# 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 $\phi$?

• 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

If $A$ is a von Neumann algebra and $\phi$ is normal, we can define $\textrm{car}\phi$ to be the least projection such that $\phi(\textrm{car}\phi)=\phi(1)$. Note that $\phi$ is faithful if and only if $\textrm{car}\phi=1$. Your condition is (in this case) equivalent to $C_{\textrm{car}\phi}=1$, where $C_{\textrm{car}\phi}$ is the least central projection above $\textrm{car} \phi$:

Proposition. If $\phi \colon A \to B$ is any completely positive map between von Neumann algebras, then

$$\forall a \in A. \bigl(\forall b \in A. \phi(b^*a^*ab)=0\bigr) \Rightarrow a=0 \qquad\qquad (*)$$ is equivalent to $C_{\textrm{car} \phi}=1$.

Proof. First some preparation. Write $p = \textrm{car} \phi$. By Zorn's lemma, find a maximal family of orthogonal projections $(q_i)_{i \in I}$ from $A$, each Murray-von Neumann weaker than $p$. Then $\sup_{i \in I} q_i = C_p$, see point 1 and 2 in proof of Theorem 28 of [1]. For each $q_i$ pick a $v_i$ such that $v_iv_i^*=q_i$ and $v_i^*v_i \leq p$. (As an example, if $A=B(l^2(\mathbb{N}))$ and $p=\left|0\right>\!\!\left<0\right|$, then one can pick $I=\mathbb{N}$, $v_i=\left|i\right>\!\!\left<0\right|$ and $q_i=\left|i\right>\!\!\left<i\right|$.)

Assume $C_p=1$. We will prove $(*)$. To this end, assume for each $a \in A$ and for each $b \in A$ we have $\phi(b^*a^*ab)=0$. In particular $\phi(v_i^* a^* a v_i)=0$. Without loss of generality, we may assume $a^*a \leq 1$ and then by the previous $v_i^* a^* a v_i \leq 1-p$. So $q_i a^*a q_i = v_iv_i^* a^* a v_iv_i^* \leq v_i (1-p) v_i^* = v_iv_i^* - v_iv_i^*= 0$. Hence $a^*a \leq 1-q_i$ for each $i \in I$. So $a^*a \leq 1- \sup_i q_i=1-C_p=0$. By the C$^*$-identity, $a=0$. Thus $(*)$ holds.

For the converse, assume $(*)$ holds. Clearly for each $b \in A$, we have $$b^* (1- C_p) b= (1-C_p) b^*b (1-C_p) \leq \| b \|^2 (1- C_p) \leq \| b \|^2 (1-p)$$ and so $\phi(b^* (1-C_p)^*(1-C_p)b)=0$. By $(*)$ now $1-C_p=0$ and so $C_p=1$, as desired. QED

[1] Paschke Dilations, Westerbaan & Westerbaan http://arxiv.org/pdf/1603.04353v1.pdf