Let $M$ be a type $II_{1}$ factor, Let $B$ is an infinite dimensional nonabelian subalgebra. Is it true that $B$ always type $II_{1}$ ?
$\begingroup$
$\endgroup$
5
-
$\begingroup$ I guess you mean von Neumann subalgebra. What if it's not a factor? If a group VN algebra take the subalgebra generated by a "small" nonabelian subgroup. $\endgroup$– YCorCommented Mar 30, 2019 at 8:49
-
$\begingroup$ yes vN subalgebra $\endgroup$– user136400Commented Mar 30, 2019 at 9:09
-
$\begingroup$ What do you mean by small group? $\endgroup$– user136400Commented Mar 30, 2019 at 10:33
-
$\begingroup$ Nothing precise, whence the quotation marks. What do you think with a nonabelian finite subgroup? $\endgroup$– YCorCommented Mar 30, 2019 at 10:47
-
$\begingroup$ Oops missed "infinite-dimensional"; but say, with a virtually abelian subgroup. $\endgroup$– YCorCommented Mar 30, 2019 at 13:48
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If $M$ is type $II_1$ then it has a tracial state, and hence so does $B$ by restriction. So if $B$ is an infinite dimensional subfactor it must be type $II_1$, by inspection of factor types. If $B$ is merely a subalgebra then it could have a type $I$ part, e.g. $\mathbb{C}p + qMq$ where $p$ and $q$ are nonzero projections with $p+q=1$.
answered Mar 30, 2019 at 12:42