# Bounded operators on the Stinespring representation space

Let $A$ be a $C^*$-algebra and let $\phi:A\to B(H)$ be a completely positive map. The Stinespring representation theorem constructs a representation of $A$ on a Hilbert space $K$, which is constructed as follows. Define $K_0=A\otimes H$ with the inner product $$\langle a\otimes v, b\otimes w\rangle_{\phi}= \langle \phi(b^*a)v,w\rangle_H,$$ extended linearly. Then $K$ is the closure of $K_0/\{z:\langle z,z\rangle_\phi=0 \}$ in the norm induced by the inner product $\langle \cdot,\cdot\rangle_\phi$.

Given a unitary linear operator $T$ on $H$ we can consider an operator $S$ on $K$ induced by $$S(a\otimes v)=a\otimes (Tv).$$

My question is under what conditions on $T$ is $S$ well-defined and bounded?

In general, I don't want to assume that $T$ commutes with the range of $\phi$.

• If it is well-defined on $K_0$ / the null space, then it is clearly bounded, because it is bounded on $K_0$, and hence would extend to a bounded operator on $K$. – Yemon Choi Mar 11 '17 at 18:47
• What have you tried so far in order to answer the question? and what level of functional analytic background should we assume? – Yemon Choi Mar 11 '17 at 18:47
• I tried to find a norm estimate, which, on elementary tensors, should go something like this: if we know that $\langle \phi(a^*a)v,v\rangle_H=1$ then $\langle \phi(a^*a) Tv, Tv\rangle_H\le C$. I don't see how to find a uniform $C$. A decent background in functional analysis is what I am willing to claim :) – user10439561 Mar 11 '17 at 18:57
• In reference to the first comment, what I don't see is why is it bounded on $K_0$. – user10439561 Mar 12 '17 at 6:30
• Suppose that $\phi$ is a nondegenerate $*$-representation. Then the map $a\otimes v\mapsto \phi(a)v$ induces an isomorphism $K\cong H$, and the formula $S(a\otimes v) := a\otimes Tv$ gives a well-defined operator precisely when $T$ commutes with $\phi(A)$. – t.c. Mar 13 '17 at 15:52

Rather than use the proof of Stinespring, I prefer to work with the uniqueness part of the result. Given $\phi$ we can find a Hilbert space $K$, a linear map $V:H\rightarrow K$ and $\pi:A\rightarrow B(K)$ a $*$-representation with $\phi(a) = V^*\pi(a)V$. Under the assumption that $\{ \pi(a)V\xi : a\in A, \xi\in H \}$ is linearly dense in $K$, the triple $(K,\pi,V)$ is unique up to unitary equivalence.

Now, a very simple example is when $A=\mathbb C$. Let $\phi(1) = R^*R$ for some $R\in B(H)$. Then we find that $K$ is the range space of $R$, and $V$ is the corestriction of $R$ to a map $H\rightarrow K$. $\pi$ is of course the trivial rep of $\mathbb C$ on $K$.

So in this setting you are asking, given $T\in B(H)$, does there exist $S_0$ on $K$ with $S_0 R = R T$. We can extend $S_0$ to $H$, so this is equivalent to asking:

Given $R,T\in B(H)$ when is there $S\in B(H)$ with $SR = RT$?

I don't see how to "solve this", unless you have some sort of "commutes" condition. (By "solve this" I mean give some sort of non-trivial characterisation.)

Given that the easiest case seems intractable, I'm not sure the general case will be possible, unless you have further conditions.

• I was indeed worried that one cannot get away without knowing that the range of $\phi$ and $T$ commute, and this is a convincing argument. – user10439561 Mar 14 '17 at 7:21