Let E be a Hilbert C*-module over some C*-algebra and let $h \in K(E)$. Due to B. Blackadar's, "K-Theory for Operator algebras" Thm. 17.11.4 for a separable C*-algebra $A$, represented by elements of $B(E)$, it is possible to construct a countable approximate unit $\{u_n\}$ contained in $C^*(h)$, such that $\{u_n\}$ is quasicentral for $A$ and $u_{n+1} u_n=u_n$. The question is: is it always possible to make $u_n$ be projectors (or, at least, idempotents). The question seems to be obvious if $E$ is just a Hilbert space, but I'm not sure for Hilbert modules.
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$\begingroup$ Nice question. Would be even nicer if you would edit the question and put dollars in around the math. $\endgroup$– RegenbogenCommented Mar 22, 2010 at 20:54
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1$\begingroup$ Personally, I think that the question is sufficiently "symbol-light" that there is no need here to hit everything with jsMath. Most readers likely to contribute to this question are IMHO capable of a certain amount of `internal LaTeX parsing' $\endgroup$– Yemon ChoiCommented Mar 22, 2010 at 21:10
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$\begingroup$ Here's a link to the Theorem you cite: books.google.com/…. $\endgroup$– Jonas MeyerCommented Mar 23, 2010 at 0:22
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1$\begingroup$ @Yemon: But we built computers so that they do our work for us! $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 23, 2010 at 0:45
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2$\begingroup$ It seems that really, for this question, Blackadar is actually following Pedersen, "C*-algebras and their automorphism groups", 3.12.14 (no Google preview). Also, h is absolutely NOT an arbitrary element in K(E): it seems that, at least, we want h to be "strictly positive": I think this is equivalent to hK(E)h being dense in K(E). $\endgroup$– Matthew DawsCommented Mar 23, 2010 at 10:51
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1 Answer
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No. In fact, K(E) need not even contain any nonzero projections. Take a (nontrivial) C*-algebra B with no nonzero projections1 and take E=B as a right module over B with inner product 〈a,b〉=a*b. Then K(E)≅B, as mentioned in Example 13.2.4 (a) in Blackadar, and proved for instance as Proposition 2.2.2 (i) in Manuilov and Troitsky. (Note that in this case K(E) also has no nonzero idempotents.)
Edit: I removed an overly complicated comment on the Hilbert space case, forgetting to take into account that h is strictly positive, and in particular positive. I added a comment on trouble that may arise even in this case.
In the case when E is a Hilbert space over B=ℂ, taking into account the fact that h is self-adjoint, C*(h) is the C*-algebra generated by a self-adjoint compact operator, and therefore the spectral projections of h provide an approximate identity {un} consisting of increasing projections. Because h is strictly positive, its range is dense, so this will be a sequence of projections converging weakly to the identity operator. However, this approximate identity need not be quasicentral for A⊆B(E). E.g., suppose you have un equal to the projection onto the span of the first n elements of an orthonormal basis. If S is the unilateral shift with respect to that basis, then ||unS−Sun|| = 1 for all n, so {un} is not quasicentral for C*(S). Pedersen uses the Hahn-Banach theorem and the axiom of choice to show the existence of a quasi-central approximate identity in the closed convex hull of {un}, but you typically will not be able to find one consisting of projections, even in the Hilbert space case.
1 E.g., take B to be C0(X) for some noncompact, locally compact, connected space X, or if you prefer simple algebras see Blackadar's 1980 paper.
answered Mar 23, 2010 at 0:05
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$\begingroup$ Thanks for the answer, though I feel sorry for asking such primitive questions. Actually I just would like to rewrite the Blackadar's theorem for my own purposes, namely to construct such $D$ that I'll have an upper estimation for $\|[D,a_j]\|$. As for the case of Hilbert space, the spectrum of the compact operator $h$ is then just a sequence of points on R tending to 0, so, as I thought, we may just use functional calculus to makefinite-dimensional projectors. $\endgroup$ Commented Mar 23, 2010 at 14:32
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$\begingroup$ Good point, that's probably a simpler way to get the approximate identity in the Hilbert space case. I haven't thought much about it, but the strict positivity implies quasicentrality of the approximate identity in this case, doesn't it? $\endgroup$ Commented Mar 23, 2010 at 16:25
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$\begingroup$ I thought about it more and realized that the answer to my question is no. I've edited my answer accordingly. $\endgroup$ Commented Mar 24, 2010 at 0:05
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$\begingroup$ Kolya Ivankov, by the way, there is no reason to apologize for your question. $\endgroup$ Commented Mar 24, 2010 at 0:16
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$\begingroup$ Maybe this is a little off topic, but Manuilov has done some work on defining generalized "eigenvalues" for compact operators on Hilbert A-modules, in case A is a von Neumann algebra. This can be found in the book by Manuilov and Troitsky on Hilbert C* modules or in the paper by Manuilov "Diagonalizing operators in Hilbert modules over C*-algebras". $\endgroup$ Commented Mar 24, 2010 at 12:08