# Extension of Spectral Scales to Unbounded Operators

Authors: M. D. Wills
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We extend the notion of a spectral scale ton-tuples of unbounded
operators affiliated with a finite von Neumann Algebra. We focus primarily
on the single-variable case and show that many of the results from the bounded
theory go through in the unbounded situation. We present the currently available material on the unbounded multivariable situation. Sufficient conditions
for a set to be a spectral scale are established. The relationship between convergence of operators and the convergence of the corresponding spectral scales
is investigated. We establish a connection between the Akemann et al. spectral scale (1999) and that of Petz (1985).


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## Body

## 1. Introduction and Preliminaries

The notion of the spectrum of a self-adjoint operator has proved to be of great interest and use in various branches of mathematics. It is natural to try and extend the notion ton-tuples of operators. In 1999, Akemann et al. came up with the notion of a spectral scale [1, page 277]. The setting is as follows. Let M be a finite von Neumann algebra equipped with a normal, faithful tracial state, τ. Elements of M can be thought of as bounded operators on some Hilbert Space, H, [2, page 308]. For a given self-adjoint b∈M the corresponding spectral scale, B, which we will define below, yields information about the spectrum of b in a nice geometric way. Many of the results can be extended to n-tuples of self-adjoint operators in M. The primary aim of this paper is to explain several of the results on spectral scales, and show how they can be extended when, instead of considering b∈M, we consider g∈M*.In Section2, we consider the single-variable case which is fairly well developed. A sequence of technical lemmas culminating in Lemma 2.10 are required before we can make significant progress in the single-variable case. We illustrate with examples. Finally, we establish sufficient conditions to guarantee that a subset of ℝ2 is a spectral scale.In Section3, we consider the geometric structure of the n-dimensional spectral scale. It turns out that there is little difficulty in generalizing from the bounded situation.In Section4, we discuss certain invariance properties of the spectral scale. Significant difficulties arise in the unbounded situation although we believe that, if Conjecture 4.5 is correct, many of the difficulties would be removed.Section5 addresses some miscellaneous results. First, we address the natural question of whether the convergence of a sequence of operators implies the convergence of the corresponding spectral scales. Second, we establish a relationship between two logically distinct objects [1, 3] which were both defined by their authors as “spectral scales”.Finally, in Section6 we outline some possible future directions of research.Let us start with some preliminary definitions.Definition 1.1.
LetH be a Hilbert space. Let M be a subalgebra of B(H). If M is closed in the weak operator topology, self-adjoint, and contains 1, then M is a von Neumann algebra [2, page 308].LetM+ denote the set of positive elements of M.Definition 1.2.
Letτ:M+→[0,∞] be a function such that for a,b,aα∈M+ and λ,μ∈[0,∞] we have:
(1.1)τ(λa+μb)=λτ(a)+μτ(b),τ(a*a)=τ(aa*),aα↑a⇒τ(aα)↑τ(a)[normal],τ(a)=0⇒a=0[faithful],τ(1)<∞[finite].
(The last equation implies that ∞∉Im(τ)).
Thenτ is a faithful, finite, normal trace on M+ [4, pages 504-5].Theorem 1.3.
Letτ be a faithful, finite, normal trace on  M+. Since any element of  M can be written as a finite linear combination of positive elements of  M,  τ can be extended to a linear functional on all of  M[5, page 309].Two projectionsp,q in M are equivalent(p~q) if there exists u∈M such that uu*=p and u*u=q. A projection p is finite if p~q≤p⇒p=q. M is finite if the projection 1 is finite [5, page 296]. Throughout, we will assume that M is finite.Further, we will assume that there exists a faithful, finite, normal traceτ of M, with τ(1)=1; that is, τ is a faithful, normal, tracial state on M.A crucial property ofτ is that “things” commute in trace—that is, although, in general ab≠ba, for a,b∈M, we do have the equality τ(ab)=τ(ba) [4, page 517].LetM1+={a∈M∣0≤a≤1}. Let (b1,b2,…,bn) be an n-tuple of self-adjoint operators in M. Let (1.2)Ψ:M→ℂn+1a↦(τ(a),τ(b1a),τ(b2a),…,τ(bna)).Definition 1.4 (see [1, page 260]).
Ψ(M1+):=Bis called the spectral scale of(b1,b2,…,bn)with respect to τ.Nowτ is normal. Further Ψ is linear and continuous with respect to the weak operator topology. Moreover, M1+ is convex and compact in the weak operator topology: τ(M1+)∈ℝ and τ(bia)=τ(a1/2bia1/2)∈ℝ for i=1,…,n. Therefore B is a compact, convex subset of ℝn+1.There have been a large number of results concerning spectral scale. Some papers on the subject include those in [1, 6, 7].In 2004, Akemann and David Sherman conjectured that, if we replaceB with the set (1.3){(τ(a),g1(a),g2(a),…,gn(a))∣a∈M1+},
where each gi∈M* is self-adjoint, we will yield similar results. This paper verifies this, and generalizes much of the first paper on spectral scales [1].Some results on “noncommutative integration” will prove useful in our exposition. We will use Nelson's 1972 [8] paper on the subject with specific theorem and page references as appropriate.In his paper, Nelson definesL1(M), the predual of M [8, Section 3, pages 112 ff.]. The duality is given by the bilinear form (a,b)↦τ(ab)=τ(ba) [8, Section 3, page 112] for a∈M and b∈L1(M). Now ba∈L1(M) [8, page 112 ff.], and Nelson shows that elements of L1(M) are closed, densely defined operators affiliated with M [8, Theorem 1, page 107, and Theorem 5, page 114]. It follows that a bounded linear functional, g∈M* can be represented by a (possibly unbounded) linear operator b affiliated with M and we get the equality g(a)=τ(ba) for every a∈M.


## 2. Spectral Scale Theory for Unbounded Operators—the Single-Variable Case

We are now prepared to discuss how the spectral scale theory generalizes. We start with the single-variable situation.Definition 2.1.
Letg∈M* be a self-adjoint linear functional. Let
(2.1)B(g)={(τ(a),g(a))∣a∈M1+}.
Then B(g) is the spectral scale of g with respect to τ.From the theory of noncommutative integration, we see thatB(g)={(τ(a),τ(ba))∣a∈M1+} for some operator b affiliated with M. Since g is self-adjoint, b too will be self-adjoint, and hence, as with the original spectral scale, our generalized spectral scale is a compact, convex subset of ℝ2.Notation 1.
We will often writeB for B(g).The following definition was suggested to the author in conversation by Akemann.Definition 2.2 (Akemann).
Ifb is bounded, we will call g an operator functional.Our main goal in this section is to show thatg is an operator functional if and only if the slopes of the lower boundary function of B are all finite. We remark that Akemann et al. have already shown the “only if” part of this statement [1, Section 1, pages 261–274]. For this reason, we may assume throughout that b is unbounded, and show that the lower boundary curve of B has, as a consequence, an infinite slope. To get there, we will need a number of preliminary results.Proposition 2.3.
B is mapped onto itself by a reflection through the point  (1/2,τ(b)/2).Proof.
Leta∈M1+. Then (τ(a),τ(ba))∈B. Therefore
(2.2)(τ(1-a),τ(b(1-a)))=(1-τ(a),τ(b)-τ(ba))∈B.
Thus the map υ:(x0,x1)↦(1-x0,τ(b)-x1) takes Ψ(a) to Ψ(1-a) and hence takes B onto itself. The fixed point of υ is (1/2,τ(b)/2), and the points (x0,x1) and (1-x0,τ(b)-x1) lie on a straight line that passes through (1/2,τ(b)/2). The straight line is given by the equation
(2.3)y-x1=(τ(b)-2x11-2x0)(x-x0).
Note also that υ2 is the identity map on ℝ2. Hence, υ(B)=B and υ reflects B through the point (1/2,τ(b)/2).For the next several results, we will need the unbounded spectral theorem for self-adjoint operators. We state it here in the functional calculus form.Theorem 2.4 (see von Neumann in [9, page 562]).
Letb be a (densely defined) self-adjoint operator in H with domain D(b). Then ∃! algebraic *-homomorphism ϕ takes bounded Borel functions on ℝ into B(H) such that the following hold.(a)
ϕ is norm continuous.(b)
Let{hn(x)}n∈ℕ be a sequence of bounded Borel functions with hn(x)→x as n→∞ for each x and |hn(x)|≤|x| for every x∈ℝ and n∈ℕ. Then for ψ∈D(b), ϕ(hn)ψ→bψ as n→∞. The convergence is in norm.(c)
Ifhn(x)→h(x) pointwisely, and the sequence {||hn||∞}n∈ℕ is bounded, then ϕ(hn)→ϕ(h) strongly.(d)
Ifbψ=λψ, then ϕ(h)ψ=h(λ)ψ.(e)
Ifh≥0, then ϕ(h)≥0.For a givenh, a bounded Borel function on ℝ, it is customary to write ϕ(h) as h(b). In other words, the “ϕ” is understood. For now it is more convenient to write ϕ explicitly.Definition 2.5.
Fors∈ℝ let ϕ(χ(-∞,s))=ps- and ϕ(χ(-∞,s])=ps+. More generally, if h is a characteristic function on a Borel subset of ℝ, then ϕ(h) is a projection; such projections are referred to as spectral projections [9, pages 234, 267].For the most part, we will only need spectral projections obtained from intervals. Note that, fors∈ℝ, ps+-ps- is nonzero on the domain of b (and hence all of H) if and only if s is an eigenvalue of b. Also, since Borel functions commute with respect to multiplication and ϕ is a homomorphism, Im(ϕ) is an Abelian subalgebra of B(H). Assume now that b is affiliated with our finite von Neumann algebra, M. In this case it turns out that Im(ϕ) is an Abelian subalgebra of M. This follows from the way that ϕ is constructed.Lemma 2.6.
Letc∈[ps-,ps+]. Then
(2.4)(b-s1)(1-c)=(b-s1)(1-ps+)≥0,(s1-b)c=(s1-b)ps-≥0.Proof.
Using the decomposition1=ps-+(ps+-ps-)+(1-ps+), we can write
(2.5)b=(b1000s000b2),c=(1000c′0000).
Hence,(2.6)(b-s1)(1-c)=(b1-s10000000b2-s1)(00001-c′0001)=(b-s1)(1-ps+).
(Of course, these equalities only make sense on the domain of b.)
For everyN>s we get:(2.7)xχ(s,N]≥sχ(s,N],ϕ(xχ(s,N])≥sϕ(χ(s,N]),limN→∞ϕ(xχ(s,N])≥limN→∞sϕ(χ(s,N]),b(1-ps+)≥s(1-ps+).
Thereforeb(1-ps+)≥s(1-ps+) and hence (b-s1)(1-ps+)≥0. The other statement in this lemma follows via a similar argument.Lemma 2.7.
Leth be a characteristic function of a bounded Borel subset of   ℝ. Then  Im(ϕ(h))⊂D(b).Proof.
Lethn be a sequence of bounded Borel functions such that limn→∞hn(x)=x for all x∈ℝ and |hn(x)|≤|x|. By the Spectral theorem,
(2.8)ϕ(hn)ψ→bψ
for every ψ∈D(b). Note that hn(x)h(x) converges to xh(x) for every x. Since xh(x):=k(x) is a bounded Borel function, ϕ(k)∈B(H). For ψ∈D(b),ξ∈H, we have
(2.9)〈bψ,ϕ(h)ξ〉=limn→∞〈ϕ(hn)ψ,ϕ(h)ξ〉=limn→∞〈ϕ(hhn)ψ,ξ〉=〈ψ,(ϕ(k))*ξ〉.
Hence ϕ(h)ξ∈D(b*) and b*ϕ(h)ξ=(ϕ(k))*ξ. Since b is self-adjoint, we have the desired result.Corollary 2.8.
The following set relations hold:(2.10)⋃t>s(pt--ps+)H⊂(D(b)∩(1-ps+)H),⋃t<s(ps--pt+)H⊂(D(b)∩(ps-)H).Lemma 2.9.
The range projection of(b-s1)(1-ps+) is  1-ps+. The range projection of  (s1-b)ps- is  ps-.Proof.
Letq be the range projection of (b-s1)(1-ps+). Let {hn}n∈ℕ be a sequence of bounded Borel functions on ℝ such that limn→∞hn(x)=xforallx∈ℝ,n∈ℕ. Let h(x)=χ(s,∞)(x). Then ϕ(h)=1-ps+. From the Spectral theorem, we have
(2.11)ϕ(hn)ψ→bψ∀ψ∈D(b),ϕ(hn)ϕ(h)ψ→b(1-ps+)ψ(byLemma2.7),ϕ(hn)(ϕ(h))2ψ→b(1-ps+)ψ,ϕ(h)ϕ(hn)ϕ(h)ψ→b(1-ps+)ψ.
Taking the limit asn→∞ on the left side, we get(2.12)(1-ps+)b(1-ps+)ψ=b(1-ps+)ψ,
and therefore
(2.13)(1-ps+)(b-s1)(1-ps+)ψ=(b-s1)(1-ps+)ψ
for every ψ∈D(b).
We have shown thatq≤1-ps+. For t>s let qt be the range projection of (b-s1)(pt--ps+). By the same reasoning as above, qt≤pt--ps+. Also,(2.14)qb(pt--ps+)=qb(1-ps+)(pt--ps+)=b(1-ps+)(pt--ps+)=b(pt--ps+).
Hence, qt≤q. Similarly, for t1≥t2>s, we have qt1≥qt2.
Nowb(pt--ps+)=ϕ(hχ(s,t))∈B(H) by Lemma 2.7. Therefore, (b-s1)(pt--ps+) is a bounded operator. In fact, (b-s1)(pt--ps+)≥0 on (b-s1)(pt--ps+)H:=Ht. We show that Ht is an invariant subspace of H under (b-s1)(pt--ps+). Suppose that η∈Ht and η⊥ Im(((b-s1)(pt--ps+))Ht). Then for every ξ∈Ht,(2.15)〈(b-s1)(pt--ps+)ξ,η〉=0,〈(b-s1)ξ,η〉=0,〈ξ,(b-s1)η〉=0,(b-s1)η=0,
since b≠s1 in Ht, and η=0.
Hence,Im((b-s1)(pt--ps+))¯=Im(qt). Thus,(2.16)qt=pt--ps+→1-ps+
as t→∞. Since qt≤q for every t,
(2.17)1-ps+=limt→∞qt≤q.
But we already know that (1-ps+)≥q and so equality holds.
The second statement in the lemma follows from an analogous proof.Lemma 2.10.
Leta∈M1+. If  a1/2(b-s1)(1-ps+)a1/2=0, then  a≤ps+.
If(1-a)1/2(s1-b)ps-(1-a)1/2=0, then a≥ps-.Proof.
Writea1/2=(a11a12a13a12*a22a23a13*a23*a33), using the decomposition
(2.18)1=ps-+(ps+-ps-)+(1-ps+).
Note that(1-ps+)ai3*=ai3* for i=1,2,3. Assume that(2.19)a1/2(b-s1)(1-ps+)a1/2=0.
The diagonal entries of a1/2(b-s1)(1-ps+)a1/2 are ai3(b2-s1)ai3* for i=1,2,3. Hence, ai3(b2-s1)ai3*=0. Thus, for all ψ, such that ai3*ψ∈D(b2), ai3(b2-s1)ai3*ψ=0. Since b2-s1≥0 on (1-ps+)H∩D(b2) which contains ai3*ψ, we get
(2.20)〈ai3(b2-s1)ai3*ψ,ψ〉=0,〈(b2-s1)1/2ai3*ψ,(b2-s1)1/2ai3*ψ〉=0.
Hence (b2-s1)1/2ai3*ψ=0 and so (b2-s1)ai3*ψ=0. Since s is not an eigenvalue of b2, a13*ψ=0. Therefore
(2.21)(1-ps+)(b2-s1)ai3*ψ=0,
and so
(2.22)(pt--ps+)(b2-s1)ai3*ψ=0
for t>s. From the Spectral theorem we can then conclude that
(2.23)(b2-s1)(pt--ps+)ai3*ψ=0.
Thus,
(2.24)(pt--ps+)ai3*ψ=0
for every ai3*ψ∈D(b2). But
(2.25)limt→∞(pt--ps+)ai3*ψ=(1-ps+)ai3*ψ=ai3*ψ,
and (pt--ps+)ai3*ψ∈D(b2) for every ψ∈H and every t>s. Hence, D(b2) is dense in Im(ai3*). Since D(b2)∩Im(ai3*)={0}, we have ai3*≡0 for i=1,2,3. Thus,
(2.26)a1/2=(a11a120a12*a220000),
so
(2.27)a=(a1/2)2≤ps+.
The other statement in the lemma follows from an analogous argument.We remark that in the original paper on spectral scales [1, Lemma 1.2, pages 262, 263], the above conclusion was obtained with a little less work, since, in that situation, b was bounded and so we did not have to worry about the domain of b. The proofs of the next several results, however, are virtually identical to the original proofs. In other words, much of the hard work has now been done.Lemma 2.11.
Fixs∈ℝ,  c∈[ps-,ps+], and  a∈M1+.  Suppose that τ(a)=τ(c).  Then the following hold:(1)
τ(ba)≥τ(bc).(2)
Ifτ(ba)=τ(bc), then a∈[ps-,ps+].(3)
Ifc=ps±, then τ(ba)=τ(bc)⇒a=c.Proof.
Note thatb(1-c)≥s(1-c) and bc(1-a)≤sc(1-a) from Lemma 2.6. Hence
(2.28)τ(b(1-c)a)=τ(a1/2b(1-c)a1/2)≥τ(a1/2s(1-c)a1/2)=τ(s(1-c)a),
since τ is faithful. Similarly, τ(bc(1-a))≥τ(sc(1-a)).
(1) We compute(2.29)τ(ba)-τ(bc)=τ(ba)-τ(bca)-τ(bc)+τ(bca)=τ(b(1-c)a)-τ(bc(1-a))≥τ(s(1-c)a)-τ(sc(1-a))=τ(sa)-τ(sc)=0,
and so τ(ba)≥τ(bc).
(2) Suppose that τ(ba)=τ(bc). Then(2.30)τ((1-c)a)=τ(a-ca)=τ(c-ca)=τ(c(1-a)).
Similarly,
(2.31)τ(b(1-c)a)=τ(bc(1-a)).
Therefore
(2.32)τ(bc(1-a))=τ(b(1-c)a)≥sτ((1-c)a)=sτ(c(1-a))≥τ(bc(1-a)),
and hence equality holds throughout. Thus
(2.33)τ(b(1-c)a)=sτ((1-c)a),(2.34)τ(bc(1-a))=sτ(c(1-a)).
From (2.33),
(2.35)τ(a1/2(b-s1)(1-c)a1/2)=0,
while from (2.34),
(2.36)τ((1-a)1/2(b-s1)c(1-a)1/2)=0.
Sinceτ is faithful and the arguments are positive, the arguments are in fact equal to zero. By Lemma 2.10, ps-≤a≤ps+.
(3)
Suppose thatc=ps± and τ(ba)=τ(bc). Then τ(a)=τ(ps±). Since a is comparable to ps±, and τ is faithful, a=ps±.We next state a theorem proved by Akemann and Pedersen [10, Theorem 2.2].Theorem 2.12.
IfM and  N are von Neumann algebras,  Ψ is a normal linear map from  M to  N, and  F a face of  Ψ(M1+), then there are unique projections  p and  q in  M with  p≤q such that  Ψ-1(F)∩M1+=[p,q] and  F=Ψ([p,q]).The following results are generalizations of the main theorems for then=1 case from the first paper on spectral scales [1, Theorems 1.5–1.7, pages 266–274]. We will introduce some new notation at this time.Notation 2.
Recall that we are assuming thatM⊂B(H) is a finite von Neumann algebra equipped with a faithful, normal, tracial state τ. The operator b is unbounded and self-adjoint on H affiliated with M obtained from a linear functional g∈M* (i.e., g(a)=τ(ba) for each a∈M). B:={(τ(a),τ(ba))∣a∈M1+} is the spectral scale of b. The lower boundary of B is given by
(2.37){(x,y)∈B∣(x,y′)∈B⇒y′≥y}.
The upper boundary of B is given by
(2.38){(x,y)∈B∣(x,y')∈B⇒y'≤y}.
The endpoints of the lower boundary are f(0) and f(1).
Letσ(b) denote the spectrum of b, and let σp be the point spectrum of b. Let f be the function on [0,1] whose graph is the lower boundary. For s,α∈ℝ, let(2.39)L(s,α)={(x0,x1)∣x1=sx0+α}.
Let L↑(s,α) be the positive half-plane determined by L(s,α).Our next result describes the faces of the lower boundary ofB. We do not include the endpoints at this time.Theorem 2.13.
(1) The zero-dimensional faces in the lower boundary of B are precisely the points of the form Ψ(ps±) for s∈σ(b). Also,
(2.40)Ψ-1(Ψ(ps±))∩M1+={ps±}.
(2) The one-dimensional faces in the lower boundary of B are the sets of the form F=Ψ([ps-,ps+]) for s∈σ(b). For each face F,(2.41)Ψ-1(F)∩M1+=[ps-,ps+].
The slope of F is s.Proof.
We have the following steps.Step 1.
We show thatΨ(ps±) are zero-dimensional faces.
Fixs∈σ(b). If a∈M1+ and τ(a)=τ(ps±), then τ(ba)≥τ(bps±) by Lemma 2.11. Hence,(2.42)Ψ(ps±)=(τ(ps±),τ(bps±))=(τ(a),τ(bps±))
is on the lower boundary of B.
But(2.43)Ψ(ps±)=Ψ(a)⇔τ(ps±)=τ(a),τ(bps±)=τ(ba)⇔ps±=a,
again by Lemma 2.11. Hence,
(2.44)Ψ-1(Ψ(ps±))∩M1+={ps±}.
We now show thatΨ(ps±) is an extreme point of B. Suppose that(2.45)Ψ(ps±)=λΨ(a1)+(1-λ)Ψ(a2)=Ψ(λa1+(1-λ)a2)
for a1,a2∈M1+,0<λ<1. Since
(2.46)Ψ-1(Ψ(ps±))∩M1+={ps±},ps±=λa1+(1-λ)a2.
Since projections are extreme points in M1+, a1=a2=ps±.Step 2.
Fors∈σp(b), Ψ([ps-,ps+]) are faces of B.
Fixs∈σp(b) so that(2.47)ps-<ps+⇒Ψ(ps-)≠Ψ(ps+).
Then
(2.48)b(ps+-ps-)=s(ps+-ps-).
If c∈[ps-,ps+], then τ(bc)≤τ(ba) for each a such that τ(a)=τ(c). Hence Ψ(c) is on the graph of f which is the lower boundary curve.
Writepλ=λps-+(1-λ)ps+forallλ∈(0,1). Then pλ∈[ps-,ps+], and Ψ(pλ) is a typical point on the line segment connecting Ψ(ps-) and Ψ(ps+). Hence graph(f) contains this line segment. The slope is(2.49)τ(bps+)-τ(bps-)τ(ps+)-τ(ps-)=τ(b(ps+-bps-))τ(ps+-ps-)=τ(s(ps+-ps-))τ(ps+-ps-)=s.
LetF denote the line segment in the graph of f that contains Ψ([ps-,ps+]) and consider the endpoints of F. By Theorem 2.12, there are projections p<q such that [p,q]=Ψ-1(F)∩M1+, and hence(2.50)p≤ps-<ps+≤q.
If p<ps-, then, since ps-<ps+≤q, Ψ(ps-) is in the interior of F, contradicting Step 1. Hence p=ps- and similarly q=ps+. Thus, Ψ([ps-,ps+]) is a line segment in graph(f) with slope s and
(2.51)Ψ-1([ps-,ps+])∩M1+=[ps-,ps+].Step 3.
We show that we have accounted for all of the graph off, except possibly the endpoints.
Fix a point(2.52)(x0,x1)=Ψ(a)=(τ(a),τ(ba)),
with 0<x0<1, and assume that Ψ(a)≠Ψ(ps±) for every s∈σ(b). Write
(2.53)r1=sup{s∈σ(b)∣τ(ps-)≤x0},r2=inf{s∈σ(b)∣τ(ps+)≥x0}.
We would like to show that r1=r2. Since σ(b) is closed, ri∈σ(b) for i=1,2.
By definition,(2.54)(pr1-)≤x0≤τ(pr2+).
Since (x0,x1)≠Ψ(ps±) for s∈σ(b),
(2.55)τ(pr1-)<x0<τ(pr2+).
If s<r2, s∈σ(b), then by definition
(2.56)τ(ps+)<x0<τ(pr2+).
Similarly, r1<s and
(2.57)s∈σ(b)⇒τ(pr1-)<x0<τ(ps-).
Suppose thatr1<r2. Then τ(pr1+)≤x0≤τ(pr2-). If(2.58)r1<s<r2
for some s∈σ(b), we would have
(2.59)τ(ps+)<x0<τ(ps-),
which is clearly false. Hence,
(2.60)(r1,r2)∩σ(b)=∅.
But then pr1+=pr2- and τ(pr1+)=τ(pr2-), which again is a contradiction. Hence,
(2.61)r1=r2:=r.
Since
(2.62)τ(pr-)<x0<τ(pr+),pr-<pr+ so r is an eigenvalue of b. From Step 2, Ψ([pr-,pr+]) is a line segment in graph(f). Hence (x0,x1) is on the interior of that line segment.Corollary 2.14.
(1) The extreme points on the upper boundary (excluding the endpoints) are precisely the points of the form F=Ψ(1-ps±) for s∈σ(b) and
(2.63)Ψ-1(1-ps±)∩M1+={1-ps±}.
(2) The line segments on the upper boundary are precisely the sets of the form F=Ψ[1-ps+,1-ps-], for s∈σp(b). The slope of F is s and(2.64)Ψ-1(F)∩M1+=[1-ps+,1-ps-].Proof.
This result is a direct consequence of applying Proposition2.3 to Theorem 2.13.Letsmin be the left endpoint of σ(b). Note that smin may be -∞. Let smax be the right endpoint of σ(b). Note that smax may be ∞.Proposition 2.15.
Ifs>smin, then the left derivative of the lower boundary function  f at  τ(ps-) exists and is given by the formula
(2.65)f-′(τ(ps-))=sup{s′∈σ(b)∣s′<s}.
If s<smax, then the right derivative of f at τ(ps+) exists and is given by the formula
(2.66)f+′(τ(ps+))=inf{s′∈σ(b)∣s′>s}.Proof.
SinceB is convex, f is a convex function, and so the left and right derivatives exist. Fix s∈ℝ with s>smin. Then σ(b)∩(-∞,s)≠∅. Define r=sup{s'∈σ(b)∣s'<s}. Since σ(b) is closed, r∈σ(b).Case 1 ((r-ϵ,r)∩σ(b)=∅ for some ϵ>0).
Ifr=s, (r-ϵ,r)∩σ(b)=(s-ϵ,s)=∅ which contradicts the choice of r. Thus r<s, and hence r is an isolated point in the spectrum, that is, r is an eigenvalue of b. Moreover, pr+=ps- and so Ψ(ps-)=Ψ(pr+) is the right-hand endpoint of a line segment in graph(f) with slope r by Theorem 2.13. Hence f-′(τ(ps-))=r.Case 2 ((r-ϵ,r)∩σ(b)≠∅ for every ϵ>0).
Choosern∈σ(b) such that rn↑r and rn≠r for n∈ℕ. Then Ψ(prn-) is on the graph of f. Furthermore, prn-→pr- in the weak-* topology. Since τ is normal, τ(prn-)→τ(pr-). Since τ is faithful, τ(prn-)≠τ(pr-). We have
(2.67)rn(pr--prn-)≤b(pr--prn-)≤r(pr--prn-)
for every n. Hence,
(2.68)τ(rn(pr--prn-))≤τ(b(pr--prn-))≤τ(r(pr--prn-))
for every n. Thus,
(2.69)rn≤f(τ(pr-))-f(τ(prn-))f(pr-)-f(prn-)≤r.
Letting n→∞ gives the desired result.The statement regarding right derivatives is proved in a similar way.Proposition 2.16.
The corners off are in one-to-one correspondence with the gaps of  σ(b), that is, the maximal bounded intervals in the complement of the spectrum.  (One is not currently concerned with unbounded maximal intervals in the complement of the spectrum, that is, those which take the form (-∞,s) or  (s,∞).)Proof.
Let(r,t) be an interior gap of the spectrum. Then for every s1,s2∈(r,t) we have ps1+=ps2-. Fix s∈(r,t). Then
(2.70)f-′(τ(ps-))=sup{s'∈σ(b)∣s'<s}<inf{r'∈σ(b)∣r'>s}=f+′(τ(ps+))=f+′(τ(ps-)).
Hence, f is not differentiable at τ(ps-), and so a gap in the spectrum corresponds to a corner. Conversely, we have already seen that f is differentiable at ps± for s∈σ(b).Proposition 2.17.
For eachs∈ℝ,
(2.71)τ((b-s1)ps-)=τ((b-s1)ps+).
The line L(s,α) is a line of support for B such that
(2.72)B⊂L↑(s,α)⇔α=τ((b-s1)ps±).
In this case, L(s,α) passes through Ψ(ps±). Moreover, one has
(2.73)Ψ-1(L(s,α))∩M1+=[ps-,ps+],Ψ([ps-,ps+])=L(s,α)∩B.Proof.
Fixs∈ℝ. If s is an eigenvalue, then
(2.74)b(ps+-ps-)=s(ps+-ps-).
Otherwise, ps-=ps+. Either way,
(2.75)(b-s1)(ps+-ps-)=0,
and so
(2.76)τ(b-s1)ps+=τ(b-s1)ps-.
Let α=τ(b-s1)ps±. Then
(2.77)-sτ(ps±)+τ(bps±)=α,
so Ψ(ps±) lies in L(s,α). We now wish to show that L(s,α) is a line of support for B. There are several cases to consider.Case 1 (s∈σp(b)).
In this situation,Ψ(ps-) and Ψ(ps+) are endpoints of a line segment in graph(f) whose slope is s. L(s,α) passes through both points and has slope s. Thus, L(s,α) contains this line segment and is tangent to f. Hence B⊂L↑(s,α).Case 2 (s∈σ(b)∖σp(b)).
Note thats is not an isolated point in σ(b). Moreover, ps-=ps+=ps. At least one of the one-sided derivatives of f takes the value s at τ(ps). Hence, B admits a line of support at Ψ(ps) with slope s. As with Case 1, the line is L(s,α) and B⊂L↑(s,α).Case 3 ((s1,s2) is an interior gap in the spectrum).
In this caseps1+=ps2-. Let α1=τ((b-s1)ps1+) and α2=τ((b-s2)ps2-). Then L(s1,α1) and L(s2,α2) are lines of support passing through Ψ(ps1+) whose slope lies between s1 and s2. If L is a line of support for B whose slope lies between s1 and s2, then L↑⊃B. But L(s,α) is such a line for s∈(s1,s2). Hence, the statement is true for any s∈(s1,s2).Case 4 (s<smin or s>smax).
Sinceb is unbounded, at least one of smin and smax has infinite magnitude. Suppose that smin is finite (and so smax must be infinite). Then smin∈σ(b). Moreover, psmin-=0. L(smin,0) is a line of support for B at Ψ(0) and b⊂L↑(smin,0) by Case 1. Suppose s<smin. Then L(s,0) is also a line of support for B at Ψ(0) and B⊂L↑(s,0).
The case fors>smax is dealt with similarly. Hence, for every s∈ℝ, L(s,α) is a line of support for B and B⊂L↑(s,α).
Conversely, for fixeds, the lines L(s,β) are all parallel as β varies over ℝ. Hence, there exists a unique β0 for which L(s,β0) is a line of support and B⊂L↑(s,β0). But L(s,α) has these properties and hence α=β0.
For the last statement, considerF=L(s,α)∩B. Then F is a face of B. Hence, F is an extreme point or a line segment on graph(f).
IfF is an extreme point, then F={Ψ(ps±)}, then s∈σ(b). Since F is an extreme point, then F={ψ(ps-)}={Ψ(ps+)}, and so(2.78)Ψ-1(F)∩M1+=Ψ-1(L(s,α))∩M1+={ps±}=[ps-,ps+].
Similarly, ifF is a line segment, then F=Ψ([ps-,ps+]) for some s∈σp(b), and so(2.79)Ψ-1(F)∩M1+=Ψ-1(L(s,α))∩M1+=[ps-,ps+].From the above results, ifsmin=-∞, then the right derivative of f(x) approaches -∞ as x↓0. If in addition smax=∞, then the left derivative of f(x) approaches ∞ as x↑1. By Proposition 2.3, the graph of the upper boundary curve of B is vertical at x=0. Hence, the only line of support at Ψ(0) is vertical. Similarly, the only line of support at Ψ(1) is vertical. Therefore, if both smin and smax are nonreal, then Ψ(0) and Ψ(1) are not corners of B.Conversely, if one ofsmin and smax is finite, then Ψ(0) and Ψ(1) are corners of B.Here the bounded and unbounded spectral scale theories do not coincide, since, in the bounded situation,Ψ(0) and Ψ(1) are always corners.In both situations, we can read spectral data of the lower boundary curve as follows(1)
1-dimensional faces correspond to eigenvalues ofb. The slope of each face is the corresponding eigenvalue.(2)
Other places where the lower boundary curve is differentiable correspond to elements of the continuous spectrum. The slope at such a given point is the corresponding element of the spectrum.(3)
Corners on the lower boundary curve correspond to gaps in the spectrum.We now exhibit two examples. In both examples, we will takeH=L2[0,1] and M=L∞[0,1]. The trace τ is integration with respect to Lebesgue measure and a(ψ)(x):=a(x)ψ(x) for a∈M, ψ∈H, and x∈[0,1]. Then M*=L1[0,1] and τ makes sense on M*.Example 2.18.
Defineb(x)=1/x almost everywhere. Then b∈M* is densely defined and self-adjoint on H. It turns out that the equation of the lower boundary function is w=f(y)=2-2(1-y). This was obtained by integrating b multiplied by appropriate characteristic functions. Observe that 1≤f'(y)≤∞ and τ(b)=2. Hence the center of B is P=(0.5,1) and we get Figures 1 and 2.Figure 1
Graph ofb for Example 2.18.Figure 2
Spectral scale forb in Example 2.18.Example 2.19.
Defineb(x)=1/x-2 for 0<x≤0.5 and b(x)=2-1/(1-x) for 0.5<x<1. Then the lower boundary curve for 0≤y≤0.5 is given by w=f(y)=-2y+2y. And b was chosen so that we would get a spectral scale that is invariant under the reflection y=0.5. Note that f'(0)=-∞, f'(0.5)=0, f(0.5)=-2+1/2, τ(b)=0, and f'(1)=∞. The resulting pictures are shown in Figures 3 and 4.Figure 3
Graph ofb for Example 2.19.Figure 4
Spectral scale forb in Example 2.19.We now examine a question posed to the author by Crandall. We start by stating the necessary properties thatU⊂ℝ2 must have in order for it to be a spectral scale for an operator functional.Definition 2.20.
Aprespectral scale is a set U contained in ℝ2 which satisfies the following properties.(i)
U is compact and convex.(ii)
(0,0)∈U and there are no other points of the form (0,y) in U.(iii)
U⊂{(x,y)∣0≤x≤1}.
Further, ifu=inf{y∣(1/2,y)∈U} and v=sup{y∣(1/2,y)∈U}, then the following are given.
(iv)
U is invariant under the reflection (a,b)↦(1-a,u+v-b).(v)
The setG:={(x,z)∈U∣(x,y)∈U⇒y≥z} is the graph of a function f on [0,1], which we will call the lower boundary curve of U.Lemma 2.21.
LetU be a prespectral scale with lower boundary curve  f. Then  f is a continuous convex function.Proof.
SinceU is closed, G⊂U. Let 0<a<b<1. Since U is convex, the line segment
(2.80){w∈ℝ2∣w=t(a,f(a))+(1-t)(b,f(b))}
is a subset of U. Now
(2.81)(ta+(1-t)b,f(ta+(1-t)b))∈G⊂U.
From the definition of G,
(2.82)tf(a)+(1-t)f(b)≥f(ta+(1-t)b).
Hence, f is convex on (0,1), and therefore continuous on (0,1) [11, pages 61, 62]. Then limx↓0f(x) and limx↑1f(x) exist as extended real numbers [12, page 116]. Since U is compact, then limx↓0f(x) is finite and limx↓0(x,f(x))∈U. Therefore, (0,limx↓0f(x))∈U. By (ii) in Definition 2.20, limx↓0f(x)=0=f(0). Applying (iv) from Definition 2.20, limx↑1f(x)=f(1). Thus, f is continuous and convex on [0,1].Sincef is convex on [0,1], the left and right derivatives of f exist for all x∈[0,1] as extended real numbers, and f is differentiable almost everywhere [12, pages 113, 114].It is easy to see that a spectral scale must be a prespectral scale: condition (i) is noted on page 3 of this paper, condition (ii) follows from Definition2.1, condition (iii) follows from the fact that τ is a state, condition (iv) follows from Proposition 2.3, and condition (v) follows from the definition of the lower boundary (Notation 2.14).Crandall asked whether a prespectral scale is automatically a spectral scale. In the next theorem, we show that the answer is yes.Theorem 2.22.
LetM=L∞[0,1] and let  τ be the Lebesgue integral on  [0,1]. Given a prespectral scale  U, there exists  g∈L1[0,1]=M* self-adjoint such that  U=B(g).Proof.
From the symmetry required forU (condition (v)) it is sufficient to examine the lower boundary curve, f, of U. The function f has the following properties (as noted in Lemma 2.21):(i)
f:[0,1]→ℝ is continuous and convex,(ii)
f(0)=0.
Let us denotefR′(t) as the right derivative of f at t∈[0,1). Similarly, denote fL′(t) as the left derivative of f at t∈(0,1]. Let g(t)=fR′(t) on [0,1) and g(1)=fL′(1). Since f is convex, then g is nondecreasing. Hence, g has at most a countable number of discontinuities. Since f is convex, then f is of bounded variation. By Exercise 14.H in [13, page 244], f is absolutely continuous. By Theorem 7.20 in [11, page 148], g∈L1[0,1]=M*, and the fundamental theorem of calculus holds. Let a∈M1+, with ∫01a(s)ds=t∈[0,1]. Since g is increasing,(2.83)∫01g(s)a(s)ds≥∫0tg(s)ds=f(t)-f(0)=f(t).
Hence, f is the lower boundary curve of B(g).


## 3. The Geometry of Spectral Scales in Higher Dimensions

This section is devoted to further generalizations of results from the original paper on spectral scales by Akemann et al. [1, Section 2, pages 276–280]. Often, with some modifications, the proofs are the same as in the original paper. Recall that H is a Hilbert space, M⊂B(H) is a finite von Neumann algebra equipped with faithful, normal, tracial state, τ.Notation 3.
In Section2 of this paper, we considered g∈M*. We now consider an n-tuple of self-adjoint linear functionals, (g1,…,gn)∈M*n. Let t=(t1,…,tn)∈ℝn∖{0}. Let gt=∑i=1ntigi. Then gt is also self-adjoint since each gi is self-adjoint and each ti is real. For each gk (k=i,t) there is an associated self-adjoint, densely defined operator in H, bk, and for every a∈M we have gk(a)=τ(bka).Definition 3.1.
LetΨ(a)=(τ(a),g1(a),…,gn(a)) for every a∈M. Let Ψt(a)=(τ(a),gt(a)) for each a∈M. Then B:=Ψ(M1+) is the spectral scale of (b1,…,bn) with respect to τ and Bt:=Ψt(M1+) is the spectral scale of gt with respect to τ.Essentially the motivation for the introduction ofgt is it allows us to reduce the n-dimensional case to the 1-dimensional case by studying “2-dimensional cross-sections” of the spectral scale. We note that bt≠∑i=1ntibi. Indeed, the right hand side may have trivial domain. However, as we will see, equality “almost” holds; that is, equality holds in trace.Defineπt(x0,x1,…,xn)=(x0,∑i=1ntixi), where x0,xi∈ℝ.Proposition 3.2.
The equalityΨt=πt∘Ψ holds.Proof.
Fora∈M we have
(3.1)πt(Ψ(a))=πt(τ(a),τ(b1a)⋯τ(bna))=πt(τ(a),g1(a),…,gn(a))=(τ(a),∑i=1ntigi(a))=(τ(a),gt(a))=(τ(a),τ(bta))=Ψt(a).Corollary 3.3.
As a consequence of this calculation,Bt=πt(B).We next introduce some additional notation.Notation 4.
Letpt,s+ be the spectral projection of bt determined by (-∞,s].
Letpt,s- be the spectral projection of bt determined by (-∞,s).
LetP(t,s,α)={(x0,x1,…,xn)∈ℝn+1∣-sx0+∑i=1ntixi=α}.
LetP↓(t,s,α)={(x0,x1,…,xn)∈ℝn+1∣-sx0+∑i=1ntixi≤α}.
LetP↑(t,s,α)={(x0,x1,…,xn)∈ℝn+1∣-sx0+∑i=1ntixi≥α}.The following results discuss the geometrical properties ofB.Proposition 3.4.
Ifx is an extreme point of  B, the n there exists a projection  p∈M, such that  Ψ(p)=x and  Ψ-1(x)∩M1+={p}. Further,  Ψ(pt,s±) is an extreme point of  B.Proof.
Fix an extreme pointx∈B. Since {x} is a face of B, by Theorem 2.12 there are unique projections, p≤q in M, such that Ψ-1(x)∩M1+=[p,q]. Thus, Ψ(p)=Ψ(q)=x and so τ(p)=τ(q). Since τ is faithful, we have that p=q, and so Ψ-1(x)∩M1+={p}.
Next, suppose thatΨ(pt,s±)=Ψ(λa1+(1-λ)a2) for some λ∈(0,1), a1,a2∈M1+. Then(3.2)Ψt(pt,s±)=πt(Ψ(pt,s±))=πt(Ψ(λa1+(1-λ)a2))=Ψt(λa1+(1-λ)a2).
Thus,
(3.3)τ(bt(pt,s±))=τ(bt(λa1+(1-λ)a2)),τ(pt,s±)=τ(λa1+(1-λ)a2).
By Lemma 2.11, pt,s±=λa1+(1-λ)a2. Since projections are extreme points in M1+, then a1=a2=pt,s±, and hence Ψ(pt,s±) is an extreme point of B.Proposition 3.5.
Suppose thatτ((bt-s1)pt,s+)=τ((bt-s1)pt,s-). Then  P(t,s,α) is a hyperplane of support for  B with
(3.4)B⊂P↑(t,s,α)⇔α=τ((bt-s1)pt,s±).
In this case, Ψ(pt,s±)∈P(t,s,α).Proof.
⇐
Letα=τ((bt-s1)pt,s±). Then(3.5)α=τ(btpt,s±)-sτ(pt,s±)=(∑i=1ntiτ(bi(pt,s±)))-sτ(pt,s)±.
ThereforeΨ(pt,s±)∈P(t,s,α). Let Lt(s,α)={(x0,x1)∣-sx0+x1=α}. By Proposition 2.17, Lt(s,α) is a line of support for Bt and Bt⊂Lt↑(s,α). Fix a∈M1+. Then Ψt(a)∈Lt↑(s,α). Hence,(3.6)α≤-sτ(a)+τ(bta)=-sτ(a)+∑i=1ntiτ(bia)
and Ψ(a)∈P↑(t,s,α).
⇒
Fixt and s, and let β vary over ℝ. The hyperplanes P(t,s,β) are all parallel and hence there exists a unique β0 such that P(t,s,β0) supports B and B⊂P↑(t,s,β0). But we have seen that α satisfies these conditions and so α=β0.Proposition 3.6.
Ifα=τ((bt-s1)pt,s±), then  F=P(t,s,α)∩B is a face of  B. Further,
(3.7)Ψ-1(P(t,s,α))∩M1+=[pt,s-,pt,s+],
and F=Ψ([pt,s-,pt,s+]).Proof.
Letα=τ((bt-s1)pt,s±). By our previous result, P(t,s,α) is a supporting hyperplane for B. Hence, F:=P(t,s,α)∩B is a face of B. By Theorem 2.12, there are unique projections p≤q in M such that Ψ-1(F)∩M1+=[p,q] and Ψ([p,q])=F. If a∈M1+ and Ψ(a)∈P(t,s,α), then Ψ(a)∈P(t,s,α)∩B, and therefore
(3.8)[p,q]=Ψ-1(F)=M1+=Ψ-1(P(t,s,α)∩B)∩M1+=Ψ-1(P(t,s,α))∩M1+.
We would like to show thatp=pt,s- and q=pt,s+.
SinceΨ(pt,s±)∈F, we have(3.9)p≤pt,s-≤pt,s+≤q.
But πt(P(t,s,α))=Lt(s,α) and πt(B)=Bt. Hence,
(3.10)Ψ([p,q])=πt(Ψ([p,q]))=πt(F)=πt(Ψ(P(t,s,α))∩B)⊂Lt(s,α)∩Bt=Ψt([pt,s-,pt,s+]).
Therefore,pt,s-≤p≤q≤pt,s+, and so pt,s-=p and pt,s+=q.Proposition 3.7.
Letβ=τ((bt-s1)(1-pt,s±)). Then  P(t,s,β) supports  B, and passes through  Ψ(1-pt,s±) and  B⊂P↓(t,s,β).Proof.
Ifγ=τ((b-t-(-s)1)p-t,-s±), then P(-t,-s,γ) is a hyperplane of support for B that contains Ψ(p-t,-s±) and B⊂P↑(-t,-s,γ). But g-t=-gt, so τ(b-t·)=-τ(bt·) and p-t,-s±=1-p-t,-s∓. Therefore,
(3.11)γ=τ((b-t-(-s)1)p-t,s±)=τ((-bt+s1)(1-pt,s∓))=-τ((bt-s1)(1-pt,s±))=-β.
Therefore,
(3.12)P(-t,-s,γ)=P(-t,-s,-β)=P(t,s,β),P↑(-t,-s,-β)=P↓(t,s,β).


## 4. Invariance Properties of the Spectral Scale

The main goal in this section is to establish the circumstances required for the spectral scale to determine (up to equivalence of tracial representations) the algebra and then-tuple, (g1,…,gn). Let N be the algebra generated by 1 and biχw(bi) where w ranges over the bounded Borel subsets of ℝ and i ranges from 1 to n.Observe thatlimk→∞(τ(biχ[-k,k](bi)a)=τ(bia)). To see this, note that χ[-k,k](bi)→1 strongly. Hence, for a fixed a∈M, χ[-k,k](bi)a→a strongly and so χ[-k,k](bi)a→a weakly. Since τ(bi·)=gi(·) is a bounded linear functional on M, we have the desired convergence.We now show that we only needN to generate the spectral scale for the n-tuple (g1,…,gn) with respect to τ. By Proposition 2.35 from [5, page 232], there exists a faithful normal projection E:M→N, with ∥E∥=1 such that τ=τ∘E. Hence for a∈M, (4.1)Ψ(a)=(τ(a),τ(b1a),…,τ(bna))=limk→∞(τ(a),τ(b1χ[-k,k](b1)a),…,τ(bnχ[-k,k](bn)a))=limk→∞(τ(E(a)),τ(b1χ[-k,k](b1)E(a)),…,τ(bnχ[-k,k](bn)E(a)))=(τ(E(a)),τ(b1E(a)),…,τ(bnE(a)))=Ψ(E(a)).
Since E is faithful and normal, Ψ(M1+)=Ψ(N1+) as desired.We now introduce additional notation and change some of the old notation.Notation 5.
LetU and V be finite von Neumann algebras equipped with faithful, normal, tracial states τU and τV, respectively. Let HU and HV be the associated Hilbert spaces obtained by the tracial Gelfand-Naimark-Segal (GNS) construction [2, pages 278, 279]. Let g1,…,gn∈U* and h1,…,hn∈V* be self-adjoint. Then there exist b1,…,bn closed, densely defined, self-adjoint operators affiliated with U such that τU(biu)=gi(u) for all u∈U. Similarly, there exist c1,…,cn closed, densely-defined, self-adjoint operators affiliated with V such that τV(civ)=hi(v) for all v∈V. Let M be the von Neumann algebra generated by 1 and biχω(bi), where i=1,…,n and ω ranges over the bounded Borel subsets of ℝ. Similarly, let N be the von Neumann algebra generated by 1 and ciχω(ci). Note that g1,…,gn∈M* and h1,…,hn∈N*. When we are concerned only with objects restricted to M and N, we will write ·M and ·N, respectively.LetB be the spectral scale for b1,…,bn relative to τM determined by ΨM and C the spectral scale for c1,…,cn relative to τN determined by ΨN. Let πM and πN be the GNS representations of M and N. Let ξM and ξN be the canonical cyclic vectors that arise from this tracial GNS construction.Definition 4.1.
Suppose that there exists a surjective unitary transformationu:HM→HN such that uξM=ξN and uπM(biχω(bi))=πN(ciχω(ci))u for i=1,…,n and all bounded Borel subsets ω of ℝ. Then the tracial representations of M and N are said to be equivalent.This definition is unsatisfying since it requires uncountably many conditions. We believe that there exists a more satisfactory definition of equivalence using thegi's and the hi's. We have not to date been able to formulate such a definition.Proposition 4.2.
Suppose thatB=C. Then there exists an isometry,  Φ, from  M* to  N* such that  Φ(gi)=hi for  i=1,…,n and  Φ(τM)=τN.Proof.
Let us temporarily denoteτM=gn+1 and τN=hn+1. For i=1,…,n+1, define Φ(gi)=hi.
We would first like to show thatΦ is well defined and can be extended linearly to the span of the gi's. Suppose one of the gi's is a linear combination of the others. Without loss of generality, g1=∑n=2∞αigi. Let v∈N1+. Since B=C, there exists u∈M1+ such that gi(u)=hi(v) for every i. Thus,(4.2)h1(v)=g1(u)=∑n=2∞αigi(u)=∑n=2∞αihi(v).
If v∈N+ is not zero, then w=(1/∥v∥)v∈N1+. There exists u∈M1+ such that gi(u)=hi(w) for every i. Therefore,
(4.3)h1(v)=∥v∥h1(w)=∥v∥g1(u)=∥v∥∑n=2∞αigi(u)=∥v∥∑n=2∞αihi(w)=∑n=2∞αihi(v).
Since any element in N is a finite linear combination of elements in N+, it follows that if g1=∑n=2∞αigi, then h1=∑n=2∞αihi. Hence, Φ is well-defined and we can therefore extend it linearly to linear combinations of the gi's and hence to all of M*.
We now show thatΦ is an isometry. For v∈M* let us denote Φ(v)=Φv. Let M1 be the set of points in M with norm 1, and let N1 be the set of points in N with norm 1. We need to show that(4.4)∥v∥=supx∈M1|v(x)|=supy∈N1|Φv(y)|=∥Φv∥.
Considerx≥0. Then x∈M1+. There exists y∈N1+ such that gi(x)=hi(y) for i=1,…,n+1, and y≠0. Let w=y/||y||, and let v=∑i=1n+1αigi. Hence,(4.5)∑i=1n+1αigi(x)=∑i=1n+1αihi(w)∥w∥,|∑i=1n+1αigi(x)|=|∑i=1n+1αihi(w)|∥w∥≤|∑i=1n+1αihi(w)|,supx∈M1∩M1+|v(x)|≤supw∈N1∩N1+|Φv(w)|.
A similar calculation shows the reverse inequality and therefore
(4.6)supx∈M1∩M1+|v(x)|=supw∈N1∩N1+|Φv(w)|.Notation 6.
Recall thatpt,s+ is the spectral projection of bt corresponding to (-∞,s] and pt,s- is the spectral projection of bt corresponding to (-∞,s). Let qt,s± denote the spectral projections of ct on the same intervals.Proposition 4.3.
The following are equivalent:(1)
B=C,(2)
Bt=Ct for t∈ℝn∖{0},(3)
τM(pt,s±)=τN(qt,s±) for s∈ℝ and t∈ℝn∖{0},(4)
τM(f(bt))=τN(f(ct)) for t∈ℝn∖{0} and f a bounded Borel function on ℝ,(5)
τM((btχω(bt))k)=τN((ctχω(ct))k) for every k∈ℕ,t∈ℝn∖{0},andω is a bounded Borel subset of ℝ.Proof.
(1)⇒(2). Consider
(4.7)Bt=πt(B)=πt(C)=Ct.
(2)⇒(1).
Suppose thatB≠C. Then (relabeling if necessary) there exists a vector x=(x0,x1,…,xn)∈B∖C. Since C is compact and convex and x∉C, there exists a hyperplane that strictly separates C from x. Thus, there exists t=(t0,t1,…,tn)∈ℝn+1, and β∈ℝ such that for y=(y0,y1,…,yn)∈C, we have(4.8)∑k=0ntkxk<β<∑k=0ntkyk.
Hence, (x0,∑k=1ntkxk)≠(y0,∑k=1ntkyk). Therefore, πt(x)≠πt(y) for every y∈C, and so πt(x)∈Bt∖Ct.
(1)⇒(3).
Fixs,t. There exists unique α such that P(t,s,α) is a hyperplane of support for B and B⊂P↑(t,s,α). Since B=C, then α has the same properties with respect to C. Hence, P(t,s,α)∩B=P(t,s,α)∩C. Therefore,(4.9)ΨM([pt,s-,pt,s+])=P(t,s,α)∩B=P(t,s,α)∩C=ΨN([qt,s-,qt,s+]),
and so τM(pt,s±)=τN(qt,s±).
(3)⇒(4).
Given (3), (4) holds when f is a characteristic function of an interval (-∞,s) or (-∞,s]. These intervals generate the Borel structure of ℝ. Now τM and τN are normal and linear. Since any bounded Borel function, f, is uniformly approximated by linear combinations of characteristic functions, (4) holds for all such f.
(4)⇒(3).
Since characteristic functions on intervals are bounded and Borel, this is immediate.
(4)⇒(2).
Takef-=χ(-∞,s) and f+=χ(-∞,s]. Then f±(bt)=pt,s±. Therefore, τU(pt,s±)=τV(qt,s±). Define g± on ℝ by g±(a)=af±(a) for a∈ℝ. g± is a bounded Borel function, and g±(bt)=btpt,s±. Hence, τM(btpt,s±)=τN(ctqt,s±) for every nonzero t∈ℝn,s∈ℝ. We have(4.10)(τM(pt,s±),τM(btpt,s±))=(τN(qt,s±),τN(ctqt,s±)).
These are the extreme points of the lower boundaries of Bt and Ct, and so the lower boundaries coincide; hence the upper boundaries coincide by Proposition 2.3, and so Bt=Ct.
(4)⇒(5).
Definef(s)=(sχω(s))k for s∈ℝ. Then f is a bounded Borel function, and hence, by assumption, τM(f(bt))=τN(f(ct)).
(5)⇒(4).
Defineh(s)=(sχω(s))k for s∈ℝ. By assumption, τM(h(bt))=τN(h(ct)), for every ω, a bounded Borel subset of ℝ, and every k∈ℕ. But the h(bt)'s are weakly dense in M, and the h(ct)'s are weakly dense in N. Also, τM and τN are normal, and so τM(f(bt))=τN(f(ct)), for f being a bounded Borel function on ℝ.Lemma 4.4.
The tracial representations ofM and  N are equivalent if and only if
(4.11)τM(ϕ(b1χω(b1),…,bnχω(bn)))=τN(ϕ(c1χω(c1),…,cnχω(cn)))
for every ϕ, a monomial in n variables, and ω, a bounded Borel subset of ℝ.Proof.
We begin by making the notation a little less cumbersome. Letbω = (b1χω(b1),…,bnχω(bn)), and let cω=(c1χω(c1),…,cnχω(cn)).
⇐
Suppose thatτM(ϕ(bω))=τN(ϕ(cω)) for every ϕ and ω.
DefineuπM(ϕ(bω))ξM=πN(ϕ(cω))ξN. Extend this definition by linearity to polynomials. Let ϕ1, ϕ2 be two such polynomials. Then for i=1,…,n we have(4.12)〈πM(biχωi(bi))πM(ϕ1(bω))ξM,πM(ϕ(bω))ξM〉=〈(πM(ϕ2(bω))*πM(biχωi(bi))πM(ϕ1(bω)))ξM,ξM〉=〈(πM((ϕ2(bω))*biχωi(bi)ϕ1(bω)))ξM,ξM〉=τM((ϕ2(bω))*biχωi(bi)ϕ1(bω))=τN((ϕ2(cω))*ciχωi(ci)ϕ1(cω))=〈πN(ciχωi(ci))πN(ϕ1(cω))ξN,πN(ϕ(cω))ξN〉=〈πN(ciχωi(ci))uπM(ϕ1(bω))ξM,uπM(ϕ(bω))ξM〉.
Such polynomials are dense in HM and HN, so u extends to a unitary transformation from HM to HN with the desired properties.
⇒
Suppose that the tracial representations ofM and N are equivalent. Then(4.13)τN(ϕ(cω))=〈πN(ϕ(cω))ξN,ξN〉=〈uπM(ϕ(bω))u*ξN,ξN〉=〈πM(ϕ(bω))ξM,ξM〉=τM(ϕ(bω)).Suppose thatM is Abelian and d, e are closed, densely defined operators affiliated with M. Then d+e and de are closable, densely defined operators whose closures are affiliated with M and M′, the set of operators affiliated with M is an Abelian *-algebra [2, pages 351, 352]. If in addition d and e are self-adjoint, then d+e and de are also self-adjoint and hence closed [14, page 536]. Further, we have(4.14)τ(bta)=gt(a)=∑i=1ntigi(a)=∑i=1ntiτ(bia)=τ(∑i=1ntibia)
for every a∈M. Thus, (4.15)τ((bt-∑i=1ntibi)a)=0,
for every a∈M. Let b+ be the positive part of bt-∑i=1ntibi and b- the negative part. Choose a=b+χω(b+), where ω is any bounded Borel set. Then a is positive since M' is commutative and (4.16)(bt-∑i=1ntibi)a=(b+)2χω(b+)≥0.
But τ((b+)2χω(b+))=0, since τ is faithful (b+)2χω(b+)=0. Thus b+=0 on its domain. Similarly b-=0, and so bt-∑i=1ntibi=0. Thus ∑i=1ntibi=bt.To proceed with the theory as given in [1, Section 3, page 281 ff.], it would be convenient if the following conjecture were true.Conjecture 4.5.
IfM and  N are Abelian and  B=C, then the tracial representations of  M and  N are equivalent.By Lemma4.4, it is enough to show that, if ϕ(x1,…,xn)=x1k1⋯xnkn denotes a monomial in the commuting variables x1,…,xn, then (4.17)τM(ϕ(b1χω(b1),…,bnχω(bn)))=τN(ϕ(c1χω(c1),…,cnχω(cn)))
for every ω, a bounded Borel subset of ℝ. By part (5) of Proposition 4.3, we know that τM((btχω(bt))k)=τN((ctχω(ct))k) for every k∈ℕ,t∈ℝn∖{0},withω being a bounded Borel subset of ℝ. Let us fix k and ω, and let P be the set of all monomials ϕ in n commuting variables such that ∑i=1nki=k. Routine computations show that(4.18)τM((btχω(bt))k)=∑ϕ∈Pt1k1⋯tnknτM(ϕ(b1,…,bn)χω(bt)).
Similarly, (4.19)τN((ctχω(ct))k)=∑ϕ∈Pt1k1⋯tnknτN(ϕ(c1,…,cn)χω(ct)).
Since (4.20)τM((btχω(bt))k)=τN((ctχω(ct))k),
then we have (4.21)∑ϕ∈Pt1k1⋯tnknτM(ϕ(b1,…,bn)χω(bt))=∑ϕ∈Pt1k1⋯tnknτN(ϕ(c1,…,cn)χω(ct)).
In the bounded case, no characteristic functions are present and so we can equate coefficients of the polynomials. Even if we could do that here, we still do not have the desired result since we want something independent of t.


## 5. Miscellaneous Results

A natural question is to ask whether convergence ofn-tuples of self-adjoint operators implies convergence of the corresponding spectral scales. Since spectral scales are compact and convex, the Hausdorff Metric is a natural metric to work with. The following definition is taken from [15, page 274].Definition 5.1.
Let(X,d) be a metric space, with A and B being nonempty subsets of X. Define d(A,B)=inf{d(a,b)∣a∈A,b∈B}. For γ>0, let us define
(5.1)Aγ={x∈X∣d({x},A)<γ},Bγ={x∈X∣d({x},B)<γ}.
Define
(5.2)dH(A,B)=inf{γ>0∣A⊂Bγ,B⊂Aγ}.
Then dH(A,B) is the Hausdorff distance between A and B.We first establish a result for the original definition of a spectral scale, that is the spectral scale from Definition1.4.Theorem 5.2.
Letm∈ℕ. Suppose that  bn→b strongly in each coordinate for  bn,b∈Mm where  bn and  b are self-adjoint (in each coordinate). If  B(bn) and  B(b) are the corresponding spectral scales, then  B(bn)→B(b) in the Hausdorff metric induced by the usual topology on  ℝm+1.Proof.
Writebjn for the jth coordinate of bn and bj0 for the jth coordinate of b. Let ϵ>0. Now B(bn)={(τ(a),τ(b1na),…,τ(bmna))∣a∈M1+} and B(b)={(τ(a),τ(b10a),…,τ(bm0a))∣a∈M1+}. Fix a j. Define cjn=bjn-bj0 and note that cjn is self-adjoint. Further, cjn→0 strongly, and so cjn2→0 strongly as well. Since τ is normal, τ(cjn2)→0. Therefore, there exists Nj∈ℕ such that n≥Nj⇒|τ(cjn2)|<ϵ2. Since τ is a weight defined on all of M [4, page 486], the map (a,b)→τ(b*a) (a,b∈M) is a positive-definite inner product on M [4, page 489]. (The map is definite because τ is faithful.) Hence, the Cauchy-Schwarz inequality applies. For a∈M1+ and n≥Nj, we have
(5.3)|τ(cjna)|=〈a,cjn〉≤[τ(a2)]1/2[τ(cjn2)]1/2[Cauchy-Schwarz]<ϵ[τ(a2)]1/2≤ϵ(∥a2∥τ(1))1/2[Hölder]≤ϵ(∥a2∥·1)1/2≤ϵ.
Let N=maxj=1,…,mNj, and fix n≥N. For a∈M1+, let
(5.4)α(a)=(τ(a),τ(b1na),…,τ(bmna)),β(a)=(τ(a),τ(b10a),…,τ(bm0a)).
Note that all points in B(bn) are of the form α(a) and all points in B(b) are of the form β(a). Then by the inequality in (5.3) we have for every a(5.5)∥(α-β)(a)∥∞=supj=1,…,m|τ(bjna)-τ(bi0a)|≤ϵ.
Hence dH(B(b),B(bn))≤2ϵ for n≥N. Since ϵ was arbitrary, the result follows.Theorem5.2 is false if we replace strong convergence with weak convergence. Sherman came up with the following example in conversation with the author of this paper, and kindly gave permission for it to be included here.Example 5.3.
LetH=L2[0,1] and M=L∞[0,1]. Then there exist bn,b∈M, self-adjoint, such that bn→b weakly, but B(bn)↛B(b) in the Hausdorff metric.Proof.
LetUn={x∈[0,1]∣ the nth binary digit of x is 0}. Let
(5.6)bn=TUn-T(Un)c,
where TW is the characteristic function on the set W. Each bn is clearly self-adjoint, and standard analysis arguments show that the sequence {bn}n∈ℕ converges weakly to 0. Now σ(bn)={1,-1} for every n. Since μ(Un)=μ((Un)c)=0.5 for every n, B(bn) is the ball in (ℝ2,∥·∥1) centered at (0.5,0) with radius 0.5 (see Figure 5). On the other hand, the spectral scale for 0 is simply the line segment {(x,0)∣0≤x≤1}. Hence, dH(B(bn),B(0))=0.5 for every n∈ℕ. (The base metric for dH is the Euclidean norm.)Figure 5
Spectral scales for Example5.3.The next result concerns the unbounded situation that we have been dealing with for most of this paper.Theorem 5.4.
Letgn=(g1n,…,gmn),  g=(g10,…,gm0) be self-adjoint  m-tuples of bounded linear functionals on our finite von Neumann algebra  M. Suppose that  gjn→gj0 in the dual norm. Then  B(gn)→B(g) in the Hausdorff metric.Proof.
Letϵ>0. Then there exists N∈ℕ such that
(5.7)n≥N⇒supj=1,…,m∥gjn-gj0∥<ϵ.
Let α∈B(g). We want to find that β∈B(gn) (for n≥N ) such that ∥α-β∥≤ϵ.
Sinceα∈B(g), then there exists a∈M1+ such that α={(τ(a),g10(a),…,gm0(a))}. Define β={(τ(a),g1n(a),…,gmn(a))}. Then(5.8)|gjn(a)-gj0(a)|=|(gjn-gj0)(a)|≤∥gjn-hj0∥<ϵ.
Note that ϵ does not depend on j. Hence taking supremums over all the j's, we have that ∥α-β∥∞<ϵ. Hence, B(g)⊂(B(gn))2ϵ. Similarly, B(gn)⊂(B(g))2ϵ. Since ϵ was arbitrary, we have convergence in the Hausdorff metric as desired.The spectral scale as given in Definition1.4 is not the only object that has been called a spectral scale. The following definition of a spectral scale was formulated by Petz.Definition 5.5 (see [3, page 74]).
LetA be a finite von Neumann algebra equipped with a faithful, normal, tracial state, τ. Let b be a self-adjoint operator affiliated with A. The Spectral Scale is defined for t∈[0,1] as follows:
(5.9)λt(b)=inf{s∣1-τ(ps+)≤t},
where ps+=χ(-∞,s](b) as before.Notation 7.
We shall call this spectral scale thePetz spectral scale. We will call the spectral scale from Definition 2.1 the AAW spectral scale.We now show how the two notions are related. To this end, we first find what valuesλt(b) can take for a given t. To this end, fix t and note that we can write (5.10)λt(b)=inf{s∣τ(ps+)≥1-t}.
There are 3 cases to consider.Case 1.
There existss0∈σ(b) such that τ(ps0+)=1-t.
In this case,s<s0⇒τ(ps+)<1-t and hence, λt(b)=s0.Case 2.
For everys∈σ(b), τ(ps+)≠1-t but there exists s0∈σ(b) such that τ(ps0+)>1-t.
Sinceσ(b) is closed, and τ is weak-* continuous, we may choose s0∈σ(b) so that s0>s1⇒τ(ps1+)<1-t. Hence, s0 is the smallest real value such that τ(ps0+)>1-t, and so s0=λt(b).Case 3.
For everys∈σ(b), τ(ps+)<1-t.Note that, in this case,t=0 and b must be an unbounded operator, with unbounded spectrum on the right. In this case, λ0(b)=∞.The following result was proposed by Pavone in conversation with the author of this paper.Proposition 5.6.
Letf be the function whose graph is the upper boundary curve of the AAW spectral scale. Then, for  t>0,  f-′(t)=λt(b).Proof.
By the rotational symmetry of the AAW spectral scale,f-′(t)=g+′(1-t) where g is the function whose graph is the lower boundary curve of the AAW spectral scale. If s0∈σ(b) and τ(ps0+)=1-t, then g+′(τ(ps0+))=s0=λt(b). If t∈(s1,s0), a gap in the spectrum, with s0,s1∈σ(b) then the right-hand derivative of g at t is s0=λt(b).Att=0, the slope of f is supσ(b). If this is a finite number smax, then λ0(b)=smax. If supσ(b)=∞, then, as we saw in Case 3 above, λ0(b)=∞. This completes our discussion of the relationship between the AAW spectral scale and the Petz spectral scale.


## 6. Future Research

A great deal of further work has been done with the spectral scale in the bounded situation. For us, the first question to ask is whether Conjecture4.5 is in fact true. If so, we believe that many of the remaining results in [1] can be extended to the unbounded case fairly readily.Additionally, we believe that the idea of a spectral scale of an unbounded operator can be used in the discussion of numerical range.Definition 6.1.
Letb be a (bounded) linear operator on H. Define
(6.1)Wk(b)={1k∑i=1k〈bxi,xi〉∣i≠j⇒〈xi,xj〉=0,〈xi,xi〉=1}.
ThenWk(b) is the k-numerical range of b. When k=1, we simply write W(b), and refer to it as the numerical range [6, page 226].We can writeb=b1+ib2, where b1 and b2 are self-adjoint and so we can define the spectral scale of b to be B(b):=B(b1,b2). It turns out that the boundary of W(b) is exactly the set of radial complex slopes on B(b) at the origin.In the unbounded situation, we start withg=g1+ig2∈M*, where M is a finite von Neumann algebra equipped with τ, a finite, faithful, normal, tracial state. We can certainly find b,bi such that g(a)=τ(ba) and gi(a)=τ(bia) for every a∈M, but (except in the Abelian case) it is not obvious that there is any relationship between b and the bi's. We define the numerical range for g by making the additional assumption in Definition 6.1 that the xi's are in the domain of B. At the moment, it is not clear that there is any relationship between W(g) and B(g1,g2). However, if we can establish some kind of relationship, it is natural to ask how much of the theory developed in [6, 7] can be extended in this context.Finally, we ask whether Theorem2.22 can be extended beyond the single-variable situation?



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