Projection lattice to operator algebra

Question

Suppose I have a lattice of projections on some (finite dimensional) Hilbert space. Can I construct an algebra of operators of which this lattice is the set of all projections in the algebra? If not, are there any extra conditions on the lattice such that in would suffice?

Answer 1

The answer in general is a big "no". Consider $M_2(\mathbb C)$ for instance. Its only C$^*$-subalgebras (which are also von Neumann, because of the finite-dimension) are $$\{0\},\ \ \mathbb C, \ \ \mathbb C\oplus\mathbb C,\ \ M_2(\mathbb C).$$

Thus the only possible lattices of projections are $$ \{0\},\ \ \{0,1\},\ \{(0,0),(1,0),(0,1),(1,1)\}, $$ and $$\tag{1} \left\{\begin{bmatrix}0&0\\0&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix}\right\}\cup\left\{\begin{bmatrix} t&e^{i\theta}\,\sqrt{1-t^2}\\ e^{-i\theta}\,\sqrt{1-t^2}&1-t\end{bmatrix}:\ {t\in[0,1],\theta\in[0,2\pi)}\right\}. $$ They are all of the form "a bunch of non-comparable elements, with a single common supremum and a common single infimum". So, if in the set $(1)$ you remove some (or many) of the projections depending on the parameters, you will still have a lattice of projections, but it won't be that of a C$^*$-algebra.

I might be wrong here, but I don't think there's an easy way to tell if a given lattice is the projection lattice of a von Neumann algebra; I know a necessary condition is that the lattice is complete, and maybe some other property is obvious. But, in general, I don't see how you could guarantee that limits and products of linear combinations of the given projections do not produce new projections.