For any $C^\ast$-algebra $A$, don't we have the isomorphism $\mathcal{H}_A \oplus \mathcal{H}_A \cong \mathcal{H}_A$? In some places, it is only mentioned when $A$ is $\sigma$-unital. But $\mathcal{H}_A \cong A \otimes l^2$, and since tensor product commutes with direct sums, we should be getting $(A \otimes l^2) \oplus (A \otimes l^2) \cong A \otimes (l^2 \oplus l^2) \cong A \otimes l^2$. What's wrong here?
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$\begingroup$ Please edit your post to include your attempts to prove this isomorphism. $\endgroup$– Anne BauvalMar 29 at 8:54
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2$\begingroup$ @AnneBauval done $\endgroup$– AnupamMar 29 at 9:30
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$\begingroup$ How do you define $\mathcal{H}_A$? $\endgroup$– AndromedaMar 29 at 10:04
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2 Answers
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There is no mistake. The result holds for general $C^*$-algebras, as you proved.
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$\begingroup$ Are you considering the fact that $\mathcal{H}_A$ need not be countable generated? $\endgroup$– AnupamMar 29 at 10:23
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$\begingroup$ @Anupam But look at your own proof: you never use assumption? $\endgroup$– J. De RoMar 29 at 10:47
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$\begingroup$ Yes, but still Hilbert modules seem a bit scary as compared to Hilbert Spaces, so was just confirming! :) $\endgroup$– AnupamMar 29 at 11:06
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Nothing wrong and your proof shows that the hypothesis $A$ is $\sigma$-unital is useless (you could complete your algebraic proof whith a little argument justifying why the isomorphism goes over to the completions).
You will find a related theorem, the "Stabilization or Absorption Theorem", in Bruce Blackadar's K-Theory for Operator Algebras: (for any $C^*$-algebra $B$),
"The theorem in this form is due to Kasparov [...]
Theorem 13.6.2. If $E$ is a countably generated Hilbert $B$-module, then $E\oplus H_B\cong H_B.$"
and
"Exercise 13.7.1 (a) Use the stabilization theorem to prove that if $B$ is $\sigma$-unital and $E$ is a countably generated full Hilbert $B$-module, then $E^\infty\cong H_B.$" ($E^\infty:=E\otimes l^2$, and "full" means $\langle E,E\rangle=B$.)
Enlightning comment below by @MaoWao: "I think the stabilization theorem is the reason why some authors state this result [$\mathcal{H}_A \oplus \mathcal{H}_A \cong \mathcal{H}_A$] only for $σ$-unital $C^∗$-algebras, because only then it is a direct consequence."
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$\begingroup$ The point is $\mathcal{H}_A$ is not countable generated if $A$ is not $\sigma$-unital $\endgroup$– AnupamMar 29 at 10:23
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$\begingroup$ I know, still it is not countably generated. In fact, $\mathcal{H}_A$ is countably generated iff $A$ is $\sigma$-unital $\endgroup$– AnupamMar 29 at 10:35
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$\begingroup$ Well $A$ itself need not be generated by countably many elements if it doesn't have a countable approximate unit. $\endgroup$– AnupamMar 29 at 10:41
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$\begingroup$ Sorry if I am being foolish, but can you give a countable generating set (according to your definition of course) which exhibits $\mathcal{H}_A$ as a countably generated $A$-module? $\endgroup$– AnupamMar 29 at 11:41
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$\begingroup$ @AnneBauval But $A$ does not need to have a unit (if it does, then $A$ is of course $\sigma$-unital). $\endgroup$– MaoWaoMar 29 at 11:51