As is well known, there exists a fibration $\mathbb{CP}^3 \to S^4$, of the four sphere by complex projective $3$space, called the Penrose twistor fibration. Does this fibration admit a "canonical" generalisation to a fibration of $S^6$ by some projective $n$space, or possibly flag manifold. What about general $2n$spheres?

$\begingroup$ The Whitney sum formula for Pontryagin classes imposes strong conditions on the fiber tangent bundle of such a fibration. Similarly, the long exact homotopy sequence and the Leray spectral sequence impose conditions. It looks like the fibers must be $\mathbb{CP}^{n3}$. Then the cohomology is determined by $c_3$ of the associated $SU(n2)$bundle. On $BSU(n2)$, does $c_3$ pair as $\pm 1$ with a generator of $\pi_6(BSU(n2)) \cong \mathbb{Z}$? If not, there is no fibration of $\mathbb{CP}^n$ by $\mathbb{CP}^{n3}$ over $S^6$ (analogous to the Penrose fibration). $\endgroup$ – Jason Starr Aug 13 '15 at 8:17
Yes, there is such a twistor fibration over each $S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic $n$plane field transverse to the fibers of the mapping. Namely, one writes $S^{2n} = \mathrm{SO}(2n{+}1)/\mathrm{SO}(2n)$ and then, using the inclusion $\mathrm{U}(n)\subset\mathrm{SO}(2n)$, one has the coset fibration $$ Z_n = \mathrm{SO}(2n{+}1)/\mathrm{U}(n)\longrightarrow\mathrm{SO}(2n{+}1)/\mathrm{SO}(2n). $$ The manifold $\mathrm{SO}(2n{+}1)/\mathrm{U}(n)$ canonically has the structure of a complex manifold and is now known as the twistor space of $S^{2n}$. Interestingly, this space was already used in 1967 by Calabi in his study of minimal $2$spheres in $S^{2n}$ (Minimal immersions of surfaces in Euclidean spheres, J. Differential Geom. 1 (1967), 111–125), where, using a quite different language, he showed that such immersions all have lifts to the twistor space as holomorphic curves that are tangent to a certain holomorphic plane field on $Z_n$.
Now, there are generalizations of this picture for each of the socalled 'inner' symmetric spaces $G/K$ where $K$ is the fixed subgroup of an involution that is an inner automorphism of $G$. The twistor fibration is of the form $G/U\to G/K$ where $U\subset K$ is a subgroup such that $K/U$ (the typical fiber of the fibration) is an Hermitian symmetric space. (There are also other kinds of twistor spaces over $G/K$ that are flag manifolds of the form $G/T$ where $T\subset K$ is a maximal torus.)
For more details on these topics, especially the generalization to the inner symmetric spaces and flag manifolds, one can, for example, have a look at my paper Lie groups and twistor spaces, Duke Mathematical Journal 52 (1985), pp. 223–261.

1$\begingroup$ Beautiful answer as always ! Sorry for not citing you in my answer. I digged a little hoping to add better references and discovered Burstall & Rawnsley's book (MR1059054), citing in particular your work. $\endgroup$ – BS. Aug 15 '15 at 10:26

$\begingroup$ A long time since I've thought about this but maybe worth saying the $(4n+2)$dimensional case differs from the $4n$dimensional: you get a pair of "twistor spaces" with an antiholomorphic isomorphism between them instead of a real structure. The case of the six sphere mentioned in the OP is especially interesting because you end up with a complex 6dimensional quadric with a real structure (the original $S^6$) and another pair of 6dimensional quadrics with the antiholomorphic map between them. I seem to remember this paper: sciencedirect.com/science/article/pii/0393044095000364 $\endgroup$ – Oliver Nash Aug 16 '15 at 20:25
One possible generalization of Penrose twistor fibration is the fibration $\pi:\mathcal{J}_M\to M$ over any oriented $2n$dimensional conformal manifold $(M,[g])$, whose fiber $\pi^{1}(x)$ is the space of orthogonal almost complex structures on $(T_xM,[g_x])$, namely the orthogonal $J_x:T_xM\to T_xM$ squaring to $\mathrm{id}$ such that the complex orientation $(e_1,Je_1,e_2,Je_2,\dots)$ is the given one.
For $n=2,3$, the fibers are projective spaces $\mathcal{J}_4=\mathbb{CP}^1$, $\mathcal{J}_6=\mathbb{CP}^3$, and for general $n$ complex projective varieties of dimension $n(n1)/2$. This is because $\mathcal{J}_{2n}$ identifies with the grassmannian of $n$dimensional totally isotropic subspaces of $\mathbb{C}^{2n}$ for any non degenerate quadratic form, e.g. $z_1 w_1+\dots+ z_n w_n$. For $n=4$, one gets $\mathcal{J}_{8}=Q_6$, the six dimensional smooth complex quadric hypersurface in $\mathbb{CP}^7$. As a real manifold, it is also the homogeneous space $SO(2n)/U(n)$, with the central $U(1)$ giving the complex structure on tangent spaces.
For $M=S^{2n}$ the round sphere, the total space $\mathcal{J}_{S^{2n}}$ identifies with $\mathcal{J}_{2n+2}$, by identifying $T_xS^{2n}\oplus\mathbb{R}^2$ with $T_xS^{2n}\oplus\mathbb{R}x\oplus\mathbb{R}e=T_x\mathbb{R}^{2n+1}\oplus\mathbb{R}e=\mathbb{R}^{2n+2}$ in the natural way, and any $J_x \in \mathcal{J}_{S^{2n}}$ to $\hat J_x\in \mathcal{J}_{2n+2}$ which coincides with $J_x$ on $T_xS^{2n}$ and sends [edit] $x$ to $e$ (and $e$ to $x$) [/edit].
So the fibration over $S^6$ is $\mathbb{CP}^3\to Q_6\to S^6$, and in general $\mathcal{J}_{2n}\to\mathcal{J}_{2n+2}\to S^{2n}$. Note that, contrary to the case $n=2$, the $n=3$ fibration has a section, meaning that $S^6$ has an orthogonal almost complex structure (noted by Ereshman, but maybe ealier).
I don't have a reference at hand, but a first guess is that it should be found in Lawson & Michelsohn "Spin Geometry" and maybe also in Harvey "Spinors and calibrations".
The link to spinors is convenient to identify $\mathcal{J}_{2n}$ for low $n$, because it identifies with the projectivization of the cone of "pure (complex) spinors" of $\mathbb{R}^{2n}$, which are all of them for $n\leq 3$, and a quadric cone for $n=4$.

$\begingroup$ For the $\mathbb{CP}^3$bundle over $S^6$ whose total space is $Q_6$, for the associated $SU(4)$bundle over $S^6$, $c_3$ equals $2$ times the Poincare dual of a point class. $\endgroup$ – Jason Starr Aug 14 '15 at 23:55

$\begingroup$ @Jason Starr : it's a long time since you posted this, but I missed it somehow. May I ask what was your point ? $\endgroup$ – BS. Oct 20 '19 at 7:43