How can we detect the existence of almost-complex structures? - MathOverflow

How can we detect the existence of almost-complex structures?

2011-04-29T16:46:39Z

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).

Answer by Sean Tilson for How can we detect the existence of almost-complex structures?

2011-04-29T17:06:18Z

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.

Answer by Francesco Polizzi for How can we detect the existence of almost-complex structures?

2011-04-29T20:51:05Z

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.

Answer by Joel Fine for How can we detect the existence of almost-complex structures?

2011-04-29T21:46:16Z

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.