Strict topology on the multiplier algebra

Question

Let $A$ be a $C^*$-algebra. Let $M(A)$ be its multiplier $C^*$-algebras. The strict topology on $M(A)$ is given by $$x_\lambda \to x \iff \forall a\in A: (\|x_\lambda a-xa\| + \|ax_\lambda - ax\| \to 0 ).$$

We can identify $M(A) \cong \mathcal{L}(A)$ where $\mathcal{L}(A)$ are the adjoinable operators of the right Hilbert $A$-module $A$ with respect to the inner product $\langle a,b \rangle:= a^*b.$ The strict topology on $\mathcal{L}(A)$ is given as follows:

$$t_\lambda \to t \iff \forall a \in A: (\|t_i(a)-t(a)\| + \|t_i^*(a) -t^*(a)\| \to 0).$$

When we identify $A$ as a subset of $\mathcal{L}(A)$ via the mapping $ab^* \mapsto \theta_{a,b}$ where $\theta_{a,b}(x) = a\langle b,x\rangle$, we obtain two notions of strict topologies on $M(A)$. I guess on bounded sets these topologies agree, but are these topologies in general equal?

Answer 1

Yes, these topologies agree basically by definition once you understand the isomorphism $M(A)\cong \mathcal L(A)$ (since $tθ_{a,b}=θ_{t(a),b}$ the canonical isomorphism $A≅ \mathcal K(A)$ takes this element to $t(a)b^*=t(ab^*)$). The isomorphism $A \to \mathcal K(A)$ is given by $a \mapsto \theta_{a_1,a_2}$ where $a=a_1a_2^\ast$ is any way of writing $a$ as a product of two elements (any element in a $C^\ast$-algebra can be written as a product of two elements, e.g. if a $a= u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast\ast}$, then $a_1 := u|a|^{1/2} \in A$ (this can be seen by approximating $t\mapsto t^{1/2}$ by polynomials) and $a = a_1 |a|^{1/2}$).