strong operator topology and $L_2$ norm

Question

In many places, I saw that SOT coincides with the norm $\tau(X^*X)^{1/2}$, $X\in M$, where $M$ is a semifinite von Neumann algebra and $\tau$ is a semifinite faithful normal trace on $M$. I want a reference about this argument but I just can not find it.

Answer 1

I'm not sure if I have seen a reference for this. I guess it's "folklore"? I will answer for $M$ finite, but the argument is basically the same for semifinite (you need to believe that you can do GNS there too).

They induce the same topology on bounded sets. Say $M\subset B(H)$. What happens is that if you do GNS for $\tau$, you get a normal faithful representation $\pi_\tau:M\to B(K)$. It follows that $\pi_\tau$ is sot-sot continuous (see, for instance, Corollary 7.1.16 in Kadison Ringrose).

So, suppose that $x_j\to0$ sot in $M$. Then $\pi_\tau(x_j)\to0$ sot in $B(K)$. Thus, $$ 0=\lim \langle \pi_\tau(x_j)\Omega,\pi_\tau(x_j)\Omega\rangle_K=\tau(x_j^*x_j). $$ Conversely, if $\tau(x_j^*x_j)\to0$ then, for any $y\in M$, $$ \|\pi_\tau(x_j)y\Omega\|^2=\langle \pi_\tau(x_j)y\Omega,\pi_\tau(x_j)y\Omega\rangle_K =\tau(y^*x_j^*x_jy)=\tau(x_jyy^*x_j^*)\leq\|y\|^2\,\tau(x_jx_j^*)\to0. $$ Now, since we are assuming that $\{x_j\}$ is bounded, we can do the following: let $\xi\in K$. Fix $\varepsilon>0$. Then there exists $y\in M$ such that $\|\xi-y\Omega\|<\varepsilon. $ Thus $$ \|\pi_\tau(x_j)\xi\|\leq\|x_j\|\,\|\xi-y\Omega\|+\|\pi_\tau(x_j)y\Omega\|\leq k\varepsilon +\|\pi_\tau(x_j)y\Omega\|. $$ Then $$ 0\leq \limsup \|\pi_\tau(x_j)\xi\|\leq k\varepsilon +\limsup \|\pi_\tau(x_j)y\Omega\|=k\varepsilon. $$ As $\varepsilon$ was arbitrary, we get $\lim \|\pi_\tau(x_j)\xi\|=0$, so $\pi_\tau(x_j)\to0$ (sot). We can apply 7.1.16 in KR to $\pi_\tau^{-1}$, and so $x_j\to0$ (sot).