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?


1 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.

  • $\begingroup$ Do you maybe have a hint how to show the linear independence? I can't seem to find it. $\endgroup$
    – user353840
    Aug 2, 2018 at 20:39
  • $\begingroup$ It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $\mathbb C^3$ are linearly independent. $\endgroup$ Aug 2, 2018 at 21:16
  • $\begingroup$ But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing? $\endgroup$
    – user353840
    Aug 2, 2018 at 21:44
  • $\begingroup$ That $p_2\leq p_1$. $\endgroup$ Aug 2, 2018 at 22:54
  • 1
    $\begingroup$ No thanks. I don't use chat. $\endgroup$ Aug 3, 2018 at 15:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .