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The result stated in the title is thoroughly standard - or that's the impression I got. I seem to remember seeing it stated somewhere in a book I was reading in the library, and then reverse-engineering a proof from the hints given.

For a preprint I'm working on, it would be preferable to give a precise citation from a "standard text", rather than spend time giving the proof "for the reader's convenience". Any suggestions?

If anyone's interested, an outline of a proof is as follows: consider an idempotent P in B(H), with H a Hilbert space, and note that we can always decompose H as an orthogonal sum with respect to which P has the block matrix form

$$ P= \left(\begin{matrix} I & R \\\\ 0 & 0 \end{matrix}\right) $$

Then it's not hard to see that conjugating $P$ by the invertible operator

$$ S= \left( \begin{matrix} I & R \\\\ 0 & I \end{matrix} \right) $$

will give

$$ E = \left(\begin{matrix} I & 0 \\\\\ 0 & 0 \end{matrix} \right) $$

Since $S= I+P-E$, it suffices to show that $E$ is in the C*-algebra generated by I and P (for then S will also lie in that algebra, and then we're done). This follows by messing around with various combinations of P, its adjoint, and their products.

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For the sake of completeness, I want to mention that "messing around" shows that $E=PP^*(1+(P-P^*)(P^*-P))^{-1}$. –  Jonas Meyer Mar 20 '10 at 1:37

1 Answer 1

up vote 3 down vote accepted

Blackadar's K-Theory for operator algebras has it, although the way it is done there is perhaps overkill if this is all you need. The result is generalized to local $C^*$-algebras, and he shows similarity by showing the stronger property of homotopy equivalence. It is Proposition 4.6.2 on page 23 of the 2nd edition (1998). (Proposition 4.3.3 shows that homotopy equivalence is stronger.)

The stronger equivalence (but just for $C^*$-algebras) is also shown in the K-theory book by Rørdam et al., Lemma 11.2.7, with a very similar proof.

Added after the first two comments:

Kaplansky's Rings of operators has another approach. Since there is no Google preview, I'll outline what is done. Theorem 26 shows that if $A$ is a unital ring with involution $*$ such that $1+x^*x$ is invertible for all $x\in A$, then for each idempotent $f\in A$ there is a projection $e\in A$ such that $fA=eA$. (The projection is obtained just as in the previous two sources, $e=ff^*(1+(f-f^*)(f^*-f))^{-1}$.) A previous exercise (4 on page 24) shows that if $f$ and $e$ are idempotents in a unital ring $A$ and $fA=eA$, then $f$ and $e$ are similar.

Here is yet another reference, older and with a different approach, but this time not from a standard text. It is Lemma 16 on page 856 of Kaplansky's "Modules over operator algebras", American Journal of Mathematics, Vol. 75, No. 4 (Oct., 1953), pp. 839-858. Of the references I have provided, this is the only one that sets out to prove precisely the result in question as opposed to something stronger or more general. The approach is to invoke the theory of polynomial identities to reduce to the case of 2-by-2 scalar matrices. This paper is famous among operator algebraists because of its introduction of what are now called Hilbert $C^*$-modules (but only over commutative $C^*$-algebras).

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Thanks - either of those references would do, although I would still prefer a more general and less specialist source. (To give some background: I first saw it, stated without proof, in a short Bull. AMS paper from the 1950s, so it predates K-theory of C*-algebras by a long way...) –  Yemon Choi Mar 3 '10 at 3:08
I don't know such a source. I've seen the same sketch you provided, but I also don't have any standard reference for that. I'll be curious to see what turns up. I understand your desire for a more general source. While K-theory is a good place to look for results about equivalence of idempotents (now that the subject exists), it does seem a bit much for just this result. Of course, no K-theory is actually used in the proofs I linked to! –  Jonas Meyer Mar 3 '10 at 3:26
Thanks Jonas for the continued efforts. I think the Kaplansky references are the closest to what I was looking for (and the place where I first saw this result quoted without proof was a short article by Montgomery, based on ... something in Kaplansky's Fields and Rings). –  Yemon Choi Mar 9 '10 at 22:30
A remark on the minimalist approach in Kaplansky's Rings of operators: Gelfand and Naimark initially took as an axiom for abstract C*-algebras that 1 + x*x is invertible for all x. It was 17 years before this was shown to follow from the other axioms, even though G & N believed it did from the start. –  Jonas Meyer Mar 30 '10 at 3:30

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