MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$ be an infinite dimensional separable complex Hilbert space. Denote by $\mathcal{B}(H)$ the C*-algebra of bounded operators on $H$, $\mathcal{K}(H)$ the ideal of compact operators on $H$, and $\mathcal{C}(H)=\mathcal{B}(H)/\mathcal{K}(H)$ the Calkin algebra. The collection of projections $\mathcal{P}(\mathcal{C}(H))$, i.e., self-adjoint idempotents, in $\mathcal{C}(H)$ form a partially ordered set in the usual way, namely $p\leq q$ if and only if $pq=p$.

It is my understanding that $(\mathcal{P}(\mathcal{C}(H)),\leq)$ does not form a lattice, i.e., meets and joins need not exist in general. (This would mean that the Calkin algebra is another example for C*-algebras with bizzarre structure of projections.)

This fact is mentioned in Hadwin's paper "Maximal Nests in the Calkin Algebra" (PAMS, 1998), and there is a sketch of a proof (credited to Weaver) in Farah's notes "Set theory and operator algebras" from the Appalachian Set Theory Workshop, however I've never quite understood this proof.

Would someone be able to explain/sketch this result?

share|cite|improve this question
up vote 7 down vote accepted

Maybe it's clearer in our original email correspondence!

From me to Ilijas, 12/31/06:

The poset of projections in any von Neumann algebra is a lattice. Is this true of C(H)? The answer seems to be no. I can give a pair of projections that looks like a counterexample, though proving this might take some work.

From Ilijas to me, 1/4/07:

Right, I can see this now. There is an example that is not too bad.

Enumerate a basis of H as e(i,j,k), i and j in omega and k in {0,1}. Let P be the projection to the closed span of all e(i,j,0). To define Q, let f(i,j) be a linear combination of e(i,j,0) and e(i,j,1) such that ||f(i,j)-e(i,j,0)||=2^{-i}. Let Q be the projection to the span of all f(i,j). Now with pi the quotient map into the Calkin algebra C(H), I claim that pi(P)/\pi(Q) does not exist in C(H). Proof: Fix a projection R' which is below both pi(P) and pi(Q). We can find R below P in the lattice of projections of B(H) such that pi(R)=R'. For each i, there is n(i) such that for all j>=n(i) the distance between e(i,j,0) and R is at least 1/2. Let S be the projection to the span of e(i,n(i),0). Then pi(R/S) is below both pi(P) and pi(Q), but it is strictly above pi(R).

Is that easier to follow?

No doubt this is also the proof that Don had in mind. (And it's also more or less what I was thinking, but this proof is Ilijas's, not mine. So I certainly don't deserve the credit here.)

share|cite|improve this answer
Nik, this does seems a bit clearer. Two questions: What is allowing us to find the n(i) as claimed? And what is meant by R/S? – Iian Smythe Jun 17 '14 at 21:14
To say that pi(R) lies below pi(Q) is to say that R is essentially contained in Q: for every $\epsilon > 0$, every unit vector in some cofinite-dimensional subspace of R has distance at most $\epsilon$ to Q. If, for some i, there were infinitely many j satisfying d(e(i,j,0), R) < 1/2, then R could not be essentially contained in Q. – Nik Weaver Jun 17 '14 at 21:31
R/S is a typo, it should be R\/S (R join S). – Nik Weaver Jun 17 '14 at 21:32
When defining f(i,j), should ||f(i,j)-e(i,j,0)|| read ||f(i,j)-e(i,j,1)||? – Aaron Tikuisis Jun 21 '14 at 13:56
@Aaron: I don't think so ... if you defined it that way then pi(P)/\pi(Q) would be 0. – Nik Weaver Jun 21 '14 at 16:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.