9

Why doesn't $\mathbb CP^2 \mathbin\# \mathbb CP^2$ admit an almost complex structure?

Michael Albanese
  • 104,029
Maciej Starostka
  • 660
  • In general, if $X$ and $Y$ are compact four-manifolds which admit almost complex structures, then $X# Y$ does not. Even more generally, if two of $X, Y, X# Y$ admit almost complex structures, the third does not. Both of these facts follow from the necessary condition stated in Jason DeVito's answer. – Michael Albanese Jun 22 '15 at 15:20

1 Answers1

11

According to the comments here, a necessary condition for a $4$-manifold to admit an almost complex structure is that $e+\tau = 0 \pmod 4$, where $e$ is the Euler characteristic and $\tau$ is the signature. (I don't know of a reference for that fact).

But $\mathbb{C}P^2\sharp \mathbb{C}P^2$ has euler Characteristic $4$ (its homology groups are that of $S^2\times S^2$) and its signature $\tau$ is $2$ (since the identity represents the bilinear form in the "usual" basis of $H^2(\mathbb{C}P^2\sharp\mathbb{C}P^2)$.)

Then $e+\tau = 6 \cong 2 \pmod 4$.

Community
  • 1
Jason DeVito - on hiatus
  • 51,620
  • If anyone happens to know of a reference for the fact that $e +\tau = 0 \pmod{4}$ for a $4$-manifold admitting an almost complex structure, I'd love to know one. – Jason DeVito - on hiatus Feb 28 '16 at 19:31
  • You can deduce this from knowing that $c_1^2$ is equivalent to $\tau$ mod 2 (and using the formula from Wu involving $c_1^2$ and $p_1$), which is ultimately a result of $c_1$ being a characteristic element of the cup-product pairing on $H^2$. I would imagine that stuff like this is found in Milnor's book on quadratic forms, or Scorpan's "Wild World" book. – Chris Gerig Sep 20 '16 at 17:46