# Why only consider Dolbeault cohomology?

On a complex manifold we have the differential operators

$$\partial:A^{p,q}\to A^{p+1,q}$$ $$\bar\partial:A^{p,q}\to A^{p,q+1}$$ which both square to zero. Hence one can define cohomology groups $$H^{p,q}_\partial=\frac{\ker\partial:A^{p,q}\to A^{p+1,q}}{\partial A^{p-1,q}}$$ $$H^{p,q}_{\bar\partial}=\frac{\ker\bar\partial:A^{p,q}\to A^{p,q+1}}{\bar\partial A^{p,q-1}}$$ But for some reason I only ever see people discussing the $H_{\bar\partial}^{p,q}$, i.e. Dolbeaut cohomology. What makes this more interesting the the $H_\partial^{p,q}$ groups?

• You should really write $A^{p,q}(X)$, $H^{p,q}_{\bar{\partial}}(X)$, etc. Apr 22, 2016 at 23:58

$\newcommand{\dd}{\partial}$To supplement the existing (excellent) answers: If $E \to M$ is a holomorphic vector bundle, the transition functions of $E$ are "constant with respect to $\bar{\dd}$", so there's a natural notion of $E$-valued Dolbeault cohomology $H_{\bar{\dd}}^{p,q}(M, E)$. Unless the transition functions of $E$ are constant, however, there is no corresponding notion of $H_{\dd}^{p,q}(M, E)$.

• Very good point I'd never noticed before!
– user98602
Apr 22, 2016 at 21:30
• But we could just as easily consider anti-holomorphic vector bundles could we not? Apr 22, 2016 at 23:27
• @MichaelAlbanese: There is no such thing. :) A composition of anti-holomorphic maps isn't anti-holomorphic, so there's no "anti-holomorphic pseudo-group". Apr 23, 2016 at 1:40
• So the 'anti-holomorphic tangent space' $T^{0,1}M$ is not an anti-holomorphic vector bundle. That's annoying. Apr 23, 2016 at 15:16
• @MichaelAlbanese: Or, if you prefer, being holomorphic is an incredibly rich, chiral property, whose "mirror image" is (surprisingly!) undistinguished from the "merely smooth". ;) Apr 23, 2016 at 16:37

Note that $H^{0,0}_{\bar{\partial}}(X) = \ker\bar{\partial} : A^{0, 0}(X) \to A^{0, 1}(X)$ is precisely the collection of holomorphic functions on $X$ which, historically, were of more interest than anti-holomorphic functions on $X$ which is given by $H^{0,0}_{\partial}(X) = \ker\partial : A^{0,0}(X) \to A^{1,0}(X)$.

Moreover, Dolbeault cohomology $H^{p,q}_{\bar{\partial}}(X)$ determines $H^{p,q}_{\partial}(X)$ and vice versa. More precisely, the maps

\begin{align*} \psi : H^{p,q}_{\bar{\partial}}(X) &\to H^{q,p}_{\partial}(X)\\ [\alpha] &\mapsto [\bar{\alpha}] \end{align*}

and

\begin{align*} \varphi : H^{q,p}_{\partial}(X) &\to H^{p,q}_{\bar{\partial}}(X)\\ [\beta] &\mapsto [\bar{\beta}] \end{align*}

are well-defined and are inverses of one another. Therefore $H^{p,q}_{\bar{\partial}}(X) \cong H^{q,p}_{\partial}(X)$.

If $X$ is compact, then Hodge Theory tells us that $H^{p,q}_{\bar{\partial}}(X) \cong \ker\Delta_{\bar{\partial}} : A^{p,q}(X) \to A^{p,q}(X)$ and $H^{p,q}_{\partial}(X) \cong \ker\Delta_{\partial} : A^{p,q}(X) \to A^{p,q}(X)$ where $\Delta_{\bar{\partial}}$ and $\Delta_{\partial}$ are the holomorphic and anti-holomorphic Laplacians respectively. If $X$ is also Kähler, then $\Delta_{\bar{\partial}} = \Delta_{\partial}$ so we see that

\begin{align*} H^{p,q}_{\bar{\partial}}(X) &\cong \ker\Delta_{\bar{\partial}} : A^{p,q}(X) \to A^{p,q}(X)\\ &= \ker\Delta_{\partial} : A^{p,q}(X) \to A^{p,q}(X)\\ &\cong H^{p,q}_{\partial}(X). \end{align*}

Combining these isomorphisms with the ones above, we see that $H^{p,q}_{\bar{\partial}}(X) \cong H^{q,p}_{\bar{\partial}}(X)$ so the Hodge numbers satisfy $h^{p,q} = h^{q,p}$. Therefore the odd Betti numbers of a compact Kähler manifold are even.

$H^{p,q}_{\partial}$ is canonically isomorphic to $H^{q,p}_{\bar \partial}$ via complex conjugation.

As for why one would prefer the latter group over the former, $H_{\bar \partial}^{p,0}$ is the set of holomorphic $p$-forms, and that's a more natural object than the antiholomorphic $q$-forms ($H_{\partial}^{0,q}$).