Let $X$ be a complex manifold, you can assume it's compact, if necessary. We have the Dolbeault complex $$0 \rightarrow \mathcal{A}^{0,0} \xrightarrow{\bar{\partial}} \mathcal{A}^{0,1} \xrightarrow{\bar{\partial}} \ldots \xrightarrow{\bar{\partial}} \mathcal{A}^{0,n} \rightarrow 0.$$ It is well known that we can consider the completion of $\mathcal{A}^{0,*}$ with respect to the $L^2$-norm induced by some metric on $X$, get a Hilbert space $H$, and perform functional calculus on the Dolbeault operator $D=\bar{\partial} + \bar{\partial}^*$, to get a Fredholm module $(\frac{D}{\sqrt{1+D^2}} , H) \in KK(C_0(X), \mathbb{C})$ in Kasparov K-group. The reference I'm using for this process is chapter 10 of Analytic K-homology, by Higson and Roe.

Now, let's say instead of working with the Hilbert space $H$, we want to work with Hilbert $C_0(X)$-modules. So I guess I need to consider the continuous sections of the bundle $\wedge ^{0,*} T^*X$ over $X$, and $\bar{\partial}$ is a densely defined (unbounded) operator acting on this space. If I understand correctly, unlike the case of Hilbert spaces, performing functional calculus for a densely defined operator on a Hilbert module is not automatic, and one needs a "regularity" condition on the operator(the operator $T$ acting on Hilbert module $E$ is regular if it is densely defined, the adjoint $T^*$ is also densely defined, and $1+T^*T$ is invertible.).

My question is if my understanding above is correct, and more importantly, if I can perform functional calculus on the Dolbeault operator or not?(i.e. is $D$ regular?) Even if one can somehow define it up to compact operators, that's good enough for me.

Thanks in advance for any help!

Edit: As Johannes pointed out, $D$ is not $C_0(X)$-linear. However, what I need is to only define functional calculus from $$\mathcal{L}_{C_0(X)}(H_X, H^{odd})/\mathcal{K}_{C_0(X)}(H_X, H^{odd}) \rightarrow \mathcal{L}_{C_0(X)}(H_X, H^{even}) / \mathcal{K}_{C_0(X)}(H_X, H^{even})$$, where in here, $H_X$, is the canonical Hilbert $C_0(X)$-module $\mathcal{l}^2(C_0(X))$ of sequences in $C_0(X)$, $\mathcal{L}_{C_0(X)}(H_1,H_2)$ is the space of $C_0(X)$-linear bounded adjointable maps from $H_1$ to $H_2$, $\mathcal{K}$ is the ideal of compact operators, and $H^{odd}$ is Hilbert $C_0(X)$-module corresponding to $\mathcal{A}^{0,odd}$ ($H^{even}$ is defined similarly).

I realize that one might still need $D$ to be $C_0(X)$-linear to define this, but is there a way to get around this issue(since we may only need $D$ to be $C_0(X)$-linear up to compact operators for "only a dense subset" for this to make sense)?


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    $\begingroup$ The problem is that D is not C(X)-linear at all. $\endgroup$ – Johannes Ebert Nov 25 '17 at 20:39
  • $\begingroup$ Thanks for pointing out the issue. I edited the question, since I needed something slightly weaker. I was wondering if the weaker statement above can be true? $\endgroup$ – Kashayar Nov 27 '17 at 1:02

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