### Around and under my talk at Fields

This week I am attending a Workshop on Developments and Technical Aspects of Free Noncommutative Functions at the Fields Institute in Toronto. Since I plan to give a chalk-talk, I cannot post my slides online (and I cannot prepare for my talk by preparing slides), so I will write here what some ideas around what I want to say in my talk, and also some ramblings I won’t have time to say in my talk.

[Several years ago I went to a conference in China and came back with the insight that in international conferences I should give a computer presentation and not a blackboard talk, because then people who cannot understand my accent can at least read the slides. It’s been almost six years since then and indeed I gave only beamer-talks since. My English has not improved over this period, I think, but I have several reasons for allowing myself to give an old fashioned lecture – the main ones are the nature of the workshop, the nature of the audience and the kind of things I have to say].

In the workshop Guy Salomon, Eli Shamovich and I will give a series of talks on our two papers (one and two). These two papers have a lot of small auxiliary results, which in usual conference talk we don’t get the chance to speak about. This workshop is a wonderful opportunity for us to highlight some of these results and the ideas behind them, which we feel might be somewhat buried in our paper and have gone unnoticed.

## Two versions of the Nullstellensatz for noncommutative functions, and their applications

#### Prologue

I want to begin by discussing something very general that we mathematicians do. I want to say something about how we give birth to problems and how they then start having a life of their own.

Suppose we have an infinite countable discrete group $G$. Then there is a very natural and well known construction of the group von Neumann algebra $L(G)$, which is an operator algebra on $\ell^2(G)$ (for the construction see, e.g., the second section of this post). The operator algebra $L(G)$ contains information on the group: for example the group has the ICC property if and only if $L(G)$ is a factor; further, $G$ is amenable if and only if $L(G)$ is hyperfinite. However, these fancy results are somewhat sophisticated, and the mathematical child is prone to ask a more basic question: do these algebras know what group they came from? if $L(G)$ and $L(H)$ are isomorphic von Neumann algebras, then are the groups $G$ and $H$ isomorphic?

Even though this is the first and most naive question that a mathematical child might ask, it very hard to solve, even for mathematical grown-ups. Notoriously, it is an open question whether $L(F_2)$ and $L(F_3)$ are isomorphic, where $F_k$ denotes the free group on $k$ generators (even though the question is open for such a basic pair of groups, there do exist known pairs of non-isomorphic groups $G$ and $H$ which give rise to isomorphic von Neumann algebras).

A mathematician might obsess over such a question for an entire lifetime. To solve this question one might introduce all kinds of strategies, involving disparate fields of mathematics. To carry out the strategies one will need to introduce new tools, and then one will have to develop and refine these tools. Soon enough, the development and refinement of the tools raise new questions, and answering these questions gives rise to the need for more new tools. Some very interesting results might be proved along the way, some very surprising applications to completely different problems are sometimes discovered. The original question was asked by the mathematical child out of curiosity and wonder more than anything else, but is never forgotten, even though it is sometimes regarded as a naive fancy and not as something that one can openly confess to be aiming at. The various technical issues that arise in analyzing the fine structure of the constructions defining the question in many cases reveal themselves as beautiful problems in their own right, filling the mathematician with new wonder and ambition.

The process described above is what I do for a living. I do not work on the free group factor problem nor on the problems in free probability that it gave birth to, nor on the problems that free probability gave birth to, etc. That example was given only for the sake of illustration.

As a mathematical child (and I still am) I asked a different question. But it is in the same spirit. If you want to get the gist of what I do – this is it. But to really tell you about my work, I need to get say more.

Sometimes I feel the need to defend the choice of my own questions. The BIG PEOPLE asked different questions – maybe those are the important questions?

We observe nature and wish to understand it. We see animals. They are beautiful, they are interesting, they excite us. We wish to understand these animals. What does it mean to understand?

One aspect of understanding is classification. We know that animal A is different from animal B, because one is standing here and the other is sitting there. So they are not the same animal. This classification is somewhat too fine to be interesting. A coarser classification might be more interesting (but not too coarse, we are not satisfied by noting that they are both “animals”).

To be able to classify in a meaningful way, we need to study the properties of the animals. For example, suppose that we have never seen African animals before, but we walk into a savanna where there are two kinds: elephants and giraffes. If we have never seen and never heard of elephants and giraffes, we do not know that these are elephants and giraffes, but maybe we are clever enough to notice that there are two distinct kinds of animals in sight. Now, different people will find different properties interesting, and this will affect their classification scheme. Some will tell the elephants and the giraffes apart by noting that one kind is grey and the other is yellow with brown spots. Some will note that one kind has a long nose (or whatever that is!) and the other has a long neck! Someone else might notice that one has four knees while the other has two knees and two elbows. They are all classifying but they are using different kinds of properties.

Anyone who is studying the world in a creative way will have to ask their own questions, and will have to define to themselves what constitutes an answer.

#### 1. The isomorphism problem for complete Pick algebras

Let $d \in \mathbb{N}$, let $\mathbb{B}_d$ be the open unit ball in $\mathbb{C}^d$. We let $H^2_d$ be the Hilbert space of all analytic functions $h : \mathbb{B}_d \to \mathbb{C}$ such that the coefficients of the power series for $h$ given by $h(z) = \sum_\alpha c_\alpha z^\alpha$ satisfies $\|h\|^2 := \sum_\alpha \frac{|\alpha|!}{\alpha!} |c_\alpha|^2 < \infty$. This space $H^2_d$ is known as the Drury-Arveson space (see, e.g., this post or this survey). Let $\mathcal{M}_d$ be the multiplier algebra of $H^2_d$, that is:

$\mathcal{M}_d = \{f : \mathbb{B}_d \to \mathbb{C} : fh \in H^2_d \,\,$ for all $h \in H^2_d\}$.

We can identify every multiplier $f \in \mathcal{M}_d$ with the multiplication operator $M_f \in B(H^2_d)$ that it gives rise to $M_f: h \mapsto fh$. In this way, $\mathcal{M}_d$ becomes an operator algebra.

Now given an analytic variety $V \subseteq \mathbb{B}_d$, we can consider $\mathcal{M}_V := \mathcal{M}_d \big|_V$ by which we mean all functions $g : V \rightarrow \mathbb{C}$, for which there exists $f \in \mathcal{M}_d$ such that $g = f\big|_V$. It can be shown that $\mathcal{M}_V$ is also a multiplier algebra, acting on the space of functions $H^2_V = H^2_d \big|_V$. It can also be shown that the algebra of multiplication operators is completely isometrically isomorphic to $\mathcal{M}_d/J_V$, where

$J_V := \{ f \in \mathcal{M}_d : f(x) = 0$ for all $x \in V\}$.

We ask the mathematical child’s question: do these algebras know the variety from which they came?

With Davidson and Ramsey we studied this question, and found the following answer:

Theorem: $\mathcal{M}_V$ and $\mathcal{M}_W$ are completely isometrically isomorphic if and only $V$ is the image of $W$ under a conformal automorphism of the ball.

However, it is more fun to ask questions in the wrong category. We ask: these algebras have a coarser structure that depends on the variety and is simply there. For example, there is the raw algebraic structure. Does the algebraic structure hold the variety in its memory? Following my work with Davidson and Ramsey, one can come up with the following statement.

“Theorem”: $\mathcal{M}_V$ and $\mathcal{M}_W$ are isomorphic (as algebras) if and only $V$ and $W$ biholomorphically equivalent via a multiplier map.

There are two problems with the above theorem.

The first one is that it only treats quotients of the form $\mathcal{M}_d/J$ where $J$ is a “radical” ideal.

The second problem is that it is false! The forward implication is proved only under additional assumptions, and we don’t really know if it is true in general (though I suspect that it is). As for the backward implication, there are counter-examples showing that there are pairs of varieties $V$ and $W$ which are multiplier biholomorphic but the algebras are non-isomorphic.

So it’s not really a theorem.

As I learned from George Elliott, when your candidate for a classification fails, there are two things you can do: 1) you can try to work in a restricted class of algebras or varieties, in hope that the invariant is a complete invariant in that setting; 2) you can refine the invariant.

We have papers dealing with restricted classes of varieties (see this paper with Kerr and McCarthy where we show that the “theorem” is true for reasonable one dimensional varieties; this paper with Davidson and Hartz studies the extent of the failure of “theorem” when we consider discs embedded in an finite dimensional ball). More recently I have been concentrating on studying the isomorphism problem through a refined invariant. The refined invariants are noncommutative (nc) varieties and their derivatives. These were described in depth in the talks by Eli Shamovich and by Guy Salomon

#### 2. The purely algebraic case

To illustrate how noncommutative varieties arise even for commutative problems, I will describe a Nullstellensatz that we obtained apropos our work on the isomorphism problem for algebras of bounded nc analytic functions. (BTW, I blogged about this in the past, and what follows has some overlap with the previous post).

Let $A = \mathbb{C}[z_1, \ldots, z_d]$ be the algebra of polynomials in $d$ commuting variables.

Let us define the zero locus $Z(J)$ as follows:

$Z(J) = \{z \in \mathbb{C}^d : p(z) = 0$ for all $p \in J\}$.

We also introduce the following notation: given $S \subseteq \mathbb{C}^d$, we write

$I(S) = \{p \in A : p(z) = 0$ for all $z \in S\}$.

The question that every mathematical child will ask immediately, is: (to what extent) can we recover $J$ from $Z(J)$?

Theorem (Hilbert’s Nullstellensatz): For every $J \triangleleft A$,

$I(Z(J)) = \sqrt{J}$

Recall that the radical of $J$ is the ideal

$rad(J) = \sqrt{J} := \{f \in A :$ there exists some $n$ such that $f^n \in J \}$.

To describe the Nullstellensatz that appears in my paper with Guy and Eli I will need to introduce some notation (after we proved it, we found that it can be dug out of a paper of Eisenbud and Hochester – but does it does not seem to be well known, at least not in our transparent formulation).

Let $M_d(n)$ denote the set of all $d$-tuples of $n \times n$ matrices. We let $M_d = \sqcup_{n=1}^\infty M_d(n)$ be the disjoint union of all $d$-tuples of $n \times n$ matrices, where $n$ runs from $1$ to $\infty$. That is, we are looking at all $d$-tuples of commuting matrices of all sizes. Elements of $M_d$ can be plugged into polynomials in noncommuting variables.

Similarly, we let $CM_d(n)$ denote the set of all commuting $d$-tuples of $n \times n$ matrices. We let $CM_d = \sqcup_{n=1}^\infty CM_d(n)$. Now we are looking at all commuting $d$-tuples of commuting matrices of all sizes. This can be considered as the “noncommutative variety” cut out in $M_d$ by the $\frac{d(d-1)}{2}$ equations (in $d$ noncommuting variables)

$Z_i Z_j - Z_j Z_i = 0$    ,    $1\leq i.

Points in $CM_d$ can be plugged into any polynomial $p \in A$

In fact, points in $CM_d$ can be naturally identified with the space of finite dimensional representations of $A$, by

$X \in CM_d \longleftrightarrow \Big(ev_X : p \mapsto p(X)\Big)$.

(We shall use the word “representation” to mean a homomorphism of an algebra or ring into $M_n(\mathbb{C})$ for some $n$).

Now, given an ideal $J \triangleleft A$, we can consider its zero set in $CM_d$:

$Z(J) = Z_{CM_d}(J) = \{X \in CM_d : p(X) = 0$ for all $p \in J\}$.

(We will omit the subscript $CM_d$ for brevity.) In the other direction, given a subset $S \subseteq CM_d$, we can define the ideal of functions that vanish on it:

$I(S) = \{p \in A : p(X) = 0$ for all $X \in S\}$.

Tautologically, for every ideal $J \triangleleft A$,

$I(Z(J)) \supseteq J$,

because every polynomial in $J$ annihilates every tuple on which every polynomial in $J$ is zero, right? The beautiful (and maybe surprising) fact is the converse.

The following formulation is taken from Corollary 11.7 from the paper “Algebras of bounded noncommutative analytic functions on subvarieties of the noncommutative unit ball” by Guy Salomon, Eli Shamovich and myself.

Theorem (free commutative Nullstellensatz): For every $J \triangleleft A$,

$J = I(Z(J))$.

For a proof, see the paper, or if you want a version for dummies (== analysts) see this blog post

The first thing to note about this theorem is that it is not trivial. One might say: “yeah sure, $A$ is finite dimensional so can guess that everything is determined by the finite dimensional representations”, but please note that this theorem fails if we replace $A$ by the (finitely generated) algebra of polynomials in $d$ non-commuting variables. It also fails in the algebra $H^\infty(\mathbb{D})$ of bounded analytic functions on the disc, even if one restricts attention to weak-* closed ideals (a setting in which $H^\infty(\mathbb{D})$ becomes a (commutative) principal ideal domain).

Consider the following theorem.

Theorem: Let $p$ be such that $\rho(p) = 0$ for every representation $\rho$ of $A$ that annihilates $J$. Then $p \in J$

In a sense, this is the correct theorem, which holds for every algebra, not just for the algebra $A = \mathbb{C}[z_1, \ldots, z_d]$. Isn’t it a better theorem?

NO! Because this theorem is too good to be wrong! It is stated in such a way that it is true automatically. Indeed, just consider the representation $A \to A/J$. By assumption, $p$ goes to zero under this representation, and hence $p \in J$

The commutative free Nullstellensatz is interesting precisely because it does not hold for all algebras, it is a special theorem true for the polynomial algebra $A$

Besides being non-trivial, the free commutative Nullstellensatz is pleasing and beautiful. Let us now see what we can do with it.

Theorem: $A/J \cong \mathbb{C}[V_{CM_d}(J)] = \{p\big|_{V_{CM_d}(J)} : p \in A\}$

Proof: The map $p \mapsto p\big|_{V_{CM_d}(J)}$ is surjective, and by the commutative free Nullstellensatz its kernel is $I(V_{CM_d}(J)) = J$

From here it is easy to show:

Theorem: $A/J \cong A/I$ iff there exists a polynomial isomorphism of varieties $G : V_{CM_d}(I) \to V_{CM_d}(J)$

Proof: The existence of such a map $G : V_{CM_d}(I) \to V_{CM_d}(J)$ implies the existence of an isomorphism $p \mapsto p \circ G$ between $A/J \cong \mathbb{C}[V_{CM_d}(J)]$ and $A/I \cong \mathbb{C}[V_{CM_d}(I)]$. Conversely, and isomorphism $\alpha : A/J \to A/I$ gives rise to a map $\alpha^* : \rho \mapsto \rho \circ \alpha$ between the spaces of finite dimensional representations. But it is easy to see that the finite dimensional representations of $A/J$ are in bijection with $V_{CM_d}(J)$ via

$\rho \longleftrightarrow (\rho(z_1), \ldots, \rho(z_d))$.

That completes the proof.

#### 3. Back to quotients of multiplier algebras

We now return to operator algebras. The ideas in the last section help us find the correct invariant for quotients of $\mathcal{M}_d = Mult(H^2_d)$ by non-radical ideals. The completely general problem is still not solved – we need to assume that the ideals are homogeneous.

For a homogeneous ideal $J \triangleleft A$, we write $\mathcal{M}_J = \mathcal{M}_d / \overline{J}^{WOT}$. We write $CB_d$ for the nc set of all commuting tuples which are also strict row contractions. Then we have the following result, which identifies our quotient algebras as algebras of nc functions on an nc variety.

Theorem: $\mathcal{M}_J \cong H^\infty(V_{CB_d}(J))$

Using this, we showed:

Theorem: $\mathcal{M}_I \cong \mathcal{M}_J$ if and only if there exists an appropriate nc analytic “isomorphism” $G : V_{CB_d}(J) \to V_{CB_d}(I)$

In my talk, the emphasis is on that an isomorphism of the algebras gives rise to a map between the varieties. To make this work, we needed to show that if $\alpha: \mathcal{M}_I \to \mathcal{M}_J$ then the induced map $\alpha^*$ is regular. The case of radical homogeneous ideals was previously solved in works with Davidson and Ramsey, and in a work of Hartz. The case of non-radical homogeneous ideals is solved by an argument that makes use of the proof of the radical case, essentially reducing things to the radical case using the following Nullstellensatz:

Theorem: Let $J \triangleleft A$ be a homogeneous ideal. There exists an $N \in \mathbb{N}$ such that for every $f \in \mathcal{M}_d$, if $f\big|_{Z(J) \cap \mathbb{B}_d} = 0$, then $f^N\big|_{V_{CB_d}(J)} = 0$