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Let $A$ and $B$ be $C^*$-algebras. Given $f \in B^*$, we can form the right slice map $$\iota \otimes f: A \otimes B \to A: a \otimes b \mapsto af(b)$$ which extends uniquely to a bounded linear map $$\iota \otimes f: M(A \otimes B) \to M(A)$$ that is strictly continuous on the unit ball.

Assume $X \in M(A \otimes B)$ satisfies $(\iota \otimes f)(X)=0$ for all $f \in A^*$. Can we conclude that $X=0?$

Attempt: When $B$ is unital, we can proceed as follows: if $a\in A$ $$0 = (\iota\otimes f)(X)a = (\iota \otimes f)(X(a \otimes 1_B))$$ for all $f \in A^*$, so since $X(a \otimes 1_B) \in A \otimes B$ we conclude that $X(a \otimes 1_B)=0$. Hence, $X(A \otimes B) = 0$ which implies $X=0$.

How to deal with the case that $B$ is non-unital?

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    $\begingroup$ Think about a representation $A \otimes B \subset B(H\otimes K)$. $\endgroup$ Nov 16, 2021 at 23:03
  • $\begingroup$ I would be interested to know what your definition of the slice map is? $\endgroup$ Nov 17, 2021 at 9:50
  • $\begingroup$ @MatthewDaws I think my post contains the definition. I first form the slice map $\iota \otimes f: A \otimes B \to A$ which is defined on pure tensors by $a \otimes b \mapsto a f(b)$. Routine arguments show that this really gives a bounded map defined on the minimal tensor product. Then, I define the slice map $\iota \otimes f: M(A \otimes B) \to M(A)$ to be the unique extension of the slice $A \otimes B \to A$ that is strictly continuous on bounded subsets. $\endgroup$
    – Andromeda
    Nov 17, 2021 at 11:11
  • $\begingroup$ Sorry, I meant, what reference (or if no reference, what argument) are you using to make, specifically, the "unique extension" which is "strictly continuous on bounded subsets". I think it is not true that any bounded linear map so extends? $\endgroup$ Nov 17, 2021 at 12:12
  • $\begingroup$ @MatthewDaws Call a linear map $f: A \to M(B)$ between $C^*$-algebras strict if it is strictly continuous on bounded subsets and norm-continuous. For such a map, it can be shown that such a unique extension exists. I guess if I look hard enough, I can find a reference. Now, the point is that the slice map $\iota \otimes f: A \otimes B \to A\subseteq M(A)$ satisfies this property. To see this, we may assume that $f$ is a state (because any functional is a linear combination of 4 states). Moreover, $f$ is strict. Then use that the tensor product of two strict c.p. maps is again strict. $\endgroup$
    – Andromeda
    Nov 17, 2021 at 12:53

1 Answer 1

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From comments, it seems that the OP is using the "abstract" definition of multipliers (compare below). A good reference is indeed the appendix of arXiv:funct-an/9707009. Let's use some remarks from there (bottom of page 38) to show that $\iota\otimes f:A\otimes B\rightarrow A$ is indeed strict. By Cohen--Hewitt factorisation, we can find $g\in B^*, c\in B$ with $f = cg$, and so $$ (\iota\otimes f)(a\otimes b) = f(b) a = g(bc) a = (\iota\otimes g)(a\otimes bc) $$ Thus if $(u_i)$ is a bounded net in $A\otimes B$ converging strictly to $0$, for $a\in A$ we have that $$ (\iota\otimes f)(u_i) a = (\iota\otimes g)(u_i(a\otimes c)) \rightarrow 0 $$ because $u_i(a\otimes c)\rightarrow 0$ in norm.

So we form the strict extension to $M(A\otimes B)$. If $(u_i)$ is a bounded net in $A\otimes B$ converging strictly to $X\in M(A\otimes B)$ then by definition of the strict extension (or by strict continuity), $$ (\iota\otimes f)(X) a = \lim_i (\iota\otimes f)(u_i) a \qquad (a\in A). $$ However, this is equal to $$ \lim_i (\iota\otimes g)(u_i(a\otimes c)) = (\iota\otimes g)(X(a\otimes c)). $$ So if $(\iota\otimes f)(X)=0$ for all $f$, then $(\iota\otimes g)(X(a\otimes c)) = 0$ for all $g,c$ and $a$, and so $X(a\otimes c)=0$ for all $a,c$ so $X=0$.

(I think of this as the "factorisation trick". Aside from CP maps, most examples of strict linear maps seem to feature some notion of "factorisation".)


Taka's comment was to use a representation of $A\otimes B$ on a Hilbert space. This is the "centraliser" picture of multipliers: if $A\subseteq\mathcal B(H), B\subseteq\mathcal B(K)$ acting non-degenerately, then $A\otimes B\subseteq\mathcal B(H\otimes K)$ non-degenerately and $$ M(A\otimes B) \cong \{ T\in\mathcal B(H\otimes K) : Tu, uT\in A\otimes B \ (u\in A\otimes B) \}. $$ This is independent of the representations chosen, so let's suppose that $f\in B^*$ is the restriction of $\omega_{\xi,\eta}$ to $B$. Then we have a natural notion of what $(\iota\otimes f)$ is acting on $M(A\otimes B)$: just the restriction of $\iota\otimes\omega_{\xi,\eta}:\mathcal B(H\otimes K) \rightarrow \mathcal B(H)$. Of course, you'd need to check that this gave the same definition as before. The required result is now obvious.

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    $\begingroup$ Matthew, you are on fire! $\endgroup$
    – Nik Weaver
    Nov 17, 2021 at 16:28

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