9
$\begingroup$

While I learn about $\partial$ and $\bar{\partial}$ operators, I had some questions about the reason why people prefer $\bar\partial$ over $\partial$. Specifically,

  1. When defining Dolbeault cohomology, one uses $\bar{\partial}$ but not $\partial$. I wonder whether there happens any problem if one define a cohomology by $\partial$. Or is this because it isn't interesting?

  2. Any textbook on complex variables say only about $\bar{\partial}$-Poincare lemma. Is there a version with $\partial$? If not, where the difference between two operators fundamentally comes?

  3. For a holomorphic vector bundle $E$, we define the operator $\bar{\partial}_E$ only. Again why don't we define $\partial_E$?

  4. Let $(E,h)$ be an hermitian holomorphic vector bundle on a compact hermitian manifold $(X,g)$. When we show that the operator $\bar\partial_E^*:=-\bar*_{E^*}\circ\bar\partial_{E^*}\circ\bar*_E$ on $A^{p,q}(X,E)$ is adjoint to $\bar\partial_E$, one uses $$\int_X\bar\partial(\alpha\wedge\bar*_E\beta)=\int_X d(\alpha\wedge\bar*_E\beta)$$ for $\alpha\in A^{p,q}(X,E)$ and $\beta\in A^{p,q+1}(X,E)$(c.f. "complex geometry" by Huybrechts, p.170). Here how we know $$\int_X\partial(\alpha\wedge\bar*_E\beta)=0$$? Again, the two operators appear to have different rules.

$\endgroup$
13
$\begingroup$
  1. On differential forms, take complex conjugate to turn $\partial$ into $\bar\partial$, and holomorphic functions into conjugate holomorphic.
  2. All of the proofs about differential forms then go through the complex conjugation effortlessly, including the Poincare lemma. We use $\bar\partial$ because we like holomorphic functions, i.e. $\bar\partial f=0$.
  3. As for vector bundles, the operator $\partial_E$ is defined only for conjugate holomorphic vector bundles, i.e. with conjugate holomorphic transition maps, because the $\partial$ operator passes through conjugate holomorphic functions: $\partial (fg)=f \partial g$ for all $g$ precisely for $f$ conjugate holomorphic, so in a conjugate holomorphic local trivialization.
  4. In local holomorphic coordinates $z^1,\dots,z^n$, count numbers of $dz^1, \dots, dz^n$ in the wedge products; you already have $n$ of them, so you get zero if you wedge in another.
$\endgroup$
  • 3
    $\begingroup$ This is all true. It might be worth stating the obvious: people care more about $\bar{\partial}$ because its kernel is the holomorphic functions. Since we care about holomorphic functions in complex analysis, the $\bar{\partial}$ complex is of more importance. $\endgroup$ – Steven Gubkin Mar 29 '18 at 16:12
  • $\begingroup$ Thanks for your guidence. So I understood that the two operators are not fundamentally different and they are just conjugate to each other. This makes me comfortable. For the last question I found that it was unrelevant to operators as the wedge product is of type (n,n-1). $\endgroup$ – Ramanasa Mar 29 '18 at 16:20
  • $\begingroup$ @AndrewMcHugh I think "passes through" means $\bar\partial(f\cdot g)=f\cdot\bar\partial g$, which holds when $\bar\partial f=0$, i.e., when $f$ is holomorphic. $\endgroup$ – Andreas Blass Apr 4 '18 at 0:48
3
$\begingroup$

If $X$ is a complex manifold and $E\to X$ is a holomorphic vector bundle, only $\bar\partial_E$ can be defined naturally, i.e., it depends only on the complex structures of $X$ and $E$. The $\partial_E$ operator cannot be defined intrinsically.

If $E$ has a flat connection, you can also define the $\partial$-operator. This is the case when $E$ is trivial, and then the $\partial$ operator you mention is the one we all know. The flat connection is $d$.

In general, the operators $\bar\partial$ and $\partial$ have different roles, even if $\partial$ is well-defined.

To answer your questions:

  1. $\partial$ is not defined, therefore it cannot be used to define some cohomology. If you have a holomorphic vector bundle over $X$, then the $\bar\partial$ complex give you some cohomology which is isomorphic to the Cech cohomology, and this a very deep result.

  2. Sure, the $\bar\partial$-Poincare lemma is a local statement, and it can be conjugated to get a statement about the $\partial$ operator. No problem here.

  3. Because $\partial_E$ cannot be defined naturally. You need some other conditions (like flatness or a Hermitian metric) in order to define the $\partial$ operator.

"Again why don't we define $\partial_E$?" Because it doesn't exist.

  1. You already have some metric since you can define the star operator. The short answer is "for bidegree reasons"
$\endgroup$

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.