q ∈ N,
−(ÃD−1)q =
−(ÃDD−2)q =
σ∈{+,−}q
− Aσ.
Proof. Since P0 ∈ OP−∞, D−1 = DD−2 mod OP−∞ and
(ÃD−1)q =
(ÃDD−2)q.
Lemma 6.7. Let A be a selfadjoint one-form, n ∈ N and q ∈ N with 2 ≤ q ≤ n and σ ∈ {−,+}q.
Then ∫
− Aσ =
− A−σ.
Proof. Let us first check that JP0 = P0J . Since DJ = εJD, we get DJP0 = 0 so JP0 =
P0JP0. Since J is an antiunitary operator, we get P0J = P0JP0 and finally, P0J = JP0. As a
consequence, we get JD2 = D2J , JDD−2 = εDD−2J , JA+J−1 = A− and JA−J−1 = A+.
In summary, JAσiJ−1 = A−σi .
The trace property of
now gives
− Aσ =
− Aσq · · ·Aσ1 =
− JAσqJ−1 · · · JAσ1J−1
− A−σq · · ·A−σ1 =
− A−σ.
Definition 6.8. In [4] has been introduced the vanishing tadpole hypothesis:
− AD−1 = 0, for all A ∈ Ω1D(A). (56)
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 ∈ N, A = L(−iAα)⊗γα = −i
l∈Zn aα,l Ul⊗γα, Aα ∈ AΘ, { aα,l }l ∈ S(Zn),
be a hermitian one-form. Then,
ApD−q =
(ǫJAJ−1)pD−q = 0 for p ≥ 0 and 1 ≤ q < n (case p = q = 1 is tadpole
hypothesis.)
(ii) If 1
Θ is diophantine, then
BD−q = 0 for 1 ≤ q < n and any B in the algebra generated
by A, [D,A], JAJ−1 and J [D,A]J−1.
Proof. (i) Let us compute ∫
− Ap(ǫJAJ−1)p′D−q.
With A = L(−iAα)⊗ γα and ǫJAJ−1 = R(iAα)⊗ γα, we get
Ap = L(−iAα1) · · ·L(−iAαp)⊗ γα1 · · · γαp
(ǫJAJ−1)p
= R(iAα′1
) · · ·R(iAα′
)⊗ γα′1 · · · γα
We note ãα,l := aα1,l1 · · · aαp,lp . Since
L(−iAα1) · · ·L(−iAαp)R(iAα′1) · · ·R(iAα′p′ )Uk = (−i)
ãα,l ãα′,l′ Ul1 · · ·UlpUkUl′
· · ·Ul′1 ,
Ul1 · · ·UlpUk = UkUl1 · · ·Ulp e−i(
i li).Θk,
we get, with
Ul,l′ := Ul1 · · ·UlpUl′
· · ·Ul′1 ,
gµ,α,α′(s, k, l, l
′) := eik.Θ
kµ1 ...kµq
|k|s+2q
ãα,l ãα′,l′ ,
′,µ := γα1 · · · γαpγα′1 · · · γα
p′γµ1 · · · γµq ,
Ap(ǫJAJ−1)p
D−q|D|−sUk ⊗ ei ∼c (−i)p ip
gµ,α,α′(s, k, l, l
′)UkUl,l′ ⊗ γα,α
′,µei.
Thus,
Ap(ǫJAJ−1)p
D−q = Res
f(s) where
f(s) : = Tr
Ap(ǫJAJ−1)p
D−q|D|−s
∼c (−i)p ip
〈Uk ⊗ ei,
gµ,α,α′(s, k, l, l
′)UkUl,l′ ⊗ γα,α
′,µei〉
∼c (−i)p ip
gµ,α,α′(s, k, l, l
′)Ul,l′
Tr(γµ,α,α
∼c (−i)p ip
gµ,α,α′(s, k, l, l
Ul,l′
Tr(γµ,α,α
It is straightforward to check that the series
k,l,l′gµ,α,α′(s, k, l, l
Ul,l′
is absolutely summable
if ℜ(s) > R for a R > 0. Thus, we can exchange the summation on k and l, l′, which gives
f(s) ∼c (−i)p ip
gµ,α,α′(s, k, l, l
Ul,l′
Tr(γµ,α,α
If we suppose now that p′ = 0, we see that,
f(s) ∼c (−i)p
′ kµ1 ...kµq
|k|s+2q
ãα,l δ
Tr(γµ,α,α
which is, by Proposition 2.16, analytic at 0. In particular, for p = q = 1, we see that
AD−1 = 0,
i.e. the vanishing tadpole hypothesis is satisfied. Similarly, if we suppose p = 0, we get
f(s) ∼c (−i)p
′ kµ1 ...kµq
|k|s+2q
ãα,l′ δPp′
i=1 l
Tr(γµ,α,α
which is holomorphic at 0.
(ii) Adapting the proof of Lemma 5.11 to our setting (taking qi = 0, and adding gamma matrices
components), we see that
− BD−q = Res
eiφ(l,l
′) δPr
1 li+l
ãα,l b̃β,l′
kµ1 ···kµq e
1 li.Θk
|k|s+2q
Tr(γ(µ,α,β))
where γ(µ,α,β) is a complicated product of gamma matrices. By Theorem 2.5 (ii), since we
suppose here that 1
Θ 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:
− AqD−q = −δq,2 4π τ
(ii) Case n = 4: with the shorthand δµ1,...,µ4 := δµ1µ2δµ3µ4 + δµ1µ3δµ2µ4 + δµ1µ4δµ2µ3 ,
− AqD−q = δq,4 π
Aα1 · · ·Aα4
Tr(γα1 · · · γα4γµ1 · · · γµ4)δµ1,...,µ4 .
Proof. (i, ii) The same computation as in Lemma 6.9 (i) (with p′ = 0, p = q = n) gives
− AnD−n = Res
(−i)n
′ kµ1 ...kµn
|k|s+2n
l∈(Zn)n
ãα,lUl1 · · ·Uln
Tr(γα1 · · · γαnγµ1 · · · γµn)
and the result follows from Proposition 2.16.
We will use few notations:
If n ∈ N, q ≥ 2, l := (l1, · · · , lq−1) ∈ (Zn)q−1, α := (α1, · · · , αq) ∈ {1, · · · , n}q, k ∈ Zn\{0},
σ ∈ {−,+}q, (ai)1≤i≤n ∈ (S(Zn))n,
lq := −
1≤j≤q−1
lj , λσ := (−i)q
j=1...q
σj , ãα,l := aα1,l1 . . . aαq ,lq ,
φσ(k, l) :=
1≤j≤q−1
(σj − σq) k.Θlj +
2≤j≤q−1
σj (l1 + . . .+ lj−1).Θlj ,
gµ(s, k, l) :=
kµ1 (k+l1)µ2 ...(k+l1+...+lq−1)µq
|k|s+2|k+l1|2...|k+l1+...+lq−1|2
with the convention
2≤j≤q−1 = 0 when q = 2, and gµ(s, k, l) = 0 whenever l̂i = −k for a
1 ≤ i ≤ q − 1.
Lemma 6.11. Let A = L(−iAα) ⊗ γα = −i
l∈Zn aα,l Ul ⊗ γα where Aα = −A∗α ∈ AΘ and
{ aα,l }l ∈ S(Zn), with n ∈ N, be a hermitian one-form, and let 2 ≤ q ≤ n, σ ∈ {−,+}q.
Then,
Aσ = Res
f(s) where
f(s) :=
l∈(Zn)q−1
φσ(k,l) gµ(s, k, l) ãα,l Tr(γ
αqγµq · · · γα1γµ1).
Proof. By definition,
Aσ = Res
f(s) where
Tr(Aσq · · ·Aσ1 |D|−s) ∼c
〈Uk ⊗ ei, |k|−s Aσq · · ·Aσ1Uk ⊗ ei〉 =: f(s).
Let r ∈ Zn and v ∈ C2m . Since A = L(−iAα)⊗ γα, and ǫJAJ−1 = R(iAα)⊗ γα, we get
+Ur ⊗ v = ADD−2Ur ⊗ v = A rµ|r|2+δr,0Ur ⊗ γ
µv = −i rµ
|r|2+δr,0
AαUr ⊗ γαγµv ,
−Ur ⊗ v = ǫJAJ−1DD−2Ur ⊗ v = ǫJAJ−1 rµ|r|2+δr,0Ur ⊗ γ
µv = i
|r|2+δr,0
UrAα ⊗ γαγµv.
With UlUr = e
Ur+l and UrUl = e
Ur+l, we obtain, for any 1 ≤ j ≤ q,
σjUr ⊗ v =
(−σj) i eσj
r.Θl rµ
|r|2+δr,0
aα,l Ur+l ⊗ γαγµv.
We now apply q times this formula to get
|k|−sAσq · · ·Aσ1Uk ⊗ ei = λσ
l∈(Zn)q
φσ(k,l) gµ(s, k, l) ãα,l Uk+
⊗ γαqγµq · · · γα1γµ1ei
φσ(k, l) := σ1 k.Θl1 + σ2 (k + l1).Θl2 + . . .+ σq (k + l1 + . . . + lq−1).Θlq.
Thus,
f(s) =
l∈(Zn)q
φσ(k,l) gµ(s, k, l) ãα,l U
Tr(γαqγµq · · · γα1γµ1)
l∈(Zn)q
φσ(k,l) gµ(s, k, l) ãα,l δ(
lj)Tr(γ
αqγµq · · · γα1γµ1)
l∈(Zn)q−1
φσ(k,l) gµ(s, k, l) ãα,l Tr(γ
αqγµq · · · γα1γµ1)
where in the last sum lq is fixed to −
1≤j≤q−1 lj and thus,
φσ(k, l) =
1≤j≤q−1
(σj − σq) k.Θlj +
2≤j≤q−1
σj (l1 + . . .+ lj−1).Θlj .
By Lemma 2.10, there exists a R > 0 such that for any s ∈ C with ℜ(s) > R, the family
φσ(k,l) gµ(s, k, l) ãα,l
(k,l)∈(Zn\{ 0 })×(Zn)q−1
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
c := 4π
Lemma 6.12. Suppose n = 4. Then, with the same hypothesis of Lemma 6.11,
(i) 1
−(A+)2 = 1
−(A−)2 = c
aα1,l aα2,−l
lα1 lα2 − δα1α2 |l|2
(ii) − 1
−(A+)3 = −1
−(A−)3 = 4c
li∈Z4
aα3,−l1−l2 a
aα1,l1 sin
l1.Θl2
(iii) 1
−(A+)4 = 1
−(A−)4 = 2c
li∈Z4
aα1,−l1−l2−l3 aα2,l3 a
aα2l1 sin
l1.Θ(l2+l3)
sin l2.Θl3
(iv) Suppose 1
Θ diophantine. Then the crossed terms in
(A+ + A−)q vanish: if C is the
set of all σ ∈ {−,+}q with 2 ≤ q ≤ 4, such that there exist i, j satisfying σi 6= σj, we have∑
Aσ = 0.
Proof. (i) Lemma 6.11 entails that
A++ = Res
l∈Zn −f(s, l) where
f(s, l) :=
′ kµ1 (k+l)µ2
|k|s+2|k+l|2
ãα,l Tr(γ
α2γµ2γα1γµ1) and ãα,l := aα1,l aα2,−l .
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 Zn(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 Zn functions. We
get, using three times (16),
|k+l|2
− 2k.l+|l|
(2k.l+|l|2)2
− (2k.l+|l|
|k|6|k+l|2
. (57)
Let us define
fα,µ(s, l) :=
′ kµ1 (k+l)µ2
|k|s+2|k+l|2
ãα,l
so that f(s, l) = fα,µ(s, l)Tr(γ
α2γµ2γα1γµ1). Equation (57) gives
fα,µ(s, l) = f1(s, l)− f2(s, l) + f3(s, l)− r(s, l)
with obvious identifications. Note that the function
r(s, l) =
′ kµ1 (k+l)µ2 (2kl+|l|
|k|s+8|k+l|2
ãα,l
is a linear combination of functions of the typeH(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 fα,µ(s, l) ∼ f1(s, l)− f2(s, l) + f3(s, l).
Let’s now compute f1(s, l).
f1(s, l) =
′ kµ1 (k+l)µ2
|k|s+4
ãα,l = ãα,l
′ kµ1kµ2
|k|s+4
Proposition 2.1 entails that s 7→
′ kµ1kµ2
|k|s+4
is holomorphic at 0. Thus, f1(s, l) satisfies (H1),
and fα,µ(s, l) ∼ −f2(s, l) + f3(s, l).
Let’s now compute f2(s, l) modulo (H1). We get, using several times Proposition 2.1,
f2(s, l) =
′ kµ1 (k+l)µ2 (2kl+|l|
|k|s+6
ãα,l =
′ (2kl)kµ1kµ2+(2kl)kµ1 lµ2+|l|
2kµ1kµ2+lµ2 |l|
|k|s+6
ãα,l
∼ 0 +
′ (2kl)kµ1 lµ2
|k|s+6
ãα,l +
′ |l|2kµ1kµ2
|k|s+6
ãα,l + 0 .
Recall that
|k|s+6
Zn(s + 4). Thus,
f2(s, l) ∼ 2lilµ2 ãα,l
Zn(s+ 4) + |l|2 ãα,l
δµ1µ2
Zn(s+ 4).
Finally, let us compute f3(s, l) modulo (H1) following the same principles:
f3(s, l) =
′ kµ1 (k+l)µ2 (2kl+|l|
|k|s+8
ãα,l
′ (2kl)2kµ1kµ2+(2kl)
2kµ1 lµ2+|l|
4kµ1kµ2+|l|
4kµ1 lµ2+(4kl)|l|
2kµ1kµ2+(4kl)|l|
2kµ1 lµ2
|k|s+8
ãα,l
∼ 4lilj
′ kikjkµ1kµ2
|k|s+8
ãα,l + 0.
In conclusion,
fα,µ(s, l) ∼ −14(2lµ1 lµ2 + |l|
2 δµ1µ2)ãα,lZn(s+ 4) + 4l
ilj ãα,l
′ kikjkµ1kµ2
|k|s+8
=: gα,µ(s, l).
Proposition (2.1) entails that Zn(s+ 4) and s 7→
′ kikjkµ1kµ2
|k|s+8
extend holomorphically in a
punctured open disk centered at 0. Thus, gα,µ(s, l) satisfies (H2) and we can apply Lemma 2.14
to get
−(A+)2 = Res
f(s, l) =
gα,µ(s, l)Tr(γ
α2γµ2γα1γµ1) =:
g(s, l).
The problem is now reduced to the computation of Res
g(s, l). Recall that Ress=0 Z4(s+4) = 2π
by (20) or (17), and
Ress=0
′ kikjklkm
|k|s+8
= (δijδlm + δilδjm + δimδjl)
Thus,
gα,µ(s, l) = −π
ãα,l (lµ1 lµ2 +
|l|2δµ1µ2).
We will use
Tr(γµ1 · · · γµ2j ) = Tr(1)
all pairings of { 1···2j }
s(P ) δµP1µP2 δµP3µP4 · · · δµP2j−1µP2j (58)
where s(P ) is the signature of the permutation P when P2m−1 < P2m for 1 ≤ m ≤ n. This gives
Tr(γα2γµ2γα1γµ1) = 2m(δα2µ2δα1µ1 − δα1α2δµ2µ1 + δα2µ1δµ2α1). (59)
Thus,
g(s, l) = −c ãα,l (lµ1 lµ2 + 12 |l|
2δµ1µ2)(δ
α2µ2δα1µ1 − δα1α2δµ2µ1 + δα2µ1δµ2α1)
= −2c ãα,l (lα1 lα2 − δα1α2 |l|2).
Finally,
−(A+)2 = 1
−(A−)2 = c
aα1,l aα2,−l
lα1 lα2 − δα1α2 |l|2
(ii) Lemma 6.11 entails that
A+++ = Res
(l1,l2)∈(Zn)2
f(s, l) where
f(s, l) :=
l1Θl2 kµ1 (k+l1)µ2 (k+
bl2)µ3
|k|s+2|k+l1|2|k+bl2|2
ãα,l Tr(γ
α3γµ3γα2γµ2γα1γµ1)
=: fα,µ(s, l)Tr(γ
α3γµ3γα2γµ2γα1γµ1),
and ãα,l := aα1,l1 aα2,l2 aα3,−bl2
with l̂2 := l1 + l2.
We use the same technique as in (i):
|k+l1|2
− 2k.l1+|l1|
(2k.l1+|l1|
|k|4|k+l1|2
|k+bl2|2
|k|2 −
2k.bl2+|bl2|
(2k.bl2+|bl2|
|k|4|k+bl2|2
and thus,
|k+l1|2|k+bl2|2
− 2k.l1
− 2k.bl2
+R(k, l) (60)
where the remain R(k, l) is a term of order at most −6 in k. Equation (60) gives
fα,µ(s, l) = f1(s, l) + r(s, l)
where r(s, l) corresponds to R(k, l). Note that the function
r(s, l) =
l1Θl2 kµ1(k+l)µ2 (k+
bl2)µ3R(k,l)
|k|s+2
ãα,l
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 fα,µ(s, l) ∼ f1(s, l).
Let us compute f1(s, l) modulo (H1)
f1(s, l) =
l1Θl2 kµ1 (k+l1)µ2 (k+
bl2)µ3
|k|s+6
ãα,l −
l1Θl2 kµ1 (k+l1)µ2 (k+
bl2)µ3 (2k.l1+2k.
|k|s+8
ãα,l
l1Θl2 kµ1kµ2
bl2µ3+kµ1kµ3 l1µ2
|k|s+6
ãα,l −
l1Θl2 kµ1kµ2kµ3 (2k.l1+2k.
|k|s+8
ãα,l
= i e
l1Θl2 ãα,l
(l1µ2δµ1µ3 + l̂2µ3δµ1µ2)
Z4(s+ 4)− 2(li1 + l̂i2)
′ kµ1kµ2kµ3ki
|k|s+8
=: gα,µ(s, l).
Since gα,µ(s, l) satisfies (H2), we can apply Lemma 2.14 to get
−(A+)3 = Res
(l1,l2)∈(Zn)2
f(s, l)
(l1,l2)∈(Zn)2
gα,µ(s, l)Tr(γ
α3γµ3γα2γµ2γα1γµ1) =:
Recall that l3 := −l1 − l2 = −l̂2. By (17) and (19),
gα,µ(s, l)i e
l1Θl2 ãα,l
2(−li1 + li3)π
(δµ1µ2δµ3i + δµ1µ3δµ2i + δµ1iδµ2µ3)
+ (l1µ2δµ1µ3 − l3µ3δµ1µ2)π
We decompose Xl in five terms: Xl = 2
l1Θl2 ãα,l (T1 + T2 + T3 + T4 + T5) where
T0 :=
(−li1 + li3)(δµνδρi + δµρδνi + δµiδνρ) + l1νδµρ − l3ρδµν ,
T1 := (δ
α3ρδα2νδα1µ − δα3ρδα2α1δµν + δα3ρδα2µδα1ν)T0,
T2 := (−δα2α3δρνδα1µ + δα2α3δα1ρδµν − δα2α3δρµδα1ν)T0,
T3 := (δ
α3νδα2ρδα1µ − δα3νδα1ρδα2µ + δα3νδρµδα1α2)T0,
T4 := (−δα1α3δα2ρδµν + δα1α3δρνδα2µ − δα1α3δρµδα2ν)T0,
T5 := (δ
α3µδα2ρδα1ν − δα3µδρνδα1α2 + δα3µδα1ρδα2ν)T0.
With the shorthand p := −l1 − 2l3, q := 2l1 + l3, r := −p− q = −l1 + l3, we compute each Ti,
and find
3T1 = δ
α1α2(2− 2m)pα3 + δα3α1qα2 − δα2α1qα3 + δα3α2qα1 + δα3α2rα1 − δα2α1rα3 + δα3α1rα2 ,
3T2 = (2
m − 2)δα2α3pα1 − 2mδα2α3qα1 − 2mδα2α3rα1 ,
3T3 = δ
α1α3pα2 − δα2α3pα1 + δα1α2pα3 + 2mδα2α1qα3 + δα3α2rα1 − δα3α1rα2 + δα1α2rα3 ,
3T4 = −δα1α32mpα2 − δα1α32mqα2 + δα1α3(2m − 2)rα2 ,
3T5 = δ
α1α3pα2 − δα1α2pα3 + δα3α2pα1 + δα3α2qα1 − δα1α2qα3 + δα3α1qα2 + (2− 2m)δα1α2rα3 .
Thus,
Xl = 2
m 2π2
l1.Θl2 ãα,l (q
α3δα1α2 + rα2δα1α3 + pα1δα2α3) (61)
−(A+)3 = i 2c (S1 + S2 + S3),
where S1, S2 and S3 correspond to respectively q
α3δα1α2 , rα2δα1α3 and pα1δα2α3 . In S1, we
permute the li variables the following way: l1 7→ l3, l2 7→ l1, l3 7→ l2. Therefore, l3.Θ l1 7→ l3.Θ l1
and q 7→ r. With a similar permutation of the αi, we see that S1 = S2. We apply the same
principles to prove that S1 = S3 (using permutation l1 7→ l2, l2 7→ l3, l3 7→ l1). Thus,
−(A+)3 = i 2c
ãα,l e
l1.Θl2 (l1 − l2)α3δα1α2 = S4 − S5,
where S4 correspond to l1 and S5 to l2. We permute the li variables in S5 the following way:
l1 7→ l2, l2 7→ l1, l3 7→ l3, with a similar permutation on the αi. Since l1.Θ l2 7→ −l1.Θ l2, we
finally get
−(A+)3 = −4c
aα1,l1 aα2,l2 aα3,−l1−l2 sin
l1.Θl2
lα31 δ
α1α2 .
(iii) Lemma 6.11 entails that
A++++ = Res
(l1,l2,l3)∈(Zn)3
fµ,α(s, l)Tr γ
µ,α where
θ := l1.Θl2 + l1.Θl3 + l2.Θl3,
Tr γµ,α := Tr(γα4γµ4γα3γµ3γα2γµ2γα1γµ1),
fµ,α(s, l) :=
θ kµ1 (k+l1)µ2 (k+
bl2)µ3 (k+
bl3)µ4
|k|s+2|k+l1|2|k+bl2|2|k+bl3|2
ãα,l,
ãα,l := aα1,l1 aα2,l2 aα3,l3 aα4,−l1−l2−l3 .
Using (16) and Corollary 2.13 successively, we find
fµ,α(s, l) ∼
θ kµ1kµ2kµ3kµ4
|k|s+2|k+l1|2|k+l1+l2|2|k+l1+l2+l3|2
ãα,l ∼
θ kµ1kµ2kµ3kµ4
|k|s+8
ãα,l.
Since the function
θ kµ1kµ2kµ3kµ4
|k|s+8
ãα,l satisfies (H2), Lemma 2.14 entails that
−(A+)4 =
(l1,l2,l3)∈(Zn)3
ãα,l Res
′ kµ1kµ2kµ3kµ4
|k|s+8
Tr γµ,α =:
Therefore, with (19), we get Xl =
ãα,l e
(A+B + C), where
A := Tr(γα4γµ4γα3γµ4γ
α2γµ2γα1γµ2),
B := Tr(γα4γµ4γα3γµ2γα2γµ4γ
α1γµ2),
C := Tr(γα4γµ4γα3γµ2γ
α2γµ2γα1γµ4).
Using successively {γµ, γν} = 2δµν and γµγµ = 2m 12m , we see that
A = C = 4 Tr(γα4γα3γα2γα1),
B = −4
Tr(γα4γα3γα1γα2) + Tr(γα4γα2γα3γα1)
Thus, A+B +C = 8 2m
δα4α3δα2α1 + δα4α1δα3α2 − 2δα4α2δα3α1
, and
ãα,l
δα4α3δα2α1 + δα4α1δα3α2 − 2δα4α2δα3α1
. (62)
By (62), we get ∫
−(A+)4 = 2c (−2T1 + T2 + T3),
where
T1 :=
l1,...,l4
aα4,l4 aα3,l3 aα2,l2 aα1,l1 e
δα4α2 δα3α1 ,
T2 :=
l1,...,l4
aα4,l4 aα3,l3 aα2,l2 aα1,l1 e
δα4α3 δα2α1 ,
T3 :=
l1,...,l4
aα4,l4 aα3,l3 aα2,l2 aα1,l1 e
δα4α1 δα3α2 .
We now proceed to the following permutations of the li variables in the T1 term : l1 7→ l2,
l2 7→ l1, l3 7→ l4, l4 7→ l3. While
i li is invariant, θ is modified : θ 7→ l2.Θl1 + l2.Θl4 + l1.Θl4.
With δ0,
in factor, we can let l4 be −l1 − l2 − l3, so that θ 7→ −θ. We also permute the αi
in the same way. Thus,
l1,...,l4
aα3,l3 aα4,l4 aα1,l1 aα2,l2 e
δα3α1 δα4α2 .
Therefore,
2T1 = 2
l1,...,l4
aα4,l4 aα3,l3 aα2,l2 aα1,l1 cos
δα4α2 δα3α1 . (63)
The same principles are applied to T2 and T3. Namely, the permutation l1 7→ l1, l2 7→ l3, l3 7→ l2,
l4 7→ l4 in T2 and the permutation l1 7→ l2, l2 7→ l3, l3 7→ l1, l4 7→ l4 in T3 (the αi variables are
permuted the same way) give
l1,...,l4
aα4,l4aα3,l3aα2,l2 aα1,l1 e
δα4α2 δα3α1 ,
l1,...,l4
aα4,l4 aα3,l3aα2,l2 aα1,l1 e
δα4α2 δα3α1
where φ := l1.Θ l2 + l1.Θ l3 − l2.Θ l3. Finally, we get
−(A+)4 = 4c
l1,...,l4
aα1,l4 aα2,l3 a
− cos θ
l1,...,l3
aα1,−l1−l2−l3 aα2,l3 a
l1.Θ(l2+l3)
sin l2.Θl3
. (64)
(iv) Suppose q = 2. By Lemma 6.11, we get
− Aσ = Res
λσfα,µ(s, l)Tr(γ
α2γµ2γα1γµ1)
where
fα,µ(s, l) :=
′ kµ1 (k+l)µ2
|k|s+2|k+l|2
eiη k.Θl ãα,l
and η := 1
(σ1 − σ2) ∈ {−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
fα,µ(s, l) ∼ f1(s, l)− f2(s, l) + f3(s, l)
where
f1(s, l) =
′ kµ1(k+l)µ2
|k|s+4
eiη k.Θl ãα,l,
f2(s, l) =
′ kµ1(k+l)µ2 (2k.l+|l|
|k|s+6
eiη k.Θl ãα,l,
f3(s, l) =
′ kµ1(k+l)µ2 (2k.l+|l|
|k|s+8
eiη k.Θl ãα,l.
Suppose that l = 0. Then f2(s, 0) = f3(s, 0) = 0 and Proposition 2.1 entails that
f1(s, 0) =
kµ1kµ2
|k|s+4
ãα,0
is holomorphic at 0 and so is fα,µ(s, 0).
Since 1
Θ is diophantine, Theorem 2.5 3 gives us the result.
Suppose q = 3. Then Lemma 6.11 implies that
− Aσ = Res
l∈(Zn)2
fµ,α(s, l) Tr(γ
µ3γα3 · · · γµ1γα1)
where
fµ,α(s, l) :=
ik.Θ(ε1l1+ε2l2)e
σ2l1.Θl2 kµ1 (k+l1)µ2 (k+l1+l2)µ3
|k|s+2|k+l1|2|k+l1+l2|2
ãα,l,
and εi :=
(σi − σ3) ∈ {−1, 0, 1}. By hypothesis (ε1, ε2) 6= (0, 0). There are six possibilities
for the values of (ε1, ε2), corresponding to the six possibilities for the values of σ: (−,−,+),
(−,+,+), (+,−,+), (+,+,−), (−,+,−), and (+,−,−). As in (ii), we see that
fµ,α(s, l) ∼
′ eik.Θ(ε1l1+ε2l2)kµ1 (k+l1)µ2 (k+
bl2)µ3
|k|s+6
′ eik.Θ(ε1l1+ε2l2)kµ1 (k+l1)µ2 (k+
bl2)µ3 (2k.l1+2k.
|k|s+8
λσ ãα,l e
σ2l1.Θl2 .
With Z := {(l1, l2) : ε1l1 + ε2l2 = 0}, Theorem 2.5 (iii) entails that
l∈(Zn)2\Z fµ,α(s, l) is
holomorphic at 0. To conclude we need to prove that
g(σ) :=
fµ,α(s, l) Tr(γ
µ3γα3 · · · γµ1γα1)
is holomorphic at 0. By definition, λσ = iσ1σ2σ3 and as a consequence, we check that
g(−,−,+) = −g(+,+,−), g(+,−,+) = −g(+,−,−), g(−,+,+) = −g(−,+,−),
which implies that
σ g(σ) = 0. The result follows.
Suppose finally that q = 4. Again, Lemma 6.11 implies that
− Aσ = Res
l∈(Zn)3
fµ,α(s, l) Tr(γ
µ4γα4 · · · γµ1γα1)
where
fµ,α(s, l) :=
i=1 εili e
(σ2l1.Θl2+σ3(l1+l2).Θl3) kµ1 (k+l1)µ2 (k+l1+l2)µ3 (k+l1+l2+l3)µ4
|k|s+2|k+l1|2|k+l1+l2|2|k+l1+l2+l3|2
ãα,l
and εi :=
(σi − σ4) ∈ {−1, 0, 1}. By hypothesis (ε1, ε2, ε3) 6= (0, 0, 0). There are fourteen pos-
sibilities for the values of (ε1, ε2, ε3), corresponding to the fourteen possibilities for the values of
σ: (−,−,−,+), (−,−,+,+), (−,+,−,+), (+,−,−,+), (−,+,+,+), (+,−,+,+), (+,+,−,+),
(+,+,+,−), (−,−,+,−), (−,+,−,−), (+,−,−,−), (−,+,+,−), (+,−,+,−) and (+,+,−,−).
As in (ii), we see that, with the shorthand θσ := σ2l1.Θl2 + σ3(l1 + l2).Θl3,
fµ,α(s, l) ∼
i=1 εili e
θσ kµ1kµ2kµ3kµ4
|k|s+8
ãα,l =: gµ,α(s, l) .
With Zσ := {(l1, l2, l3) :
i=1 εili = 0}, Theorem 2.5 (iii), the series
l∈(Zn)3\Zσ
fµ,α(s, l) is
holomorphic at 0. To conclude, we need to prove that
g(σ) :=
gµ,α(s, l) Tr(γ
µ4γα4 · · · γµ1γα1) = 0.
Let C be the set of the fourteen values of σ and C7 be the set of the seven first values of σ given
above. Lemma 6.7 implies ∑
g(σ) = 2
g(σ).
Thus, in the following, we restrict to these seven values. Let us note Fµ(s) :=
kµ1kµ2kµ3kµ4
|k|s+8
so that
g(σ) = Res
Fµ(s)λσ
θσ ãα,l Tr(γ
µ4γα4 · · · γµ1γα1).
Recall from (62) that
Fµ(s)Tr(γ
µ4γα4 · · · γµ1γα1) = 2c
δα4α3δα2α1 + δα4α1δα3α2 − 2δα4α2δα3α1
As a consequence, we get, with ãα,l := aα1,l1 · · · aα4,l4 ,
g(σ) = 2cλσ
l∈(Zn)4
θσ ãα,l δP4
i=1 li,0
i=1 εili,0
δα4α3δα2α1 + δα4α1δα3α2 − 2δα4α2δα3α1
=: 2cλσ(T1 + T2 − 2T3).
We proceed to the following change of variable in T1: l1 7→ l1, l2 7→ l3, l3 7→ l2, l4 7→ l4. Thus,
we get θσ 7→ ψσ := σ2l1.Θl3 + σ3(l1 + l3).Θl2, and
i=1 εili 7→ ε1l1 + ε3l2 + ε2l3 =: uσ(l). With
a similar permutation on the αi, we get
l∈(Zn)4
ψσ ãα,l δP4
i=1 li,0
δε1l1+ε3l2+ε2l3,0 δ
α4α2δα3α1 .
We proceed to the following change of variable in T2: l1 7→ l2, l2 7→ l3, l3 7→ l1, l4 7→ l4. Thus,
we get θσ 7→ φσ := σ2l2.Θl3 + σ3(l2 + l3).Θl1, and
i=1 εili 7→ ε3l1 + ε1l2 + ε2l3 =: vσ(l). After
a similar permutation on the αi, we get
l∈(Zn)4
φσ ãα,l δP4
i=1 li,0
δε3l1+ε1l2+ε2l3,0 δ
α4α2δα3α1 .
Finally, we proceed to the following change of variable in T3: l1 7→ l2, l2 7→ l1, l3 7→ l4, l4 7→ l3.
Thus, we get θσ 7→ −θσ, and
i=1 εili 7→ (ε2− ε3)l1+(ε1− ε3)l2− ε3l3 =: wσ(l). With a similar
permutation on the αi, we get
l∈(Zn)4
θσ ãα,l δP4
i=1 li,0
δ(ε2−ε3)l1+(ε1−ε3)l2−ε3l3,0δ
α4α2δα3α1 .
As a consequence, we get
g(σ) = 2c
l∈(Zn)4
Kσ(l1, l2, l3) ãα,l δP4
i=1 li,0
δα4α2δα3α1 ,
where Kσ(l1, l2, l3) = λσ
ψσ δuσ(l),0 + e
φσ δvσ(l),0 − e
θσ δP3
i=1 εili,0
θσ δwσ(l),0
The computation of Kσ(l1, l2, l3) for the seven values of σ yields
K−−++(l1, l2, l3) = δl1+l3,0 + δl2+l3,0 − δl1+l2,0 − δl1+l2,0,
K−+−+(l1, l2, l3) = δl1+l2,0 + δl1+l2,0 − δl1+l3,0 − δl1+l3,0,
K−−++(l1, l2, l3) = δl2+l3,0 + δl1+l3,0 − δl2+l3,0 − δl2+l3,0,
K−−−+(l1, l2, l3) = −
l1.Θl2δP3
i=1 li,0
l2.Θl1δP3
i=1 li,0
l2.Θl1δP3
i=1 li,0
l1.Θl2δl3,0
K−+++(l1, l2, l3) = −
l3.Θl2δl1,0 + e
l3.Θl1δl2,0 − e
l2.Θl3δl1,0 − e
l3.Θl1δl2,0
K+−++(l1, l2, l3) = −
l1.Θl2δl3,0 + e
l2.Θl1δl3,0 − e
l1.Θl3δl2,0 − e
l3.Θl2δl1,0
K++−+(l1, l2, l3) = −
l1.Θl3δl2,0 + e
l2.Θl3δl1,0 − e
l1.Θl2δl3,0 − e
l2.Θl1δP3
i=1 li,0
Thus, ∑
Kσ(l1, l2, l3) = 2i(δl3,0 − δP3
i=1 li,0
) sin l1.Θl2
and ∑
g(σ) = i4c
l∈(Zn)4
(δl3,0 − δP3
i=1 li,0
) sin l1.Θl2
ãα,l δP4
i=1 li,0
δα4α2δα3α1 .
The following change of variables: l1 7→ l2, l1 7→ l2, l3 7→ l4, l4 7→ l3 gives
l∈(Zn)4
1 li,0
sin l1.Θl2
ãα,l δP4
1 li,0
δα4α2δα3α1 = −
l∈(Zn)4
δl3,0 sin
l1.Θl2
ãα,l δP4
1 li,0
δα4α2δα3α1
g(σ) = i8c
l∈(Zn)4
δl3,0 sin
l1.Θl2
ãα,l δP4
1 li,0
δα4α2δα3α1 .
Finally, the change of variables: l2 7→ l4, l4 7→ l2 gives
l∈(Zn)4
δl3,0 sin
l1.Θl2
ãα,l δP4
1 li,0
δα4α2δα3α1 = −
l∈(Zn)4
δl3,0 sin
l1.Θl2
ãα,l δP4
1 li,0
δα4α2δα3α1
which entails that
g(σ) = 0.
Lemma 6.13. Suppose n = 4 and 1
Θ diophantine. For any self-adjoint one-form A,
ζDA(0) − ζD(0) = −c τ(Fα1,α2Fα1α2).
Proof. By (34) and Lemma 6.6 we get
ζDA(0) − ζD(0) =
(−1)q
σ∈{+,−}q
− Aσ.
By Lemma 6.12 (iv), we see that the crossed terms all vanish. Thus, with Lemma 6.7, we get
ζDA(0)− ζD(0) = 2
(−1)q
−(A+)q. (65)
By definition,
Fα1α2 = i
aα2,k kα1 − aα1,k kα2
aα1,k aα2,l [Uk, Ul]
(aα2,k kα1 − aα1,k kα2)− 2
aα1,k−l aα2,l sin(
τ(Fα1α2F
α1α2) =
α1, α2=1
(aα2,k kα1 − aα1,k kα2)− 2
l′∈Z4
aα1,k−l′ aα2,l′ sin(
k.Θl′
(aα2,−k kα1 − aα1,−k kα2)− 2
l”∈Z4
aα1,−k−l” aα2,l” sin(
k.Θl”
One checks that the term in aq of τ(Fα1α2F
α1α2) corresponds to the term
(A+)q given by
Lemma 6.12. For q = 2, this is
l∈Z4, α1, α2
aα1,l aα2,−l
lα1 lα2 − δα1α2 |l|2
For q = 3, we compute the crossed terms:
k,k′,l
(aα2,k kα1 − aα1,k kα2) a
Uk[Uk′ , l] + [Uk′ , Ul]Uk
which gives the following a3-term in τ(Fα1α2F
α1α2)
aα3,−l1−l2 a
aα1,l1 sin
l1.Θl2
For q = 4, this is
aα1,−l1−l2−l3 aα2,l3 a
l1.Θ(l2+l3)
sin l2.Θl3
which corresponds to the term
(A+)4. We get finally,
(−1)q
−(A+)q = − c
τ(Fα1,α2F
α1α2). (66)
Equations (65) and (66) yield the result.
Lemma 6.14. Suppose n = 2. Then, with the same hypothesis as in Lemma 6.11,
−(A+)2 =
−(A−)2 = 0.
(ii) Suppose 1
Θ diophantine. Then
− A+A− =
− A−A+ = 0.
Proof. (i) Lemma 6.11 entails that
A++ = Res
l∈Z2 −f(s, l) where
f(s, l) :=
kµ1 (k+l)µ2
|k|s+2|k+l|2
ãα,l Tr(γ
α2γµ2γα1γµ1) =: fµ,α(s, l)Tr(γ
α2γµ2γα1γµ1)
and ãα,l := aα1,l aα2,−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)
fµ,α(s, l) =
′ kµ1 (k+l)µ2
|k|s+4
ãα,l −
′ kµ1 (k+l)µ2 (2k.l+|l|
|k|s+4|k+l|2
ãα,l.
and the function r(s, l) :=
kµ1 (k+l)µ2 (2k.l+|l|
|k|s+4|k+l|2
ãα,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
fµ,α(s, l) ∼
′ kµ1 (k+l)µ2
|k|s+4
ãα,l ∼
′ kµ1kµ2
|k|s+4
ãα,l
Note that the function (s, l) 7→ hµ,α(s, l) :=
kµ1kµ2
|k|s+4
ãα,l satisfies (H2). Thus, Lemma 2.14
yields
f(s, l) =
hµ,α(s, l)Tr(γ
α2γµ2γα1γµ1).
By Proposition 2.16, we get Res
hµ,α(s, l) = δµ1µ2 π ãα,l. Therefore,
− A++ = −π
ãα,l Tr(γ
α2γµγα1γµ) = 0
according to (59).
(ii) By Lemma 6.11, we obtain that
A−+ = Res
l∈Z2 λσfα,µ(s, l)Tr(γ
α2γµ2γα1γµ1) where
λσ = −(−i)2 = 1 and
fα,µ(s, l) :=
′ kµ1 (k+l)µ2
|k|s+2|k+l|2
eiη k.Θl ãα,l
and η := 1
(σ1 − σ2) = −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
fα,µ(s, l) ∼
′ kµ1 (k+l)µ2
|k|s+4
eiη k.Θl ãα,l := gα,µ(s, l) .
Since 1
Θ is diophantine, the functions s 7→
l∈Z2\{0} gα,µ(s, l) are holomorphic at s = 0 by
Theorem 2.5 3. As a consequence,
− A−+ = Res
gα,µ(s, 0)Tr(γ
α2γµ2γα1γµ1) = Res
′ kµ1kµ2
|k|s+4
ãα,0 Tr(γ
α2γµ2γα1γµ1).
Recall from Proposition 2.1 that Ress=0
|k|s+4
= δij π. Thus, again with (59),
− A−+ = ãα,0 π Tr(γα2γµγα1γµ) = 0.
Lemma 6.15. Suppose n = 2 and 1
Θ diophantine. For any self-adjoint one-form A,
ζDA(0) − ζD(0) = 0.
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 1
Θ diophantine. Then for any self-adjoint 1-form A and
σ ∈ {−,+}q with 2 ≤ q ≤ n, ∫
− Aσ = 0 .
Proof. Since Aσ ∈ Ψ1(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,
ζDA(0)− ζD(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
S(DA,Φ,Λ) = 4πΦ2Λ2 +Φ(0) ζDA(0) +O(Λ−2),
where Φ2 =
Φ(t) dt. By Lemma 6.15, ζDA(0) − ζD(0) = 0 and from Proposition 5.4,
ζD(0) = 0, so we get the result.
(ii) Similarly, S(DA,Φ,Λ) = 8π2 Φ4Λ4+Φ(0) ζDA(0)+O(Λ−2) with Φ4 = 12
Φ(t) t dt. Lemma
6.13 implies that ζDA(0)−ζD(0) = −c τ(FµνFµν) and by Proposition 5.4, ζDA(0) = −c τ(FµνFµν)
leading to the result.
(iii) is a direct consequence of (5), Propositions 5.4, 5.5, and Corollary 6.17.
A Appendix
A.1 Proof of Lemma 3.3
(i) We have |D|T |D|−1 = T + δ(T )|D|−1 and |D|−1T |D| = T − |D|−1δ(T ). A recurrence
proves that for any k ∈ N, |D|kT |D|−k =
δq(T )|D|−q and we get |D|−kT |D|k =
q=0(−1)q
|D|−qδq(T ).
As a consequence, since T , |D|−q and δq(T ) are in OP 0 for any q ∈ N, for any k ∈ Z,
|D|kT |D|−k ∈ OP 0. Let us fix p ∈ N0 and define Fp(s) := δp(|D|sT |D|−s) for s ∈ C. Since
for k ∈ Z, Fp(k) is bounded, a complex interpolation proves that Fp(s) is bounded, which gives
|D|sT |D|−s ∈ OP 0.
(ii) Let T ∈ OPα and T ′ ∈ OP β. Thus, T |D|−α, T ′|D|−β are in OP 0. By (i) we get
|D|βT |D|−α|D|−β ∈ OP 0, so T ′|D|−β|D|βT |D|−β−α ∈ OP 0. Thus, T ′T |D|−(α+β) ∈ OP 0.
(iii) For T ∈ OPα, |D|α−β and T |D|−α are in OP 0, thus T |D|−β = T |D|−α|D|α−β ∈ OP 0.
(iv) follows from δ(OP 0) ⊆ OP 0.
(v) Since ∇(T ) = δ(T )|D|+ |D|δ(T )− [P0 , T ], the result follows from (ii), (iv) and the fact that
P0 is in OP
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 ′ ∈
Ψ(A). There exist d, d′ ∈ Z such that for any N ∈ N, N > |d|+ |d′|, there exist P,P ′ in D(A),
p, p′ ∈ N0, R ∈ OP−N−d
, R′ ∈ OP−N−d such that T = PD−2p + R, T ′ = P ′D−2p′ + R′,
PD−2p ∈ OP d and P ′D−2p′ ∈ OP d′ .
Thus, TT ′ = PD−2pP ′D−2p
+RP ′D−2p
+ PD−2pR′ +RR′.
We also have RP ′D−2p
′ ∈ OP−N−d′+d′ = OP−N and similarly, PD−2pR′ ∈ OP−N . Since
RR′ ∈ OP−2N , we get
TT ′ ∼ PD−2pP ′D−2p′ mod OP−N .
If p = 0, then TT ′ ∼ QD−2p′ mod OP−N where Q = PP ′ ∈ D(A) and QD−2p′ ∈ OP d+d′ .
Suppose p 6= 0. A recurrence proves that for any q ∈ N0,
D−2P ′ ∼
(−1)k∇k(P ′)D−2k−2 + (−1)q+1D−2∇q+1(P ′)D−2q−2 mod OP−∞ .
By Lemma 3.3 (v), the remainder is in OP d
′+2p′−q−3, since P ′ ∈ OP d′+2p′ . Another recurrence
gives for any q ∈ N0,
D−2pP ′ ∼
k1,··· ,kp=0
(−1)|k|1∇|k|1(P ′)D−2|k|1−2p mod OP d′+2p′−q−1−2p.
Thus, with qN = N + d+ d
′ − 1,
TT ′ ∼
k1,··· ,kp=0
(−1)|k|1P∇|k|1(P ′)D−2|k|1−2(p+p′) mod OP−N .
The last sum can be written QND
−2rN where rN := p qN + (p + p
′). Since QN ∈ D(A) and
−2rN ∈ OP d+d′ , the result follows.
A.3 Proof of Proposition 3.11
Let P ∈ OP k1 , Q ∈ OP k2 ∈ Ψ(A). With [Q, |D|−s] =
Q− σ−s(Q)
|D|−s and the equivalence
Q− σ−s(Q) ∼ −
r=1 g(−s, r) εr(Q) mod OP−N−1+k2 , we get
P [Q, |D|−s] ∼ −
g(−s, r)Pεr(Q)|D|−s mod OP−N−1+k1+k2−ℜ(s)
which gives, if we choose N = n+ k1 + k2,
P [Q, |D|−s]
n+k1+k2∑
g(−s, r)Tr
Pεr(Q)|D|−s
By hypothesis s 7→ Tr
Pεr(Q)|D|−s
has only simple poles. Thus, since s = 0 is a zero of the
analytic function s 7→ g(−s, r) for any r ≥ 1, we have Res
g(−s, r) Tr
Pεr(Q)|D|−s
= 0, which
entails that Res
P [Q, |D|−s]
= 0 and thus
− PQ = Res
P |D|−sQ
When s ∈ C with ℜ(s) > 2max(k1 + n + 1, k2), the operator P |D|−s/2 is trace-class while
|D|−s/2Q is bounded, so Tr
P |D|−sQ
|D|−s/2QP |D|−s/2
σ−s/2(QP )|D|−s
Thus, using (29) again,
P |D|−sQ
− QP +
n+k1+k2∑
g(−s/2, r)Tr
εr(QP )|D|−s
As before, for any r ≥ 1, Res
g(−s/2, r)Tr
εr(QP )|D|−s
= 0 since g(0, r) = 0 and the spectral
triple is simple. Finally,
P |D|−sQ
− QP.
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.
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	Introduction
	Residues of series and integral, holomorphic continuation, etc
	Residues of series and integral
	Holomorphy of certain series
	Proof of Lemma ?? for i=1:
	Proof of Lemma ?? for i=0:
	Proof of item (i.2) of Theorem ??:
	Proof of item (iii) of Theorem ??:
	Commutation between sum and residue
	Computation of residues of zeta functions
	Meromorphic continuation of a class of zeta functions
	A family of polynomials
	Residues of a class of zeta functions
	Noncommutative integration on a simple spectral triple
	Kernel dimension
	Pseudodifferential operators
	Zeta functions and dimension spectrum
	The noncommutative integral   
	Residues of DA for a spectral triple with simple dimension spectrum
	The noncommutative torus
	Notations
	Kernels and dimension spectrum
	Noncommutative integral computations
	The spectral action
	Computations of   
	Even dimensional case
	Odd dimensional case
	Proof of the main result
	Appendix
	Proof of Lemma ??
	Proof of Lemma ??
	Proof of Proposition ??
