Operator topologies on direct sum of von Neumann algebras

Question

I'm trying to understand the topological properties of infinite direct sums of von Neumann algebras (particularly in the strong operator topology).

If I take the direct sum $$ M = \bigoplus_{\lambda \in \Lambda}M_{\lambda} $$ of von Neumann algebras, I understand that algebraically one is looking at operators $\{A_{\lambda}\}_{\lambda \in \Lambda}$ such that $\sup \|A_{\lambda} \| < \infty$. Is there any easy way to understand the operator topologies on $M$ in terms of the $M_{\lambda}$? Is this just the same as taking the product topology of the $M_{\lambda}$, or something else entirely? What about the case of taking the direct sum of finite-dimensional von Neumann algebras, where the norm topology and strong operator topology are the same? How does the product topology relate (if at all) to the norm/ SOT on $M$? (in my particular case, I am actually more interested in the SOT on $U(M)$ and can I can relate it to each $U(M_{\lambda})$)

I tried considering $M$ acting on a Hilbert space and found that $\oplus A_{\lambda_i}$ convergences to $\oplus A_{\lambda}$ in SOT iff each $A_{\lambda_i} \to A_{\lambda}$ in SOT for all $\lambda \in \Lambda$ which seems to imply to me that we do have the product topology for SOT. However for norm topology we have $\oplus A_{\lambda_i} \to \oplus A_{\lambda}$ iff $\sup_{\lambda \in \Lambda} \| A_{\lambda_i} - A_{\lambda} \| \to 0$ which seems like it would be something different from the product topology. Is this correct?

Answer 1

You have to be careful with how to formulate things. For example, the strong topology (as well as the weak topology) is not inherent to the von Neumann algebra itself, but rather to the way it is represented on a Hilbert space. So, when you say the strong topology on $M=\bigoplus_\lambda M_\lambda$ is the product topology, you actually think of $M$ as being represented in a particular way (depending on the exact faithful unital representations of the $M_\lambda$).

What is inherent to the von Neumann algebra are the $\sigma$-topologies ($\sigma$-weak, $\sigma$-strong and $\sigma$-strong$^*$) and there everything works as you would expect (convergence in $M$ is the same as convergence of the factors), however you represent things.

Answer 2

One thing that should be clarified: Even for the intrinsic topologies like $\sigma$-strong etc., the topology on the direct product is not the product topology. For example, take $\ell^\infty$, which is the direct product of countably many copies of $\mathbb C$. The product topology on $\ell^\infty$ is nothing but the topology of pointwise convergence.

The sequence $(n\delta_n)_n$ converges to $0$ in the product topology: For every $k\in\mathbb N$ we have $n\delta_n(k)=0$ for $n>k$. On the other hand, $(n\delta_n)$ cannot converge to $0$ in the $\sigma$-strong topology because the sequence is unbounded (by the uniform boundedness principle, every strongly convergent sequence is bounded). You can produce similar examples in the general case as soon as infinitely many $M_\lambda$'s are non-zero.

What is true is that the $\sigma$-strong topology on the unit ball $(M)_1$ coincides with the product topology induced by the $\sigma$-strong topologies on $(M_\lambda)_1$. This is not hard to see once you realize that if $M_\lambda$ is faithfully represented on $H_\lambda$, then $M$ is faithfully represented on $\bigoplus_\lambda H_\lambda$ in the natural way.