The definitions I am using are:

Def: a projection $e \in M$, von Neumann algebra, is said to be purely infinite, if given any projection $p \in Z(M) \subset M$, then $pe$ is finite if and only if $pe=0$.

Def: A von Neumann algebra $M$ is said to be purely infinite, if I (the identity) is a purely infinite projection.

Fact: A projection $e$ is purely infinite if and only if $e=e_1 + e_2$ with $e_1e_2=0$ and $e_1 \sim e_2 \sim e$ (where $\sim$ is the usual equivalence relation on projections).

Now, let $M=B(\mathcal{H})$ be the von Neumann algebra of bounded operators on the separable Hilbert space $\mathcal{H}$, then we have that this is a purely infinite algebra, that is because we can imagine the identity operator as an "infinite dimensional diagonal matrix" with $1$ on the diagonal, and the it is easy to find two orthogonal projection as in the "Fact" above.

Now, what I can't understand is why the identity doesn't have no nontrivial finite subprojections? Namely: why the obvious rank-one projections (i.e. the projections on the $i-$th components of the base) are not a counter-example to the fact that the identity is purely infinite?


1 Answer 1


Because the definition requires cutting it with central projections. And $B(H)$ is a factor, so the only nonzero projection is the identity itself.

What makes a projection not properly infinite is that you cannot do halving. For instance, consider $ M=\mathbb C\oplus B(H)$, and take the identity $1\oplus I$. You cannot halve this projection, because you would need to halve $1\in\mathbb C$.

  • $\begingroup$ Oh, that is crystal clear, thank you Martin! $\endgroup$
    – HaroldF
    Jan 31, 2019 at 23:49

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