In this [question](https://mathoverflow.net/questions/202956/a-conjecture-about-parallelizable-generalized-spheres), we have discussed how the following bundle:


$E\_{d} = TS^{d}\oplus \Lambda^2 T^{\ast}S^{d}$


is always trivial, where $S^{d}$ is the $d$-dimensional standard sphere. Now, let us take $d=6$. The six-sphere $S^{6}$ is an example of a strict nearly-Kahler homogeneous manifold which indeed is not parallelizable, although as we have said $E\_{6}$ is. The natural question then is:


Is the vector bundle $W = TM\_{6}\oplus \Lambda^{2} T^{\ast}M\_{6}$ trivial, where $M\_{6}$ is any homogeneous strict, nearly-Kahler manifold?


Notice that there are only four strict nearly-Kahler homogeneous manifolds, namely:


$S^{6},\qquad Sp(2)/(SU(2)\times U(1)), \qquad SU(3)/(U(1)\times U(1)), \qquad S^{3}\times S^{3}.$ 


For $S^{6}$ we already know that the answer is affirmative, and in fact, since $S^{3}$ is parallelizable, then the $W$ corresponding to $S^{3}\times S^{3}$ is also trivial. So there are only two cases left to check.


Thanks.



Parallelizable nearly-Kahler manifolds
The answer is 'no'. Already, it is not true for $M = \mathrm{Sp}(2)/\bigl(\mathrm{SU}(2)\times\mathrm{U}(1)\bigr)$, which is known to be diffeomorphic to $\mathbb{CP}^3$.


To see this, note that the first Pontrjagin class of $T\mathbb{CP}^3$ satisfies
$$
p\_1(T\mathbb{CP}^3) = 8u^2\in H^4(\mathbb{CP}^3,\mathbb{R}),
$$
where $u\in H^2(\mathbb{CP}^3,\mathbb{R})\simeq\mathbb{R}$ is the standard generator. By the usual formula, this implies that
$$
p\_1\bigl(\Lambda^2(T\mathbb{CP}^3)\bigr) = 32u^2,
$$
so that
$$
p\_1\bigl(T\mathbb{CP}^3\oplus\Lambda^2(T\mathbb{CP}^3)\bigr) = 40 u^2\not=0.
$$
In particular, $T\mathbb{CP}^3\oplus\Lambda^2(T\mathbb{CP}^3)$ is not a trivial bundle over $\mathbb{CP}^3$.


