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I know the technical difference between an almost complex manifold and a complex manifold, namely in the former the almost complex structure $J$ may not be integrable while in the later it is. However, in an almost complex manifold one can do many things that can be done in a complex manifold: namely one can still make the Dolbeault decomposition of tensors, and also of the exterior derivative as well as define holomorphic functions. I wonder what are the key differences between the two scenarios, namely what things can be done in a complex manifold that cannot be done in an almost complex manifold because of $J$ being non-integrable.

By the way, could someone recommend me a good book about complex differential geometry?

Thanks.

Michael Albanese
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Bilateral
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  • Are you sure you have the Dolbeaut complex? As for the book, I found Wells - Differential Analysis on Complex Manifolds useful a while ago. – orangeskid Feb 16 '15 at 00:40
  • Yes, you can check for example Anna's book

    http://www.math.ist.utl.pt/~acannas/Books/lsg.pdf

     – Bilateral Feb 16 '15 at 00:47
  • If you'll forgive a tangential "public service comment" on terminology: It would be more logical if we all spoke of "complex manifolds" (admitting an endomorphism field $J$ with $J^{2} = -I$) and "holomorphic manifolds" (complex manifolds for which $J$ is integrable), the way we already do for vector bundles. But for this changeover to happen, students (and working mathematicians) have to embrace the terminology. :) – Andrew D. Hwang Apr 14 '15 at 21:53
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    You don’t have the Dolbeault complex for an almost complex manifold, but you still have a bigraded cdga with differential $d$ which splits into four graded pieces, $d = \bar\mu + \bar\partial+ \partial + \mu$, where the bidegrees are (-1,2), (0,1), (1,0), (2,-1) respectively. The complex structure is integrable iff $\mu = 0$. The operator $\bar\partial$ doesn’t necessarily square to zero in the almost complex case, but $\bar \mu$ does, and $\bar\partial$ descends to a map on $\bar \mu$ cohomology which does square to 0. Recent work of Cirici and Wilson studies this and a general Frolicher ss – Aleksandar Milivojević Nov 18 '18 at 17:48

1 Answers1

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Here are some things which make sense on complex manifolds, but not almost complex manifolds:

  • Complex coordinates $(z_1, \ldots, z_n)$. In particular, vectors like $\frac{\partial}{\partial z_k}$ only make sense on a complex manifold.

  • On complex manifolds, we have $\overline{\partial}^2 = 0$. This gives rise to the Dolbeault complex $$\Omega^{p,0}(M) \xrightarrow{\overline{\partial}} \Omega^{p,1}(M) \xrightarrow{\overline{\partial}} \cdots$$ and therefore to the notion of Dolbeault cohomology $H^{p,q}_{\overline{\partial}}(M)$. That is, Dolbeault cohomology does not make sense on an almost complex manifold. In particular, note that we need $\overline{\partial}^2 = 0$ to ensure that we actually have $$ \overline{\partial}\text{-exact} \implies \overline{\partial}\text{-closed.}$$ For this reason, it's not really worth having notions of "$\overline{\partial}$-exact" or "$\overline{\partial}$-closed" on a manifold which is merely almost complex.

  • For reasons as in the previous point, the $\overline{\partial}$-Poincaré Lemma and Local $i\partial \overline{\partial}$-Lemma only work on complex manifolds.

  • Holomorphic vector bundles require both the total space and the base to be complex manifolds. Note that we need holomorphic vector bundles to talk about the Chern connection.

There are probably one or two more points that I'm missing (e.g., regarding Hodge stars and Laplacians (when a Hermitian metric is present)), but I think the above should be the main differences.

As for textbook recommendations, I like these two:

  • Andrei Moroianu, "Lectures on Kahler Geometry."

  • Daniel Huybrechts, "Complex Geometry."

Jesse Madnick
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    Thanks. About the first point, I guess you mean that the changes of coordinates on overlapping open sets are holomorphic, because just local complex coordinates can be defined on any almost-complex manifold (and the corresponding $\frac{\partial}{\partial z}$. Of course in that case changes of coordinates do not preserve the local holomorphicity condition. – Bilateral Apr 14 '15 at 23:15
  • @Bilateral: Yes, that's exactly right, thanks. Sorry, I was being careless. – Jesse Madnick Apr 14 '15 at 23:17