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Any smooth $k$-manifold $M$ comes with a well-defined map $f:M\rightarrow BGL_{k}(\mathbb{R})$ (up to homotopy) classifying its tangent bundle. Since $GL_{k}(\mathbb{R})$ deformation-retracts onto $O_k$, then $BGL_{k}(\mathbb{R})\simeq BO_k$, which is a cute way (though it's certainly overkill) of proving that every smooth manifold admits a Riemannian metric. An almost-complex structure, on the other hand, is equivalent to a reduction of the structure group from $GL_{2n}(\mathbb{R})$ to $GL_n(\mathbb{C})$, which is the same as asking for a lift of the classifying map through $BU_n\simeq BGL_n(\mathbb{C})\rightarrow BGL_{2n}(\mathbb{R})$.

Can we detect the nonexistence of a lift entirely using characteristic classes? If not, what else goes into the classification?

I'd imagine these don't suffice themselves. I know that $w_{2n}(TM) \equiv_2 c_n(TM)$, so this holds in the universal case $H^\ast(BO_{2n};\mathbb{Z}/2) \rightarrow H^\ast(BU_n;\mathbb{Z}/2)$. And certainly there are necessary conditions like $w_1(TM)=0$ (which of course just means that $TM$ is an orientable bundle, which is the same as asking that $M$ be an orientable manifold). But I have no idea of what sufficient conditions would look like. I've heard that this problem is indeed solved. Maybe it takes some characteristic class & cohomology operation gymnastics, or maybe it even needs extraordinary characteristic classes. Or maybe there's yet another ingredient in the classification?

Edit: Apparently I misquoted my source, and this is only known stably (which makes sense, in light of Joel's answer and Tom's comments on it).

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Since this is a lifting problem, there are the standard cohomological obstructions (as in Hatcher Section 4.3). But that's not really an answer to the question you asked. –  Dan Ramras Apr 29 '11 at 17:05
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3 Answers

up vote 10 down vote accepted

Edit: Now updated to include reference and slightly more general result. Edit 2: Includes remark about integrability.

Similar to Francesco Polizzi's answer, there is the following Theorem concerning 6-manifolds.

A closed oriented 6-dimensional manifold $X$ without 2-torsion in $H^3(X,\mathbb{Z})$ admits an almost complex structure. There is a 1-1 correspondence between almost complex structures on $X$ and the integral lifts $W \in H^2(X, \mathbb{Z})$ of $w_2(X)$. The Chern classes of the almost complex structure corresponding to $W$ are given by $c_1 = W$ and $c_2 = (W^2 - p_1(X))/2$.

In fact, a necessary and sufficient condition for the existence of an almost complex structure is that $w_2(X)$ maps to zero under the Bockstein map $H^2(X,\mathbb{Z}_2) \to H^3(X,\mathbb{Z})$.

I think the reason for results such as this and the one mentioned by Francesco is the following. To find an almost complex structure amounts to finding a section of a bundle over $X$ with fibre $F_n=SO(2n)/U(n)$. The obstructions to such a section existing lie in the homology groups $H^{k+1}(X, \pi_k(F_n))$. When $n$ is small I would guess we can compute these homotopy groups and so have a good understanding of the obstructions. For example, in the case mentioned above, n=3, $F_n = \mathbb{CP}^3$ and so the only non-trivial homotopy group which concerns us is $\pi_2 \cong \mathbb{Z}$. This is what leads to the above necessary and sufficient condition concerning 2-torsion. On the other hand when $n$ is large I don't know what $F_n$ looks like, let alone its homotopy groups...

For the proof of the above mentioned result see the article "Cubic forms and complex 3-folds" by Okonek and Van de Ven. (I highly recommend this article, it's full of interesting facts about almost complex and complex 3-folds.)

It is worth pointing out that in real dimension 6 or higher there is no known obstruction to the existence of an integrable complex structure. In other words, there is no known example of a manifold of dimension 6 or higher which has an almost complex structure, but not a genuine complex structure. By the classification of compact complex surfaces, those 4-manifolds admitting integrable complex structures are well understood.

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A couple of comments about the spaces $O_{2n}/U_n$ that Joel Fine mentions: 1. $O_{2n+2}/U_{n+1}\cong O_{2n+1}/U_n$ is a bundle over $O_{2n+1}/O_{2n}=S^{2n}$ with fiber $O_{2n}/U_n$, making $O_{2n+2}/U_{n+1}$ a kind of twisted product of the spheres $S^0$, $S^2$, $S^4$, ... , $S^{2n}$ (although this doesn't give you a lot of insight right away into the homotopy groups). –  Tom Goodwillie Apr 30 '11 at 12:29
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2. The limiting space $O/U$ is something whose homotopy groups are known. It is part of the story of Bott periodicity: it is in fact homotopy equivalent to $\Omega O$. So the obstruction to putting a <i>stable</i> complex structure on a given real vector bundle is some kind of real $K$-theory class. –  Tom Goodwillie Apr 30 '11 at 12:29
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Thanks! I like the homotopy theory in here. By way, just to clarify: when you say there is no known obstruction ..., do you mean that there are no almost-complex manifolds that are known not to admit complex structures? The Nijenhuis tensor is supposed to tell us about any particular almost-complex structure that we happen to care about, if I understand correctly. –  Aaron Mazel-Gee May 3 '11 at 8:45
    
Yep that's right: there is no known example of a manifold of dimension 6 or higher which admits an almost complex structure but does not admit a complex structure. I agree my original wording was very poor, so I've changed it so it actually makes sense! –  Joel Fine May 3 '11 at 20:44
    
@Tom For obstruction theory we need $\pi_k(O_2n/U_n)$ only for $k<n$. Aren't they in the stable range already? (So unstable problem seems to be equivalent to the stable one?..) –  Grigory M Jun 26 '11 at 10:05
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If $M$ is a $4$-manifold, the existence of an almost-complex structure can be often detected by using the following result due to Wu:

Theorem. A $4$-manifold $M$ admits an almost-complex structure $J$ if and only if there exists $h \in H^2(M, \mathbb{Z})$ such that

$h^2=3 \sigma(X)+2 \chi(X) \quad \textrm{and} \quad h \equiv w_2(X) (\ \textrm{mod} \ 2).$

In this case $h=c_1(M, J).$

See Gompf-Stipsicz, "4-manifolds and Kirby calculus", p. 30 for more details.

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This is more of a comment:

Maybe you have already thought of this, or maybe there is something I am missing. Or perhaps this isn't the answer you want.

Suppose $M$ is a smooth (finite-dimensional) compact manifold (without boundary). Let $f: M \to BO$ be the map that classifies it's stable normal bundle (or "equivalently" it's tangent bundle). Then an almost complex structure is a lift of $f$ over $i: BU \to BO$ to $\widetilde{f} : M \to BU$. So to see if we can get such a lift can't we just compose with $f$ with $j: BO \to Cof(i)$? Perhaps this is easier said than done. But we can apply homtopy and see what we get.

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Is that right? Even if you get a complex structure on the stable normal bundle, this may not sit as a complex subbundle of the tangent bundle of $\mathbb{R}^N$, and then you'd be stuck. In any case, I'm confused: why do you prefer to work with the normal bundle anyways? –  Aaron Mazel-Gee Apr 29 '11 at 20:51
    
I am pretty sure that is just the one people work with anyway. In the references I have seen, or can recall (I would be happy for counterexamples) structures on spaces are on the stable normal bundle, or at least almost complex structures are. I am not entirely sure why people work with it instead of the tangent bundle, one possible reason is the Pontryagin-Thom construction. Also, the stable normal bundle data is "equivalent" to tangent bundle data. I apologize I can not make any of this significantly more precise. –  Sean Tilson Apr 30 '11 at 3:12
    
Well, knowing the stable normal bundle is equivalent to knowing the stable tangent bundle since they're inverses in $KO(X)$. But it seems like it would be an equivalently hard problem to destabilize information about the stable tangent bundle, perhaps similarly to how it's hard to figure out the spheres of origin for stable homotopy classes of maps between spheres. –  Aaron Mazel-Gee Apr 30 '11 at 18:16
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