# strict topology on multiplier algebras

Suppose $$A$$ is a $$C^*$$ algebra,$$M(A)$$ is the multiplier algebra.If $$S$$ is a subset of $$M(A)$$ which is compact for the strict topology on $$M(A)$$,is $$S$$ also a subset of $$M(M(A))$$ which is compact for the strict topology on $$M(M(A))$$?

• If $B$ is a unital algebra then $M(B) = B$. So surely $M(M(A)) = M(A)$? Or did you mean something else by this notation? – Matthew Daws Jan 10 at 20:18

So, if I understand the question, you are asking this: Let $$S$$ be a subset of $$M(A)$$ which is compact for the strict topology. (The strict topology is such that a net $$(x_i)$$ converges to $$x$$ exactly when $$x_ia\rightarrow xa, ax_i\rightarrow ax$$ in norm, for each $$a\in A$$). Now set $$B=M(A)$$, a unital algebra, and consider $$M(B)$$, which is just $$B$$. We also consider the strict on $$M(B)=B$$, which is just the norm topology (let $$a=1$$ in the above description). Is $$S \subseteq B$$ also compact in the strict (=norm) topology.
Let $$S$$ be a subset of $$M(A)$$ which is compact for the strict topology. Is $$S$$ also compact for the norm topology on $$M(A)$$?
As you might expect, the answer is "no". For a counter-example, let $$A=K(H)$$ the compact operators on a separable Hilbert space with basis $$(e_n)$$, and let $$t_n(\xi) = (\xi | e_n) e_n$$ be the rank-one projection onto the span of $$(e_n)$$. Then $$t_n =t_n^*$$ and $$t_n(\xi)\rightarrow 0$$ for each $$\xi\in H$$, and $$(t_n)$$ is bounded. It follows that $$t_n\rightarrow 0$$ in the strict topology on $$M(A) = B(H)$$. Set $$S = \{ t_n : n\geq 1 \} \cup \{0\}.$$ It follows that $$S$$ is compact in $$M(A)$$. However, as a subset of $$B(H)$$ with the norm topology, $$S$$ is not compact.
• @mathrookie The strict topology on a unital C$^*$-algebra is the norm topology, and $M(A)$ is unital, whether $A$ is or not. I am a bit baffled by your second sentence, because the whole point of Matthew Daws's answer is to show that the strict topology on $M(A)$ and $M(M(A))$ are different. – Robert Furber Jan 11 at 11:02
• @Robert Furber，You mean the strict topology on $M(M(A))$ is the norm topology on $M(A)$?Does there exist a sufficent and necessary condition such that every strictly compact subset of $M(A)$ is also a strictly compact of $M(M(A))$? – mathrookie Jan 11 at 16:36