Let $A$ be a (unital) $C^*$-algebra and $X,Y$ right Hilbert $A$-modules which are finitely generated and projective. It seems to be well-known that if $T: X \to Y$ is an $A$-linear map, then $T$ is necessarily adjointable. I could not find a proof though. Can someone give a reference or proof of this little fact?
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$\begingroup$ Maybe you could add definitions of "finitely generated" and "projective" (just in case you mean something different from the pure-algebra case)? And, perhaps, some justification for the "well-known" claim? $\endgroup$– Matthew DawsCommented Dec 7, 2022 at 12:07
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$\begingroup$ @MatthewDaws I'm not entirely sure what the exact definition is of either terms actually. If I would guess: finitely generated as Hilbert module (i.e. you need the closure of the $A$-span of finitely many element of the Hilbert module) and projective means that it is a direct summand of a free Hilbert module $\bigoplus_{i\in I} A = \ell^2(I)\otimes A$ for some index set $I$. I'm not sure if the latter definition is equivalent with the categorical definition of projectivity (in the category of say, Hilbert modules and adjointable maps). $\endgroup$– AndromedaCommented Dec 7, 2022 at 17:16
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$\begingroup$ So there must be some reason why you believe that this result is true, even though you are not quite sure of the relevant definititions (!) I'm guessing you read this somewhere. It might help others to answer your question if you could give a reference to where you read the result ?? $\endgroup$– Matthew DawsCommented Dec 8, 2022 at 11:48
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2$\begingroup$ @MatthewDaws Apologies. It is from the proof of theorem 6.23 in arxiv.org/pdf/1604.00159.pdf But these notes do not define the terminology. $\endgroup$– AndromedaCommented Dec 8, 2022 at 12:52
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2 Answers
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Looking at the proof of Lemma 6.21 in the notes of de Commer that you're reading (https://arxiv.org/pdf/1604.00159.pdf, per the comments), it seems like the relevant property of the modules is that of having compact identity operator. Suppose that a Hilbert $A$-module $X$ has this property. Since finite sums of the form $\sum_{i} |x_i\rangle\,\langle y_i|$ are dense in the compact operators, some such sum must be very close to the identity, and hence invertible. Let $K=\left( \sum_i |x_i\rangle\,\langle y_i|\right)^{-1}$. Then for every Hilbert module $Y$ and every $A$-linear map $T:X\to Y$ we have \begin{equation*} T = T\circ \operatorname{id}_{X} = \sum_i |TKx_i\rangle\,\langle y_i|, \end{equation*} which is compact and therefore adjointable.
To tie this to the actual question as asked: for Hilbert modules over unital $C^*$-algebras, the property of having compact identity operator is equivalent to being finitely generated and projective in the purely algebraic sense, and it is also equivalent to being the image of an orthogonal projection on the Hilbert module $A^n$ for some $n$. These equivalences are explained in Wegge-Olsen's book on $K$-theory.
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1$\begingroup$ Thanks! For those interested, it is theorem 15.4.2 and remarks following this in the book that is relevant here. $\endgroup$ Commented Dec 8, 2022 at 23:41
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$\begingroup$ @t.c. Per your last paragraph: I don't think you don't need unitality of the $C^*$-algebra for this. $\endgroup$– J. De RoCommented Dec 30, 2022 at 11:57
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This is a comment on the definition of being "finitely generated".
There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert $C([0,1])$-module as a maximal ideal of that $C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.
