Measurable operators and the asymptotics of heat kernels and zeta functions

0.21 MB·English

by  Alan Carey

Measurable operators and the asymptotics of heat
kernels and zeta functions
Alan Carey, Fedor Sukochev
2
Mathematical SciencesInstitute, Australian National University,Canberra ACT, 0200,
1 Australia
0
2 School of Mathematics and Statistics, Universityof New South Wales, Sydney, 2052,
Australia.
n
a
J

7
1

Abstract
]
A
In this note we answer some questions inspired by the introduction in [6, 7],
F
by Alain Connes, of the notion of measurable operators using Dixmier traces.
.
h These questions concernthe relationshipofmeasurability to the asymptoticsof
t ζ functions and heat kernels. The answers have remained elusive for some 15
a
−
m years 1.
[ Keywords: Dixmier traces, heat kernels, measurable elements, generalized

1 limits, Ces`aro operator.
v
2000 MSC: Primary: 58B34, 46L52
1
6
6
3 1. Introduction and Preliminaries
.
1
0 In [7] (see also [6]) Alain Connes described in part the relationship between
2 Dixmier traces, heat kernel asymptotics and the behaviour of ζ functions at
1 their leading singularity. In that discussion he introduced the not−ion of a mea-
:
v surable operator. Subsequently these notions have arisen in other contexts
i and interest has been generated in obtaining a comprehensive picture of how
X
they are related. The present authors were forced to confront these ideas in
r
a their attempts to develop tools for semifinite noncommutative geometry in [3],
[5]. Similar issues arise also in [2]. In addition, after discussions with many
colleagues, it became clear to us that, for applications, extensions of [7, 3]
were needed. There has been considerable progress in the last few years in
[2, 4, 14, 16, 17, 22, 23, 24, 26, 27]. In this note we provide the final answer to
two of the outstanding questions.
∗CorrespondingAuthor
Email addresses: [e-mail] (AlanCarey),[e-mail]
(FedorSukochev)
1Wededicate thispapertothememoryofNigelKalton.
Preprint submitted toElsevier January 19, 2012
As we have done previously in [3], [5] we will work in the generality of
semifinite von Neumann algebras although even for the more standard case of
the bounded operators on Hilbert space the results of this paper are new. Let
be a von Neumann algebra equipped with a faithful normal semifinite trace
M
τ.ForeveryoperatorA ,letE (s, )denotethespectralmeasureof A,
|A|
∈M ∞ | |
then its distribution function d and rearrangement µ(A) are defined by the
A
following formulas:
d (s)=τ(E (s, )), s>0
A |A|
∞
µ(t,A)=inf s: d (s) t , t>0.
A
{ ≤ }
The following sets of operatorsfrom are widely used in noncommutative
M
geometry (see [7, 2, 3, 4, 5, 16, 17, 27]). The readershould be awareof the fact
that the notation we are using is not that of [7].
1 t
= A : sup µ(s,A)ds<
M1,∞ { ∈M log(1+t)Z ∞}
t∈(0,∞) 0
and
= A : suptµ(t,A)< .
1,∞
L { ∈M ∞}
t>0
Equipped with the norm
1 t
A := sup µ(s,A)ds
k kM1,∞ log(1+t)Z
t∈(0,∞) 0
the first set is an example of a Marcinkiewicz operator space. The second
set is the so-called weak space, which is a linear (non-closed) subspace in
1
L
( , ). Recall also that is not dense in with respect to
M1,∞ k·kM1,∞ L1,∞ M1,∞
the norm (see e.g. [13, Lemma 5.5 in Ch.II.7]).
k·kM1,∞
We need the (multiplicative) Cesarooperatoracting on the space L (0, )
∞
∞
of all essentially bounded Lebesgue measurable functions given by the formula
1 ν ds
(Mx)(ν)= x(s) . (1)
log(ν)Z s
1
If A , then it follows from [4, Lemma 5.1] that
1,∞
∈L
sup 1τ(e−(λA)−1)< . (2)
λ ∞
λ
The inequality (2) does not necessarilyhold for A (see [4, Example on
1,∞
∈M
p. 274]). It is implicitly proved in [4] (see also [27, Theorem 40 and Corollary
41] where a much stronger result is established) that
supM(λ 1τ(e−(λA)−1))< (3)
→ λ ∞
λ
foreveryA wherethenotationisashorthandfortakingthesupremum
1,∞
of the funct∈ionMobtained from applying M to λ 1τ(e−(λA)−1).
→ λ
This note is motivated by the following two questions (that we will com-
pletely answer here).
2
Question 1. Suppose that A + is such that the limit
∈L1,∞
lim 1τ(e−(λA)−1)
λ→∞λ
exists. What information is then available on the distribution function of the
operator A?
Question 2. Suppose that A + is such that the following limit
∈M1,∞
lim M(λ 1τ(e−(λA)−1))
λ→∞ → λ
exists. What information is then available on the distribution function of the
operator A?
NoteweareagainusinganobviousshorthandnotationinQuestion2. These
questions are in fact related to somewhat similar matters studied in [4] (there
theyarecalledquestionsAandB)fortheζ-functionandtheconnectionbetween
them can be established via the theory of Dixmier traces (see e.g. [9, 7, 3, 2, 5,
17, 16, 14, 24, 26]). Very briefly, we now recallsome basic definitions from that
theory.
First, a positive normalised functional on a unital von Neumann algebra is
calledastateandanystateonthealgebraL (0, )iscalledageneralisedlimit
∞
∞
if it vanishes on every function with compact support.
Second, given s > 0, a dilation operator σ : L (0, ) L (0, ) is
s ∞ ∞
∞ → ∞
definedbysetting (σ x)(t)=x(t/s). Ageneralisedlimit ω is saidtobe dilation
s
invariant if ω σ =ω for every s>0.
s
◦
Third,ifω isanarbitrarydilationinvariantgeneralisedlimitthenaDixmier
trace τ on is defined (see [14, Definition 9]) by the formula
ω 1,∞
M
1 t
τ (A):=ω(t µ(s,A)ds), A + .
ω → log(1+t)Z ∈M1,∞
0
Recall that a positive linear functional ϕ on is called fully symmetric if
1,∞
M
for all 0 A,B such that
1,∞
≤ ∈M
t t
µ(s,B)ds µ(s,A)ds, for all t>0,
Z ≤Z
0 0
wehaveϕ(B) ϕ(A). InthisnotethelargestpossibleclassofDixmiertraces,
≤
namely the class
:= τ : ω is a dilation invariant generalized limit
ω
D { }
ofallDixmiertracesisneeded. (Seefurtherpossibilitiesin[7,17,5,16,26,24]).
Thatthisclassisnaturalisconfirmedbythefactthat coincideswiththeclass
D
of all fully symmetric singular functionals on .
1,∞
M
More precisely, the following assertion follows from [14, Theorem 11] if we
set the function denoted by ψ in that theorem to be ψ(t)=log(1+t).
3
Theorem 1. For every fully symmetric functional ϕ on , there exists a
1,∞
M
dilation invariant generalised limit ω such that the Dixmier trace τ =ϕ.
ω
Now we establish the notation for, and background to, our main theorem.
Let ω be an arbitrary dilation invariant generalised limit. A heat kernel
functional ξ is defined (see [27, Sections 1 and 5]) by the formula
ω
ξ (A):=(ω M)(λ 1τ(e−(λA)−1)), A + .
ω ◦ → λ ∈M1,∞
See [7, 3, 4, 27]) for the reasons for this particular form of the definition of the
heat kernel functional.
Let γ be an arbitrary generalised limit. The ζ-function residue (associated
with γ) is defined (see [27, Section 1] and also [7, 3]) by the formula
1
ζ (A):=γ(r τ(A1+1/r)).
γ
→ r
Evidence that the zeta andheatkernelfunctionals areclosely relatedcomes
from the following theorems, proved in [27].
Theorem 2. [27, Theorem 8]. For every generalised limit γ, the ζ-function
residue ζ is a fully symmetric functional on .
γ 1,∞
M
Theorem 3. [27, Theorem 22]. For every dilation invariant generalised limit
ω, the heat kernel functional ξ is a fully symmetric functional on .
ω 1,∞
M
Theorem 4. [27, Theorem 31]. For every fully symmetric functional ϕ on
, there exists a dilation invariant generalised limit ω such that the heat
1,∞
M
kernel functional ξ =ϕ.
ω
Remark 5. In fact, it is proved in [27, Theorem 31] and [27, Lemma 20] that
foreveryfullysymmetricfunctionalϕon ,thereexistsadilationinvariant
1,∞
M
generalised limit ω such that for every q >0,
(ω M)(λ 1τ(e−(λA)−q))=Γ(1+1/q)ϕ
◦ → λ
.
InviewofTheorems1to4,itisnaturaltoaskwhetherthe equalityτ =ξ
ω ω
holds for an arbitrary dilation invariant generalised limit ω. This is however
notthe case,see[27, Theorem37]whereexamplesofω’saregivenforwhichwe
have τ =ξ .
ω ω
6
Finally, we come to one ofthe majornew notionsintroducedin this context
in [7] (see also [6]) and generalised in [17, 5, 16, 24, 26]:
Definition 6. The operator A is said to be measurable if and only if
1,∞
∈ M
the set τ (A): τ consists of a single point.
ω ω
{ ∈D}
4
Inviewofthepreviouslycitedresultsandcounter-examplesourmainresult,
whichwenowstate,isnotentirelyexpected. ItanswersQuestion2andcomple-
ments and extends earlier results in [7, 3, 4]. It also provides a very short new
proof of the main result from [17] (see the proof of the implication (i) = (ii)
⇒
below).
Theorem 7. Let A be a positive operator. The following conditions
1,∞
∈ M
are equivalent.
(i) The operator A is measurable.
(ii) The limit lim 1 tµ(s,A)ds exists.
t→∞ log(1+t) 0
(iii) The limit lim M(λ R1τ(e−(λA)−1)) exists.
λ→∞ → λ
(iv) The limit lim sτ(A1+s) exists.
s→0
Furthermore, if any of the conditions (i)-(iv) above holds, then we have the
coincidence of the three limits
lim 1 tµ(s,A)ds= lim M(λ 1τ(e−(λA)−1))= limsτ(A1+s)
t→∞log(1+t)Z0 λ→∞ → λ s→0
with the value given by τ (A): τ .
ω ω
{ ∈D}
Remark 8. Let a = τ (A) for every τ in (i) in Theorem 7. Let a , a
1 ω ω 2 3
∈ D
and a be thelimits in (ii), (iii)and (iv)(respectively) in Theorem 7. For every
4
1 i,j 4, we show in the proof of Theorem 7 that (i) = (j) and a =a .
i j
≤ ≤ ⇒
The results should be seen in the generalcontext of the continuing study of
the notion of measurable operators introduced in [6, 7] and further elaborated
in [17, 5, 16]. The main interest remains in the following areas: (i) comparing
various modifications of this notion with respect to various subsets of Dixmier
traces(asarulewithadditionalpropertiesofinvariance),(ii)findingconvenient
descriptionsofthesetofself-adjointmeasurableoperators,and(iii)determining
when a given self-adjoint measurable operator is Tauberian. We remark that
this current note is related to progress on these directions which will appear
in [24, 26], where it is shown that not every self-adjoint measurable operator
is necessarily Tauberian (which is in stark contrast with the case of positive
operators). It will also be shown in [26] that the notion of measurability as
originallyintroducedbyConnesin[6]anditsversionconsideredin[17]actually
coincide.
Acknowledgements. ThisresearchwassupportedbytheAustralianResearch
Council. The authors thank Dima Zanin for numerous discussions and help in
the preparationofthis article. The ideato useLemma11belowbelongsto him
andthe usageof this lemma has significantlysimplified our originalproofs. We
also thank Bruno Iochum for many discussions on the issues surrounding the
resultsofthisnote. ThefirstnamedauthorthankstheAlexandervonHumboldt
Stiftung and colleagues at the University of Mu¨nster.
5
2. Proof of the main result
The following lemma is well-known. The proof can be found in e.g. [12,
Section 6.8].
Lemma 9. Let z L (0, ) be a positive differentiable function. If tz′(t)
∞
∈ ∞ ≥
const for every t>0, then the following implication holds
1 t
lim z(s)ds=C = lim z(t)=C.
t→∞ t Z ⇒t→∞
0
The following lemma is also well-known. Due to the lack of a suitable refer-
ence we provide a short proof for convenience of the reader.
Lemma 10. Let x L (0, ) and let a R. The following conditions are
∞
∈ ∞ ∈
equivalent.
1. We have the bounds
liminfx(t) a limsupx(t).
t→∞ ≤ ≤ t→∞
2. There exists a generalised limit γ such that γ(x)=a.
Proof. Theimplication(2) (1)followsimmediatelyfromthedefinitionofthe
→
generalised limit.
In orderto provethe implication (1) (2), define a functional γ on R+xR
→
by setting γ(α+βx)=α+βa. Clearly,
γ(z) limsupz(t), z R+xR.
≤ ∈
t→∞
The assertion follows now from the Hahn-Banach theorem.
Our next lemma plays an important role in the proof of our main result.
Lemma 11. Let z be a positive locally integrable function on (0, ). If Mz
∞ ∈
L (0, ), then we have
∞
∞
lim(M2z)(t)=C = lim(Mz)(t)=C.
t→∞ ⇒t→∞
Proof. Set x=(Mz) exp. We have
◦
1 t du 1 log(t)
(M2z)(t)= (Mz)(u) = x(s)ds,
log(t)Z u log(t)Z
1 0
where we used the substitution u=es in the second equality. By the assump-
tion, we have
1 t
lim x(s)ds =C.
t→∞ t Z
0
6
Letusnowverifythatthefunctiont tx′(t)satisfiestheassumptionofLemma
→
9. We have
1 et ds 1 et ds
tx′(t)=t( z(s) )′ = z(s) +z(et).
t Z s −t Z s
0 0
Since z is positive, we have tx′(t) (Mz)(et) and since Mz L (0, ), we
∞
≥ − ∈ ∞
conclude tx′(t) const. By Lemma 9, we have lim x(t) = C and hence
t→∞
≥
lim (Mz)(t)=C.
t→∞
The following remark is well known and can be found in e.g. [7].
Remark12. Foreverygeneralisedlimitγ,thestateγ M isadilationinvariant
◦
generalised limit.
With these preliminary results in hand we come to the proof of our main
result.
Proof. (Of Theorem 7.) First, the implication (ii)= (i) follows from the defi-
⇒
nition of τ . Next, the implication (i)= (ii) was first proved in [17, Theorem
ω
⇒
6.6](seealso[5]). Weprovidehereanew(veryshortandstraightforward)proof.
Let
C :=τ (A), for all τ .
ω ω
∈D
In particular, by Remark 12, we have τ (A) = C for every generalised limit
γ◦M
γ. That is, we have the equality
1 t
(γ M)(t µ(s,A)ds)=C,
◦ → log(1+t)Z
0
which, due to Lemma 10, guarantees
1 t
lim M(t µ(s,A)ds)=C. (4)
t→∞ → log(1+t)Z
0
Setz(t):=tµ(t,A). Observethatz isapositivemeasurable,butnotnecessarily
bounded function. However, since A , the function
1,∞
∈M
1 t
t (Mz)(t)= µ(s,A)ds
→ log(t)Z
1
is bounded. Thus, Mz L (0, ) and obviously
∞
∈ ∞
1 t
lim(Mz)(t) µ(s,A)=0. (5)
t→∞ − log(1+t)Z
0
Combining (4) and (5), and using the (obvious) fact that lim (My)(t) = 0
t→∞
whenever y L (0, ) satisfies lim y(t)=0, we infer that
∞ t→∞
∈ ∞
lim(M2z)(t)=C.
t→∞
7
By Lemma 11, we obtain from the preceding equality
lim(Mz)(t)=C
t→∞
and the proof of the implication is completed by referring to (5).
(iii) = (i). Let C be the limit in (iii). By definition of ξ , we have
ω
⇒
ξ (A)=C for every dilation invariant generalisedlimit ω. By Theorems 1 and
ω
4, the class coincides with the class of all heat kernel functionals and so,
D
we also have τ (A) = C for every τ and the proof of the implication is
ω ω
∈ D
completed.
(i) = (iii). Suppose that τ (A) = C for every τ . Then the same
ω ω
⇒ ∈ D
argumentasaboveshowsthatξ (A)=C foreverydilationinvariantgeneralised
ω
limit ω. In particular, due to Remark 12, we have ξ (A) = C for every
γ◦M
generalised limit γ. That is,
(γ M2)(λ 1τ(e−(λA)−1))=C.
◦ → λ
It follows from Lemma 10 that
lim M2(λ 1τ(e−(λA)−1))=C.
λ→∞ → λ
Due to (3), we know that the mapping λ M(λ 1τ(e−(λA)−1)) is bounded
→ → λ
and therefore the proof of the implication is completed by invoking Lemma 11.
(i) = (iv). Suppose that τ (A) = C for every τ . It follows from
ω ω
⇒ ∈ D
Theorems 2 and 1 that the class of all ζ function residues is a subclass of .
− D
Hence, for every generalised limit γ, we have
1
γ(t τ(A1+1/t))=C.
→ t
An appeal to Lemma 11 completes the proof of the implication.
Finally, the implication (iv)= (i) is established in [3, Theorem 3.1].
⇒
Our methods have a further interesting consequence. Repeating the ar-
gument (i) = (iii) in Theorem 7 verbatim (and using Remark 5 instead of
⇒
Theorem 4), we obtain the following result.
Proposition 13. For every measurable positive operator A and every
1,∞
∈M
q >0, we have
lim M(λ 1τ(e−(λA)−q))=Γ(1+ 1)limsτ(A1+s).
λ→∞ → λ q s→0
3. Answering Question 1
The following corollary answers Question 1. Its proof immediately follows
from the implication (iii)= (ii) established in Theorem 7.
⇒
8
Corollary 14. Let A be a positive operator. If the limit
1,∞
∈L
lim 1τ(e−(λA)−1)
λ→∞λ
exists then so does the limit
1 t
lim µ(s,A)ds.
t→∞log(1+t)Z
0
The following example shows that the converse to Corollary 14 does not
hold.
Example 15. There exists a positive operator A such that
1,∞
∈L
1 t
lim µ(s,A)ds=0 (6)
t→∞log(1+t)Z
0
and
limsup 1τ(e−(λA)−1)>0. (7)
λ
t→∞
Proof. Define a positive operator A by setting
s−1, s (een,neen), n 1
∈ ≥
µ(s,A)=e−en+1, s (neen,een+1), n 1
 ∈ ≥
e−e, s (0,ee).
∈

For every n 1, we have
≥
een+1 n keek eek+1
µ(s,A)ds=1+  µ(s,A)ds+ µ(s,A)ds=
Z Z Z
0 kX=1 eek keek
 
n
=1+ log(k)+1 (k+1)e−(e−1)ek =n+log(n!)+O(1)=O(nlog(n)).
X(cid:16) − (cid:17)
k=1
Here, the last equality follows from Stirling’s formula
n n
n!=√2πn e12θn, 0<θ <1.
(cid:16)e(cid:17)
For every t>e, let ν =ν(t)=[log(log(t))]. It follows that
t eeν+1
µ(s,A)ds µ(s,A)ds=O(νlog(ν))=o(log(t)),
Z ≤Z
0 0
which yields (6).
On the other hand, we have
1τ(e−(λA)−1) 1 ∞ neen e−λ−1sds= ∞ e−λ−1een e−nλ−1een.
λ ≥ λnX=1Zeen nX=1 −
9
For a given n N, set λ=een. It follows that
∈
1τ(e−(λA)−1) e−λ−1een e−nλ−1een =e−1 e−n.
λ ≥ − −
Therefore,
limsup 1τ(e−(λA)−1) e−1,
λ ≥
λ→∞
yielding (7).
This example has a further interesting consequence.
Corollary 16. The limit
lim 1τ(e−(λA)−1) (8)
t→∞λ
does not exist and hence we cannot omit M in Theorem 7.
Proof. Suppose that the limit in (8) exists and is equal to c. Then, obviously
lim M(λ 1τ(e−(λA)−1))=c
λ→∞ → λ
and by Theorem 7, we obtain that
1 t
lim µ(s,A)ds=c.
t→∞log(1+t)Z
0
It follows from (6) that c =0. Thus, we should then have that the limit in (8)
is 0. However,the latter contradicts (7).
Finally we see thatthis example demonstratesthatwe arenotableto claim
any meromorphic continuation property for the zeta function on the basis of
our results to this point.
Lemma 17. For the operator A constructed in Example 15, the ζ-function
s τ(A1+s) does not have a pole or a removable singularity at 0.
→
Proof. Assume the contrary, that is, the ζ-function admits an analytic con-
tinuation into the punctured neighborhood of 0 and has an n-th order pole
there. By Theorem 7 and (6), we have lim sτ(A1+s) = 0. Therefore, we
s→0
have lim snτ(A1+s)=0, which contradicts the assumption. The ζ-function
s→0
s τ(A1+s) does not have a removable singularity at 0 because A / (that
1
→ ∈ L
is the limit lim τ(A1+s) does not exist).
s→0
Itis importantto observethatwearealsoina positionto answeranalogues
of Questions 1 and 2 in the case of arbitrary operators from (not neces-
1,∞
M
sarily positive). For brevity, we state and prove such analogues for self-adjoint
operators.
10