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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?

flag	asked Apr 25 2012 at 19:56 Sergio A. Yuhjtman 309●1●8

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As proven below, the answer is no. In fact, the rest of the "weak" topologies are the weak$^\ast$ topology (in the sense of Banach spaces, coming from the unique predual of $M$). Since $M$ is not reflexive, these topologies are different. – Steven Deprez Apr 26 2012 at 13:33

Matthew Daws · Accepted Answer · 2012-04-25 20:14:57Z UTC

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.

Weak topology restricted to the unitary group of a von Neumann algebra

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