Let $M$ be a smooth homogeneous $G$space for a Lie group $G$, and let $J$ be a $G$invariant almostcomplex structure for $M$. Do there exist succinct sufficient (and necessary) conditions for $J$ to be integrable? Besides the six sphere, what other examples of a nonintegrable invariant almostcomplex structure for a smooth homogeneous space are there? Do there exist nonintegrable almostcomplex structures for any flag manifolds?

$\begingroup$ When you give flag manifolds as examples of homogeneous spaces, do you mean homogeneous for $G$ the compact group? (I'm pretty sure you are.) $\endgroup$– Allen KnutsonNov 19, 2011 at 21:10

$\begingroup$ Yes, that's what I mean $\endgroup$– Ago SzekeresNov 19, 2011 at 22:53
3 Answers
Oops, I was thinking of a Lie group. Edit: On a Lie group, it is pretty easy to test for integrability. Take a basis of complex linear left invariant 1forms (i.e. left translate from a choice of such 1forms on the Lie algebra). Then compute their exterior derivatives. You have an integrable almost complex structure if and only if the (0,2) parts of the exterior derivatives all vanish. To see this, you express the exterior derivatives in linear combinations of the wedge products of the original 1forms and their complex conjugates. So examples are easy to check, and only require the differential familiar from Lie algebra cohomology, i.e. a pure Lie algebra calculation.
On a homogeneous space $G/H$ the computation is a little trickier. You take $\omega=g^{1}dg$, the left invariant Maurer Cartan form on $G$, and then $\omega+\mathfrak{h}$ is semibasic for the quotient map $G \to G/H$ and splits into a complex linear part and a conjugate linear part on each tangent space of $G/H$. Let $\eta$ be the complex linear part. Pick a splitting of $\mathfrak{g}$ into $\mathfrak{g}/\mathfrak{h}$ and some complement, identified with $\mathfrak{h}$, and let $\Omega$ be the projection of $\omega$ to complement. The equation $d \omega + (1/2)[\omega,\omega]=0$ gives an equation $d \eta =  \Omega \wedge \eta + a \eta \wedge \eta + b \eta \wedge \bar{\eta} + c \bar{\eta} \wedge \bar{\eta}$. The Nijenhuis tensor vanishes just when $c=0$. I am pretty sure that the generic invariant almost complex structure on a flag manifold (invariant under the compact form of the automorphism group) is not integrable, but I would have to check.

$\begingroup$ Thanks for your answer Ben. But to have a basis of leftinvariant wouldn't my homogeneous need to be parallelizable? Or do you mean on the Lie group? In which case I don't understand your answer. $\endgroup$ Nov 19, 2011 at 22:59
Presumably, you are looking for something other than "the Nijenhaus tensor, possibly simplified by the action of $(\rho_{G})_*$"?
In the local case, there have been a few studies constructing families of almostcomplex manifolds in low dimensions, along with conditions for varying amounts of notquiteintegrability. You might look at http://projecteuclid.org/euclid.ajm/1175789046 (Robert Bryant's AJM paper on almostcomplex 6manifolds and the references therein). However, this only addresses the local question and even then would require refinement to study the case of homogeneous or flag manifolds. I suspect very little is known otherwise.
As for examples: A nearly Kähler manifold, which is not Kähler, always has a nonintegrable almost complex structure. You mention $S^6$, there are three more compact simply connected homogoneous nearly Kähler manifolds on the spaces: $\mathbb{CP}^3,S^3\times S^3$ and also on the flag manifold $\mathrm{SU}(3)/{\mathbb{T}^2}$. You can also consider a compact semisimple Lie group $G$ with maximal torus $T$. Then $\mathfrak{g}\otimes \mathbb{C}$ splits as a sum of $\mathfrak{t}^2\otimes \mathbb{C}$ and a sum of root spaces. A $G$invariant almost complex structure on $G/T$ is then same as an appropriate choice of a subset of root spaces. The integrability is then read off from the commutator relationships between the root spaces.