5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Mateusz Wasilewski Mar 21 at 12:18
  • $\begingroup$ 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.) $\endgroup$ – condexp Mar 21 at 12:30
  • 1
    $\begingroup$ 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. $\endgroup$ – Mateusz Wasilewski Mar 21 at 12:57
5
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.