If $A$ is unital C$^*$-algebra, is it true that the multiplier algebra of $A \otimes \mathcal{K} $ is $ A \otimes \mathcal{B}(\mathcal{H})$? Where $\mathcal{K}$ is C$^*$-algebra of compact operators on the Hilbert space $\mathcal{H}$.


If $A=C_0(X)$ and $B$ is a $C^\ast$-algebra then $M(A\otimes B)$ is the set of strictly continuous functions $\beta X\to M(B)$, where $\beta$ stands for Stone-Čech compactification.

If you take $X$ to be compact and $B=\mathcal{K}$ then we are in your setting.

But $C(X)\otimes\mathcal{B(H)}$ is the set of norm-continuos functions $\beta X=X\to\mathcal{B(H)}$. The strict topology is the $\sigma$-strong-$^\ast$ topology on $\mathcal{B(H)}$, which is different form the norm topology. This should answer your question in the negative.

  • $\begingroup$ What space $X$ do you have in mind? (It must be infinite, of course.) $\endgroup$ – Yemon Choi May 25 '16 at 15:06
  • 1
    $\begingroup$ The one-point compactification of the natural numbers should work. We can use this space to index a sequence converging in the strict topology of bounded operators, but not converging in norm. $\endgroup$ – vap May 25 '16 at 16:15

The fact stated in the answer by vap is proven in the paper "Multipliers of C*-algebras" by Akemann, Pedersen and Tomiyama (see Theorem 3.3, I guess). Moreover, they prove in Theorem 3.8 that multiplier algebras are not very well behaved with respect to minimal tensor products:

Let $A$ and $B$ be $C^*$-algebras and assume that $B$ has a countable approximate unit, but no unit (think of $\mathcal{K}$ here in your case) and that $A$ is infinite dimensional. Then
$$ M(A) \otimes M(B) \subsetneq M(A \otimes B) $$ where the tensor product is the minimal one.

So, in particular, for any infinite dimensional unital $C^*$-algebra $A$, the tensor product $A \otimes \mathcal{B}(\mathcal{H})$ is always a proper subalgebra of $M(A \otimes \mathcal{K})$.


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