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.

• $S^d$ has a metric and orientation, which gives $T\simeq T^*$ and $\Lambda^kT\simeq\Lambda^{d-k}T$. Thus, the claim is that $T\oplus\Lambda^2T$ is trivial. As others have said, this is certainly true stably. I would guess that it is not true unstably, but that is just a guess. – Neil Strickland Apr 15 '15 at 7:07
• @NeilStrickland: Thanks Neil. What means "stably" in this context? – Bilateral Apr 15 '15 at 10:35

3 Answers

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 $SO(d+1)$.

• Thanks for the answer. What about the opposite question? namely which orientable Riemannian manifolds $M$ have the corresponding $E_{d} = TM_{d}\oplus \Lambda^{2} T^{\ast}M_{d}$ trivial? For example: $S^{6}$ is a prototypical example of irreducible nearly Kahler manifold. It is true that $E_{6}$ is trivial when taking $M_{6}$ to be any irreducible nearly-Kahler manifold? – Bilateral Apr 15 '15 at 12:17
• By the way, does this answer depend on whether $S^{d}$ is the standard or the squashed sphere? – Bilateral Apr 15 '15 at 13:57
• While the argument I gave is based on the fact that $S^d$ is a space-form, the statement about triviality of the bundle is invariant under diffeomorphisms. So if by "squashed sphere" you mean the sphere endowed with a different metric, the statement holds there, too. I have no idea whether there is a relation to nearly K\"ahler structures, but I would rather doubt it. – Andreas Cap Apr 15 '15 at 18:00
• I am not sure about the relation to nearly Kahler. But there is an analog of the result for any symmetric space, and I think this gives a refinement for $S^6$. Namely, $TS^6\oplus\mathfrak{su}(TS^6)$ is a trivial bundle (where in the second summand you take skew-Hermitian rather than skew symmetric maps). This is related to viewing $S^6$ as a homogeneous space of $G_2$. – Andreas Cap Apr 22 '15 at 5:56
• (ttoo small to edit): istance -> instance, anlogous -> analogous – ypercubeᵀᴹ Nov 26 '16 at 0:16

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 $T\oplus\Lambda^2T$.

• would the same proof work for manifolds different than $S^{d}$? – Bilateral Apr 17 '15 at 12:48
• @Neil is there a short proof as above to prove that $TS^{n}\oplus TS^{n}$ is trivial? – Ali Taghavi Apr 20 '15 at 15:28

Yes, every sphere (and even every homotopy sphere) is stably parallelizable. See page 79 in this paper by Tim Lance.