# Why are ultraweak *-homomorphisms the `right' morphisms for von Neumann algebras (and say, not ultrastrong)?

Both the ultraweak and ultrastrong topologies are intrinsic topologies in the sense that the image of a continuous (unital) $$*$$-homomorphism between von Neumann algebras (in either topology) is a von Neumann subalgebra of the target von Neumann algebra, unlike say just norm-continuous $$*$$-homomorphisms. One may then speak of the category of von Neumann algebras with morphisms as ultrastrong $$*$$-homomorphisms.

Why do we predominantly think of the ultraweak topology as the intrinsic one when presumably there could be many more topologies that are intrinsic in the above sense? I understand that the ultraweak topology is the weak-topology coming from the pre-dual and hence quite natural to study. But is there a guiding logical or category-theoretic principle that tells us to make this choice?

Thank you.

• Ultrastrongly continuous $\ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $\ast$-homomorphisms is potentially broader; might be the same, I'm not sure. – Mateusz Wasilewski Mar 21 '19 at 12:18
• The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($\mathcal{T}$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $\mathcal{T} - \mathcal{T}$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.) – condexp Mar 21 '19 at 12:30
• Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space. – Mateusz Wasilewski Mar 21 '19 at 12:57

## 2 Answers

A $$*$$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).

A von Neumann algebra is a $$C^*$$-algebra $$A$$ that admits a predual, i.e., a Banach space $$A_*$$ such that there is an isomorphism $$A\to(A_*)^*$$.

A morphism of von Neumann algebras is a morphism of $$C^*$$-algebras $$A\to B$$ that admits a predual, i.e., a morphism of Banach spaces $$B_*\to A_*$$ such that $$(A_*)^*\to (B_*)^*$$ is isomorphic to $$A\to B$$.

The weak topology induced by the predual on $$A$$ is precisely the ultraweak topology, and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.