# Properly infinite projection/von Neumann algebra

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?

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 $$1\in\mathbb C$$.