The multiplier algebra $M(A)$ of a $C^*$-algebra $A$ can be identified with the idealizer of $A$ in $A^{**}$. I am wondering if the same identification extends to the fourth dual (seen as the universal enveloping von Neumann algebra of the universal enveloping von Neumann algebra of $A$).
In other words, if $m \in A^{****}$ is such that $mA, Am \subset A$, is it true that $m \in A^{**}$ (under the standard identification of subalgebras $A \subset A^{**} \subset A^{****}$)?
It feels to me that this should follow from the universal property of $M(A)$, but I can't quite get things to work.
Thanks!
This is false, but it will be true if you tweak the question.
Let $H$ be a separable infinite-dimensional Hilbert space. In the following we let $K(H)$ be the compact operators on $H$, $B(H)$ the bounded operators on $H$, and $C(H) = B(H)/K(H)$ the calkin algebra. The counter-example will be $A= K(H)$. In fact, $K(H)^{**} = B(H)$ (canonically), and $B(H)^{**} = K(H)^{**} \oplus C(H)^{**} = B(H) \oplus C(H)^{\ast\ast}$ (also canonically). The inclusion of $K(H)$ into $K(H)^{****} = B(H)\oplus C(H)^{\ast\ast}$ is therefore simply the inclusion into the left direct summand. Hence $K(H)$ is an ideal in $K(H)^{\ast\ast\ast\ast}$, so the idealiser is all of $K(H)^{\ast\ast\ast\ast}$ (which is much larger than $M(K(H)) = B(H)$, for instance, $B(H)$ is $\sigma$-finite, but $K(H)^{\ast\ast\ast \ast}$ isn't since it contains $C(H)$).
What goes wrong is that $A \to A^{\ast\ast\ast\ast}$ isn't non-degenerate (there is a non-zero projection in $A^{\ast\ast\ast\ast}$ which annihilates $A$). Howevever, if you let $p$ be the support projection of $A$ inside $A^{\ast\ast\ast\ast}$, then $M(A)$ is the idealiser of $A$ in $pA^{\ast\ast\ast\ast}p$ since this is true however you embed $A$ into a von Neumann algebra $N$ in a non-degenerate way (this result goes back to the Akemann-Pedersen-Tomiyama paper from the 70's).