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Hi,

Given a $\sigma$-unital $C^*$-algebra $A$ and a full Hilbert $A$-module $E$, is it possible to find an approximate unit $ \{\epsilon_i\}, i\in I$ in $A$ such that each $\epsilon_i$ is of the form $< e_i,e_i>_A$, where $e_i \in E, \forall i\in I$? If not, what are the conditions on $E$ and $A$ for which this might be possible?

Thanks in advance.

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  • $\begingroup$ Not sure why this should have a "noncommutative geometry" tag, since we're asking about abstract Hilbert A-modules divorced from any particular geometric construction... Perfectly fine question, though. $\endgroup$
    – Yemon Choi
    Feb 4 '10 at 2:24
  • $\begingroup$ I agree with Yemon, but I see that it was already removed once. $\endgroup$ Feb 4 '10 at 3:40
  • $\begingroup$ On the topic of the question, although I only wish I knew a counterexample, I suspect the answer is no. It is true that there is an approximate unit consisting of finite sums of $<e_i,e_i>$'s. $\endgroup$ Feb 4 '10 at 4:10
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    $\begingroup$ But I don't see any index theorem being used... this is just like claiming that every theorem about C*-algebras is noncommutative topology (GN-thm). Is it not possible to do functional analysis any more without it being noncommutative widgetry? :( $\endgroup$
    – Yemon Choi
    Feb 4 '10 at 9:49
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    $\begingroup$ Indrava, I'll admit it's not a good reason, but it is because standard references like Lance and Manuilov & Troitsky, which develop some of the theory of full modules over sigma-unital C*-algebras, include Brown's result I mentioned above (using finite sums) without mentioning this problem. As I said, I even am not satisfied with my answer, and I would like to think more about this. $\endgroup$ Feb 4 '10 at 16:13
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No, there is not always such an approximate unit. (This will be easier to formulate in terms of left modules, and with inner products linear in the first entry. The warning seems necessary due to the common convention in C*-module theory to do the opposite.)

Example

Let $A=B(\mathbb{C}^2)$ (linear operators), $E=\mathbb{C}^2$, with the module action given by operators acting on vectors (on the left) and the $A$-valued inner product of $x$ and $y$ in $E$ given by $<x,y>_A(z)=<z,y>_\mathbb{C}x$. Then $<x,y>_A$ has rank at most one for all $x$ and $y$, so no such approximate identity exists. $\square$


Any finite dimensional Hilbert space with dimension at least 2 would give a slight modification of this example. Or, let $E=H$ be a separable, infinite dimensional Hilbert space, and let $A=\mathcal{K}(H)$ be the algebra of compact operators on $H$. (Added: Note that fullness follows from the fact that the span of the range of the inner product is the set of finite rank operators.) Or, given a C*-algebra $B$, one could form $H_B=B\oplus B\oplus\ldots$ with its usual right $B$-module structure and consider the analogous construction with $A=\mathcal{K}(H_B)$ and $E=H_B$. If $B$ is $\sigma$-unital, then so is $A$, but there will be no approximate identity of the desired form.

I do not have anything useful to say about formulating sufficient conditions for such an approximate identity to exist, but this simple example shows that lack of existence is common.

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  • $\begingroup$ @Jonas: is that example a full Hilbert $A$-module? $\endgroup$
    – Yemon Choi
    Feb 5 '10 at 3:59
  • $\begingroup$ Absolutely. The span of the range of the inner product is the finite rank operators, which for the finite dimensional case is $A$, and in the infinite dimensional case is dense in $A$. $\endgroup$ Feb 5 '10 at 4:01
  • $\begingroup$ @Jonas: yes, of course it is. Nice example. (I was vaguely wondering if there might be some polarization trick - perhaps that could be done if $A$ is commutative? - but your example shows I wasn't thinking about things properly.) $\endgroup$
    – Yemon Choi
    Feb 5 '10 at 4:05
  • $\begingroup$ Thanks, the counterexample is great. Sorry for my late reply. $\endgroup$ Feb 10 '10 at 13:32
  • $\begingroup$ You're welcome, and don't worry about the time. This example arises in Morita equivalence of C*-algebras. It yields an equivalence bimodule between $\mathcal{K}(H)$ and $\mathbb{C}$ $\endgroup$ Feb 10 '10 at 15:52
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I haven't properly thought about it, but it seems to me that for countably generated modules a positive answer follows from proposition 1.6 in http://arxiv.org/abs/math/0301259

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  • $\begingroup$ Unfortunately incorrect, but that is an interesting reference. $\endgroup$ Feb 5 '10 at 6:23

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