4
$\begingroup$

For a von Neumann algebra $\mathcal{A} \subseteq \mathcal{B(H)}$ where $\mathcal{B(H)}$ is the space of all bounded linear operators on the Hilbert space $\mathcal{H}$, there is a Banach space $ \mathcal{A}_∗$, called the predual of $\mathcal{A}$, such that the Banach dual of $\mathcal{A}_*$ coincides with $\mathcal{A}$ with norm topology, whereas the weak-$∗$ topology coincides with the ultra-weak topology of $\mathcal{A}$. In other words, predual of a von Neumann algebra is the Banach space of all ultraweakly continuous linear functionals on $\mathcal{A}$ In fact, Sakai showed that a von Neumann algebra can be characterized in the class of $C^∗$-algebras by this property of having predual.

Since there is an isometrically isomorphism, $\mathcal{B_1(H)^*} \cong \mathcal{B(H)}$ under the map $A \longrightarrow tr(.A)$. Predual of $\mathcal{B(H)}$ is $\mathcal{B_1(H)}$, the space of trace-class operators. That is ultra-weak topology on $\mathcal{B(H)}$ is the topology such that all the trace class operators are continuous as seminorms. For a von Neumann algebra $\mathcal{A} \subseteq \mathcal{B(H)}$ ultra-weak topology on $\mathcal{A}$ can be seen as subspace- topology induced from $\mathcal{B(H)}$. But I am rather interested in explicit description of ultra-weak topology in terms of semi-norms, that is the Banach space of all ultraweakly continuous linear functionals on $\mathcal{A}$. If there is some isomorphism between predual $\mathcal{A}_*$ and Banach space containing the space of trace-class operators $\mathcal{B_1(H)}$?

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ Yes, if ${\mathcal A}$ is a subspace of ${\mathcal B}(H)$ then ${\mathcal A}_*$ is a quotient of ${\mathcal B}(H)_*$. You can identify ${\mathcal A}_*$ with ${\mathcal B}_1(H)/{\mathcal I}$ where ${\mathcal I} = \{A \in {\mathcal B}_1(H): {\rm tr}(AB) = 0$ for all $B \in {\mathcal A}\}$. $\endgroup$
    – Nik Weaver
    Commented May 14, 2014 at 13:59
  • $\begingroup$ @NikWeaver any references for the above identification? and one thing which still confusing me is, since for a Banach spaces $A, B$ , $ A \subseteq B$ will imply $ B^* \subseteq A^*$. so shouldn't $\mathcal{B_1(H)} \subseteq \mathcal{A_*}$ $\endgroup$
    – preetinder
    Commented May 14, 2014 at 16:47
  • 1
    $\begingroup$ You're mistaken, $A \subseteq B$ implies that $A^*$ is a quotient of $B^*$. Consult any standard text on functional analysis, for instance A Course in Functional Analysis by Conway. $\endgroup$
    – Nik Weaver
    Commented May 14, 2014 at 23:42
  • 1
    $\begingroup$ Oh, it's in my book Measure Theory and Functional Analysis too ... $\endgroup$
    – Nik Weaver
    Commented May 14, 2014 at 23:42

0

Reset to default

You must log in to answer this question.