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Lel $B$ be a Banach algebra and give $B^{**}$ one of the Arens products in order to make it a Banach algebra. Then the canonical embedding $\kappa\colon B\to B^{\ast\ast}$ is a homomorphic embedding wrt to either Arens product.

Suppose $M$ is a maximal abelian subalgebra of $B$. Can infer from this that $\overline{\kappa(M)}^{w^*}$ is a maximal abelian subalgbra of $B^{\ast\ast}$? I suspect it need not be the case... How about when $B$ is Arens regular, that is, when the Arens products in $B^{\ast\ast}$ coincide?

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Yes, thanks for spotting this. – Cleft Feb 17 '13 at 21:00
up vote 7 down vote accepted

It need not be so. The simplest example would be the Toesplitz algebra $\cal T$. It is the ${\rm C}^\ast$-algebra generated by the unilateral shift $S$ on $\ell_2{\bf N}$, $Se_n=e_{n+1}$. The algebra ${\cal T}$ contains the ideal $\cal K$ of the compact operators, and the quotient is $\pi\colon{\cal T} \to {\cal T}/{\cal K} \cong C({\bf T})$. The diagonal subalgebra $D:=\ell_\infty{\bf N} \cap {\cal T}$ is a masa of $\cal T$, which coincides with $c_0{\bf N}+{\bf C}1$. This follows from the facts that $(c_0{\bf N})'\cap{\bf B}(\ell_2{\bf N})=\ell_\infty{\bf N}$ and that $\langle xe_n,e_n\rangle$ converges for every $x\in{\cal T}$ (to $\int_{\bf T} \pi(x)\ dm$). One has $$\bigl( D^{\ast\ast}\subset {\cal T}^{\ast\ast} \bigr) \cong \bigl( (c_0{\bf N})^{\ast\ast}\oplus {\bf C}1\subset {\cal K}^{\ast\ast} \oplus C({\bf T})^{\ast\ast} \bigr)$$ and $D^{\ast\ast}$ is not a masa.

In fact, I think the weak*-closure of a masa is never a masa in the second dual $B^{\ast\ast}$ unless the situation is very degenerate, e.g., $c_0 \subset {\cal K}$ or the masa has finite index. This is because every quasi-equivalence class of a *-representation $\pi$ of $B$ has a corresponding central projection $c([\pi])$ in $B^{\ast\ast}$, which satisfies $\pi(B)'' \cong B^{\ast\ast}c([\pi])$. That $\overline{M}^{w\ast}$ contains them all means that the quasi-equivalence class of a *-representation $\pi$ of $B$ is uniquely determined by its restriction $\pi|_M$ to $M$. This is rather rare.

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