# Ideals in Factors

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?

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?

We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2=\{x\in M | tr(x^\ast x)<\infty\}$$ and the trace class operators: $$I_1=\{x\in M | tr(|x|)<\infty\}=I_2^\ast I_2 =\left\{\sum^n_{i=1} x_i^\ast y_i | x_i, y_i\in I_2\right\}.$$ 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 $N\otimes L^j(H)$?

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## 2 Answers

Blackadar in his textbook on operator algebras gives a complete classification of norm-closed ideals in factors. See Proposition III.1.7.11.

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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).

And I think it is correct that $I_j$ is the norm closure of $N\otimes L^j(H)$, $j=1,2$.

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