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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1$\begingroup$ 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. $\endgroup$– Steven DeprezCommented Apr 26, 2012 at 13:33
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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.
answered Apr 25, 2012 at 20:14