Two $C^*$-algebras with the same multiplier algebra

Question

Is it possible for two non-isomorphic $C^*$-algebras $A$ and $B$ to have the same multiplier algebra? If so, what is a simple example?

Remark: I am thinking that one example might be: $A$ is the Roe algebra on $L^2(\mathbb{R})$ and $B$ is the uniform Roe algebra on the same space. But if this is true, then there must be a simpler example.

Added later: As I mentioned in the comments to Martin's answer, I would be interested in an example where neither $A$ nor $B$ is the multiplier algebra of the other.

Answer 1

Let me just introduce a basic fact, and then produce an example.

Fact. For any pair of C*-algebras $A_1,A_2$, we have that $\mathcal{M}(A_1)\oplus\mathcal{M}(A_2) = \mathcal{M}(A_1\oplus A_2)$. (This is actually true more generally for infinite sums, see this discussion.)

An example that suits your request would be something like the following. Choose your favourite separable, infinite dimensional Hilbert space $H$ and denote by $B(H)$ and $K(H)$ the C*-algebras of bounded linear operators on $H$ and its ideal of compact operators, respectively. Then consider the ideals of $B(H) \oplus B(H)$ given by

• $A = B(H) \oplus K(H)$, and
• $B = K(H) \oplus K(H)$.

We can now use the fact above to conclude that $\mathcal{M}(A) = B(H)\oplus B(H) = \mathcal{M}(B)$ even though $A \not\cong B$.

Answer 2

You can take any non-unital C$^*$-algebra $A$ and its multiplier algebra $B$.

So, for instance, here are some pairs that foot the bill:

• $A=K(H)$, $B=B(H)$ (for infinite-dimensional $H$)

• $A=c_0(\mathbb N)$, $B=\ell^\infty(\mathbb N)$.