Consider the weak topology in a von Neumann algebra (weak in the sense of Banach spaces).


Does this topology coincide with the rest of the "weak" topologies when restricted to the unitary group?



Weak topology restricted to the unitary group of a von Neumann algebra
No. Consider the algebra $\ell^\infty$. Let $w$ be an ultrafilter and define $\mu\in(\ell^\infty)^\*$ by $\mu((x\_n)) = \lim\_{n\rightarrow w} x\_n$. Now consider the sequence $x^{(n)}$ in the unitary group of $\ell^\infty$, defined by
$$x^{(n)}\_m = \begin{cases} 1 &: m\leq n, \\ -1 &:m>n. \end{cases}$$
So $\mu(x^{(n)}) = -1$ for all $n$; thus $x^{(n)}$ does not converge weakly to $1$.


However, for any $a=(a\_n) \in \ell^1$, we have that $a(x^{(n)}) = \sum\_{k\leq n} a\_k - \sum\_{k>n} a\_k \rightarrow a(1) = \sum\_k a\_k$. So $x^{(n)} \rightarrow 1$ in the $\sigma$-weak topology.


