Complex structure on $\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2$

Question

I know the chern classes-related theorem that states that $\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2$ ($k$ times) has no almost complex structure (hence no complex structure) if and only if $k$ is even.

I also know that $\mathbb{C}\mathbb{P}^2\# \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ has no complex structure, hence the almost complex ones you define on it are not integrable.

Where can I find a proof of this proposition? How do I see that the connected sum of three (or five, or seven, or every odd number) copies of $\mathbb{C}\mathbb{P}^2$ doesn't admit a complex structure?

related: Almost complex structure on $\mathbb CP^2 \mathbin# \mathbb CP^2$ — Grigory M, May 08 '14 at 17:34
related: How can we detect the existence of almost-complex structures? @ MO — Grigory M, May 08 '14 at 17:35
yeah thanks, that is the theorem about Chern classes I was referring to, if anyone wondered :) — AnalysisStudent0414, May 08 '14 at 17:35
@mezhang Connected sum (the notation is absolutely standard) — Grigory M, May 08 '14 at 19:32

Michael Albanese · Answer 1 · 2017-04-12T01:28:52.440

If you just want to know the answer for the standard smooth structure, you can prove this result by using Seiberg-Witten invariants.

Suppose $k\mathbb{CP}^2$ (the connected sum of $k$ copies of $\mathbb{CP}^2$) admits a complex structure for some $k \geq 2$.

First of all, a compact complex surface is Kähler if and only if its first Betti number is even. As $k\mathbb{CP}^2$ is simply connected, if it admits a complex structure, it is Kähler and hence $k\mathbb{CP}^2$ is symplectic. As $b^+(k\mathbb{CP}^2) = k \geq 2$, it has a non-trivial Seiberg-Witten invariant.

However, by a theorem of Taubes, if $M = M_1\# M_2$ and $b^+(M_1), b^+(M_2) \geq 1$, then $M$ has no non-trivial Seiberg-Witten invariants. Alternatively, as $\mathbb{CP}^2$ admits a positive scalar curvature metric (e.g. the Fubini-Study metric), so does $k\mathbb{CP}^2$ by a result of Gromov and Lawson, but such a space must have trivial Seiberg-Witten invariants.

Therefore, $k\mathbb{CP}^2$ does not admit a complex structure for any $k \geq 2$.

score 4 · Accepted Answer · answered May 11 '14 at 23:12

This is a corollary of a theorem on page 376 in "Compact complex surfaces" by Barth, Hulek, Peters and van de Ven:

If $X$ is a simply-connected compact complex surface with definite intersection form $q$ then $X$ is $CP^2$.

Note that in your case, $q$ has the identity matrix, hence, is positive-definite.

Complex structure on $\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2$

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