Why doesn't $\mathbb CP^2 \mathbin\# \mathbb CP^2$ admit an almost complex structure?
Almost complex structure on $\mathbb CP^2 \mathbin\# \mathbb CP^2$
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differential-geometry
algebraic-topology
complex-geometry
almost-complex
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0In 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. – 2015-06-22
1 Answers
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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 $e+\tau = 6 \cong 2 \pmod 4$.
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0You 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. – 2016-09-20