Showing that $B_0(H)^{**}= B(H)$

Question

Let $H$ be a Hilbert space. Consider the $C^*$-algebra of compact operators $B_0(H)$ and its double enveloping von Neumann algebra $B_0(H)^{**}$.

I want to show that $B_0(H)^{**}= B(H)$ as $C^*$-algebras.

Though I didn't check the details, we can probably proceed as follows:

Given a $C^*$-algebra $A$, there exists a unique $W^*$-algebra structure on the Banach space dual $A^{**}$ such that the canonical inclusion $i: A \hookrightarrow A^{**}$ becomes a $*$-homomorphism with weak$^*$-dense range.

Then, using that we can canonically identify $B_0(H)^{**}=B(H)$ as Banach spaces in a way such that the canonical inclusion maps agree under the identifications, it follows that $B_0(H)^{**}=B(H)$ as $C^*$-algebras.

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A possible other approach is the following: We check that the identity map $\pi: B_0(H) \hookrightarrow B(H)$ is a universal representation. Since the enveloping von Neumann algebra can be abstractly defined as the bicommutant of the image of a universal representation, we will be done once again.

Again, I didn't check all the details but at first sight this also seems to work.

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Are there other ways to see the isomorphism $B_0(H)^{**}= B(H)$ on the $C^*$-level? For example, if we follow the GNS-construction to canonically define the universal representation of $B_0(H)$ and take the von Neumann algebra generated by its image, is it clear that we get a $C^*$-algebra isomorphic to $B(H)$?

Thanks in advance.

Answer 1

The "universal" in the universal representation means the following (Takesaki I, Definition III.2.3):

$\pi:A\to B(H)$ is universal if for every representation $\rho:A\to B(K)$ there exists a $\sigma$-weak continuous $*$-homomorphism $\tilde\rho:\pi(A)''\to\rho(A)''$ such that $\rho=\tilde\rho\circ\pi$ on $A$. The universal representation is unique in the sense that the von Neumann algebra $\pi(A)''$ is uniquely determined up to isomorphism.

Consider $A=B_0(H)$ and $\pi$ the identity. Given a representation $\rho:B_0(H)\to B(K)$, we want to show that it can be extended to $\tilde\rho:B(H)\to B(K)$. The extension both exists and is unique due to the fact that $B_0(H)$ is an ideal. Indeed, by shrinking $K$ is needed we may assume that $\rho(B_0(H))K$ is dense. Then one defines $$\tag1 \tilde\rho(a)(\rho(b)k):=\rho(ab)k,\qquad a\in B(H),\quad b\in B_0(H), \quad k\in K. $$ It is not hard to check that this is well-defined, bounded (so it extends to acting in all of $K=\overline{\rho(A)K}$) and that it is indeed a representation. Any other representation $B(H)\to B(K)$ that agrees with $\rho$ on $B_0(H)$ will satisfy $(1)$, which gives uniqueness.

If we show that $\tilde\rho$ is $\sigma$-weak continuous then we can conclude that the identity representation is universal.

To show $\sigma$-weak continuity (that is, normality), it is enough to check that our map is sot countinuous on bounded sets. So suppose that $\{a_j\}\subset B(H)$, that $\|a_j\|\leq1$ for all $j$, and that $a_j\to0$ sot. Given $b$ compact, $b=\lim b_n$ with $b_n$ finite-rank. The key fact is that $a_jb_n\to0$ in norm. Then $$ \|a_jb\|\leq\|a_j(b-b_n)\|+\|a_jb_n\|\leq\|b-b_n\|+\|a_jb_n\|, $$ which implies $$ \limsup_j\|a_jb\|\leq\|b-b_n\|. $$ As this works for all $n$, we get that $a_jb\to0$. By $(1)$, $$ \tilde\rho(a_j)\rho(b)k=\rho(a_jb)k\to0. $$ For arbitrary $k\in K$, there exists a sequence $\{\rho(b_n)k_n\}\subset K$ with $\rho(b_n)k_n\to k$. Then $$ \|\tilde\rho(a_j)k\|\leq\|\tilde\rho(a_j)(k-\rho(b_n)k_n)\|+\|\tilde\rho(a_j)\rho(b_n)k_n\| \leq\|k-\rho(b_n)k_n\|+\|\tilde\rho(a_j)\rho(b_n)k_n\|. $$ Then $$ \limsup_j\|\tilde\rho(a_j)k\|\leq\|k-\rho(b_n)k_n\| $$ and we get that $\tilde\rho(a_j)k\to0$, and so $\tilde\rho(a_j)\to0$ sot, making $\tilde\rho$ normal.

Answer 2

It seems to be relatively unknown (for example I don't think it's in Takesaki's books), but the product on the double dual of a $C^\ast$-algebra is explicit:

For $\omega\in A^\ast$ and $\phi\in A^{\ast\ast}$ define $\omega_\phi\in A^{\ast}$ by $$ \omega_\phi(a)=\phi(a\omega). $$ Then the Arens product of $\phi,\psi\in A^{\ast\ast}$ is defined as $$ \phi\psi(\omega)=\psi(\omega_\phi). $$ (In fact, for a general normed algebra, there are two Arens products, but they coincide for a $C^\ast$-algebra).

Now if $\omega=\mathrm{tr}(\cdot\rho)$ for a trace-class operator $\rho$ and $\phi\in B_0(H)^{\ast\ast}$ the functional associated with $T\in B(H)$, then $$ \omega_\phi(a)=\phi(a\omega)=\phi(\mathrm{tr}(\cdot a\rho))=\mathrm{tr}(Ta\rho)=\mathrm{tr}(a\rho T), $$ that is, $\omega_\phi=\mathrm{tr}(\cdot\rho T)$.

Consequently, if $\phi,\psi\in B_0(H)^{\ast\ast}$ are the functionals associated with $S,T\in B(H)$, respectively, then $$ \phi\psi(\omega)=\psi(\omega_\phi)=\psi(\mathrm{tr}(\cdot\rho S))=\mathrm{tr}(T\rho S)=\mathrm{tr}(ST\rho). $$ Thus, $\phi\psi\in B_0(H)^{\ast\ast}$ is the functional associated with $ST$. That means that the isomorphism $B(H)\to B_0(H)^{\ast\ast}$ also preserves the product. For the involution this is easy to see, so that it is in fact a $\ast$-isomorphism.