# Does $\Bbb{CP}^{2n} \mathbin{\#} \Bbb{CP}^{2n}$ ever support an almost complex structure?

This question has now been crossposted to MathOverflow, in the hopes that it reaches a larger audience there.

$\Bbb{CP}^{2n+1} \mathbin{\#} \Bbb{CP}^{2n+1}$ supports a complex structure: $\Bbb{CP}^{2n+1}$ has an orientation-reversing diffeomorphism (complex conjugation!), so this is diffeomorphic to the blowup of $\Bbb{CP}^{2n+1}$ at one point.

On the other hand, $\Bbb{CP}^2 \mathbin{\#} \Bbb{CP}^2$ does not even support an almost complex structure: Noether's formula demands that its first Chern class $c_1^2 = 2\chi + 3\sigma = 14$, but if $c_1 = ax_1 + bx_2$ (where $x_1, x_2$ generate $H^2$, $x_1^2 = x_2^2$ is the positive generator of $H^4$, and $x_1x_2 = 0$), then $c_1^2 = a^2 + b^2$, and you cannot write $14$ as a sum of two squares.

Using a higher-dimensional facsimile of the same proof, I wrote down a proof here that $\Bbb{CP}^4 \mathbin{\#} \Bbb{CP}^4$ does not admit an almost complex structure. The computations using any similar argument would, no doubt, become absurd if I increased the dimension any more.

Can any $\Bbb{CP}^{2n} \mathbin{\#} \Bbb{CP}^{2n}$ support an almost complex structure?

• In the answer you link to, you use the eight-dimensional version of the Wu theorem. Is there a known $4n$-dimensional version of the Wu theorem? I didn't know about the eight-dimensional version prior to reading your answer. Aug 31 '15 at 21:31
• @MichaelAlbanese: I don't know of one. The article works as follows: the first three formulae are equivalent to the existence of a stable complex structure; this may be an easier problem (but $\Bbb{CP}^4 \# \Bbb{CP}^4$ does support a stable ACS), and one can promote a stable ACS to an actual ACS iff $c_n(X) = \chi(X)$; and the last formula is precisely this statement. This only appears to work when $M$ is $8n$-dimensional. Perhaps one could follow a similar, but much more difficult, approach for arbitrary $8n$-dimensional manifolds.
– user98602
Aug 31 '15 at 21:37
• E: the first formula is equivalent to the existence of a stable ACS; the next two are formulae that its first and third chern classes must satisfy, solely for use in the last formula.
– user98602
Aug 31 '15 at 21:56
• It anyways has stable complex structure, doesnt it? Feb 18 '17 at 8:15
• @Rick_Student Yeah, but that doesn't really convince me of much either way.
– user98602
Feb 18 '17 at 20:23

## 1 Answer

The question has been answered in the crosspost to mathoverflow by Panagiotis Konstantis. I copy this answer here to close the question.

The $m$-fold connected sum $m\# {\mathbb{CP}}^{2n}$ admits an almost complex structure if and only if $m$ is odd, as we show in our recent [preprint][1]. (By the way, thanks to Mike for this interesting question, which motivated us to write the paper!)

Here's a brief summary of the proof's idea. Our main tool is a result by Sutherland resp. Thomas from the 60s which tells us when a stable almost complex structure is induced by an honest almost complex structure: this is the case iff its top Chern class equals the Euler class of the manifold.

As the connected sum of manifolds admitting a stable almost complex structure admits one as well, we certainly have stable almost complex structures on $m\# {\mathbb{CP}}^{2n}$ at our disposal, and we can understand the full set of all such structures by explicitly determining the kernel of the reduction map from complex to real K-theory. We then compute the top Chern class of all these structures: luckily for us, it turns out that in order to show the non-existence of almost complex structures for even $m$, it suffices to compute its value modulo 4 and compare it to the Euler characteristic of $m\# > {\mathbb{CP}}^{2n}$. For odd $m$, we explicitly find a stable almost complex structure for which the criterion above is satisfied.

• As is pointed out in the paper, the fact that $m\mathbb{CP}^{2n}$ does not admit an almost complex structure for $m$ even (in particular, $m = 2$) follows from a result of Hirzebruch which states that the Euler characteristic and signature of a closed almost complex manifold $M$ of dimension $4n$ satisfies $\chi(M) \equiv (-1)^n\sigma(M) \bmod 4$. This is proved using the Hirzebruch $\chi_y$ genus, see here for example. Sep 10 '18 at 14:17