For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from the usual anti-holomorphic derivative, that acts on $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ so as to give a complex $$ \Gamma^{\infty}(E) \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,1)}(M) \overset{\overline{\partial}}{\to} \cdots \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,n)}(M)\overset{\overline{\partial}}{\to} 0. $$

As I prefer global constructions, I began to wonder how one would construct this globally. To apply id $\otimes \overline{\partial}$ to $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ is of course not well-defined since we are tensoring over ${C^{\infty}(M)}$. So what can one do . . .?

P.S. Does such a complex exist in the purely real case, and if not, then why not?

  • 1
    $\begingroup$ Locally $\overline{\partial}_E$ "looks like" $\overline{\partial}$, but this is because you have chosen a holomorphic trivialisation of $E$. As a global operator $\overline{\partial}_E$ is not something that you can get from $\overline{\partial}$, i.e., from the complex structure on $M$. It is determined by (and determines) the holomorphic structure on $E$. There may be many such (or none) on the same complex vector bundle. $\endgroup$ – Peter Dalakov Aug 1 '12 at 19:36

To get the Dolbeault complex, you need a choice of holomorphic structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty_M\otimes_{\mathcal{O}_M}\mathcal{E}$ and $\overline{\partial}_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$ is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying $$ \overline{\partial}_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}_E(\sigma),$$ for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that $\left. \overline{\partial}_E\right| _{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$. You can then extend the Dolbeault operator to $\overline{\partial}_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$, $\overline{\partial}_E^2=0$, by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution $$ 0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$ The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.

You cannot do any of this without the holomorphic structure. Differently put, you need the total space of $E$ to be a complex manifold and the projection $E\to M$ to be holomorphic. You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.


The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure $\overline{\partial}_E$ is non-trivial. It can be phrased as follows:

$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that $D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.

  • $\begingroup$ So in the Kahler case all holomorphic structures on $E$ induce the same set of holomorphic sections? $\endgroup$ – Jean Delinez Aug 1 '12 at 21:00
  • 2
    $\begingroup$ There can be many inequivalent holomorphic structures even in the Kaehler case. For example, if $M$ is a Riemann surface of genus $g>1$, then the trivial line bundle admits a whole $g$-dimensional torus of non-isomorphic holomorphic structures, $\textrm{Pic}^0(M)$. $\endgroup$ – Peter Dalakov Aug 1 '12 at 21:30

There is an analogous construction in the purely real case. The analogue of the sheaf of homolorphic functions is the sheaf $LC$ of locally constant functions. A locally free $LC$-module is equivalent to a local system, and hence to a vector bundle $V$ with integrable connection $\nabla: V \to V \otimes \Omega^1$. One then has the exact sequence of sheaves:

$$LC(V) \to C^{\infty}(V) \to C^{\infty}(V) \otimes \Omega^1 \to C^{\infty}(V) \otimes \Omega^2 \to \cdots \to C^{\infty}(V) \otimes \Omega^n.$$

Here $LC(V)$ is the sheaf of locally constant sections of the local system $V$, and $C^{\infty}(V)$ is the sheaf of smooth sections. Just as in the complex case, one can then show that the sheaf cohomology $H^q(X, V)$ is isomorphic to the deRham cohomology $$\mathrm{Ker}(C^{\infty}(V) \otimes \Omega^q \to C^{\infty}(V) \otimes \Omega^{q+1}) / \mathrm{Im} (C^{\infty}(V) \otimes \Omega^{q-1} \to C^{\infty}(V) \otimes \Omega^{q}).$$

In particular, if $V$ is the trivial local system, this is the isomorphism between sheaf cohomology and deRham cohomology.

Regarding your question about constructing $\bar{\partial}$ in a coordinate free way: Surprisingly, I don't know how to do this. But I can give you a coordinate free axiomatization of $\bar{\partial}$. Let $\mathcal{O}$ be the sheaf of holomorphic functions and $C^{\infty}$ the sheaf of smooth functions. Let $M$ be a locally free $\mathcal{O}$ sheaf and let $C^{\infty}(M)$ be $M \otimes_{\mathcal{O}} C^{\infty}$.

$\bar{\partial}: C^{\infty}(M) \to > C^{\infty}(M) \otimes \Omega^{0,1}$ is the unique $\bar{\partial}$ connection annihilating $M$.

Once you have used coordinates to check that this connection exists and is unique, you can probably prove anything you want about it faster from this description than from the coordinate formulae.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.