Minimal projections and Type II von Neumann Algebras.

Question

Let $M \subseteq B(H)$ be a type $II_1$ factor. Can it contain a minimal projection? If it can't, what would go wrong? I assume something about the trace being faithful?

Answer 1

No, it can't. The argument depends on your definition of "II$_1$-factor".

Let us assume that a II$_1$-factor is an infinite-dimensional von Neumann factor with a faithful trace. Suppose that $p$ is a non-zero minimal projection. Because in a factor all projections are comparable, $p\preceq q$ for any other nonzero projection $q$. We deduce that $\tau(p)\leq\tau(q)$ for any other projection $q$, and that any projection with trace $\tau(p)$ is minimal.

Now, also because we are in a factor, there exists $q\leq1-p$ with $q\sim p$. Then $1-p-q$ is a projection, which is either zero or $p\preceq 1-p-q$. Then we would be able to find another projection $r$, equivalent to $p$, with $r\leq1-p-q$. Continuing this way and using Zorn's Lemma, we get a family of pairwise equivalent projections $\{p_j\}$, all minimal, with $\sum p_j=1$.

This family cannot be infinite, because then the identity would be infinite and there wouldn't be a finite trace. So $1=p_1+\cdots+p_n$, with $p_1,\ldots,p_n$ minimal.

Now, as $p_j$ is minimal, $p_jMp_j=\mathbb C\,p_j$. It is not hard to see that this implies that $M$ is isomorphic to $M_n(\mathbb C)$, so finite-dimensional. This contradiction shows that no minimal projection can exist.