# Direct sum of standard Hilbert A-modules

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?

• Please edit your post to include your attempts to prove this isomorphism. Mar 29 at 8:54
• @AnneBauval done Mar 29 at 9:30
• How do you define $\mathcal{H}_A$? Mar 29 at 10:04

There is no mistake. The result holds for general $$C^*$$-algebras, as you proved.

• Are you considering the fact that $\mathcal{H}_A$ need not be countable generated? Mar 29 at 10:23
• @Anupam But look at your own proof: you never use assumption? Mar 29 at 10:47
• Yes, but still Hilbert modules seem a bit scary as compared to Hilbert Spaces, so was just confirming! :) Mar 29 at 11:06

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."

• The point is $\mathcal{H}_A$ is not countable generated if $A$ is not $\sigma$-unital Mar 29 at 10:23
• I know, still it is not countably generated. In fact, $\mathcal{H}_A$ is countably generated iff $A$ is $\sigma$-unital Mar 29 at 10:35
• Well $A$ itself need not be generated by countably many elements if it doesn't have a countable approximate unit. Mar 29 at 10:41
• 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? Mar 29 at 11:41
• @AnneBauval But $A$ does not need to have a unit (if it does, then $A$ is of course $\sigma$-unital). Mar 29 at 11:51