### Topics in Operator Theory, Lecture 10: hyperrigidity

In this lecture we discuss the notion of hyperrigidity, which was introduced by Arveson in his paper The noncommutative Choquet boundary II: Hyperrigidity, shortly after he proved the existence of boundary representations (and hence the C*-envelope) for separable operator systems. Most of the results and the examples that we will discuss in this lecture come from that paper, and we will certainly not be able to cover everything in that paper. In the last section of this post I will put some links concerning a result of Kennedy and myself which connects hyperrigidity to the Arveson’s essential normality conjecture.

#### 1. Background and definition of hyperrigidity

Recall that the in Bernstein’s proof of the Weierstrass polynomial approximation theorem, one associates with every continuous function $f : [0,1] \to [0,1]$ a Bernstein polynomial

$B_nf(x) = \sum_{k=0}^n f\left(\frac{k}{n}\right) {n \choose k}x^k(1-x)^{n-k}$.

The operators $B_n : C[0,1] \to C[0,1]$ are clearly linear, positive and unital. It can be shown that $B_n x = x$ and $B_n x^2 = x^2 + \frac{x - x^2}{n}$. Therefore

(*) $B_n f \to f$ uniformly for every $f \in span\{x,x^2\}$.

To prove Weierstrass’ approximation theorem, one needs to show that $B_n f \to f$ uniformly for all $f \in C[0,1]$. One can give a non-probabilistic proof of this fact, using just that $B_n$ is a sequence of positive and unital maps satisfying (*) (see Chapter 10.3 in Davidson and Donsig’s book “Real Analysis and Applications“).

In fact, Korovkin proved that given a sequence $\phi_n$ of positive linear and unital operators on $C[0,1]$ such that $\phi_n(f) \to f$ uniformly for $f \in span \{x,x^2\}$, then $\phi_n(f) \to f$ uniformly for all $f \in C[0,1]$. This implies that the generating set $\{x,x^2\}$ has a very “rigid” hold on the (closed) unital algebra that it generates, $C[0,1]$.

The above discussion should serve as background and motivation for the following definition.

Definition 1: Let $G$ be a generating subset of a C*-algebra $B$. We say that $G$ is hyperrigid in $B$ if for every faithful nondegenerate representation $B \subseteq B(H)$ and every sequence $\phi_n$ of UCP maps,

$\forall a \in G . \phi_n(a) \to a \Longrightarrow \forall b \in B . \phi_n(b) \to b$.

In the above definition I forced myself to overcome my pedantic self and use Arveson’s notation $B \subseteq B(H)$ for different faithful (and nondegenerate) representations of $B$. What we really mean is that $B$ is a certain fixed C*-algebra (either abstract or represented faithfully on some space, that’s not the point here) and that for every faithful and nondegenerate representation $\pi : B \to B(H)$and sequence of UCP maps $\phi_n : B(H) \to B(H)$, the condition

(*) $\phi_n(\pi(a)) \to \pi(a)$ for all $a \in G$

implies the consequence

(**) $\phi_n(\pi(b)) \to \pi(b)$ for all $b \in B$.

Conventions: Unless emphasized otherwise, our C*-algebras will be unital and all the representations will be nondegenerate (hence unital). In his paper Arveson discussed the nonunital setting to some extent, and time permitting I will touch upon this briefly in class (for a additional discussion of hyperrigidity in the context of nonunital algebras see this paper by Guy Salomon). We also note that a generating set $G$ is hyperrigid if and only if the operator system that it generates is. Although it is more precise to say that a set of generators $G$ (or the operator algebra that it generates $S$) is “hyperrigid in B“, we will sometimes just say that it is “hyperrigid“.

In Arveson’s paper it is always assumed that the generating set is finite or countably infinite, and that all Hilbert spaces are separable. I think the reason is that at the time, the existence of boundary representations was known only for separable operator systems. I will not make any countability assumptions here, I think they are not needed now that we know that boundary representations always exist (I will be grateful if an alert reader finds a gap in these notes). On the other hand, we may always stay within the realm of separable Hilbert spaces if our C*-algebras are separable – this is left as an exercise.

#### 2. Characterization of hyperrigidity

The definition of hyperrigidity (Definition 1) has an approximation theoretic flavor. The following theorem connects hyperrigidity to the unique extension property, which we studied in previous lectures.

Theorem 2: Let $S$ be an operator system generating a C*-algebra $B$. The following conditions are equivalent.

1. $S$ is hyperrigid in $B$.
2. For every nondegenerate representation $\pi : B \to B(H)$ and every sequence $\phi_n \in UCP(B,B(H))$, if $\phi_n(a) \to \pi(a)$ for every $a \in S$, then $\phi_n(b) \to \pi(b)$ for all $b \in B$.
3. For every nondegenerate representation $\pi : A \to B(H)$, the UCP map $\pi\big|_S$ has the unique extension property.
4. For every unital C*-algebra $\tilde{B}$, every unital $*$-homomorphism $\theta : B \to \tilde{B}$ and every UCP map $\phi : \tilde{B} \to \tilde{B}$

$\phi(x) = x$ for all $x \in \theta(S)$ $\quad$ $\,\implies\,$ $\quad$ $\phi(x) = x$ for all $x \in \theta(B)$.

Proof: 1 $\implies$ 2: Condition 2 is very similar in appearance to the definition of hyperrigidity given by “(*) implies (**)” in the previous section. Indeed, it follows readily by assuming that $B$ is represented faithfully as $A \subseteq B(K)$, and then considering the faithful representation

$a \hookrightarrow \pi(a) \oplus a \in B(H \oplus K)$

If $\phi_n(a) \to \pi(a)$ for all $a \in S$, then $\phi_n(a) \oplus a \to \pi(a) \oplus a$ for all $a \in S$. By the definition of hyperrigidity (summoned only after invoking Arveson’s extension theorem, so that we will be discussing UCP maps on $B(H \oplus K)$), we find that

$\phi_n(b) \oplus b \to \pi(b) \oplus b$ for all $b \in B$,

which implies that $\phi_n(b) \to \pi(b)$. Note how we really needed the assumption of hyperrigidity to address every faithful nondegenerate representation of $B$.

2 $\implies$ 3: This follows by taking $\phi_n = \pi$ for all $n$.

3 $\implies$ 4: Let $\tilde{B}, \theta$ and $\phi$ be as in Condition 4. Represent $\tilde{B}$ faithfully (and non-degenerately) on a Hilbert space as $\tilde{B} \subseteq B(H)$, and extend $\phi$ to a UCP map $\phi : B(H) \to B(H)$. Then $\theta$ can be considered as representation of $B$ on $H$. If $\phi$ does not fix $\theta(B)$, then then $\phi \circ \theta$ is an extension of $\theta\big|_S$ which is different from $\theta$, in contradiction to Condition 3.

4 $\implies$ 1: Suppose that $B \subseteq B(H)$ faithfully and non-degenerately, and let $\phi_n : B(H) \to B(H)$ be a series of UCP maps. Assume, as in the definition of hyperrigidity, that

$\phi_n(a) \to a$ for all $a \in S$.

Assuming that Condition 4 holds, we wish to prove that

$\phi_n(b) \to b$ for all $b \in B$.

Construct the C*-algebra $D = \ell^\infty(B(H))$ of bounded sequences with values in $B(H)$ with the obvious structure, and consider the UCP map $\psi : D \to D$ given by

$\psi(a_1, a_2, \ldots) = (\phi_1(a_1), \phi_2(a_2), \ldots)$.

If $C = c_0(B(H))$ is the ideal of all sequence tending to zero, then $\psi(C) \subseteq C$. Write $\tilde{B} = D/C$ and so $\psi$ induces a UCP map $\phi : \tilde{B} \to \tilde{B}$. By defining a *-homomorphism $\theta : B \to \tilde{B}$ by

$\theta(b) = (b,b,b, \ldots) + C$

we are precisely in the situation of Condition 4. It follows that $\phi$ fixes $\theta(B)$, and this is the same as $\phi_n(b) \to b$ for all $b \in B$.

That concludes the proof.

The following is a simple corollary that is worth recording:

Corollary 3: Let $S$ be a hyperrigid operator system in $B = C^*(S)$. Then $B$ is the C*-envelope of $S$.

Proof: Since for every representation $\pi$, the restriction $\pi\big|_{S}$ has the UEP, the Shilov ideal – which we know is the intersection of all the kernels of boundary representations – is trivial.

Corollary 4: Let $S$ be a hyperrigid operator system in $B$, let $I \triangleleft B$ be an ideal, and let $q: B \to B/I$ be the quotient map. Then $q(S)$ is hyperrigid in $B/I$.

Proof: Condition 2 in the theorem is preserved under taking quotients.

It is important to note that the converse to Corollary 3 does not hold: an operator system with trivial Shilov ideal in the C*-algebra it generates need not be hyperrigid. Moreover, in the context of Corollary 4, we note that having trivial Shilov ideal is not preserved by quotients. Here is an example (due to Davidson; personal communication) that illustrates both of these statements.

Example 1: Let $\{e_n\}_{n=1}^\infty$ be an orthonormal basis for a Hilbert space $H$, and let $\{t_n\}_{n=1}^\infty$ be a sequence of complex numbers such that $t_1 = 2, t_2 = 0$ and $\{t_n\}_{n=3}^\infty$ is a dense sequence in $\mathbb{D}$. Define $T e_n = t_n e_n + \frac{1}{n}e_{n+1}$ and let $A$ be the unital (norm closed) operator algebra generated by $T$. (We are suddenly discussing an operator algebra and not an operator system, but we can always pass from $A$ to $S:=A+A^*$ and back).

One can check that $B = C^*(A)$ is an irreducible operator algebra containing the compacts. The Calkin map $q$ is not isometric on $A$, so by the boundary theorem, the Shilov ideal is trivial and $B = C^*_e(A)$.

On the other hand, $q(T)$ is a normal operator with spectrum $\overline{\mathbb{D}}$. It follows that $q(B) \cong C(\overline{\mathbb{D}})$ and $q(A)$ is the disc algebra. Thus, after passing to the quotient, the Shilov boundary ideal is not trivial.

This example shows that – unlike hyperrigidity – trivial Shilov ideal is a property that does not pass to quotients. It also shows that a trivial Shilov boundary does not imply hyperrigidity (why?).

By Theorem 2, if $S \subseteq B = C^*(S)$ is hyperrigid, then, in particular, every irreducible representation is a boundary representation. Arveson conjectured that the converse also holds true.

Arveson’s Hyperrigidity Conjecture: $S$ is hyperrigid in $B = C^*(S)$ if and only if every irreducible representation of $B$ is a boundary representation for $S$.

#### 3. Examples

Example 2: Suppose that $T \in B(H)$ is a selfadjoint operator with at least three points in the spectrum, and let $B$ be the (unital, as always) C*-algebra generated by $T$. We will show that:

1. $\{I,T,T^2\}$ is hyperrigid in $B$.
2. $\{I,T\}$ is not hyperrigid in $B$.

(The assumption on the spectrum is no biggy: if the spectrum has two or less points, then $span \{I,T\}$ is the C*-algebra generated by $T$, and of course it is hyperrigid.)

Note that if we let $T = M_x$ be the multiplication operator by the identity function $x$ on $H = L^2[0,1]$, and if we identify $C[0,1]$ as a C*-subalgebra of $B(H)$, then we obtain a strengthened version of Korovkin’s theorem that we mentioned in the first section. Indeed, in Korovkin’s theorem, the conclusion $\phi_n(f) \to f$ for all $f \in C[0,1]$ follows from this convergence only of elements $f \in \{1,x,x^2\}$ only for sequences of (completely) positive maps $\phi_n : C[0,1] \to C[0,1]$. Hyperrigidity shows that $\phi_n(f) \to f$ follows from the same assumption, but now for any sequence of UCP maps $\phi_n : C[0,1] \to B(H)$.

Let us first show that $S = span\{1,T\}$ is not hyperrigid. Suppose that $a = \min \sigma(T)$, $b = \max \sigma(T)$, and $c$ is a point in the spectrum that lies strictly between $a$ and $b$. Then point evaluation $\delta_c$ at $c$ is representation of $C(\sigma(T)) \cong B$. However, $\delta_c\big|_S$ does not have the unique extension property, since $\delta_c\big|_S$ is a convex combination of $\delta_a$ and $\delta_b$.

Now let $S = span \{1,T,T^2\}$. To show that $S$ is hyperrigid, we need to show that for every nondegenerate $\pi : B \to B(K)$, the UCP map $\pi\big|_S$ has the UEP.

To this end, suppose that $\phi : B \to B(K)$ is a UCP map that extends $\pi\big|_S$. This just means that $\phi(T) = \pi(T)$ and $\phi(T^2) = \pi(T^2) = \pi(T)^2 = \phi(T)^2$. We have to show that $\phi$ is multiplicative – this will imply that $\phi = \pi$.

Let $\phi(b) = P_K \rho(b) \big|_K$ be a Stinespring representation of $\phi$, where $\rho : B \to B(L)$ is a *-representation and $L$ is a space containing $K$. We write $P = P_K$, and compute:

$P \rho(T) (I-P)\rho(T)P = P\rho(T^2)P - P\rho(T)P\rho(T)P = \phi(T^2) - \phi(T)^2 = 0$.

This means that $(I-P)\rho(T)P = 0$, or put differently, that $K$ is invariant under $\rho(T)$. It follows that $K$ is a reducing subspace for $B$, and so $\phi = P_K \rho(\cdot)\big|_K$ is multiplicative, as required.

It is worth noting that under the assumption above on $T$, Arveson proved that if $\{I,T,f(T)\}$ is hyperrigid (where $f$ is continuous on the spectrum of $T$) then $f$ must be either strictly convex or strictly concave. He also proved that the converse holds, assuming that the Hyperrigidity Conjecture is true.

Example 3: Let $u_1, \ldots, u_n$ be isometries generating a unital C*-algebra $B$. The generating set

$G = \{u_1, \ldots, u_n, u_1 u_1^* + \cdots + u_n u_n^*\}$.

is hyperrigid in $B$.

In particular, if $u_1, \ldots, u_n$ are Cuntz isometries (i.e., isometries with pairwise orthogonal ranges, such that $u_1u_1^* + \cdots u_n u_n^* = 1$), then it is well known that they generate the Cuntz algebra $\mathcal{O}_n$. Then we have that the operator system generated by $u_1, \ldots, u_n$ is hyperrigid in $\mathcal{O}_n$.

Let’s prove the claimed hyperrigidity in the special case of Cuntz isometries.; the general case is based on the same idea but slightly more tedious. Write $S$ for the operator system generated by $u_1, \ldots, u_n$. Let $\pi : B = \mathcal{O}_n \to B(H)$ be a (nondegenerate, as always) representation. Let $\phi: B \to B(H)$ be a UCP map such that $\phi\big|_S = \pi\big|_S$, and let $\rho: B \to B(K)$ be a representation that is a dilation of $\phi$. Define $V_k = \rho(u_k)$ for $k=1, \ldots, n$. Then for every $k$,

$V_k = \begin{pmatrix} \pi(u_k) & x_k \\ y_k & z_k \end{pmatrix}$.

$\rho$ is a *-representation, so

$I = V_k^* V_k = \begin{pmatrix} \pi(u_k^*u_k) + y_k^*y_k & * \\ * & * \end{pmatrix}$.

Comparing the (1,1)-entry, we find $I_H = \pi(u_k^*u_k) + y_k^*y_k$. Since $\pi(u_k^*u_k) = I$, we must have that $y_k^* y_k = 0$, so that $y_k = 0$. On the other hand

$\sum_k V_k V_k^* = \begin{pmatrix} \sum \pi(u_ku_k^*) + \sum x_k x_k^* & * \\* & * \end{pmatrix}$.

As before, this implies that $\sum_k x_k x_k^* = 0$, which implies that $x_k = 0$ for all $k$.

We conclude that $\rho(u_1), \ldots, \rho(u_n)$ all reduce $H$, and (since these operators generate $B$) it follows that $H$ is a reducing subspace for $\rho$. As a consequence, $\phi$ is its own minimal Stinespring dilation, so it is multiplicative. Hence $\phi = \pi$, as required.

#### 4. The hyperrigidity conjecture for C*-algebras with countable spectrum

Arveson established his hyperrigidity conjecture for the special case of countable spectrum. Recall that the spectrum of a C*-algebra $B$ is the set $\hat{B}$ of all unitary equivalence classes of irreducible representations (if you are the kind of person who -like me – always worries about such things, let me tell you that it is OK to use the word “set” for $\hat{B}$. Hint: we are speaking about irreducible representations, and the dimension of Hilbert space on which an irreducible representation of $B$ can live is restricted by the cardinality of $B$).  A C*-algebra $B$ is said to have countable spectrum if $\hat{B}$ is countable, in other words, if $B$ only has a countable number of pairwise unitary inequivalent irreducible representations.

Theorem 5: Let $S$ be an operator system such that the generated C*-algebra $B = C^*(S)$ has countable spectrum. If every $\pi \in \hat{B}$ is a boundary representation, then $S$ is hyperrigid in $B$.

Proof: Assume that $bd_S B = \hat{B}$. To prove that $S$ is hyperrigid, we need to prove that for every representation $\pi:B \to B(H)$, the UCP map $\pi\big|_S$ has the UEP. But every representation of $B$ is the direct sum of irreducible representations (here we are using the fact that $\hat{B}$ is countable, and some non-trivial facts from the representation theory of (type I) C*-algebras). Since the direct sum of UCP maps with the UEP has the UEP (yet another fact that requires proof, but not as deep as the previous fact we used), it follows that $\pi\big|_S$ has the UEP.

#### 5. Connection to Arveson’s Essential Normality Conjecture

In this section I wish to discuss the connection between the notion of hyperrigidity and another conjecture of Arveson – the essential normality conjecture.

The problem of essential normality can take place in many spaces, but I like to view in the Drury-Arveson space $H^2_d$. See this old post (mostly Section 1) and this old post (mostly Sections 2 and 3) where I already discussed this space. In this old post (Section 1) I discuss the essential normality problem.

In class I plan to discuss the paper “Essential normality, essential norms, and hyperrigidity by Kennedy and myself (here is a link to an arxiv version; here is a link to the corrigendum). I wrote about this problem and about this paper a few times before (for example, when announcing the preprint), so I with the above pointers and links in place, we end these notes!

Thanks for listening! You all get a grade 100 in the course!