### Topics in Operator Theory, Lecture 9: the boundary theorem

In this post, we come back to boundary representations and the C*-envelope, prove an important theorem, and see some examples. It is interesting to note that the theory has interesting consequences even for operators on finite dimensional spaces. Here is a link to a very interesting paper by Farenick giving an exposition of Arveson’s boundary theorem in the setting of operators on finite dimensional spaces.

#### 1. The implementation theorem

Theorem (the implementation theorem): Let $S_i$ be an operator system in $B_i = C^*(S_i)$, for $i=1,2$. If the Shilov ideal of both $S_i$ is trivial, then every unital and completely isometric map of $S_1$ onto $S_2$ is implemented by a *-isomorphism.

Proof: If the Shilov boundary ideals are both trivial, then $B_1, B_2$ are the C*-envelopes of $S_1, S_2$, respectively. By the universal property of the C*-envelope, the exist surjective *-homomorphisms $\pi : B_1 \to B_2$ and $\rho : S_2 \to S_1$ such that

$\pi(a) = \phi(a)$ and $\rho(\phi(a)) = a$

for all $a \in S_1$. We find that $\rho \circ \pi$ is a *-homomorphism of $B_1$ onto itself which fixes $S_1$. Hence it is the identity, and $\pi$ must be a *-isomorphism.

Example: Suppose that $A,B \in M_n$ are irreducible. If

$\|I\otimes X + A \otimes Y\| = \|I \otimes X + B \otimes Y\|$

for all $X,Y \in M_n$, then $A$ and $B$ are unitarily equivalent. In other words, the unitary equivalence class of an irreducible operator is completely determined by the “structure” of the unital operator space generated by the operator. This follows from the implementation theorem because $M_n$ is simple.

In fact, the same argument works for a pair of irreducible compact operators, since the C*-algebra generated by an irreducible compact operator is the algebra of compact operators, which is simple.

The assumption that the operator is irreducible is obviously required, since $A$, $A \oplus 0$ and $A \oplus A$ (for example) all generate completely isometric operator systems.

#### 2. The boundary theorem

Recall that a set $\mathcal{S} \subseteq B(H)$ of operators on a Hilbert space $H$ is said to be irreducible if there is no nontrivial subspace $M \subset H$ that jointly reduces all the operators in $\mathcal{S}$. If $\mathcal{S}$ is selfadjoint, then it is irreducible if and only there is no joint proper invariant subspace. An operator space, system or algebra is said to be irreducible if it is irreducible as a set.

If $K(H)$ denotes the algebra of compact operators on $B(H)$, then the quotient map $q: B(H) \to B(H) / K(H)$ is called the Calkin map. If a C*-algebra $B$ contains $K(H)$, then we abuse notation and use the same symbol $q$ to denote the quotient $q: B \to B/K(H)$, and we also call $q$ the Calkin map.

The C*-algebra of compacts $K(H)$ is evidently irreducible, and whenever an operator system $S$ is such that the C*-algebra $K(H) \subseteq C^*(S)$, then $S$ is also irreducible.

Let $S$ be an operator system. It is of interest to determine when $S$ has trivial Shilov boundary ideal in $C^*(S)$. Here is one easy criterion: if the identity map is a boundary representation, then the Shilov boundary ideal (being the intersection of the kernels of all boundary representations) must be trivial. The converse is also true when $C^*(S)$ contains the compacts (I leave this as an exercise).

The following is Arveson’s boundary theorem, which gives a usable criterion for when the identity is a boundary representation. We first present it as presented in the second “Subalgebras” paper.

Theorem (the boundary theorem): Let $S \subseteq B = C^*(S)$ be an operator system in $B(H)$ such that $K(H) \subseteq B$. The identity map is a boundary representation for $S$ in $B$ if and only if the restriction of the Calkin map to $S$ is not completely isometric.

The following relatively-simple-but-ingenious proof that we will present is due to Davidson. Arveson’s original proof was much more involved, and used stuff from the theory of von Neumann algebras (it is always beautiful to see proofs that make use of tools somewhat unrelated to the problem at hand, but it is always best – especially when teaching – to be able to explain things using the simplest and most relevant tools). A recent preprint by Hasegawa and Ueda presents another proof. Below we will give another very simple proof, that became available only rather recently, after it was proved that there always exist many boundary representations.

Before the proof, note that the assumption that $K(H) \subseteq B$ implies that $S$ is irreducible. If $S$ is assumed irreducible, then if the quotient map is not isometric on $S$ it means that $K(H) \subseteq B$. Indeed, $q$ is not isometric on $B$, whence it cannot be injective, so $B$ contains a compact operator. By irreducibility, $B$ contains all compact operators.

Davidson’s proof: Suppose first that $q$ is completely isometric on $S$, then we can define a completely isometric and unital map

$q(a) \in q(S) \mapsto a \in S \subseteq B(H)$,

which, by Arveson’s extension theorem, extends to a UCP map $\psi : B(H)/K(H) \to B(H)$. Now, $\psi \circ q : B \to B(H)$ is a UCP map that fixes $S$ but annihilates $K(H)$, so it is not the identity representation. Hence, the identity $id_S$ has more than one UCP extension to $B$.

For the converse, we assume that $q$ is not isometric, and prove that this implies that every UCP extension of $id_S$ to $B(H)$ is the identity. The assumption that $q$ is not isometric, is stronger than the assumption that $q$ is merely not completely isometric, and we leave it to the reader to explain why there is no loss of generality in this assumption.

So suppose that $q$ is not isometric, and let $\phi : B \to B(H)$ be a UCP extension of $id_S$ to $B(H)$. We need to show that $\phi$ is the identity map.

Let $(\pi,V,K)$ be the minimal Stinespring dilation of $\phi$. We will make use of basic facts regarding representations of C*-algebras containing an ideal. The representation $\pi : B \to B(K)$ breaks up into a direct sum on $K = K_a \oplus K_s$ as

$\pi = \pi_a \oplus \pi_s$

(here the subscript “a” stands for “analytic” and the subscript “s” stands for “singular”), such that $\pi_s$ annihilates the compacts and $\pi_a$ is a direct sum of the identity representation.

To obtain the split, one defines $K_a = [\pi(K(H)) K]$ and proceeds from there.

The explanation why $\pi_a$ is a direct sum of the identity representation has two main parts:

1. Every non-degenerate representation of the compacts breaks up into a direct sum of representations unitarily equivalent to the identity representation. We have explained this in class for representations of $M_n$ on finite dimensional spaces; here the explanation is similar (making use of matrix units, or rank one operators), and needs to throw in a Zorn Lemma argument.
2. By the above, $\pi_a\big|_{K(H)} = \oplus \pi_i$ where every summand is equivalent to the identity. Now one observes that every $\pi$ extends uniquely to $B$ by the formula
$\pi(b) \pi_i(k) h_i = \pi(b) \pi(k) h_i = \pi(bk) h_i = \pi_i(bk) h_i$.

For more details, confer either Davdison’s “C*-Algbegras by Example” or Arveson’s “An Invitation to C*-Algebras”.

Now let $T \in S$ such that $\|q(T)\| < \|T\|$. If $|T| = (T^*T)^{1/2}$, then $\|T\| = \||T|\|$, and we find that $q(|T|) < \||T|\|$. By considering the spectral projection of $[q(|T|)+\epsilon, \||T|\|]$, which must be compact and hence finite dimensional, we obtain a nonzero finite dimensional subspace

$E = \{h \in H : \|Th\| = \|T\| \|h\| \} = SpecProj\{\|T\|\}$.

Now we carry out a little computation. Fix $h \in E$, and write $Vh = (\oplus_i x_i) \oplus x_s$. Then

$\|T h\|^2 = \|\phi(T)h\|^2 = \|V^*\pi(T)Vh\|^2 = \|V^* ( \oplus_i Tx_i \oplus \pi_s(T) x_s\|^2$

$\leq \sum_i \|T x_i\|^2 + \|\pi_s(T) x_s\| \leq \|T\|^2 ( \sum\|x_i\|^2 + \|x_s\|^2) = \|T\|^2 \|h\|^2$.

Since $\|Th\| = \|T\| \|h\|$, we must have equality throughout. Using the fact that $\|\pi_s(T)\|<\|T\|$ (because $\|q(T)\| < \|T\|$ and $\pi_s(T) = \pi_s(T+k)$ for every $k \in K(H)$), we obtain, first that $VE \subseteq H_a$, and second, that $VE \subseteq \oplus_i E$.

We have found a nonzero finite dimensional space $E$ such that $VE \subseteq \oplus_i E$. Interestingly, at this point we can forget about the non-isometricness of $q$, we discard $T$, and the proof proceeds in an ingenious way.

Let $N \subseteq E$ be a minimal nonzero subspace with the property that $VN \subseteq \oplus_i N$. We define

$\Gamma = \{b \in B : V b h = \pi(b) Vh$ for all $h \in N \}$.

Note that $\Gamma$ is a closed subspace that contains the identity.

Claim: If $a \in S$ and $b \in \Gamma$, then $ab \in \Gamma$.

Assuming the claim for the moment (and in order to immediately justify this definition), let’s see how it implies that $\phi$ must be the identity. Clearly, if the claim holds, then $\Gamma = B = C^*(S)$. If $a,b \in B$, then for all $h \in N$,

$ab h = V^*V ab h = V^* \pi(ab) V h = V^* \pi(a) \pi(b) V h = V^* \pi(a) V bh = \phi(a) bh$.

But don’t forget that $B$ is irreducible, so elements of the form $bh$, where $b \in B$ and $h \in N$ must span the whole space. It follows that $\phi(a) = a$, and this concludes the proof, modulo the claim.

To prove the claim, we fix $a \in S$ and $b \in \Gamma$. Define

$N_0 = \{h \in N : \|abh \| = \|(ab)\big|_N\| \|h\|\}$.

$N_0$ is a nonzero subspace of $N$ (for the same reasons that $E$ defined above is a subspace). For every $h \in N_0$, we have

$\|abh\| = \|\phi(a) b h\| = \|V^*\pi(a) V b h\| = \|V^*\pi(a) \pi(b) V h\| =$

$= \|V^*\pi_a(ab)V h\| \leq \|\oplus_i (ab)\big|_N V h\| \leq \|(ab)\big|_N\| \|h\| = \|abh\|$.

Since we have equality throughout, it must be that $VN_0 \subseteq \oplus_i N_0$, and by minimality $N_0 = N$. The inequalities also give

$\|\pi(ab) Vh\| = \|V^* \pi(ab) Vh\| = \|VV^* \pi(ab) h\|$

so $\pi(ab)h$ must be in the range of $V$ ($=$ the range of the projection $V V^*$), and, using that $b \in \Gamma$, we find that

$\pi(ab) Vh = VV^* \pi(ab) Vh = VV^* \pi(a) \pi(b) V h = VV^* \pi(a) V b h =$

$= V \phi(a) b h = V ab h$,

and this shows that $ab \in \Gamma$, as claimed. The proof of the boundary theorem is now complete.

Modern simple proof: (Thanks to Adam Dor-On, Satish Pandey and Marina Prokhorova for pointing my attention to this).

We first recall that every representation $\pi$ of $B$ has the form

$\pi = \pi_a \oplus \pi_s$,

so that $\pi$ is a sum of identity representations and $\pi_s$ annihilates the compacts. By Davidson and Kennedy’s theorem, for every $a \in M_n(S)$,

$\|a\| = \max\{\pi^{(n)}(a) : \pi$ is a boundary rep. $\}$.

Suppose that $id$ is not a boundary representation. Then the above maximum is attained at a singular representation. If follows that $q$ is completely isometric, since every singular representation factors through the quotient $B/K(H)$.

Conversely, suppose that $id$ is a boundary representation. Then the Shilov boundary – which is the intersection of all boundary ideals, is trivial. It follows that the compacts do not constitute a boundary ideal, and so the $q$ is not completely isometric.

#### 3. Some examples

Example: Let $S_1, \ldots, S_d$ be the shift on symmetric Fock space, and let $A_d$ be the unital operator algebra that the shift generates. Note that $A_d$ is a commutative operator algebra, while $C^*(A_d)$ is not commutative and contains the compacts (showing this for $d>1$ requires some work). Using the boundary theorem, we can see that

1. The case $d = 1$: The Calkin map is isometric. It follows that the Shilov ideal of $A_1$ is equal to the compacts, so the C*-envelope is $C^*(A_1)/K(H) = C(\mathbb{T})$.
2. The case $d >2$: The Calkin map is not completely isometric, as one can see by considering the row $[S^*_1, \ldots, S^*_d]$, which has norm at least

$\|\sum_{i=1}^d S_i^* S_i 1\| = d > 1$.

On the other hand, one can show that $(S_1, \ldots, S_d)$ is essentially normal, so

$\|q(\sum S_i^*S_i)\| = \|q(\sum S_i S_i^*)\| = \|q(I)\| = 1$.

This shows that $q$ is not completely isometric. It follows that the identity is a boundary representation, and $C^*_e(A_d) = C^*(A_1)$, and in particular it is non-commutative.