A conjecture about parallelizable generalized spheres Let $S^{d}$ denote the standard $d$-dimensional sphere. I heard from a physicist that from physical arguments they have been able to show that the vector bundle:
$E_{d} =  TS^{d}\oplus \Lambda ^{d-2}T^{\ast}S^{d}$
is topologically trivial, meaning that it is parallelizable. The convention is that $\Lambda ^{0}T^{\ast}S^{2} = \mathbb{R}$ and $E_{1} = TS^{1}$. They call this bundles $E_{d}$ "generalized spheres".
Is this claim true? For $d=2$ it can be easily checked that it is correct.
Edit: What about the opposite question? namely which orientable, $d$-dimensional, Riemannian manifolds $M_{d}$ have the corresponding $E_{d} = TM_{d}\oplus \Lambda^{2} T^{\ast}M_{d}$ trivial? For example: 
$S^{6}$ is the prototypical example of irreducible nearly Kahler manifold. Is it true that $E_{6}$ is trivial when taking $M_{6}$ to be any irreducible nearly-Kahler manifold?
$S^{7}$ is an example of nearly parallel $G_{2}$-manifold. Is it true that $E_{7}$ is trivial when taking $M_{7}$ to be any nearly-parallel $G_{2}$ manifold?  
Thanks.
 A: Actually, this is simpler than I thought.  The map 
$$ (u,v) \mapsto x\wedge u + v $$
gives an isomorphism
$$ T_xS^n \oplus \Lambda^2 T_xS^n \to \Lambda^2(\mathbb{R}^{n+1}), $$
and this trivialises the bundle $T\oplus\Lambda^2T$.
A: It is true in general that $E_d$ is trivial. As remarked by Neil Strickland above, this boils down to showing that $TS^d\oplus\Lambda^2 TS^d$ is always a trivial bundle. To see this, represent $S^d$ as $SO(d+1)/SO(d)$. Then all bundles in question are homogeneous vector bundles (and the isomorphisms used in the comment are compatible with the homogeneous structure). These homogeneous bundles are classified by representations of $SO(d)$ and the representation in question is $\mathbb R^d\oplus\Lambda^2\mathbb R^d\cong \mathbb R^d\oplus\mathfrak{so}(d)$. But as a representation of $SO(d)$, this is isomrophic to the restriction of the adjoint representation $\mathfrak{so}(d+1)$ of $SO(d+1)$. This implies that $TS^d\oplus\Lambda^2 TS^d$ actually is an associated bundle to the the principal bundle $SO(d+1)\times_{SO(d)}SO(d+1)\to SO(d+1)/SO(d)$ and this principal bundle is easily seen to be trivial (mapping $(g,g')\to (gSO(d),gg')$ factorises to an isomorphism from the associated bundle to $(SO(d+1)/SO(d))\times SO(d+1)$). 
This actually is an istance of a so-called tractor bundle over the homogeneous model of a Cartan geometry. There is an anlogous result associated to any representation of $SO(d+1)$. 
A: Yes, every sphere (and even every homotopy sphere) is stably parallelizable. See page 79 in this paper by Tim Lance.
