Timeline for How well do we know relative commutants in $L(\mathbb{F}_\infty)$?

6 events

when toggle format	what		by	license	comment
Aug 6, 2014 at 4:23	comment	added	Sebastien Palcoux		Typo: $\langle M ,N \rangle$ should be replaced by $\langle M ,e_N \rangle$.
Aug 1, 2014 at 17:37	comment	added	Jesse Peterson		@SébastienPalcoux: If $N \subset M$ is an inclusion of ${\rm II}_1$ factors, then $N' \cap \mathcal B(L^2(M))$ is anti-isomorphic to the basic construction $\langle M, N \rangle = (JNJ)' \cap \mathcal B(L^2(M))$. This is always a semi-finite factor, and is finite if and only if $N$ is a finite index subfactor of $M$. In the case of free products, $N \subset N * B$ is finite index only in the case $B = \mathbb C$.
Jul 31, 2014 at 17:44	comment	added	Sebastien Palcoux		@JessePeterson: If $N_i$ are ${\rm II}_1$ factors, is it true that $N_1 \subset B(L^2(N_1 * N_2, \tau))$ and its commutant in $B(L^2(N_1 * N_2, \tau))$, are also ${\rm II}_1$ factors ?
Aug 4, 2012 at 23:53	vote	accept	Ollie
Aug 4, 2012 at 22:25	comment	added	Jesse Peterson		I should remark that the same proof shows that for any non-principal ultrafilter $\omega$ on $\mathbb N$, the asymptotic commutant $N_1' \cap (N_1 * N_2)^\omega$ is trivial if and only if $N_1$ is a non-amenable ${\rm II}_1$ factor. (A ${\rm II}_1$ factor is amenable if and only if the coarse bimodule contains almost central vectors.)
Aug 4, 2012 at 22:08	history	answered	Jesse Peterson	CC BY-SA 3.0