Topology on a $II_1$-factor is induced by the $\|.\|_2$-norm

Question

Let $(M,\tau)$ be a $II_1$-factor, i.e., $\tau$ is a faithful normal state on $M$ and $M$ has a trivial center.

The $\|\cdot\|_2$-norm induced by $\tau$, i.e., $\|a\|^2=\tau(a^*a)$ for $a\in M$ induced the $\sigma$-strong topology on the norm bounded subsets of $M$.

I am a bit confused by the notions of $*$-strong topology and the $\sigma$-strong topology. Is it true that a net $a_i\in M$ converges to $a\in M$ in $*$-SOT if and only if $\|a_i-a\|_{\varphi}\xrightarrow{i\to\infty} 0$ for every faithful normal state $\varphi$ on $M$? If $(M,\tau)$ is a $II_1$-factor, then the topology is induced by the $\|.\|_2$-norm associated with $\tau$?

Thank you for the help!

Answer 1

I will answer from last to first.

1. Assume that $(M,\tau)$ is a $II_1$-factor. $\tau$ is a faithful normal state on $M$. Let $(\pi_{\tau}, \mathcal{H}_{\tau}, \xi_{\tau})$ be the associated GNS construction. To briefly recall, $\mathcal{H}_{\tau}$ is the completion of $M$ with respect to the inner product $\langle \widehat{x}, \widehat{y}\rangle_{\tau}=\tau(y^*x)$. $\pi_{\tau}: M\to \mathbb{B}(\mathcal{H}_{\tau})$ is defined by $\pi_{\tau}(x)(\widehat{y})=\widehat{xy}$. $\xi_{\tau}=\widehat{1}$. Note that when I view $M$ as a subset of $\mathcal{H}_{\tau}$, I denote the elements of $M$ by putting a hat $\widehat{}$ on the top. Since $\tau$ is a normal state, $\pi_{\tau}$ is a normal representation, and $\pi_{\tau}(M)$ is a von Neumann algebra which is identified with $M$ by the faithfulness of $\tau$. Under this identification, assume that $\pi_{\tau}(x_i)\xrightarrow{SOT}\pi_{\tau}(x)$ on the unit ball of $M$. This means that $\pi_{\tau}(x_i)\xi_{\tau}\xrightarrow{\|.\|}\pi_{\tau}(x)\xi_{\tau}$ which is equivalent to saying that $\widehat{x_i}\to \widehat{x}$ in the norm induced by $\langle\cdot,\cdot\rangle_{\tau}$. This is equivalent to $$\|x_i-x\|_{\tau}=\tau\left((x_i-x)^*(x_i-x)\right)=\langle \widehat{x_i}-\widehat{x},\widehat{x_i}-\widehat{x}\rangle_{\tau}\xrightarrow{i\to\infty}0,$$ On the other hand, assume that $\|x_i-x\|_{\tau}\to0$. For any $\widehat{y}\in M$, $$\left\|\left(\pi_{\tau}(x_i)-\pi_{\tau}(x)\right)\widehat{y}\right\|_{\tau}=\left\|\left(\pi_{\tau}(x_i-x)\right)\widehat{y}\right\|_{\tau}\le \|x_i-x\|_{\tau}\|y\|_{\infty}\xrightarrow{i\to\infty}0.$$ By the density of $M$ inside $\mathcal{H}_{\tau}$, we see that

$$\left\|\left(\pi_{\tau}(x_i)-\pi_{\tau}(x)\right)\xi\right\|_{\tau}\xrightarrow{i\to\infty}0,~\xi\in\mathcal{H}_{\tau}.$$

2. Now, assume that $a_i\to a$ in $*$-SOT. Since $*$-SOT is stronger than SOT, $a_i\to a$ in SOT. Arguing similarly as in the (1), using the associated GNS construction $(\pi_{\varphi}, \mathcal{H}_{\varphi}, \xi_{\varphi})$, we see that $\|a_i-a\|_{\varphi}\xrightarrow{i\to\infty}0$ for every faithful normal state $\varphi$. Assume that $\|a_i-a\|_{\varphi}\to 0$ for every normal faithful state $\varphi$ on $M$. This implies that $a_i\to a$ in SOT. Using the Kaplansky density theorem, we can assume $\|a_i\|\le \|a\|$. On the ball $B(0, \|a\|)$, the $*$-SOT and SOT coincide and therefore, $a_i\to a$ in $*$-SOT.

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3. The $*$-strong topology ($*$-SOT) is given by the seminorms of the form $$T\to \|T\xi\|\text{ and } T\to \|T^*\xi\|,~T\in\mathbb{B}(H), \xi\in H.$$

On the other hand, the $\sigma$-strong topology (or ultra-strong topology) is induced by the seminorms $$T\to \sqrt{\sum_n\|T\xi_n\|^2}, ~(\xi_n)\subset H \text{ such that }\sum_n\|\xi_n\|^2<\infty$$

Assume that $T_m\to T$ in $*$-SOT. We want to show that $T_m\to T$ in $\sigma$-strong topology. Let $(\xi_n)\subset H$ be such that $\sum_n\|\xi_n\|^2<\infty$. Since $T_m\xrightarrow{*-SOT}T$, $T_m\xi\to T\xi$ for every $\xi\in H$. In particular, $\sup_m\|T_m\xi\|<\infty$ for every $\xi\in H$. Using the uniform boundedness principle, we see that $\sup_m\|T_m\| <\infty$. Let's call it $K$, which is a positive number. Let $\epsilon>0$ be given. Then, there exists $n_0\in\mathbb{N}$ such that $\sum_{n> n_0}\|\xi_n\|^2<\frac{\epsilon}{2(K+\|T\|)^2}$. We now observe that $$\sum_n\|(T_m-T)\xi_n\|^2\le \sum_{n\le n_0}\|(T_m-T)\xi_n\|^2+ (K+\|T\|)^2\sum_{n>n_0}\|\xi_n\|^2\le \sum_{n\le n_0}\|(T_m- T)\xi_n\|^2+\frac{\epsilon}{2} $$ Now, we can find $m_0\in\mathbb{N}$ such that $\forall m\ge m_0$, $$\|(T_m-T)\xi_n\|^2<\frac{\epsilon}{2},~\forall n\le n_0$$ The claim follows.