Let $\mathcal{R}$ denote the algebra of $\mathbb{N}\times\mathbb{N}$-matrices $a_{ij}$ satisfying $$\sup_{i,j}i^k j^l |a_{ij}| <\infty$$ for all $k,l>0$. By picking an orthonormal basis for $l^2(\mathbb{N})$, we can identify $\mathcal{R}$ with a dense subalgebra of the compact operators $\mathcal{K}=\mathcal{K}(l^2(\mathbb{N}))$. Note that $\mathcal{R}$ is a Frechet algebra with respect to the topology determined by the family of seminorms $$q_{k,l}(a_{ij})= \begin{cases} \sup_{i,j}i^k j^l |a_{ij}|&\text{if }k,l>0,\\ \|a_{ij}\|_{\mathcal{B}(l^2(\mathbb{N}))}&\text{if }k,l=0. \end{cases} $$
Now let $A$ be a $C^*$-algebra, and form the tensor product $A\otimes\mathcal{K}$, noting that $\mathcal{K}$ is nuclear. I would like to make sense of the statement that $$"A\otimes\mathcal{R}\subseteq A\otimes\mathcal{K}."$$ It's not quite clear to me what this means. More precisely:
- What tensor product should we take on the left?
- The answer to this question makes me wonder whether this inclusion of tensor products holds.
But perhaps there are some nice properties that make this possible. Let me ask the following:
Question 1: What type(s) of tensor product would make the above inclusion make sense?
Question 2: If $A$ is represented faithfully on $\mathcal{B}(H)$ for some Hilbert space $H$, then does $A\otimes\mathcal{R}$ have a natural representation on $\mathcal{B}(H\otimes l^2(\mathbb{N}))$?