### Matthew Kennedy awarded CMS 2012 Doctoral Prize

One of the nicest things that happen on math blogs is that people write expository posts on other people’s work. Tim Gowers and Terry Tao have set a fine example in their expositions of the works of Fields Medalists or Abel Prize laureates. These are among the most interesting and important posts out there, I think.

Recently, at the Canadian Mathematical Society’s Winter Meeting, Matthew Kennedy was awarded the CMS 2012 Doctoral Prize (at the meeting several other prizes were awarded by the CMS as you can see on the meeting’s homepage). See here for the media release, and here for a description of the prize and a list of past recipients.

Matt gave a plenary lecture at the CMS meeting surveying (some of) his work. Here are the slides, which are certainly worth looking at (thanks to Matt for allowing me to post them). Note that on the last slide there are photos of two gentlemen with no captions; these are Ken Davidson (on the left, Matt’s PhD supervisor) and Heydar Radjavi (Matt’s undergraduate research supervisor).

In this post I will describe a couple of Matt’s first really big results, one of which didn’t make it into his talk. These are the existence of wandering vectors for (certain) free semigroup algebras, and the reflexitivity of free semigroup algebras. These result appeared in the tour-de-force paper “Wandering vectors and the reflexivity of free semigroup algebras”; here are links to arxiv, mathscinet, and the official version in Crelle

#### 1. Free semigroup algebras

Let $H$ be a Hilbert space. A row isometry  is an n-tuple $V_1, \ldots, V_n$ isometries which have mutually orthogonal ranges (n here is allowed to be any integer, but let us assume that $n \geq 2$). Thus we have the relations $V_i^* V_j = 0$ if $i \neq j$ and $V_i^* V_i = I$ for all i. The reason for the terminology is that a tuple $V_1, \ldots, V_n$ is a row isometry if and only if the row operator $V = [V_1, \ldots, V_n] : H \oplus \ldots \oplus H \rightarrow H$, defined on the sum of n copies of $H$ by $V(h_1, \ldots, h_n) = \sum V_i h_i$, is isometric.

At first sight, it might not be clear if there is a rich supply of row isometries. Let me start by describing the canonical example.

Let $F^n_+$ be the free semigroup generated by n generators $1,\ldots, n$. Let $H$ be a Hilbert space with orthonormal basis $\{e_{w}\}_{w\in F^n_+}$ parametrized by $F^n_+$. We define n operators $L_1, \ldots, L_n$ on $H$ by

$L_i e_w = e_{iw} \quad , \quad w \in F^n_+ \,\, , \,\, i=1, \ldots, n .$

It is easy to see that $L_1, \ldots, L_n$ is a row isometry. The unital norm–closed algebra $\mathcal{A}_n$ that it generates is called the noncommutative disc algebra and was first introduced and studied by Gelu Popescu. The unital weak–operator–closed algebra that is generated by $L_1, \ldots, L_n$ is denoted $\mathcal{L}_n$ and is referred to as the left regular free semigroup algebra or as the noncommutative analytic Toeplitz algebra (also first introduced and studied by Popescu). The C*–algebra that this row isometry generates is called known as the Cuntz–Toeplitz algebra.

It is well known that the C*–algebra generated by a row isometry is *–isomorphic to the Cuntz algebra if $\sum V_i V_i^* = I$, and is *–isomorphic to the Cuntz–Toeplitz algebra otherwise — no other C*–algebras can be generated by a row isometry.

It is less known that, no matter how the particular isometries are chosen, the norm–closed unital algebra generated by $V_1, \ldots, V_n$ is isometrically isomorphic to the noncommutative disc algebra $\mathcal{A}_n$ (this was proved by Popescu in this paper).

It might therefore come as a surprise that the weak–operator–closed unital algebras generated by different row isometries can be totally different; in a sense, they can be as different from one another as weak–operator closed algebras can be. More on this below, but for now I’ll just state that such an algebra does not have to be even algebraically isomorphic to $\mathcal{L}_n$.

Thus, the (non–self–adjoint) weak–operator–closed algebras generated by a row isometry contains more information on the row–isometry than the C*–algebra or norm–closed algebra generated by the row isometries. Since this is the situation, and since row isometries have been of growing interest for several decades now, it makes sense to study these weak–operator–closed algebras.

free semigroup algebra  is a weak–operator–closed unital algebra generated by a row isometry. This class of operator algebras was introduced and studied by Kenneth Davidson and David Pitts in these two papers, and in a series of subsequent papers. Let me mention also this paper from 2001 by Davidson, Katsoulis and Pitts, where a general structure theorem for free semi–group algebras is obtained. This paper ended with some interesting and important open problems. Here is one of these open problems (two more will be described below).

Problem 1: Is there a row isometry such that the free semigroup algebra it generates is a von–Neumann algebra?

Charles Read solved this problem by showing, actually, that $B(H)$ is a free semigroup algebra! There are many ways to see that $\mathcal{L}_n$ is different in many ways from $B(H)$. From a spatial perspective, $\mathcal{L}_n$ has joint eigenvectors, $\mathcal{L}_n$ has no non–scalar normal elements, $\mathcal{L}_n$ a gigantic commutant, $\mathcal{L}_n$ has no non–zero compacts, etc. From an intrinsic perspective these two algebras are not even algebraically isomorphic let alone isometrically isomorphic; for example, one has many ideals and the other has only one, one has no proper projections, etc. One may consult the above mentioned papers for more details.

#### 2. Reflexivity and wandering vectors

Given an operator algebra $\mathcal{A} \subseteq B(H)$, its lattice of invariant subspaces is defined

$Lat(\mathcal{A}) = \{F \subseteq H : \forall T \in \mathcal{A} . TF \subseteq F\} .$

And given a family $\mathcal{F}$ of subspaces of $H$, the algebra corresponding to $\mathcal{F}$ is defined

$Alg(\mathcal{F}) = \{T \in B(H) : \forall F \in \mathcal{F} . TF \subseteq F \} .$

An operator algebra $\mathcal{A}$ is said to be reflexive if $\mathcal{A} = Alg(Lat(\mathcal{A}))$. The notion of reflexivity grew out of problems related to the invariant subspace problem, and was introduced by Radjavi and Rosenthal in 1969 in this paper (the terminology was suggested by Halmos). The problem of whether or not an algebra is reflexive, or satisfies some quantified version of reflexivity, has been a problem of central interest in operator algebras. One of the open problems that Davidson, Katsoulis and Pitts posed at the end of their 2001 paper was

Problem 2: Is every free semigroup algebra reflexive?

Say that a free semigroup algebra is of type L if it is algebraically isomorphic to $\mathcal{L}_n$ (I am perhaps using outdated terminology. Today type L algebras are also said to be analytic, which is perhaps more suggestive). Davidson and Pitts observed early on that $\mathcal{L}_n$ is reflexive. By the structural results of Davdison, Katsoulis and Pitts the above problem can be reduced to the problem of whether or not every free semigroup algebra of type L is reflexive. The problem is that algebras of type L can behave spatially very differently from $\mathcal{L}_n$, so at first it is not clear how being algebraically isomorphic to $\mathcal{L}_n$ helps solve the problem of reflexivity — which is a spatial problem. (Remark: by spatial we mean something that has to do with the particular representation of the algebra on a certain space. So invariant subspaces, eigenvectors and so forth are spatial notions, whereas the ideal structure is an algebraic notion). It was not clear at first, and it did not become any clearer later, and the problem of whether or not every type L algebra is reflexive remained open since 2001 until Matt solved in 2011 (I am citing year of publication in both cases).

It turned out that this problem is equivalent to a seemingly unrelated problem. Let $\mathcal{A}$ be a free semigroup algebra on $H$ generated by the row isometry $V_1, \ldots, V_n$. A unit vector $u \in H$ is said to be a wandering vector for $\mathcal{A}$ if the set

$\{V_{i_1} \cdots V_{i_k} u \big| i_1 i_2 \cdots i_k \in F^n_+ \}$

is orthonormal. For example, in the left regular free semigroup algebra every one of the basis vectors $e_w$, $w \in F^n_+$, is a wandering vector.

Since the left regular representation is the canonical one, it is at first hard to imagine that a free semigroup algebra can have no wandering vector. But this can indeed happen: indeed, if there is a wandering vector then there are many nontrivial invariant subspaces, while on the other hand $B(H)$ — a free semigroup algebra — has none.

But Davidson, Katsoulis and Pitts conjectured that every free semigroup algebra of type L has a wandering vector. This was their third open problem posed at the end of their 2001 paper.

Problem 3: Does very free semigroup algebra of type L have a wandering vector?

In 2005 Davidson, Li and Pitts proved that if a type L algebra has a wandering vector then it is reflexive.

#### 3. What Kennedy did

In his 2011 paper “Wandering vectors and the reflexivity of free semigroup algebras” Kennedy solved the above Problems 2 and 3. The main (in my opinion) technical achievement was proving that every type L algebra has a wandering vector. From what I wrote above this implies in turn (by results of other researchers) that every type L algebra is reflexive and therefore that every free semigroup is reflexive (he obtained some other and sharper results which I shall not discuss). I will end this post by discussing the main ingredient that went into his proof that a wandering vector exists.

Let $V_1, \ldots, V_n$ be a row isometry. How does one produce a wandering vector? You might try to begin with a unit vector $u$ in the range of $V_1$. Then for all $i=1, \ldots, n$, $V_i u$ is orthogonal to $V_j u$ for $j \neq i$ and also to $u$ if $i \geq 2$ (because the ranges are orthogonal). But there is no reason why $u$ would be orthogonal to $V_1 u$. If you go and try to fix this in some direct way then you will probably not get very far. Kennedy’s approach is quite indirect, using what he calls “dual algebra techniques”, which he attributes to Bercovici, Foias, Pearcy, and others, and traces back to S. Brown’s proof of the existence of invariant subspaces for subnormal operators.

Definition: Let $\mathcal{S}$ be weak*–closed subspace of $B(H)$. $\mathcal{S}$ is said to have property $\mathbb{A}_1$ if for every weak*–continuous functional $\phi$ on $\mathcal{S}$, there exist $x,y \in H$ such that for all $T \in \mathcal{S}$

$\phi(T) = (Tx,y) .$

Property $\mathbb{A}_1$ can be thought of heuristically as measure of how small the predual of a subspace is. This is a truly non–trivial property to possess (and therefore very useful); for example, $B(H)$ does not have property $\mathbb{A}_1$. On the other hand, the infinite amplification of any weak* closed subspace does have property $\mathbb{A}_1$ (why?).

Kennedy proved that every type L algebra has property $\mathbb{A}_1$ (he actually proved that it has property $\mathbb{A}_1(1)$, which is a sharper version). A thick part of his paper was dedicated to this, and the argument is quite delicate and hard. I will not go into the details of that. Instead, I want to explain how property $\mathbb{A}_1$ implies the existence of a wandering vector, which is much easier to explain and shows how property $\mathbb{A}_1$ can be useful (but it must be understood that in order to see the heart of the proof you will really have to read Kennedy’s paper).

Theorem (Kennedy): Every type L free semigroup algebra has a wandering vector.

Proof: Let $V_1, \ldots, V_n$ be a row isometry on $H$ and let $\mathcal{A}$ be the free semigroup algebra that they generate. Let $\mathcal{A}_0$ be the weak*–closed ideal that is generated by $V_1, \ldots, V_n$. The canonical algebraic isomorphism between $\mathcal{A}$ and $\mathcal{L}_n$ sending $V_i$ to $L_i$ is known to be weak*–continuous (but this is not trivial). Therefore, $\mathcal{A}_0$ is a nontrivial ideal which does not contain $I := I_H$. By the Hahn–Banach Theorem, there is a weak*–continuous functional $\phi$ on $\mathcal{A}$ which annihilates $\mathcal{A}_0$ and maps $I$ to $1$. Since $\mathcal{A}$ has property $\mathbb{A}_1$, there are $x, y \in H$ such that

$\phi(T) = (Tx,y)$

for all $T \in \mathcal{A}$. We may assume that $y \in \mathcal{A} x$, because if it is not then we replace it with its projection on that space. On the other hand, we have that $(V_{i_1} \cdots V_{i_k} x, y) = 0$ for all $i_1 i_2 \cdots i_k \in F^n_+$, therefore $y \perp \mathcal{A}_0 x$. Thus $y \in \mathcal{A}x \ominus \mathcal{A}_0 x$ (the orthogonal complement of $\mathcal{A}_0 x$ inside $\mathcal{A} x$).

Let $u = y/\|y\|$. Then $u$ is the wandering vector that we are looking for. Indeed, for every $i_1, \ldots, i_k$, the vector $V_{i_1} \cdots V_{i_k} u$ is in $\mathcal{A}_0 u \subseteq \mathcal{A}_0 \mathcal{A} x = \mathcal{A}_0 x$, and therefore orthogonal to $u$, thus

$(V_{i_1} \cdots V_{i_k} u, u) = 0 .$

From here, using the pair–wise orthogonal ranges assumption, it easily follows that for every two distinct words $i_1 i_2 \cdots i_k$ and $j_1 j_2 \cdots j_m$ in $F^n_+$,  $(V_{i_1} \cdots V_{i_k} u, V_{j_1} \cdots V_{j_m} u) = 0$. That shows that $u$ is a wandering vector and completes the proof.

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