minimal projections in finite dimensional von Neumann algebras

Question

The algebras I'm working with are defined as follows

Let $\mathcal{H}$ be a Hilbert space of finite dimension and denote by $\mathcal{B}(\mathcal{H})$ the bounded operators on $\mathcal{H}$. A matrix algebra $\mathcal{M}$ is a *-subalgebra of $\mathcal{B}(\mathcal{H})$.

Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras?

I need to prove the following:

The set of minimal projections in a commutative matrix algebra $\mathcal{M}$ is a finite set $\{p_1,\dots,p_k \}$ of pairwise orthogonal projections.

Note that I am working with the following definitions:

A projection $p$ is called minimal if there are no non-zero projections $q$ such that $q < p$, where $q < p$ means $q \mathcal{H} \subset p \mathcal{H}$ (strictly) and $\leq$ is defined similarly.

My proof is as follows. First I show that $\mathcal{M}$ contains minimal projections. Since $\mathcal{M}$ is finite-dimensional, we define $rank(p) = \dim p \mathcal{H}$. Because for any $p$ this is a finite number, we can either find $q \in \mathcal{M}$ such that $q < p$ or $p$ is minimal. Since $\mathcal{M}$ contains a unit, we know that $\mathcal{M}$ also has minimal projections. Now let $p$ and $q$ be two minimal projections. Because $\mathcal{M}$ is commutative, $pq$ is also a projection and $pq \leq p$ and $qp \leq q$. Because $p$ and $q$ are minimal, we then have that either $pq = 0$ or $p=(pq=qp=)q$, so $p$ and $q$ are orthogonal. I still need to show that there are only finitely many and intuitively, this seems natural because of the orthogonality and the finite-dimensionality of $\mathcal{M}$, but I can't seem to put my finger on the exact argument.

My question is this: The way I show that there are minimal projections in $\mathcal{M}$, is that correct? Because I know that there are von Neumann-algebras without minimal projections, but then, I think, you should be able to costruct an infinite sequence $(p_1,\dots)$ so that $p_{i+1} \leq p_i$, otherwise there would be a minimal projection. This is probably way beyond what I'm capable of, but is this true?

Answer 1

Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.

Your argument is correct, but there is no need to use the inclusion $\mathcal M\subset B(\mathcal H)$. Simply if $\mathcal M$ has no minimal projections, you would have a strict sequence $p_1\geq p_2\geq\cdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,\ldots $ and show that they are linearly independent.

Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $\mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.

And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-p\ne0$, then there is a minimal projection $q\leq I-p$, contradicting the maximality.