Ideals in Factors - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:38:40Z http://mathoverflow.net/feeds/question/605 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/605/ideals-in-factors Ideals in Factors Dave Penneys 2009-10-15T15:58:51Z 2010-04-18T18:47:55Z <p>One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($\dim(H)=\infty$, $H$ separable) have been studied extensively. What about ideals in $II_\infty$ factors?</p> <p>One might think, since every $II_\infty$ factor $M$ can be written as $N\overline{\otimes} B(H)$ for $N$ a $II_1$ factor, if $I\subset B(H)$ is an ideal, then $N\otimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $N\otimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure? </p> <p>We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2=\{x\in M | tr(x^\ast x)&lt;\infty\}$$ and the trace class operators: $$I_1=\{x\in M | tr(|x|)&lt;\infty\}=I_2^\ast I_2 =\left\{\sum^n_{i=1} x_i^\ast y_i | x_i, y_i\in I_2\right\}.$$ What is the relation of $I_j$ to $N\otimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $N\otimes L^j(H)$?</p> http://mathoverflow.net/questions/605/ideals-in-factors/1258#1258 Answer by Dmitri Pavlov for Ideals in Factors Dmitri Pavlov 2009-10-19T18:42:47Z 2009-10-19T18:42:47Z <p>Blackadar in his textbook on operator algebras gives a complete classification of norm-closed ideals in factors. See Proposition III.1.7.11.</p> http://mathoverflow.net/questions/605/ideals-in-factors/21770#21770 Answer by Martin Argerami for Ideals in Factors Martin Argerami 2010-04-18T18:47:55Z 2010-04-18T18:47:55Z <p>Regarding your first question, $N\otimes I$ is the subalgebra of "diagonal" operators, which is already norm-closed, and it is by no means an ideal. The ideal you are looking for is the ideal generated by the finite projections, which is $N\otimes K(H)$ (that is, the norm closure of the set of "finite matrices" with entries in N). </p> <p>And I think it is correct that $I_j$ is the norm closure of $N\otimes L^j(H)$, $j=1,2$. </p>