Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How do I prove this lemma?

Lemma: Suppose that $X$ is a Hilbert $A$-module. For each $x\in X$ there exists a unique $y\in X$ such that $x=y \langle y,y \rangle $.

I don't know if this is needed but we already know Cohen's factorization theorem that states If $A$ is a Banach algebra with a bounded left or right approximate identity, then for all $a\in A$ there exist $b,c \in A$ such that $a=bc$.

share|cite|improve this question
If $\langle x,x\rangle$ is invertible you can set $y=x\langle x,x\rangle^{1/3}$ – Norbert Jun 26 '12 at 21:26
@Norbert: You mean $-1/3$, right? – Jonas Meyer Jun 26 '12 at 21:49
@JonasMeyer Yes, you are right. Do you have any ideas for the general case? – Norbert Jun 26 '12 at 21:50
@Norbert: I cheated, i.e., I knew where to look it up. It is Proposition 2.31 in Raeburn and Williams's Morita equivalence and continuous trace $C^*$-algebras. They cite the article at the following link as the source of the proof they present (link to large pdf, where the result is found on the sixth page, a.k.a. page 145):… – Jonas Meyer Jun 26 '12 at 21:54
@JonasMeyer Thanks! – Norbert Jun 26 '12 at 22:06
up vote 3 down vote accepted

If you have access to the book Morita equivalence and continuous trace C*-algebras by I. Raeburn and D. Williams, you will find this proved there as Proposition 2.31 on page 21 (the proof is on the next page). In this 6 MB pdf file of the article "Déformations de C*-algèbres de Hopf" by E. Blanchard, the result is on the sixth page (labeled as page 145 in the publication).

share|cite|improve this answer
Thanks @t.b. for improving the book reference. – Jonas Meyer Jul 3 '12 at 3:34

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.