Let $M\subset B(H)$ be a finite von Neumann algebra with a faithful normal trace $\tau$. There is a norm $\|.\|_\tau$ on $M$ given by $\sqrt{\tau(xx^*)}$. How to show the $\|.\|_\tau$-topology coincides with the strong operator topology on $M_1$, i.e. the operator-norm unit ball?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ This has been answered in the negative with an explicit counter example but there is a second philiosophical reaon to doubt the statement which I mention since it is a general reason to be wary of using the strong topology--involution is not continuous. It is usually better to use $s^\ast$, its symmetrisation. Even better is the finest lc topology which agrees with the latter on the ball since it is complete and has the same convergent sequences--it is even the Mackey topology with respect to the natural duality. $\endgroup$– bathalf15320Commented Feb 11, 2021 at 10:03
-
1$\begingroup$ @bathalf15320 Nik seems to have retracted his counterexample. Also, I seem to recall that in the tracial vN case things are better than for the unit ball of B(H) itself, but I may have misremembered $\endgroup$– Yemon ChoiCommented Feb 11, 2021 at 14:14
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
7
Some errors identified in the comments by Mateusz and Matt — will try to fix these later, although I believe that the overall strategy outlined here can be made to work.
It's late here so I haven't checked through the details, but I think the following is an outline of one possible approach. I'm leaving it up as an incomplete answer because that seemed a bit clearer than offering cryptic suggestions in comments.
The restriction to $M_1$ of the SOT on B(H) coincides with the restriction to $M_1$ of the ultraweak topology on $M$, which in turn is intrinsic to $M$ (i.e. independent of the choice of faithful normal unital representation $M\to B(H)$.
Therefore it suffices to prove the claim when $H=L^2(M,\tau)$ and $M\to B(H)$ is the realization of $M$ in so-called "standard form", i.e. the GNS rep corresponding to the faithful tracial state $\tau$. In this realization, $H$ has a cyclic and separating vector for $M$, which we'll denote by $\Omega$; then $\Vert x\Vert_\tau$ is equal to ${\Vert x\Omega\Vert}_H$.
In particular, a net in $M_1$ that converges in SOT must converge in $\tau$-norm. Conversely, given a net $(x_i)$ in $M_1$ that converges in $\tau$-norm, I think a Cauchy--Schwarz argument then shows that $(x_iy\Omega)$ converges in $H$ for each $y\in M$. But since $\Omega$ is a cyclic vector, a 3-epsilon argument (this is where it's vital that our net is bounded in the norm of $M$!) shows that $(x_i \eta)$ converges in $H$ for each $\eta\in H$, which is what it means to converge SOT.
answered Feb 11, 2021 at 4:25
-
3$\begingroup$ Yemon, I think you meant the ultrastrong topology in your first bullet point, not the ultraweak one? $\endgroup$ Commented Feb 11, 2021 at 9:02
-
2$\begingroup$ In think, in step 3, you need to take $y$ in the commutant of $M$. As $\Omega$ is separating, this gives you a total subset of $L^2(M,\tau)$. I am looking at Prop 2.7.7 in Evington's thesis: theses.gla.ac.uk/8650/1/2018EvingtonPhD.pdf $\endgroup$ Commented Feb 11, 2021 at 11:42
-
1$\begingroup$ How about this for step 3: let $(x_i)$ be a net in $M_1$ that converges to 0 in $\tau$-norm. We want to show it converges strongly to 0. Since $M$ is dense in $L^2(M,\tau)$ and $(x_i)$ is bounded, it's enough to show that $\|x_i y\|_\tau \to 0$ for all $y \in M$. But $\|x_i y\|_\tau^2 = \tau(x_i y y^*x_i^*) \leq \|y\|^2\tau(x_ix_i^*) = \|y\|^2\|x_i\|^2_\tau \to 0$. $\endgroup$ Commented Feb 11, 2021 at 23:48
-
2$\begingroup$ @NikWeaver, that works, but the argument with the commutant can also be used in the non-tracial setting: the L^2-topology induced by a faithful normal state and the SOT agree on bounded subsets. $\endgroup$ Commented Feb 12, 2021 at 13:13
-
1$\begingroup$ @YemonChoi The ultrastrong topology is also independent of the representation since $x_i\to x$ ultrastrongly iff $(x-x_i)^\ast (x-x_i)\to 0$ ultraweakly. $\endgroup$– MaoWaoCommented Feb 13, 2021 at 6:29