2
$\begingroup$

The Sherman Takeda theorem says that the double dual $A^{\ast \ast}$ of a $C^{\ast}$ algebra $A$ can be identified as a Banach space with $A''$, the enveloping von Neumann algebra of $A$, i.e. the weak operator topology closure of $\pi_{U}(A)$ in $B(\mathcal{H}_{U})$, where $\pi_{U}$ is the universal representation of $A$ on the Hilbert space $\mathcal{H}_{U}$.

From the von Neumann density theorem we have that the weak operator topology closure of $\pi_{U}(A)$ in $B(\mathcal{H}_{U})$ is its double commutant $\pi_{U}(A)^{''}$.

Does this imply that for a unital $C^{\ast}$-subalgebra $A$ of $B(\mathcal{H})$, $A^{\ast \ast}=A^{''}$?

In particular, for a compact Hausdorff space $X$, are both $C(X)^{\ast \ast}$ and $C(X)^{''}$ equal to $L^{\infty}(X,\mu)$ for appropriate choice of $\mu$?

$\endgroup$
2
$\begingroup$

No. When you consider $A\subset B(H)$ you are considering one representation. And the universal representation is obtained by the direct sum of all the (classes of) representations of the algebra.

There are probably simpler examples than this one (separable in particular), but this is one that comes to my mind. Consider a II$_1$-factor $M$; it is a simple non-separable C$^*$-algebra. We can take an irreducible representation $\pi:M\to B(H)$, and so we end up with $A=\pi(M)$ irreducible, i.e. $A''=B(H)$. And we cannot have $A^{**}\simeq B(H)$, since it is has many ideals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.