### Topics in Operator Theory, Lecture 7: boundary representations

In this lecture we will present a proof that boundary representations exist in abundance, following Davidson and Kennedy’s breakthrough paper. Davidson and Kennedy’s paper was in the spirit of Arveson’s paper from 1969, and followed Arveson’s solution in the separable case from 2007. (BTW, I wrote about Davidson and Kennedy’s solution in a an old blog post).

#### 1. The unique extension property and maximal representations

Recall the definition of a boundary representation.

Our setting will be of an operator system $S$ contained in a C*-algebra $B = C^*(S)$. Recall that earlier we discussed the situation of a unital operator algebra $A \subseteq B = C^*(A)$, and later we extended our attention to unital operator spaces. In this post we will consider only operator systems, but there will be no loss of generality (because every unital completely contractive map $A \mapsto B(H)$ extends to a unique unital completely positive map $S: A + A^* \to B(H)$, and vice versa).

Definition 1: Let $A \subseteq B = C^*(A)$ be a unital operator space. A boundary representation for $A$ is an irreducible representation $\pi : B \to B(H)$ such that the only UCP map $B \to B(H)$ that extends $\pi|_A$ is $\pi$ itself.

The kernel of the above definition can also make sense when $\pi$ is not irreducible. Some authors have used the term “boundary representation” in this more general sense. We make the following definition:

Definition 2: Let $S \subseteq B = C^*(S)$ be an operator system. A UCP map $\phi : S \to B(H)$ is said to have the unique extension property (UEP for short) if

1. There exists a unique UCP map $\pi : B \to B(H)$ such that $\pi\big|_S = \phi$
2. $\pi$ is a *-representation.

Clearly, an irreducible *-representation $\pi : B \to B(H)$ is a boundary representation for $S$ if and only if $\pi\big|_S$ has the UEP.

Remark: One might imagine a situation where a UCP map $\phi : S \to B(H)$ has a unique UCP extension $\tilde{\phi} : B \to B(H)$, but $\tilde{\phi}$ is not necessarily a *-representation. It might have made sense to say that this kind of map has the UEP, but the notion has been defined as above (bby Arveson), and we stick with this definition. The alternative notion that we suggest has not been explored much.

How does one prove that a UCP map has the UEP? A very interesting solution, due to Dritschel and McCullough, and inspired by earlier work of Muhly and Solel, is the observation that the UEP is equivalent to a certain extremal property.

To define this extremal property, we need the notion of the dilation order on UCP maps. If $\phi : S \to B(H)$ and $\psi : S \to B(K)$, then we say that $\psi$ is a dilation of $\phi$, and that $\phi$ is a compression of $\psi$, if $H \subseteq K$ and $\phi(a) = P_H \psi(a) \big|_H$, for all $a \in S$. We then write $\phi \leq \psi$

A dilation $\psi \geq \phi$ is said to be trivial if there is a UCP map $\rho : S \to B(K \ominus H)$ such that $\psi = \phi \oplus \rho$

Definition 3: Let $S \subseteq B = C^*(S)$ be an operator system. A UCP map $\phi : S \to B(H)$ is said to be maximal if every UCP dilation of $\phi$ is a trivial dilation.

The following proposition connects between maximal maps and ones with the UEP.

Proposition 4: A UCP map $\phi : S \to B(H)$ is maximal if and only if it has the unique extension property.

Proof: Let $\phi : S \to B(H)$ be a maximal UCP map. We wish to prove that $\phi$ has the UEP. Let $\pi : B \to B(H)$ be a UCP extension of $\phi$ to $B = C^*(S)$. Let $\sigma : B \to B(K)$ be a minimal Stinespring dilation of $\pi$. But then $\sigma\big|_S$ is a dilation of $\phi$, so by maximality, $\sigma = \phi \oplus \rho$, so $H$ is invariant under $\sigma(S)$. By minimality of the Stinespring representation, $K = [\sigma(C^*(S)) H] = [C^*(\sigma(S)) H] = H$. It follows that $K = H$, that is $\sigma = \pi$, so $\pi$ is already multiplicative and $\phi$ has the UEP.

Conversely, assume that $\phi : S \to B(H)$ has the UEP. Suppose that $\psi : S \to B(K)$ is a dilation. Our goal is to prove that $\psi$ is a trivial dilation. Without loss of generality, we may assume that $\psi$ is a minimal dilation, in the sense that $[C^*(\psi(S)) H] = K$ (because if it wasn’t, the space on the LHS would be reducing for $\psi(S)$ and we could restrict attention to there).

Now let $\tilde{\psi} : B \to B(K)$ be an extension of $\psi$ to $B$, which we know exists, thanks to Arveson’s extension theorem. Now $B \ni a \mapsto P_H \tilde{\psi}(a)\big|_H$ is a UCP map from $B$ to $H$ which restricts to $\phi$, so by the UEP, the map $B \ni a \mapsto P_H \tilde{\psi}(a)\big|_H$ is multiplicative. For every $a \in B$, we have (using the Kadison-Schwarz inequality)

$P_H \tilde{\psi}(a)^* P_H \tilde{\psi}(a) \big|_H = P_H \tilde{\psi}(a^*a) \big|_H \geq P_H \tilde{\psi}(a)^* \tilde{\psi}(a)\big|_H$

Write $X = (I_K - P_H) \tilde{\psi}(a)\big|_H$. Then, after moving all the terms to one side, the above inequality can be rewritten as $X^* X \leq 0$, which means that $X = 0$, and consequently $H$ is invariant under $\tilde{\psi(a)}$ for all $a \in B$. Therefore $K = [C^*(\psi(S))H] = H$. We have shown that every minimal dilation has to be really trivial, and consequently every dilation must be trivial.

Dritschel and McCullough proved (essentially) that every UCP map of an operator system $S$ can be dilated to a maximal UCP map. When one begins with a completely isometric and UCP map (say, the identity representation of the operator system $S$), then the maximal dilation has the UEP, and the unique extension is a *-representation whose image is $C^*_e(S)$. This is enough to prove the existence of the C*-envelope and the Shilov ideal, but it does not settle the existence of boundary representations.

Davidson and Kennedy used the idea in Dritschel and McCullough’s proof to show the existence of sufficiently many boundary representation by dilating certain UCP maps to maximal dilations. This will be our topic in the next few sections. In the final section we will use the existence of boundary representations to prove the existence of the C*-envelope and the Shilov ideal.

#### 2. Existence of boundary representations – the main idea

By Proposition 4, we know that every maximal UCP map $\phi : S \to B(H)$ extends to a *-representation $\pi$ of $B = C^*(S)$ on $H$. To be a boundary representation, $\pi$ must be irreducible. How can we tell, looking at $\phi$, whether $\pi$ will turn out to be irreducible? The following notion, introduced by by Arveson in 1969, was shown by Davidson and Kennedy more than four decades later to be of key importance.

Definition 5: A CP map $\phi : S \to B(H)$ is said to be pure if the only CP map $\psi : S \to B(H)$ satisfying $0 \leq_{cp} \psi \leq_{cp} \phi$ (that is, $\psi$ and $\phi - \psi$ are CP) is a scalar multiple of $\phi$, i.e., $\psi = t \phi$

Examples: The identity map of of the $2 \times 2$ matrices $M_2$ is pure. The identity map of the diagonal $2 \times 2$ matrices $D_2$ is not pure.

Proposition 6: Every pure maximal UCP map $\phi : S \to B(H)$ extends to a boundary representation.

Proof: As we noted before Definition 5, Proposition 4 implies that every maximal $\phi$ extends to a *-representation $\pi$. We need to show that the pureness assumption implies irreducibility of $\pi$. So let’s show that if $\pi$ is not irreducible, then $\phi$ is not pure.

If $\pi$ is not irreducible, there is a subspace $M \subseteq H$ that reduces $\pi(B)$. Define $\psi(a) = P_M \phi(a) \big|_M$ for all $a \in S$. Then $\psi$ is clearly CP, and (using the fact that $P_M$ commutes with $\phi(a)$ for all $a \in C^*(S)$) we verify that $\phi - \psi$ is CP. But $\psi(1) = P_M$ is a non-trivial projection and not a scalar multiple of $I_H = \phi(1)$. Thus $\phi$ is not pure, and the proof is complete.

The simple Proposition is the key to Kennedy and Davidson’s approach to showing that there exist sufficiently many boundary representations. There are two steps that now need to be carried out:

1. Show that there exist sufficiently many pure UCP maps to completely norm $S$
2. Show that every pure UCP map can be dilated to to a maximal pure UCP map.

We will carry out these steps in the next sections.

#### 3. Dilation pure UCP maps to maximal pure UCP maps

The following “local” version of maximality facilitates the proof of the main result of this section.

Definition 7: Let $S$ be a an operator system. A UCP map $\phi : S \to B(H)$ is said to be maximal at $(a,h) \in S \times H$ if for every dilation $\psi \geq \phi$, we have $\phi(a) h = \psi(a) h$.

It is easy to see that $\phi$ is maximal at $(a,h)$ if and only if $\|\psi(a)h\| = \|\phi(a)h\|$ for every dilation $\psi \geq \phi$. Moreover, using that $S$ is selfadjoint, we see that $\phi$ is maximal if and only if it is maximal at $(a,h)$ for all $(a,h) \in S \times H$; it even suffices to know maximality at a dense subset of $S \times H$

Lemma 8: Let $\phi : S \to B(H)$ be a UCP map and $(a,h) \in S \times H$. Then $\phi$ can be dilated to a UCP map $\psi : S \to B(H \oplus \mathbb{C})$ such that $\psi$ is maximal at $(a,h)$

Proof: Consider the supremum

$\sup\{\|\rho(a)h\| : \rho \geq \phi\}$

The supremum is finite, since $a$ and $h$ are held fixed, and $\rho$ is assumed UCP, so in particular it is contractive. If we replace every $\rho$ in the supremum with its compression to the space spanned by $H$ and $\rho(a)h$, then the same supremum is achieved. So we may just consider

$\sup\{\|\rho(a)h\| : \phi \leq \rho : S \to B(H \oplus \mathbb{C})\}$

By compactness of $UCP(S,B(H \oplus \mathbb{C}))$ in the BW topology, the supremum is attained at some $\psi \geq \phi$, which is the sought after dilation.

The real tricky part, and a key to Davidson and Kennedy’s approach, is to show that a pure map can be dilated to a pure one which is maximal at a given pair:

Proposition 9: Let $\phi : S \to B(H)$ be a pure UCP map and $(a,h) \in S \times H$. Then for every $(a,h) \in S \times H$, $\phi$ can be dilated to a pure UCP map $\psi : S \to B(H \oplus \mathbb{C})$ such that $\psi$ is maximal at $(a,h)$

Proof: Fix $(a,h)$ as in the statement of the proposition. Define

$Y = \{\psi \in UCP(S,B(H \oplus \mathbb{C})) : \psi \geq \phi\}$

and

$X = \{\psi \in Y : \psi (a)h = \phi(a)h \oplus \eta \}$

where

$\eta = \sqrt{\sup\{\|\psi(a)h\|^2 : \psi \in Y\} - \|\phi(a)h\|^2}$.

By the lemma $X$ is non-empty, and it is clear that it is a BW-compact and convex face of $Y$. Thus, $X$ has an extreme point $\psi_0$, which is also an extreme point of $Y$

It is clear that $\psi_0$ is maximal at $(a,h)$. The point is that this extreme point $\psi_0$ must be a pure. The details will be left out of the lecture, and I refer the interested student to take a look at the proof of Lemma 2.3 in the Davidson-Kennedy paper.

Now, the existence of a pure maximal dilation follows by a transfinite induction argument.

Theorem 10: Let $\phi : S \to B(H)$ be a pure UCP map. Then $\phi$ can be dilated to a pure and maximal UCP map.

Proof: As mentioned above, this is achieved using a transfinite induction; see the proof of Theorem 2.4 in the Davidson-Kennedy for full details. For brevity, we’ll just give the case of separable $S$ and separable $H$

Let $\{(a_n,h_n)\}_n$ be a dense sequence in $S \times H$, and write $\phi_0 = \phi$. By the previous proposition, we can find inductively, for every $n=1,2,\ldots$, a pure dilation $\phi_n$ of $\phi_{n-1} \geq \phi_0$ which is maximal at $(a_1, h_1), \ldots, (a_n,h_n)$. If we let $H_n$ be the space such that $\phi_n : S \to B(H_n)$, then we can define the direct limit of these spaces

$K_0 = \overline{\cup_n H_n}$

We can define a UCP map $\psi_0 : S \to B(K_0)$ uniquely by insisting that $\psi_0(a) = \lim_n P_{H_n} \phi_n(a) \big|_{H_n}$. It is not hard to see that $\psi_0$ is pure and maximal on $S \times H$

Are we done? No, we’re not done. To be maximal, $\psi_0$ must be maximal on $S \times K_0$. We therefore repeat the procedure above and obtain a pure UCP dilation $\psi_1 \geq \psi_0 \geq \phi_0 = \phi$, which is maximal on $S \times K_0$. Repeating this infinitely many times, we find a sequence $\psi_n \in UCP(S, B(K_n))$ such that $\psi_n$ is maximal on $S \times K_{n-1}$. Finally, defining

$K = \overline{\cup_n K_n}$

and $\psi(a) = \lim_n P_{K_n}\psi_n(a)\big|_{K_n}$, we have a pure UCP map $\psi \in UCP(S,B(K))$ that dilates $\phi$ and is maximal on $S \times K$. This completes the proof.

#### 4. Abundance of pure UCP maps and boundary representations

In the first version of their paper, Davidson and Kennedy proved the following theorem.

Theorem 11: Let $S$ be an operator system. Then there are sufficiently many pure UCP maps of $S$, in the sense that for all $n$ and all $[a_{ij}] \in M_n(S)$,

(*) $\|[a_{ij}]\} = \sup\{\|\phi_n([a_{ij}]) : \phi$ is a pure UCP map $\}$

Their proof uses a characterization of pure UCP maps due to Farenick, the notion of matrix convex sets and matrix extreme points, and a Krein-Milman type theorem by Webster and Winkler. We will turn to the proof of Theorem 11 in the next lectures, since it takes time to introduce these notions.

Now if (*) holds, then dilating every pure UCP map to a maximal one (as we have done in the previous section), and then extending the maximal dilation to a boundary representation, one obtains that there are sufficiently many boundary representations, in the sense that for all $n$ and all $[a_{ij}] \in M_n(S)$

(**) $\|[a_{ij}]\| = \sup\{\|\pi_n([a_{ij}]) \| : \pi$ is a boundary representation of $S \}$

Theorem 12: Let $S \subseteq C^*(S) =B$ be an operator system. Then there are sufficiently many boundary representations for $S$ relative to $B$

In the final version of their paper, Davidson and Kennedy present also a quicker route (which they attribute to Craig Kleski) to obtaining (**). But I had some problems filling in the details. Bonus points to the student who will explain this to me!

#### 5. Existence of the Shilov boundary and the C*-envelope

Proposition 13: (Invariance principle). Let $S_i \subseteq C^*(S_i)$ , $i = 1,2$, be two operator systems, and suppose that $\theta: S_1 \to S_2$ is a unital completely isometric map of $S_1$ onto $S_2$. Then a map $\phi \in UCP(S_1,B(H))$ has the UEP if and only if $\phi\circ \theta^{-1}$ has the UEP. In particular, the boundary representations of $S_1$ and $S_2$ are in one-to-one correspondence.

Proof: Since the UEP is equivalent to maximality (Proposition 4), this is quite easy to see. Indeed, if $\psi$ is a dilation of $\phi$, then $\psi \circ \theta^{-1}$ is a dilation of $\phi \circ \theta^{-1}$, and they both have the same “shape” (i.e., break up as a direct sum or not).

We can finally show that the Shilov boundary and the C*-envelope exist.

Theorem 14: Let $S \subseteq C^*(S) = B$ be an operator system. There exists a largest boundary ideal $J$. Moreover, the quotient $C^*_e(S) := B/J$ together with the completely isometric unital quotient map $q\big|_S : S \to B/J$ have the following universal property: for every unital completely isometric map $i : S \to i(S) = S' \subseteq C^*(S') = B'$, there exits a surjective *-homomorphism $\alpha : B' \to B/J$ such that $q\big|_S = \alpha \circ i$.

Proof: Let $\pi : B \to B(H)$ be the direct sum of all boundary representations. We shall prove first that $J:= ker \pi$ is the Shilov ideal. Because of (**), $\pi$ is completely isometric on $S$, and since $B/J \cong \pi(B)$, the quotient map $q_J$ is completely isometric on $S$. Thus $J$ is a boundary ideal.

Let $I \triangleleft B$ be a boundary ideal. Our goal is to prove that $I \subseteq J$. Now, the quotient map $q_I$ is completely isometric on $S$. The map $\psi : q_I(a) \mapsto \pi(a)$ ($a \in S$) is a well defined UCIS – hence also UCP – map from $q_I(S)$ onto $\pi$. The direct sum of maps with UEP also has UEP, and thus $\pi$ has UEP. By the previous proposition, $\psi$ extends uniquely to a representation $\sigma : B/I \to \pi(B) \subseteq B(H)$. In other words, we obtain the factorization $\sigma \circ q_I = \pi$, and it follows that $I = \ker q_I$ must be contained in $J = \ker \pi$. This shows that $J$ is the Shilov ideal.

The existence of the C*-envelope follows from the invariance principle, similarly to the above. I’ll go through the details, but the readers should at least stop here and see that they can do it themselves.

Consider the map $\psi : S' \to B(H)$ given by $\psi = \pi \circ i^{-1}$. By the invariance principle, $\psi$ has the UEP, so it extends to a *-representation $\sigma$ onto $C^*(\pi(S)) = \pi(B)$. Composing $\sigma$ with the *-isomorphism $\rho: \pi(B) \to B/J$, we obtain a surjective *-homomorphism $\alpha = \rho \circ \sigma$, and we are almost done. To finish, one needs to check that $q(a) = \alpha(i(a))$ for all $a \in S$, which readily follows from the constructions.