### Topics in Operator Theory, Lecture 8: matrix convexity

In this lecture we will encounter the notion of matrix convexity. Matrix convexity is an active area of research today, and an important tool in noncommutative analysis. We will define matrix convex sets, and we will see that closed matrix convex sets have matrix extreme points which play a role similar to extreme points in analysis. As an example of a matrix convex set, we will study the set of all matrix states. We will use these notions to outline the proof that there are sufficiently many pure UCP maps, something that was left open from the previous lecture.

#### 1. Matrix convex sets

Let $V$ be a vector space (over the complex numbers, to be concrete). We can form the space of $m \times n$ matrices $M_{m,n}(V)$ with entries in $V$. If $m \leq m', n \leq n'$, then $M_{m,n}(V)$ embeds in $M_{m',n'}(V)$ in a natural way (actually, there are several). It follows that usually, it suffices to consider only the spaces of square matrices, i.e., those of the form $M_n(V) := M_{n,n}(V)$.

If $V \subseteq B(H)$ is an operator space, then we have already discussed how one obtains a natural norm and order structure on $M_n(V)$. Although usually we are interested in the case where $V$ is an operator space, this will play a small role in this lecture.

The action of the scalars on $V$ extends naturally to left and right actions of the matrices $M_n$ on $M_n(V)$, by the usual formulas of matrix multiplication.

A special case of interest is when $V$ is a finite dimensional space. Fixing a basis, we have an identification $V \cong \mathbb{C}^d$. This gives rise to a natural identification

$M_n(V) \cong M_n(\mathbb{C}^d) \cong M_n(\mathbb{C})^d$

that is, just the space of all $d$-tuples of $n \times n$ matrices. The left action of $M_n$ on $M_n(V) \cong M_n^d$ then becomes

$A \cdot (X_1, \ldots, X_d) = (AX_1, \ldots, AX_d)$

(as one easily checks) and similarly for the right action. Given $A \in M_{m,n}$, we have an action by conjugation $M_m(V) \to M_n(V)$ given by

$X \in M_m(V) \mapsto A^* \cdot X \cdot A$

If $\phi : V \to W$, then we obtain as usual a map $\phi_n : M_n(V) \to M_n(W)$ by operating entry-wise. If we view $M_n(V) = V \otimes M_n$, then this map is just $\phi \otimes id$. The complementary situation also arises: if $\phi : M_n \to W$ is a linear map, we can consider $id \otimes \phi$, which we will usually just write as $\phi$. For example, if $\phi : M_n \to M_n$ is given by multiplying from the left by a matrix $A \in M_n$, then for $X \in M_n(V)$ is

$\phi(X) = id \otimes \phi (X) = A \cdot X$

Conjugation by a matrix is also an operation of this form.

Finally, one last operation: given $X \in M_m(V)$ and $Y \in M_n(V)$, we can always form the direct sum

$X \oplus Y = \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix} \in M_{m+n}(V)$

Definition 1: A matrix convex set (in $V$ or over $V)$ is a disjoint union of $K = \sqcup K_n \subseteq \mathbb{M}(V):= \sqcup M_n(V)$ which is closed under direct sums and conjugation with isometric scalar matrices.

Being closed under direct sums just means that $X \oplus Y \in K_{m+n}$ whenever $X \in K_m, Y \in K_n$. Being closed under conjugation with isometric scalar matrices, means that whenever $A \in M_{m,n}$ satisfies $A^* A = I_n$ and $X \in K_m$, then $A^* X A \in M_n(V)$

The following exercise contains two additional equivalent definitions of matrix convex sets.

Exercise A: $K = \sqcup_n K_n \subseteq \mathbb{M}(V)$. Prove that the following are equivalent:

1. $K$ is matrix convex.
2. $K$ is closed under direct sums, and for every UCP map $\phi : M_n \to M_m$, and every $X \in K_n$, we have that $\phi(X) = id \otimes \phi(X) \in K_m$
3. For every scalar matrices $A_i \in M_{n_i,n}$, $i=1, \ldots, k$, which satisfy $\sum_i A_i^* A_i = I_n$, and for every $X^{(i)} \in M_{n_i}(V)$, $i = 1, \ldots, k$, we have that $\sum_i A_i^* X^{(i)} A_i \in K_n$. (This can be summarized by saying that $K$ is closed under matrix convex combinations).

We can equip every $M_n^d$ with the operator (row) norm $\|X\| = \|\sum X_i X_i^* \|^{1/2}$. A matrix convex set $K$ is said to be closed if $K_n$ is closed for every $n$, and it is said to be bounded if there exists $M$ such that $\|X\| \leq M$ for every $X \in K$

Example: Let $A \in B(H)^d$. The matrix range of $A$ is the set $K = \sqcup_n K_n$, where

$K_n = \{(\phi(A_1), \ldots, \phi(A_d)) : \phi \in UCP(B(H), M_n) \}$.

It is not hard to show that the matrix range of $A$ is a closed and bounded matrix convex set. In fact, every closed an bounded matrix convex set over $\mathbb{C}^d$ arises this way.

Example: Simple examples of matrix convex sets (over $\mathbb{C}$) are the matrix intervals, those matrix convex sets of the form

$\sqcup_n [\alpha I_n, \beta I_n]$,

for some $\alpha \leq \beta$, where $[\alpha I_n, \beta I_n]$ is the set of all selfadjoint $X \in M_n$ such that $\alpha I_n \leq X \leq \beta I_n$

Example: If $V$ is an operator space, then the “matrix ball” given by $K = \sqcup_n K_n$, where $K_n = \{X \in M_n(V): \|X\| \leq 1\}$, form a matrix convex set. Similarly, if $V$ is an operator system, then the “matrix cone” $P = \sqcup_n P_n$, where $P_n = \{X \in M_n(V) : X \geq 0\}$, is matrix convex.

Example: As a final example, suppose that $V$ is an operator system. Then the sets

$\mathcal{UCP}(V) = \sqcup_n UCP(V,M_n)$

where $UCP(V,M_n)$, is a matrix convex set (over the vector space $V' = B(V,\mathbb{C})$) which is often referred to as the matrix state space. Similarly if $W$ is another operator space or system, using the identification $M_n(B(V,W)) = B(V,M_n(W))$, we can discuss matrix convex sets $\mathcal{CB}(V,W)$ or $\mathcal{UCP}(V,W)$

#### 2. Effros and Winkler’s separation theorem

If $V$ is a *-vector space (meaning just that is has an involution defined on it), then $M_n(V)$ is also a *-vector space, with involution given by $[v_{ij}]^* = [v_{ji}^*]$. This allows us to define selfadjoint elements, as well as the real and imaginary parts of an element $X \in M_n(V)$. To be explicit, we define

$Re X = \frac{1}{2}(X+X^*)$ and $Im X = \frac{1}{2i}(X - X^*)$

Suppose now that $V$ is a locally convex topological vector space, and that $V'$ is its dual. (In typical situations, $V = V' = \mathbb{C}^d$ with the usual pairing). A set matrix convex set $K$ over $V$ is said to be weakly closed if it is closed with respect to the weak topology (generated by $V'$ on $V$).

Continuous linear maps $V \to M_n$ are in one-to-one correspondence with matrices of continuous linear functionals in $M_n(V')$. Effros and Winkler proved the following Hahn-Banach type theorem (see Section 5 in their paper):

Theorem 2: Let $V$ be as above, and let $K$ be a weakly closed matrix convex set over $V$ such that $0 \in K$. If $X \in M_n(V) \setminus K_n$, then there exists a continuous linear map $\phi: V \to M_n$, such that:

1. $Re \phi_n(X) \nleq I_{n\times n}$, and
2. $Re \phi_m(Y) \leq I_{m \times n}$ for all $m$ and all $Y \in K_m$

Just as Arveson’s extension theorem is a Hahn-Banach-type extension theorem, this theorem should be considered as a Hahn-Banach-type separation theorem. It is completely analogous to the fact that if $K$ is a weakly closed and convex set in $V$ which contains $0$, and if $x \notin K$, then there exists a continuous functional $f \in V'$ that separates $x$ from $K$ in the sense that $Re f(x) > 1$ and $Re f(y) \leq 1$ for all $y \in K$. The assumption that $0 \in K$ corresponds to the normalization appearing in the inequality $f(y) \leq 1$; this is obviously no big deal because we can always translate the set, and this translates the value of the functional. The assumption $0 \in K$ is no loss in the noncommutative case as well, since matrix convex sets always contain scalar points.

Example: If $V = \mathbb{C}^d$, then we identify $V'$ with $\mathbb{C}^d$ in the natural way: $b = (b_1, \ldots, b_d) \in V' = \mathbb{C}^d$ acts on $V = \mathbb{C}^d$ by $\langle a, b \rangle = a_1 b_1 + \ldots a_d b_d$. A linear map $\phi : V \to M_k$ can be identified with a tuple $B = (B_1, \ldots, B_d) \in M_k^d$, where

$\phi(a) = \sum_i a_i B_i$

The ampliation $\phi_n : M_n(V) \to M_n(M_k) = M_{nk}$ then takes the form

$\phi_n(A) = \sum_i A_i \otimes B_i$

In this setting it is convenient to discuss a bipolar theorem

Here I talked about minimal and maximal matrix convex sets, and about polar duality. (See Sections 2 and 3 in this paper for more details.)

We talked about examples: $Wmax([a,b]) = Wmin([a,b])$, and $Wmax(\overline{\mathbb{D}}) \neq Wmin(\overline{\mathbb{D}})$.

#### 3. Matrix extreme points

Definition 3: A matrix convex combination $X = \sum V_i^* X^{(i)} V_i$ is said to be proper if all $V_i$s are surjective.

It is clear why this definition is important: if $V_i$ has a proper range, then $V_i^* X^{(i)} V_i$ will vanish for certain $X^{(i)}$, and will contribute notion to the matrix convex combination.

Definition 4: A point $X$ in a matrix convex set $K$ is said to be a matrix extreme point if whenever $X = \sum V_i^* X^{(i)} V_i$ is a proper matrix convex combination, then $V_i$ are all invertible (se that $n_i = n$ for all $i$) and $X^{(i)}$ are all unitarily equivalent to $X$. We denote the set of all matrix extreme points of $K$ by $\partial K$.

It is easy to see that every matrix extreme point in $\partial K_n$ is an extreme point (in the usual sense) of $K_n$ in the linear space $M_n(V)$.

Example: $\partial (\sqcup_n [aI_n, bI_n]) = \{a,b\}$.

Example: For every matrix convex set $K$, $\partial K_1 = ext(K_1)$.

Definition 5: The closed matrix convex hull $\overline{co}(S)$ of a set $S \subseteq \mathbb{M}(V)$ is the smallest closed matrix convex set containing $S$.

The matrix convex hull $\overline{co}(S)$ is also just the closure of the set of all matrix convex combinations of elements from $S$.

The following theorem of Webster and Winkler is a noncommutative version of the Krein-Milman theorem (here is a link to their paper).

Theorem 6 (Webster-Winkler Krein-Milman theorem): Let $K \subseteq \mathbb{M}(V)$ be a bounded and closed matrix convex set. Then $\overline{co}(\partial K) = K$ (in particular, the set of matrix extreme points is not empty.

Proof: First, $\partial K$ is not empty, because $\partial K_1 = ext(K_1) \neq \emptyset$. Let us write $L = \overline{co}(\partial K)$. Then, clearly, $L \subseteq K$.

We wish to prove the converse inclusion. For this, we assume without loss of generality that $0 \in L$.

Recall how the proof of the Krein-Milman theorem roughly goes: if there is a point $X \in K \setminus \overline{co}(ext(K))$ then we can separate it from $L$ with a linear functional, say $Re f(X) > 1$ and $Re f(Y) \leq 1$ for all $Y \in L$. Then the set of points $Z \in K$ for which $Re f(Z) = Re f(X)$ is itself a weakly compact convex set, and so by the non-emptiness part of KM, it has an extreme points, which must be an extreme point of $K$ – a contradiction.

We can’t use this proof since the matrix extreme points might come from different levels (and the set of matrix extreme points in a certain level might indeed be empty – no contradiction). Thus we will separate a point $X \in K \setminus L$ from all levels of $L$ using a “matrix functionals” as in Theorem 2, and then we will try to reduce the situation somehow to the classical Krein-Milman theorem to derive a contradiction. The details are as follows.

Assume that $X \in K_n \setminus L_n$. By Theorem 2, there exists a continuous linear map $\phi: V \to M_n$, such that:

1. $Re \phi_n(X) \nleq I_{n\times n}$, and
2. $Re \phi_m(Y) \leq I_{m \times n}$ for all $m$ and all $Y \in L_m$

We define a linear functional $F : M_n(V) \to \mathbb{C}$ as we did in Notes 5 (but forgetting the normalization), by

$F([a_{ij}]) = \sum_{i,j} \phi(a_{ij})_{ij}$

or, equivalently,

$F(a \otimes T) = Trace(\phi(a)^t T)$.

A little bit of algebra shows that for every $Y \in M_m(V)$ and every pair of matrices $A,B \in M_{m,n}$, we have $A^* Y B \in M_n(V)$ and

$F(A^*Y B) = \langle \phi_m(Y)B, A \rangle$,

where in the inner product $A$ and $B$ are considered as elements of $\mathbb{C}^{m \times n}$, simply by taking the $m$ rows and putting them one after the other in one long $mn \times 1$ column (this is like deciding to identify $M_{m,n} = \mathbb{C}^n \otimes \mathbb{C}^m$).

Now, define the set

$\Delta_n(K) = \{A^* Y A : Y \in K_m, m \in \mathbb{N}, A \in M_{m,n}, \|A\|_2 = 1\}$.

(Here we write $\|A\|_2$ for the Frobenius norm $\sqrt{\sum_{i,j}|a_{ij}|^2}$).

First, we note that $\Delta_n(K)$ is convex (in $M_n(V)$). Indeed, if $A_i^* Y_i A_i \in \Delta_n(K)$ and $t \in (0,1)$, then we can write

$t A_1^*Y_1 A_1 + (1-t) A_2^* Y_2 A_2 = A^* Y A$,

where $A = \begin{pmatrix} t^{1/2} A_1 \\ (1-t)^{1/2} A_2 \end{pmatrix}$ and $Y = Y_1 \oplus Y_2$ satisfy all the requirements. This shows that $\Delta_n(K)$ is convex.

Next, we note that one can use in the above definition only $A$s that are surjective, and in particular we can use only values of $m \leq n$:

$\Delta_n(K) = \{A^* Y A : Y \in K_m, m \leq n, A \in M_{m,n}, \|A\|_2 = 1\}$.

We leave the proof of this as an exercise. From this fact it follows that $\Delta_n(K)$ is closed, too.

We shall need the following technical claim:

Claim: If $Z \in \Delta_n(K)$ is an extreme point, then there exists a matrix extreme point $Y \in \partial K_m$ for some $m$ such that $Z = A^* Y A$ (where $A$ is as in the definition of $\Delta_n(K)$).

Proof of claim: As noted above, we can choose $m \leq n$, $Y \in K_m$ and $A \in M_{m,n}$ which is surjective, so that

$Z = A^* Y A$.

We shall prove that $Y$ is a matrix extreme point in $K$. Suppose that $Y$ is given as a proper matrix convex combination:

$Y = \sum_{i=1}^k V_i^* X^{(i)} V_i$,

with $X^{(i)} \in K_{n_i}$ and $V_i \in M_{n_i,m}$ for all $i=1, \ldots, k$.

Set $t_i = \|V_i A\|^2$. Note that

$\sum t_i = \sum \|V_i A\|_2^2 = \sum Trace(A^* V_i^* V_i A) = Trace(A^* \sum V_i^* V_i A)$

$= Trace(A^* A) = \|A\|^2_2 = 1$.

Also, $t_i \neq 0$, because $V_iA$ is surjective (as a product of such), so non-zero. We therefore find $Z$ as a nontrivial convex combination (in the usual sense) of elements in $\Delta_n(X)$:

$Z = A^* Y A = \sum t_i (A^* V_i^*)/t_i^{1/2} X^{(i)} (V_i A_i)/ t_i^{1/2}$.

Now, since $Z$ was assumed to be an extreme point in $\Delta_n(X)$, we find that

$A^* Y A = (A^* V_i^*)/t_i^{1/2} X^{(i)} (V_i A)/ t_i^{1/2}$

for all $i$. Conjugating with the right inverse of $A$, we have that

(*) $Y = V_i^*/t_i^{1/2} X^{(i)} V_i / t_i^{1/2}$

for all $i = 1, \ldots, k$. We would like to deduce from (*) that $Y$ is unitarily equivalent to $X^{(i)}$ for every $i$. To achieve this, I will use some trickery I came up with, which I myself find somewhat dubious (see Webster and Winkler’s paper (and also the remark below) for a different solution).

First, I will assume that $V$ is an operator system. It is true that every weakly closed and bounded matrix convex set is “matrix affine homeomorphic” to a matrix convex set in an operator system, but since we need this theorem only for operator systems in the first place, let’s just grab this assumption without question.

Next, I will assume that $V$ is finite dimensional, say $V = \mathbb{C}^d$. I can do this, because there are only finitely many elements in $V$ that appear in (*) for $i=1, \ldots, k$ so we may as well replace $V$ by the operator subsystem that these elements generate.

Finally, I will use without proof the (easy) fact mentioned above in the first section, that every closed and bounded matrix convex set in $\mathbb{C}^d$ is the matrix range $\mathcal{W}(A)$ of some tuple $A = (A_1, \ldots, A_d) \in B(H)^d$. With this in mind, we have that there is a fixed $A \in B(H)^d$ such that

$K_n = \{\phi(A) : \phi \in UCP(B(H), M_n) \}$.

We go one step further, and throw in the identity in the tuple $A_1, \ldots, A_d$ even if it wasn’t there (say $A_1 = I$), noting that this doesn’t change the geometry of the problem. Dubious, isn’t it?

Now we have $Y = \psi(A)$ for $\psi \in UCP(B(H), M_m)$, and $X^{(i)} = \phi_i(A)$ for $\phi_i \in UCP(B(H), M_{n_i})$. The equation (*) now implies that

$\psi(\cdot) = \gamma_i^* \phi_i(\cdot) \gamma_i$

as maps on the operator system generated by $A_1, \ldots, A_d$, for every $i$, where $\gamma_i = V_i / t_i^{1/2}$. Since $\psi$ and $\phi_i$ are all UCP maps, we can plug in the identity (say $I = A_1$) and find that $\gamma_i^* \gamma_i = I_m$, so these $\gamma_i$s are isometries. But $\gamma_i$ are all surjective, hence they are unitaries, and the proof of the claim is complete.

So now we know that given any extreme point $Z \in \Delta_n(K)$ then there exists a matrix extreme point $Y \in \partial K_m$ for some $m$ such that $Z = A^* Y A$. Therefore

$Re F(Z) = Re F(A^* Y A) = \langle Re \phi_m(Y)A, A \rangle \leq \langle A, A \rangle = \|A\|_2^2 = 1$,

where we used the fact that $Y \in L$ and the assumption on $\phi$.

Since this is true for every extreme point of $\Delta_n(K)$, we find by the usual Krein-Milman theorem

$Re F \big|_{\Delta_n(K)} \leq 1$.

Now let $X \in K_n$ be from the beginning of the proof, such that $Re \phi_n(X) \nleq I_{n\times n}$. So let $D \in M_n = \mathbb{C}^n \otimes \mathbb{C}^n$ s.t.

$Re F( D^* X D) = Re \langle \phi_n(X) D, D \rangle > 1$.

This contradicts what we just found out. So no such $X \in K \setminus L$ exists, and the proof is complete.

Discussion: Recall ordered vector spaces, convex sets $K$, and the affine functions on convex sets $A(K)$. The NC generalization is operator systems, matrix convex sets, and the matrix state space.

#### 4. Existence of pure UCP maps

We are almost ready to prove the following theorem, which completes our work from the previous notes about the existence of boundary representations.

Theorem 7: Let $S \subseteq B = C^*(S)$ be an operator system. Then there exist sufficiently many pure UCP maps, in the sense that for every $n$ and every $[a_{ij}] \in M_n(S)$,

$\|[a_{ij}]\| = \sup\{\|[\phi(a_{ij})]\| : \phi : S \to B(H)$ is pure $\}$.

Before the proof, let us recall that $\phi$ is pure if whenever $\psi$ is CP and $\phi - \psi$ is CP, then $\psi = t \phi$ for some $t \in [0,1]$.

Proof: It is easy to see that

$\|[a_{ij}]\| = \sup\{\|[\phi(a_{ij})]\| : \phi \in UCP(S,B(H))$ and $H$ is finite dimensional $\}$.

Indeed, simply consider compressions. In other words (or should I say, using a different notation)

$\|[a_{ij}]\| = \sup\{\|[\phi(a_{ij})]\| : \phi \in \mathcal{UCP}(S)\}$.

Now, it is easy to see that matrix convex combinations are norm decreasing, hence thanks to Theorem 6

$\|[a_{ij}]\| = \sup\{\|[\phi(a_{ij})]\| : \phi \in \partial \mathcal{UCP}(S)\}$.

The proof will therefore be complete if the pure UCP maps in $\mathcal{UCP}(S)$ were the same as the matrix extreme points. This is in fact true (it is due to Farenick, here is a link to the relevant paper). Thus, the Theorem 9 below – which says that pure points are matrix extreme points in $\mathcal{UCP}(S)$ – completes this proof.

Remark: There is an alternative proof of the existence of pure UCP maps in Davidson and Kennedy’s paper which avoids using matrix convexity (hence avoids using Webster and Winkler’s result as well as Farenick’s). The reader s will have to fill in some details on their own.

#### 5. Characterizations of pure UCP maps

To prove Farenick’s theorem on pure UCP maps of an operator system, we shall first need to understand better pure UCP maps of C*-algebras. This task has already been carried out by Arveson in Section 1.4 of his first “Subalgebras” paper. It is interesting to note that in the beginning of that section Arveson wrote: “The results in the later parts of this section go somewhat beyond our immediate needs in this paper; we feel, however, that these results may be interesting, and that they will prove useful in future developments”.

Theorem 8: Let $B$ be a C*-algebra, and let $\phi: B \to B(H)$ be a UCP map with minimal Stinespring representation $(\pi,K,V)$. Then $\phi$ is pure if and only if $\pi$ is irreducible.

Remark: Compare this with Proposition 6 in Lecture 7.

Proof: Define

$[0, \phi] = \{\psi \in CP(B,B(H)) : 0 \leq_{cp} \psi \leq_{cp} \phi\}$.

If $T \in \pi(B)'$ is positive, then can define a CP map $\phi_T \in CP(B,B(H))$ by

$\phi_T(a) = V^* T \pi(a) V$.

By the definition of the Stinespring representation, $\phi_I = \phi$. The first part of the proof consists of proving that $[0,\phi]$ is in an order preserving one-to-one correspondence with $\{ T \in \pi(B)' : 0 \leq T \leq I\}$, via the assignment $T \mapsto \phi_T$.

Now, if $\phi \in UCP(B,B(H))$ is pure, then every $\psi \in [0,\phi]$ has the form $t\phi$ for some $t \in [0,1]$; it follows that every $T \in \pi(B)'$ satisfying $0 \leq T \leq I$ must be of the form $tI$. This implies that $\pi(B)' = \mathbb{C}I$, whence $\pi$ is irreducible.

Working the argument backwards gives the converse, and the proof is complete.

Theorem 9: For every operator system $S \subseteq B = C^*(S)$, a UCP map into $M_n$ is pure if and only if it is a matrix extreme point of $\mathcal{UCP}(S)$. Moreover, every pure UCP map $\phi : S \to M_n$ extends to a pure UCP map $\tilde{\phi} : B \to M_n$.

Proof: We shall outline the proof only in the direction which we need for the existence of boundary representations (that is, we’ll argue that matrix extreme UCP maps are pure). The converse is much simpler, is similar to an argument carried in the proof of Theorem 6, and is left as an exercise.

Assume that $\phi : S \to M_n$ is a matrix extreme point in $\mathcal{UCP}(S)$. It is required to prove that $\phi$ is pure, and that it extends to a pure UCP map $\tilde{\phi} : B \to M_n$. This will prove the “moreover” part of the theorem (thanks to the implication that we have left unproven).

Since $\phi$ is pure, its range $\phi(S)$ must be an irreducible set of operators on $\mathbb{C}^n$. Indeed, otherwise, if $q \in \phi(S)'$ is a nontrivial projection, then $\phi = q \phi + (1-q) \phi$ would be a direct sum of matrix states, in contradiction to matrix-extremality of $\phi$ (direct sums are matrix convex combinations).

Now define

$K_\phi = \{\Phi \in UCP(B, M_n) : \Phi\big|_S = \phi\}$.

This set is weakly compact and convex, so it has extreme points (by the ordinary Krein-Milman theorem). It turns out that every extreme point in $\Phi \in K_\phi$ is actually a pure UCP map. To show this, we fix $\Phi$, and consider its minimal Stinespring representation $(\pi, K,V)$. By Theorem 8, $\Phi$ is pure if and only if $\pi(B)' = \mathbb{C}I$. To show this, it suffices to prove that every strictly positive element $H \in \pi(B)'$ is a scalar multiple of the identity. Now, the fact that $\Phi$ is an extreme point in $UCP(B,M_n)$ implies that the compression map $T \mapsto V^* T V$ is injective on $\pi(B)'$ (I am omitting the proof, which involves using the definitions).

To summarize where we are in the proof: to prove that the map $\Phi \in K_\phi$ that we fixed (an extreme point that extends $\phi$), we now take a strictly positive $H \in \pi(B)'$, and we aim to prove that $V^* H V$ is a scalar. Clearly there is no loss of generality in assuming that $I - H$ is also strictly positive.

We obtain a decomposition of $\Phi$ as a sum

$\Phi = V^* \pi(\cdot)H V + V^* \pi(\cdot)(I-H) V = \psi_1 + \psi_2$.

Put $g_i = \psi_i(I)$ and $\theta_i = g_i^{-1/2}\psi_i(\cdot) g_i^{-1/2}$. Then we have

$\Phi = g_1^{1/2} \theta_1(\cdot) g_1^{1/2} + g_2^{1/2} \theta_2(\cdot) g_2^{1/2}$,

and once we restrict the domain to $S$ we obtain

$\phi = g_1^{1/2} \theta_1(\cdot) g_1^{1/2} + g_2^{1/2} \theta_2(\cdot) g_2^{1/2}$.

We have that $g_1 + g_2 = \Phi(I) = I$, so this gives $\phi$ as a matrix convex combination of $\theta_1, \theta_2$. Since $\phi$ is assumed to be a matrix extreme point, we obtain $\theta_i = u_i^* \phi(\cdot) u_i$ for some unitary $u_i$. Writing $w_i = u_i g_i^{1/2}$, we find that

$\phi = w_1^* \phi w_1 + w_2^* \phi w_2$.

This means that CP map $\Psi : a \mapsto w_1^* a w_1 + w_2^* a w_2$ agrees with the identity map on the irreducible set of operators $\phi(S)$. Recall that we proved several weeks ago that the multiplicative domain of a CP map is a C*-algebra. It follows therefore that $\Psi : a \mapsto w_1^* a w_1 + w_2^* a w_2$ is the identity CP map. By the uniqueness part of Choi’s theorem, $w_1, w_2$ must be in the linear span of $I$ (because $a \mapsto I^*a I$ is the canonical Choi-Kraus representation of the identity), and so $w_1, w_2$ are scalars. We find that $g_i = w_i^* w_i$ is a scalar, and since $g_1 = V^*HV$, we find that $H$ is a scalar, as required.

If you don’t like the multiplicative domain argument, here is a nicer one: the operator system $T = \phi(S) \subseteq M_n$ is irreducible, hence $C^*(T) = M_n$. But $M_n$ is simple, so $C^*_e(T) = M_n$. Now, by the general theory of the C*-envelope, the UCIS map $\Psi : T \to T$ must extend uniquely to a *-isomorphisms between the C*-envelopes, hence $\Psi$ uniquely extends to the identity map (roughly speaking). We will return to such considerations in more detail in the next week of lectures, when we discuss the “boundary theorem”.

Summary: Up to here, we proved that every matrix extreme point $\phi \in \mathcal{UCP}(S)$ extends to a pure map in $\mathcal{UCP}(B)$. In particular, we proved that a matrix extreme point in $\mathcal{UCP}(B)$ is pure. It remains to show that $\phi$ itself is pure. This requires some work, but since the key ideas are similar, we finish the outline of the proof at this point.