In the case of bounded operators on a Hilbert space $\mathcal{H}$, $L(\mathcal{H})$, there are multiple descriptions of the $\sigma$-strong-* topology, namely:

1) As given by seminorms $p_{\phi},~p_{\phi}^{*}$, indexed by normal linear functionals on $L(\mathcal{H})$, and defined on operators $T\in L(\mathcal{H})$ by $p_{\phi}(T)=\phi(T^{*}T)^{1/2}$ and $p_{\phi}^{*}(T)=\phi(TT^{*})^{1/2}$.

2) As given by semi-norms $p_{\xi_{i}}(T),~p_{\xi_{i}}^{*}(T)$, indexed by sequences $\lbrace \xi_{i}\rbrace_{i\in\mathbb{N}}$ such that $\sum_{i} ||\xi_{i}||\leq \infty$, and defined on operators $T\in L(\mathcal{H})$ by $p_{\xi_{i}}(T)= \sum_{i\in\mathbb{N}}||T\xi_{i}||$ and $p_{\xi_{i}}^{*}(T)=p_{\xi_{i}}(T^{*})$.

3) The strict topology coming from the strict topology when $L(\mathcal{H})$ is identified with the multiplier algebra of the compact operators $C(\mathcal{H})$. More precissely given by the seminorms $p_{K},~p_{K}^{*}$, indexed by operators $K\in C(\mathcal{H})$, and defined on operators $T\in L(\mathcal{H})$ by $p_{K}(T)=||KT||$ and $p_{K}^{*}(T)=||TK||$.

The first description is the usual definition of the $\sigma$-strong-* topology on any W*-algebra, the second is the usual definition when the W*-algebra is faithfully represented on a Hilbert space.

The question I now like to ask, is: "Is there for any W*-algebra $A$ an ideal $I$ such that $M(I)=A$ and the corresponding strict topology coincides with the $\sigma$-strong-* topology?"

May be the standard form of a Von Neumann algebra can be of help, but I am not very into Tomita-Takesaki theory and I am hoping to avoid that.

  • $\begingroup$ I tweaked the formatting slightly (often one needs to enclose math-mode expressions inside a pair of ` symbols, see the "How to write math" blurb in the right-hand sidebar). $\endgroup$
    – Yemon Choi
    Aug 12, 2010 at 18:11

3 Answers 3


You might want to make clear what you mean by "ideal". If it is a von Neumann ideal, then the answer is simply no because in that case you would have $M(I)=I$.

Now, in the motivating example the ideal is an essential C*-ideal. But even then the general answer to the question has to be no. If you consider a II$_1$-factor, then it is simple as a C*-algebra, so the only nonzero ideal is $A$ itself. Of course $M(A)=A$, but the strict topology cannot agree with the $\sigma^*$-strong: because the ideal contains the identity, the strict topology in this case agrees with the norm topology.

Regarding Dmitri's answer, in a II$_\infty$-factor $M$ the "compact" ideal $K$ (norm closure of the span of the finite projections) is an essential ideal, and so $M(K)=M$ by standard results. But I don't immediately see that the strict topology in this case agrees with the $\sigma^*$-strong on bounded sets (because the key of the argument for $K(H)\subset B(H)$ is that for a finite projection $p\in B(H)$, $\|xp\|$ can be calculated over a finite-dimensional subspace).


There is a notion of compact element for any W*-algebra, namely, the two-sided ideal of compact elements is the norm-closure of the two-sided ideal of finite-rank elements, the latter being defined as the set of all elements whose source (or target) projection is finite. Suppose N=P⊕Q⊕R is the decomposition of N into its finite part P, properly infinite semifinite part Q, and non-semifinite part R. Then finite-rank(N)=P⊕finite-rank(Q) and compact(N)=P⊕compact(Q). Thus the only interesting case apart from type I_∞ case is the type II_∞ case.

Although I am not sure about this, my guess is that one should be able to prove that M(compact(N)) is homeomorphically isomorphic to N for arbitrary semifinite W*-algebra.


I don't believe that (3) characterises the $\sigma$-strong* topology. Here's what I think is a counter-example.

Let $H=\ell^2$. Consider $\ell^\infty$ acting on $H$ in the obvious way, so $c_0$ also acts on $H$ as compact operators. Let $K$ be the compact operator induced by $(1/\sqrt k)_{k\geq 1}\in c_0$. I'm going to build a net $(x_\beta)$ in $\ell^\infty\subseteq\mathcal B(H)$ such that $x_\beta\rightarrow 0$ $\sigma$-strong* but with $\|x_\beta K\|\geq 1$ for all $\beta$, so that $(x_\beta)$ does not tend strictly to 0.

To do this, observe that the $\sigma$-strong* topology, restricted to $\ell^\infty$, is given by seminorms of the form $x=(x_k) \mapsto \sum_k |x_k|^2 |b_k|$ where $(b_k)\in\ell^1$. So, given $b^{(1)},\cdots,b^{(n)}\in\ell^1$, let $c_k = \sum_{j=1}^n b^{(j)}_k$. Then $(c_k)\in\ell^1$, and if $\sum_k |x_k|^2 |c_k|$ is small, then certainly $\sum_k |x_k|^2 |b^{(j)}_k|$ is small for each $j$.

So it suffices to show that for each $(c_k)\in\ell^1$ and $\epsilon>0$, we can find $x\in\ell^\infty$ with $\sum_k |x_k|^2|c_k|<\epsilon$, but with $|x_k|/\sqrt k\geq 1$ for some $k$. We can do this by setting $x_k=0$ unless $k=N$ in which case $x_N = \sqrt N$. This follows, as if we cannot do this, then $N|c_N|\geq\epsilon$ for all $N$, and so $\sum_k |c_k| \geq \epsilon\sum_k 1/k = \infty$, a contradiction.

Did you mean to restrict to bounded sets?? Anyway, I wonder what happens for, say, $L^\infty([0,1])$; I'd be surprised if this contained non-trivial ideals which were its own multiplier algebra.

  • 1
    $\begingroup$ I think that you are correct that the assertion has to be "restricted to bounded set". As for your last sentence, I'm not completely sure what you mean: any W*-ideal will be its own multiplier algebra. $\endgroup$ Aug 15, 2010 at 20:49

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