Coincidence of two topology on a bounded subset of a finite von Neumann algebra 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?
 A: 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.

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*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.
