## topologies on U(H)

There are many topologies on the algebra $B(H)$ of bounded operators on Hilbert space:
the weak, strong, ultraweak (also called σ-weak), ultrastrong (also called σ-strong), and some more...

Luckily, the weak and strong topologies agree when restricted to $U(H)\subset B(H)$.
Similarly, the ultraweak and ultrastrong topologies agree on $U(H)$.

Is it true that the weak and ultraweak topologies agree when restricted to $U(H)$?

Definitions:
A generalized sequence $a_i$ is weakly, strongly, ultraweakly, ultrastrongly convergent if:
• $\langle a_i\xi,\eta\rangle\to\langle a\xi,\eta\rangle\qquad \forall \xi,\eta\in H$
• $a_i\xi\to a\xi\qquad \forall \xi\in H$
• $\langle (a_i\otimes 1)\xi,\eta\rangle\to\langle (a\otimes 1)\xi,\eta\rangle\qquad \forall \xi,\eta\in H\otimes \ell^2(\mathbb N)$
• $(a_i\otimes 1)\xi\to (a\otimes 1)\xi\qquad \forall \xi\in H\otimes \ell^2(\mathbb N)$,
respectively.
Here, $H\otimes \ell^2(\mathbb N)$ denotes the Hilbert space tensor product of $H$ and $\ell^2(\mathbb N)$.

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I am not sure how you interpret the tensor. Could you please elaborate this point? – Marc Palm May 21 2011 at 19:40
Thank you pm: it's the Hilbert space completion of the algebraic tensor product. I edited the question. – André Henriques May 21 2011 at 19:46
Okay, I did not look carefully enough ξ is an element of the tensor, I thought there was some identification l2(N)⊗H≅H going on, nevermind that one. But writing $\xi = \xi_1 \otimes \xi_2$, do we not get $< (a \otimes 1) (\xi_1 \otimes \xi_2), (\eta_1 \otimes \eta_2)>= <a \xi_1, \eta_1><\xi_2, \eta_2>$, see en.wikipedia.org/wiki/…. But that would just be scalling? – Marc Palm May 21 2011 at 20:02
@pm: You're right. But don't forget that an element of $H_1\otimes H_2$ is typically not of the form $\xi_1\otimes \xi_2$. – André Henriques May 21 2011 at 21:40

The weak and ultraweak topologies coincide on bounded subsets of $B(H)$: see section 3.5 in Pedersen's book "C*-algebras and their automorphism groups".
@Marcel: The unit ball is a bounded subset of $B(H)$, so you're not really saying anything different. – André Henriques May 21 2011 at 21:43