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?
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Sign up to join this communityConsider 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?
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.