# Multipliers and corners of $C^*$-algebras

Let $$A$$ and $$B$$ be $$C^*$$-algebras. Suppose that there exists a projection $$p$$ in $$\mathcal{M}(B)$$, the multiplier algebra of $$B$$, such that $$A=pBp$$. That is, $$A$$ is a corner of $$B$$.

Question: Is it true that the corner $$p(\mathcal{M}(B))p$$ contains the multiplier algebra of $$A$$ (which we view as a subalgebra of $$B$$)? In other words, does every multiplier of the corner come from the corner of the multiplier algebra?

Yes. We have $$M(A) = M(pBp) = pBp \subseteq pM(B)p.$$ The second equality holds since $$pBp$$ is a unital C*-algebra.

Corrected version: There is an obvious map $$\iota \colon pBp \hookrightarrow p M(B)p$$. This inclusion extends to the multiplieralgebra $$\bar \iota \colon M(pBp) \to pM(B)p$$. The extension exists since $$\iota$$ is non-degenerate. Indeed, if $$(e_n)$$ is an approximate unit for $$B$$, then $$pe_np \to p$$ strictly, which is the unit of $$pM(B)p$$. Furthermore, it is clear that $$\bar \iota$$ is still injective.

• Why must $pBp$ be unital if we start off with $B$ non-unital? Jun 10, 2019 at 14:22
• Ah okay, I made a mistake. It is not the case that that $pBp$ is strictly closed in $M(B)$, e.g. $\mathbb K = 1_{B(H)} \mathbb K 1_{B(H)}$. Jun 10, 2019 at 14:49
• I have corrected the proof. Thanks for pointing out. Jun 10, 2019 at 14:56