International Journal of Mathematics and Mathematical Sciences

Volume 2010 (2010), Article ID 713563, 33 pages

http://dx.doi.org/10.1155/2010/713563

## Extension of Spectral Scales to Unbounded Operators

Department of Mathematics, Weber State University, Ogden, UT 84404, USA

Received 30 May 2010; Accepted 14 June 2010

Academic Editor: Palle E. Jorgensen

Copyright © 2010 M. D. Wills. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We extend the notion of a spectral scale to *n*-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).

#### 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 to -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 be a finite von Neumann algebra equipped with a normal, faithful tracial state, . Elements of can be thought of as bounded operators on some Hilbert Space, , [2, page 308]. For a given self-adjoint the corresponding spectral scale, , which we will define below, yields information about the spectrum of in a nice geometric way. Many of the results can be extended to -tuples of self-adjoint operators in . 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 , we consider .

In Section 2, 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 is a spectral scale.

In Section 3, we consider the geometric structure of the -dimensional spectral scale. It turns out that there is little difficulty in generalizing from the bounded situation.

In Section 4, 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.

Section 5 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 Section 6 we outline some possible future directions of research.

Let us start with some preliminary definitions.

*Definition 1.1. *Let be a Hilbert space. Let be a subalgebra of . If is closed in the weak operator topology, self-adjoint, and contains 1, then is a *von Neumann algebra* [2, page 308].

Let denote the set of positive elements of .

*Definition 1.2. *Let be a function such that for and we have:
(The last equation implies that ).

Then is a *faithful, finite, normal trace* on [4, pages 504-5].

Theorem 1.3. *Let be a faithful, finite, normal trace on . Since any element of can be written as a finite linear combination of positive elements of , can be extended to a linear functional on all of [5, page 309].*

Two projections in are *equivalent * if there exists such that and . A projection is *finite* if . is *finite* if the projection is finite [5, page 296]. Throughout, we will assume that is finite.

Further, we will assume that there exists a faithful, finite, normal trace of , with ; that is, is a *faithful, normal, tracial state* on .

A crucial property of is that “things” commute in trace—that is, although, in general , for , we do have the equality [4, page 517].

Let . Let be an -tuple of self-adjoint operators in . Let

*Definition 1.4 (see [1, page 260]). *is called the *spectral scale* ofwith respect to *. *

Now is normal. Further is linear and continuous with respect to the weak operator topology. Moreover, is convex and compact in the weak operator topology: and for . Therefore is a compact, convex subset of .

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 replace with the set where each 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 defines , the predual of [8, Section , pages 112 ff.]. The duality is given by the bilinear form [8, Section , page 112] for and . Now [8, page 112 ff.], and Nelson shows that elements of are closed, densely defined operators affiliated with [8, Theorem , page 107, and Theorem , page 114]. It follows that a bounded linear functional, can be represented by a (possibly unbounded) linear operator affiliated with and we get the equality for every .

#### 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. *Let be a self-adjoint linear functional. Let
Then is the *spectral scale* of with respect to .

From the theory of noncommutative integration, we see that for some operator affiliated with . Since is self-adjoint, too will be self-adjoint, and hence, as with the original spectral scale, our generalized spectral scale is a compact, convex subset of .

*Notation 1. *We will often write for .

The following definition was suggested to the author in conversation by Akemann.

*Definition 2.2 (Akemann). *If is bounded, we will call an *operator functional*.

Our main goal in this section is to show that is an operator functional if and only if the slopes of the lower boundary function of are all finite. We remark that Akemann et al. have already shown the “only if” part of this statement [1, Section , pages 261–274]. For this reason, we may assume throughout that is unbounded, and show that the lower boundary curve of has, as a consequence, an infinite slope. To get there, we will need a number of preliminary results.

Proposition 2.3. * is mapped onto itself by a reflection through the point .*

*Proof. *Let . Then . Therefore
Thus the map takes to and hence takes onto itself. The fixed point of is , and the points and lie on a straight line that passes through . The straight line is given by the equation
Note also that is the identity map on . Hence, and reflects through the point .

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]). *Let be a (densely defined) self-adjoint operator in with domain . Then algebraic -homomorphism takes bounded Borel functions on into such that the following hold.*(a)* is norm continuous.*(b)*Let be a sequence of bounded Borel functions with as for each and for every and . Then for , as . The convergence is in norm.*(c)*If pointwisely, and the sequence is bounded, then strongly.*(d)*If , then .*(e)*If , then .*

For a given , a bounded Borel function on , it is customary to write as . In other words, the “” is understood. For now it is more convenient to write explicitly.

*Definition 2.5. *For let and . More generally, if is a characteristic function on a Borel subset of , then 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, for , is nonzero on the domain of (and hence all of ) if and only if is an eigenvalue of . Also, since Borel functions commute with respect to multiplication and is a homomorphism, Im is an Abelian subalgebra of . Assume now that is affiliated with our finite von Neumann algebra, . In this case it turns out that Im is an Abelian subalgebra of . This follows from the way that is constructed.

Lemma 2.6. * Let . Then
*

*Proof. *Using the decomposition , we can write

Hence,
(Of course, these equalities only make sense on the domain of .)

For every we get:

Therefore and hence . The other statement in this lemma follows via a similar argument.

Lemma 2.7. * Let be a characteristic function of a bounded Borel subset of . Then . *

*Proof. *Let be a sequence of bounded Borel functions such that for all and . By the Spectral theorem,
for every . Note that converges to for every . Since is a bounded Borel function, . For , we have
Hence and . Since is self-adjoint, we have the desired result.

Corollary 2.8. *The following set relations hold:
*

Lemma 2.9. *The range projection of is . The range projection of is . *

*Proof. *Let be the range projection of . Let be a sequence of bounded Borel functions on such that . Let . Then . From the Spectral theorem, we have

Taking the limit as on the left side, we get
and therefore
for every .

We have shown that . For let be the range projection of . By the same reasoning as above, . Also,
Hence, . Similarly, for , we have .

Now by Lemma 2.7. Therefore, is a bounded operator. In fact, on . We show that is an invariant subspace of under . Suppose that and Im. Then for every ,
since in , and .

Hence, Im. Thus,
as Since for every ,
But we already know that and so equality holds.

The second statement in the lemma follows from an analogous proof.

Lemma 2.10. *Let . If , then .**If , then .*

*Proof. *Write , using the decomposition

Note that for . Assume that
The diagonal entries of are for . Hence, . Thus, for all , such that , . Since on which contains , we get
Hence and so . Since is not an eigenvalue of , . Therefore
and so
for . From the Spectral theorem we can then conclude that
Thus,
for every . But
and for every and every . Hence, is dense in Im. Since Im, we have for . Thus,
so
The other statement in the lemma follows from an analogous argument.

We remark that in the original paper on spectral scales [1, Lemma , pages 262, 263], the above conclusion was obtained with a little less work, since, in that situation, was bounded and so we did not have to worry about the domain of . 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. * Fix , , and . Suppose that . Then the following hold:*()*.*()*If , then .*()*If , then . *

*Proof. *Note that and from Lemma 2.6. Hence
since is faithful. Similarly, .

() We compute
and so .

() Suppose that . Then
Similarly,
Therefore
and hence equality holds throughout. Thus
From (2.33),
while from (2.34),

Since is faithful and the arguments are positive, the arguments are in fact equal to zero. By Lemma 2.10, .

()Suppose that and . Then . Since is comparable to , and is faithful, .

We next state a theorem proved by Akemann and Pedersen [10, Theorem ].

Theorem 2.12. * If and are von Neumann algebras, is a normal linear map from to , and a face of , then there are unique projections and in with such that and .*

The following results are generalizations of the main theorems for the 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 that is a finite von Neumann algebra equipped with a faithful, normal, tracial state . The operator is unbounded and self-adjoint on affiliated with obtained from a linear functional (i.e., for each ). is the spectral scale of . The *lower boundary* of is given by
The *upper boundary* of is given by
The endpoints of the lower boundary are and .

Let denote the spectrum of , and let be the point spectrum of . Let be the function on whose graph is the lower boundary. For , let
Let be the positive half-plane determined by .

Our next result describes the faces of the lower boundary of . We do not include the endpoints at this time.

Theorem 2.13. * The zero-dimensional faces in the lower boundary of are precisely the points of the form for . Also,
** The one-dimensional faces in the lower boundary of are the sets of the form for . For each face ,**
The slope of is . *

*Proof. *We have the following steps.*Step 1. *We show that are zero-dimensional faces.

Fix . If and , then by Lemma 2.11. Hence,
is on the lower boundary of .

But
again by Lemma 2.11. Hence,

We now show that is an extreme point of . Suppose that
for . Since
Since projections are extreme points in , .*Step 2. *For , are faces of .

Fix so that
Then
If , then for each such that . Hence is on the graph of which is the lower boundary curve.

Write . Then , and is a typical point on the line segment connecting and . Hence graph contains this line segment. The slope is

Let denote the line segment in the graph of that contains and consider the endpoints of . By Theorem 2.12, there are projections such that , and hence
If , then, since , is in the interior of , contradicting Step 1. Hence and similarly . Thus, is a line segment in graph with slope and
*Step 3. *We show that we have accounted for all of the graph of , except possibly the endpoints.

Fix a point
with , and assume that for every . Write
We would like to show that . Since is closed, for .

By definition,
Since for ,
If , , then by definition
Similarly, and

Suppose that . Then . If
for some , we would have
which is clearly false. Hence,
But then and , which again is a contradiction. Hence,
Since
so is an eigenvalue of . From Step 2, is a line segment in graph. Hence is on the interior of that line segment.

Corollary 2.14. * The extreme points on the upper boundary (excluding the endpoints) are precisely the points of the form for and
** The line segments on the upper boundary are precisely the sets of the form , for . The slope of is and*

*Proof. *This result is a direct consequence of applying Proposition 2.3 to Theorem 2.13.

Let be the left endpoint of . Note that may be . Let be the right endpoint of . Note that may be .

Proposition 2.15. *If , then the left derivative of the lower boundary function at exists and is given by the formula
**
If , then the right derivative of at exists and is given by the formula
*

*Proof. *Since is convex, is a convex function, and so the left and right derivatives exist. Fix with . Then . Define . Since is closed, .*Case 1 ( for some ). *If , which contradicts the choice of . Thus , and hence is an isolated point in the spectrum, that is, is an eigenvalue of . Moreover, and so is the right-hand endpoint of a line segment in graph with slope by Theorem 2.13. Hence .*Case 2 ( for every ). *Choose such that and for . Then is on the graph of . Furthermore, in the weak- topology. Since is normal, . Since is faithful, . We have
for every . Hence,
for every . Thus,
Letting gives the desired result.The statement regarding right derivatives is proved in a similar way.

Proposition 2.16. *The corners of are in one-to-one correspondence with the gaps of , 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 or .)*

*Proof. *Let be an interior gap of the spectrum. Then for every we have . Fix . Then
Hence, is not differentiable at , and so a gap in the spectrum corresponds to a corner. Conversely, we have already seen that is differentiable at for .

Proposition 2.17. *For each ,
**
The line is a line of support for such that
**
In this case, passes through . Moreover, one has
*

*Proof. *Fix . If is an eigenvalue, then
Otherwise, . Either way,
and so
Let . Then
so lies in . We now wish to show that is a line of support for . There are several cases to consider.*Case 1 (). *In this situation, and are endpoints of a line segment in graph whose slope is . passes through both points and has slope . Thus, contains this line segment and is tangent to . Hence .*Case 2 (). *Note that is not an isolated point in . Moreover, . At least one of the one-sided derivatives of takes the value at . Hence, admits a line of support at with slope . As with Case 1, the line is and .*Case 3 ( is an interior gap in the spectrum). *In this case . Let and . Then and are lines of support passing through whose slope lies between and . If is a line of support for whose slope lies between and , then . But is such a line for . Hence, the statement is true for any .*Case 4 ( or ). *Since is unbounded, at least one of and has infinite magnitude. Suppose that is finite (and so must be infinite). Then . Moreover, . is a line of support for at and by Case 1. Suppose . Then is also a line of support for at and .

The case for is dealt with similarly. Hence, for every , is a line of support for and .

Conversely, for fixed , the lines are all parallel as varies over . Hence, there exists a unique for which is a line of support and . But has these properties and hence .

For the last statement, consider . Then is a face of . Hence, is an extreme point or a line segment on graph.

If is an extreme point, then , then . Since is an extreme point, then , and so

Similarly, if is a line segment, then for some , and so

From the above results, if , then the right derivative of approaches as . If in addition , then the left derivative of approaches as . By Proposition 2.3, the graph of the upper boundary curve of is vertical at . Hence, the only line of support at is vertical. Similarly, the only line of support at is vertical. Therefore, if both and are nonreal, then and are not corners of .

Conversely, if one of and is finite, then and are corners of .

Here the bounded and unbounded spectral scale theories do not coincide, since, in the bounded situation, and are always corners.

In both situations, we can read spectral data of the lower boundary curve as follows()1-dimensional faces correspond to eigenvalues of . The slope of each face is the corresponding eigenvalue.()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.()Corners on the lower boundary curve correspond to gaps in the spectrum.

We now exhibit two examples. In both examples, we will take and . The trace is integration with respect to Lebesgue measure and for , , and . Then and makes sense on .

*Example 2.18. *Define almost everywhere. Then is densely defined and self-adjoint on . It turns out that the equation of the lower boundary function is . This was obtained by integrating multiplied by appropriate characteristic functions. Observe that and . Hence the center of is and we get Figures 1 and 2.

*Example 2.19. *Define for and for . Then the lower boundary curve for is given by . And was chosen so that we would get a spectral scale that is invariant under the reflection . Note that , , , , and . The resulting pictures are shown in Figures 3 and 4.

We now examine a question posed to the author by Crandall. We start by stating the necessary properties that must have in order for it to be a spectral scale for an operator functional.

*Definition 2.20. *A *prespectral scale* is a set contained in which satisfies the following properties.(i) is compact and convex.(ii) and there are no other points of the form in .(iii).

Further, if and , then the following are given.

(iv) is invariant under the reflection .(v)The set is the graph of a function on [], which we will call the *lower boundary curve* of .

Lemma 2.21. * Let be a prespectral scale with lower boundary curve . Then is a continuous convex function. *

*Proof. *Since is closed, . Let . Since is convex, the line segment
is a subset of . Now
From the definition of ,
Hence, is convex on , and therefore continuous on [11, pages 61, 62]. Then and exist as extended real numbers [12, page 116]. Since is compact, then is finite and . Therefore, . By (ii) in Definition 2.20, . Applying (iv) from Definition 2.20, . Thus, is continuous and convex on .

Since is convex on , the left and right derivatives of exist for all as extended real numbers, and 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 Definition 2.1, condition (iii) follows from the fact that 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. *Let and let be the Lebesgue integral on . Given a prespectral scale , there exists self-adjoint such that .*

*Proof. *From the symmetry required for (condition (v)) it is sufficient to examine the lower boundary curve, , of . The function has the following properties (as noted in Lemma 2.21):(i) is continuous and convex,(ii).

Let us denote as the right derivative of at . Similarly, denote as the left derivative of at . Let on [0,1) and . Since is convex, then is nondecreasing. Hence, has at most a countable number of discontinuities. Since is convex, then is of bounded variation. By Exercise .H in [13, page 244], is absolutely continuous. By Theorem in [11, page 148], , and the fundamental theorem of calculus holds. Let , with . Since is increasing,
Hence, is the lower boundary curve of .

#### 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 , pages 276–280]. Often, with some modifications, the proofs are the same as in the original paper. Recall that is a Hilbert space, is a finite von Neumann algebra equipped with faithful, normal, tracial state, .

*Notation 3. *In Section 2 of this paper, we considered . We now consider an -tuple of self-adjoint linear functionals, . Let . Let . Then is also self-adjoint since each is self-adjoint and each is real. For each ( there is an associated self-adjoint, densely defined operator in , , and for every we have .

*Definition 3.1. *Let for every . Let for each . Then is the *spectral scale* of with respect to and is the *spectral scale* of with respect to .

Essentially the motivation for the introduction of is it allows us to reduce the -dimensional case to the 1-dimensional case by studying “2-dimensional cross-sections” of the spectral scale. We note that . Indeed, the right hand side may have trivial domain. However, as we will see, equality “almost” holds; that is, equality holds in trace.

Define , where .

Proposition 3.2. *The equality holds. *

*Proof. *For we have

Corollary 3.3. *As a consequence of this calculation, . *

We next introduce some additional notation.

*Notation 4. *Let be the spectral projection of determined by .

Let be the spectral projection of determined by .

Let .

Let .

Let .

The following results discuss the geometrical properties of .

Proposition 3.4. *If is an extreme point of , the n there exists a projection , such that and . Further, is an extreme point of . *

*Proof. *Fix an extreme point . Since is a face of , by Theorem 2.12 there are unique projections, in , such that . Thus, and so . Since is faithful, we have that , and so .

Next, suppose that for some , . Then