# When does Stinespring dilation yield a faithful representation?

Let $A$ be a $C^*$-algebra, $H$ a Hilbert space, $\phi: A \to B(H)$ a completely positive map. The Stinespring construction yields a triple $(K, V, \pi)$ where $K$ is a Hilbert space, $V: H \to K$ a bounded linear map with $\|V\|^2 = \|\phi(1)\|$, and $\pi: A \to B(K)$ a *-representation satisfying $V^* \pi(\cdot) V = \phi(\cdot)$. The triple $(K, V, \pi)$ becomes unique up to unitary equivalence once we impose the minimality condition $\overline{\pi(A) VH} = K$.

I'm interested in what conditions on $\phi$ are equivalent to (or at least imply) the faithfulness of the representation $\pi$. Here are a few observations:

• Clearly, faithfulness of $\phi$ implies faithfulness of $\pi$, since any operator with a nonzero compression is perforce nonzero.
• The converse fails; for instance, if $\phi$ is a vector state on $M_2(\mathbb{C})$ (so $H$ is one-dimensional), then $\phi$ is not faithful, but Stinespring (=GNS) yields the identity representation.
• It is possible for $\pi$ not to be faithful. For instance, if $A = C(X)$ for a compact Hausdorff space $X$, if $\phi$ is the state induced by a probability measure supported on a proper subset $E$ (so again $H$ is one-dimensional), and if $f \in A$ is supported on $X \setminus E$, then $\pi(f) = 0$. Similarly, if $A = A_1 \oplus A_2$ and $\phi = \phi_1 \oplus 0$, then $\pi(0,a_2) = 0$.

In general, just by following through the details of the Stinespring construction, it looks like $\pi(a) = 0$ is equivalent to $$\forall b \in A: \, \phi(b^* a^* ab) = 0$$ so that faithfulness of $\pi$ would be equivalent to the statement $$\forall a \in A: \Big( \, \forall b \in A: \, \phi(b^* a^* ab) = 0 \Big) \Rightarrow a =0$$ or equivalently, of course, $$\forall a \in A: \, a \neq 0 \Rightarrow \exists b \in A: \, \phi(b^* a^* ab) \neq 0.$$ Is there a nicer way to say this, or a way to relate it to named/more easily understood properties of $\phi$?

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You might want to rephrase you question to make it about a minimal Stinespring dilation. Otherwise, the same $\phi$ can have different $\pi$s with different properties. –  Martin Argerami Mar 27 '12 at 17:45
Two comments: firstly, you form $K$ as the completion of $A\otimes H$ for a pre-inner product $(a\otimes\xi|b\otimes\eta) = (\phi(b^*a)\xi|\eta)$. So I get a more complicated condition than the one you have. Secondly-- can you give an answer when $\phi$ is just a state? It strikes me that there is no particularly "nice" answer even in this case... –  Matthew Daws Mar 28 '12 at 13:18
@MatthewDaws Silly me, forgetting about mixed tensors! I think the formula I wrote down is correct for a state, so you're right, even in that case there may not be a simple answer. –  Dave Gaebler Mar 29 '12 at 16:10