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I have a (basic?) question in topology.

Question 1. Is it possible to characterise compact $4$-manifolds $M^4$, such that almost complex structures on $M^4$ are uniquely defined up to homotopy by their first Chern classes? Or is there at least a large class of $4$-manifolds where this is true? Maybe there is a reference?

Recall the theorem of Wu. Denote by $\tau$ the signature of $M^4$ and by $e$ the Euler characteristics. Then for any element $c\in H^2(M^4,\mathbb Z)$ with $c^2=3\tau+2e$ and such that $c\; {\rm mod} \;\mathbb Z_2= w_2\in H^2(M^4,\mathbb Z_2)$ there is an almost complex structure $J$ on $M^4$, such that $c_1(M,J)=c$ (in particular, as Paul says $e+\tau=0\; {\rm mod} \;4$). However this $J$ might be non-unique, up to homotopy among almost complex structures with $c_1=c$, as the following example shows.

Example. Let $M^4=S^1\times S^3$, then the tangent bundle is trivial, so almost complex structures on $M^4$ can be identified up to homotopy with homotopy classes of maps $M^4\to S^2$ (see page 11 in The geometry of Four-manifolds Donaldson Kronheimer). Maps $S^3\to S^2$ can have different Hopf invariants, but $c_1=0$ since $S^1\times S^3$ has no second cohomology...

Added Question 2. Is there at least one manifold that satisfies condition of Question 1?

In McDuff-Salamon (footnote on page 120) it is written, that if $M^4$ is spin then there are precisely two homotopy classes of $J$ with given $c_1$ (this is said to be related to $\pi_4(S^2)=\mathbb Z_2$). But since $S^1\times S^3$ is spin this statement from McDuff-Salamon seem to contradict to my conclusion (that for $S^1\times S^3$ homotopy classes are can have different Hopf invariants). So, where is the mistake?...:)

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If $M$ admits a almost complex structure, $e+\tau=0$ mod 4 ($e$=Euler characteristic, $\tau$=signature).So eg $CP^2\# CP^2\# CP^2$ doesnt admit an almost complex structure, even though $c=(3,3,1)$ satisfies $c^2=19= 2e+3\tau$. Also,if I′m not mistaken,$[S^1\times S^3, S^2]=\pi_3(S^2)\oplus \pi_4(S^2)$: the cofibration sequence $$S^1 v S^3\to S^1\times S^3\to S^4 $$ gives the exact sequence $[S^4,S^2]\to [S^1\times S^3,S^2]\to [S^1 v S^3,S^2]$, and $[S^1 v S^3, S^2]=\pi_2S^2 $ (I think), so the Hopf invariant misses the $Z/2=\pi_4(S^2)$. –  Paul Mar 29 '11 at 2:50
    
(make that $\pi_3S^2$ in the last sentence) –  Paul Mar 29 '11 at 2:52
    
Paul $5+3=0$ mod $4$, so your first sentence is wrong – –  Dmitri Mar 29 '11 at 9:34
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Dmitri: You're right. make that $4 CP^2$. –  Paul Mar 29 '11 at 11:37
    
Paul, thanks! I it seems that your are right with $e+\tau=0$ and $4CP^2$ does not have an almost complex structure. I'll change this –  Dmitri Mar 29 '11 at 12:28
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1 Answer 1

up vote 14 down vote accepted

This can be answered by obstruction theory for the fibration $$ F=SO(4)/U(2) \to BU(2) \to BSO(4) $$ where the fibre is actually a 2-sphere: $F=S^2$. Start with the tangent bundle of an oriented 4-manifold $M$ and ask for existence respectively uniqueness of a lift of its Gauss-map $M\to BSO(4)$, that is, existence respectively uniqueness for an almost complex structure on $M$.

The obstructions for existence lie in $H^{i+1}(M;\pi_i(F))$ and the obstructions for uniqueness lie in $H^{i}(M;\pi_i(F))$. Assume for simplicity that $M$ is simply connected and closed and let $X$ be its 2- (and hence) 3-skeleton. The same obstruction theory tells you that the tangent bundle of $X$ has a complex structure and that it's uniquely determined by the first Chern-class. If such a structure extends to $M$, there is a single uniqueness obstruction in $$ H^{4}(M,X;\pi_4(F)) = \pi_4(S^2) = Z/2$$ By naturality, this obstruction is realized by a second almost complex structure on $M$ (with the same first Chern-class) if and only if the quotient map $M \to M/X=S^4$ induces a nontrivial map on cohomotopy: $$ \pi_4(S^2) = [S^4,S^2] \to [M,S^2] $$ It is well known that this map is nontrivial if and only if $M$ is spin, see for example the cohomotopy preprint with Kirby and Melvin on my homepage.

This explains the footnote in McDuff-Salamon (where they probably assume that $M$ is simply connected, otherwise there is another uniqueness obstruction in $H^3(M)$ that you found for $M = S^1 \times S^3$). It also answers your question for simply connected almost complex manifolds $M$ as follows:

The first Chern-class characterizes almost complex structures on $M$ if and only if $M$ is not spin. Good examples are those Hirzebruch surfaces which are nontrivial $S^2$-bundles over $S^2$.

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Thanks a lot, this is very helpful! –  Dmitri Aug 21 '11 at 9:12
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