Let $B$ be a $C^*$-algebra and $E$ a right Hilbert-$B$-Module. If I take an approximate unit $(e_i)_i$ in $B$ then for all $x \in E$ we have $x\cdot e_i \rightarrow x$ in $E$ by a simple calculation. I think by Cohens factorization Theorem we should have $E \cdot B:=\{x \cdot b | x \in E, b \in B\}=E$. But on the Wikipedia page it only says that $E \cdot B$ is dense. Now I am a little worried that I got that wrong. I would be grateful if somebody could clarify that for me. Thank you
Hilbert-Modules and Cohens factorization
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functional-analysis
operator-algebras
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0You have that there exist $e_i\in B$ so that $x\cdot e_i\to x$ for any $x$, so $x$ lies in the closure of $E\cdot B$. The previous argument does not imply that $x$ itself lies in $E\cdot B$. – 2017-01-17
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0I know, but that's exactly what Cohens Theorems says, is't it? See: https://en.wikipedia.org/wiki/Cohen%E2%80%93Hewitt_factorization_theorem – 2017-01-17
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0In that case I would check whether the implication $x\cdot e_i\to x$ is correct, I can right now only see that $x\cdot e_i \cdot f\to x\cdot f$ for every $f\in B$, but not how this implies $x\cdot e_i$ converges at all. – 2017-01-18
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0@s.harp: I think if you expand $\langle x-x\cdot e_i,x-x\cdot e_i\rangle$ using the properties of the $B$-\valued inner product it leads to the conclusion that $\|x-x\cdot e_i\|\to 0$ fof all $x\in E$. – 2017-03-19
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Theorem: Let B be a $C^∗$-algebra, X a right Banach B-module. Then XB is a closed submodule of X. This is Theorem II.5.3.7 in Blackadar, Operator Algebras: Theory of C*-Algebras and Von Neumann Algebras, Volume 13, page 90, (it refers to 4.1 in Pedersen, Gert K. Factorization in $C^∗$-algebras. Exposition. Math. 16 (1998), no. 2, 145–156. which I cannot access)It is also theorem 4.6.4 in $C^*$-algebras and Finite-dimensional Approximations By Nathanial Patrick Brown, Narutaka Ozawa.