# Generalising the Penrose Twistor Fibration

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?

• 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 $\mathbb{CP}^{n-3}$. Then the cohomology is determined by $c_3$ of the associated $SU(n-2)$-bundle. On $BSU(n-2)$, does $c_3$ pair as $\pm 1$ with a generator of $\pi_6(BSU(n-2)) \cong \mathbb{Z}$? If not, there is no fibration of $\mathbb{CP}^n$ by $\mathbb{CP}^{n-3}$ over $S^6$ (analogous to the Penrose fibration). Aug 13, 2015 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 so-called '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.

• 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.
– BS.
Aug 15, 2015 at 10:26
• 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 anti-holomorphic 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 6-dimensional quadric with a real structure (the original $S^6$) and another pair of 6-dimensional quadrics with the anti-holomorphic map between them. I seem to remember this paper: sciencedirect.com/science/article/pii/0393044095000364 Aug 16, 2015 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 $$n(n-1)/2$$. 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  $$x$$ 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 $$n=4$$.

• 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. Aug 14, 2015 at 23:55
• @Jason Starr : it's a long time since you posted this, but I missed it somehow. May I ask what was your point ?
– BS.
Oct 20, 2019 at 7:43