What does it mean for two compatible observables to be a "coarse-graining" of a third?

Question

In reading about quantum contextuality, I've encountered the statement that

if [A,B] = 0, then there exists another observable C such that the spectral projections of A and B are a coarse-graining of those of C and, thus, measuring C allows one to infer the result of both, A and B, a property called joint measurability.

See, for example, p. 4 of the accepted Rev. Mod. Phys. article on Kochen-Specker contextuality. Similarly in a lecture by Cabello (min. 17:39), he states that

coarse-grainings of two different (and incompatible) measurements represent the same observable.

It's been ages I've had algebra in college and I don't have any recollections of the notion of coarse-graining from any of my graduate quantum mechanics courses.

To me, if two observables A and B commute, it just means that they share the same set of eigenvectors. I don't see why there would be a third observable C whose eigenvectors would be more "fine-grained". What does that even mean? (My wild guess is that C's eigenvector would be a superset of the eigenvectors of A and B?)

I've clearly missed some important result from algebra and thus fail to see how it relates to compatibility (and ultimately to contextuality). What have I missed?

Answer 1

We say that an observable $B$ is a coarse-graining of observable $A$ if every eigenbasis of $A$ is an eigenbasis of $B$.

You are wrong thinking that commuting observables share « the same��» set of eigenvectors, or at least, not precise enough.

Two self-adjoint operators $A$ and $B$ commute if an only if there exists a common eigenbasis of them. But it is not difficult to build two commuting orthogonal projections and an eigenbasis of one that is not an eigenbasis of the other.