# Is the connected sum of complex manifolds also complex?

Let $M$ and $N$ be real manifolds of dimension $n$ which happen to admit complex structures (so that necessarily $n=2k$ and both are orientable). Then does their connected sum $M\# N$ also admit a complex structure?

This is true for $n=2$, because every oriented $2$-dimensional topological manifold admits a complex structure. However, this operation doesn't seem to be compatible with the information given by the structure; in particular we would want $M\sqcup N$ to at least be almost-complex cobordant to $M\# N$, or that these two manifolds should have the same Chern numbers. We have $$c_1[M\sqcup N]=\langle c_1(\tau M\sqcup\tau N),[M\sqcup N]\rangle = \langle c_1(\tau M),[M]\rangle + \langle c_1(\tau N),[N]\rangle = c_1[M]+c_1[N]$$ But then recall that the first Chern class of a complex rank 1 bundle $V$ is the Euler class of its realification, and so the first Chern number of a complex 1-manifold is its Euler characteristic. Thus $$c_1[M\# N] = \chi(M\#N)=\chi(M)+\chi(N)-2 \neq c_1[M] + c_1[N]$$ So it seems like it's possible that the fact that complex $1$-manifolds are closed under connected sum could be a coincidence.

In higher dimensions there are no complex structures on $S^{2k}$ (except MAYBE $S^6$). But, in the positive direction, there ARE these complex Calabi-Echkmann manifolds homeomorphic to $S^{2k+1}\times S^{2l+1}$. So does anyone know if, for example, the connected sum of $g$ copies of $S^5\times S^5$ admits a complex structure for any $g>1$?

• This is a very late question. But what is almost complex cobordant? The cobordism will be odd dimensional so there is no almost complex structure on it. Dec 22, 2022 at 1:57

$X = \Bbb{CP}^4 \#\Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.

Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $\Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;\Bbb Z)$ as $x_1, x_2$.

1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.

2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.

3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $\Bbb{CP}^2 \vee \Bbb{CP}^2$. Because $X_4 \hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X \to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f \vee f$, where $f$ is the restriction of the classifying map of $\Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(\Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$ Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.

Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2\ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).

Then because $\chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+\ell n) + 16(m+n) - (2k+1)^4 - (2\ell+1)^4 + 10(2k+1)^2 + 10(2\ell+1)^2 - 50.$$

Simplifying we get $$80 = 32(km + \ell n - k^3 - \ell^3 + k^2 + \ell^2) + 16(k+\ell - k^4 - \ell^4 +m +n)$$ Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 \equiv 16(k+\ell - k^4- \ell^4) \mod 32$$ But $k+\ell$ is odd iff $k^4 + \ell^4$ is odd, so the right side is $0 \mod 32$. This is a contradiction, as desired.

That this was so much work suggests that this is, uh, the wrong approach to prove that $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.

• You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice. Sep 23, 2015 at 21:25
• @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
– user98602
Sep 23, 2015 at 22:09

EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $$K3 \# \overline{K3}$$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $$46$$, and $$\chi + \sigma$$ is always divisible by $$4$$ on an almost complex 4-manifold.

I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $$M \# N$$ can support a complex structure when $$M$$ and $$N$$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.

Let $$M$$ and $$N$$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $$M \# \overline N$$ has a natural complex structure - just line up the copies of $$\Bbb C^n \setminus \{0\}$$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.

So, if $$N$$ supports a complex structure, does $$\overline N$$? This would mean that $$M \# N$$ supports a complex structure. If $$\text{dim}(N) = 4n+2$$, then this is true: if $$J$$ is your (integrable almost) complex structure, then the complex conjugate $$\overline J$$ gives a complex structure on $$\overline N$$. The reason this doesn't work in dimension $$4n$$ is because $$\overline J$$ induces the same orientation as $$J$$!

So I wouldn't quite call it a fluke. It's true for every $$4n+2$$-dimensional complex manifold, not just $$2$$-dimensional ones. But it's not true in every dimension.

For 4-manifolds, it is a theorem of Wu that $$M$$ admits an almost complex structure with $$c_1(J) = c \in H^2(M;\Bbb Z)$$ if and only if $$c$$ reduces to $$w_2$$ mod 2, and $$c^2 = 3\sigma + 2\chi$$. You can prove using this criterion that a most two of $$M, N$$, and $$M \# N$$ can admit almost complex structures; in particular, $$\Bbb{CP}^2 \# \Bbb{CP}^2$$ cannot admit an almost complex structure.

I don't know about higher dimensions $$4n$$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $$> 2$$) examples of almost complex manifolds that do not admit a complex structure.

• Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure. Aug 27, 2015 at 18:10
• Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $\#^g S^5\times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$. Aug 28, 2015 at 1:29
• @William The connected sum will admit an almost complex structure if and only if g is congruent to 1 mod 24; this follows from Theorem 2 in "Almost complex structures on (n-1)-connected 2n-manifolds" by Huijun Yang sciencedirect.com/science/article/pii/S0166864111006080 Apr 26, 2020 at 21:06