How can I show $\mathbb{TS^{5}}$ is not trivial?

Question

I want to ask if anyone has a hint how to solve this problem via elementary differential topology methods not involving topological $K$-theory like $H$-spaces, $J$-homomorphism, etc. Even if we use $K$-theory then we know $$\overline{K}(\mathbb{S}^{5})=0$$via Bott periodicity. This implies $K(\mathbb{S}^{5})=\mathbb{Z}$. But this does not help me to know whether the tangent bundle itself is trivial. It is not clear to me what elementary differential topology tools I should use as the instructor hinted $\mathbb{S}(\mathbb{TS}^{5})$ has the same homotopy groups as $\mathbb{S}^{5}\times \mathbb{S}^{4}$. I feel I need the heavy machinary of characteristic classes, but I do not know how to make it work in this case from the definitions.

A really elementary way to think about it is via the clutching construction. Then the tangent bundle is given by a transition class in $\pi_{4}(SL(4,\mathbb{R})$. The later one deformation retracts to $\pi_{4}(SO(4,\mathbb{R}))$ if we give it a metric or limit to orthogonal transformations. Since we know $SO(4,\mathbb{R})\cong \mathbb{S}^{3}\times \mathbb{RP}_{3}$, it seems we can work it out directly if we know $\pi_{4}(\mathbb{S}^{3})$ and $\pi_{4}(\mathbb{RP}_{3})$, which should be $\mathbb{Z}$ in both cases. But the fact $$\pi_{4}(SO(4,\mathbb{R}))=\mathbb{Z}\oplus \mathbb{Z}$$ tell us very little about the map classifying $\mathbb{TS^{5}}$. While it seems "clear" to me that $\mathbb{TS}^{2}$ is twice of the generator class of $\pi_{1}(SO(2,\mathbb{R}))$, it is not clear how the same geometric intuition may be of help in this case.

There is an almost identical post earlier with no conclusive answer. So I am not sure if this fit into forum guidelines.

Maybe you can try showing that there are no sets of linearly independent global sections of the bundle. — smiley06, Apr 21 '13 at 06:23
Nothing personal but I think this is too elementary to be useful at here. — Bombyx mori, Apr 21 '13 at 06:38
Stiefel Whitney class can be used to show that $S^n$ is parallelizable implies that $n$ is power of 2. See Milnor, Characteristic classes for example. — , Apr 21 '13 at 06:55
@Sanchez: I believe you confused $\mathbb{RP}^{n}$ with $\mathbb{S}^{n}$. The parallelizablability of spheres is more difficult. — Bombyx mori, Apr 21 '13 at 07:07
@Sanchez: Thanks for the effort. I did not read M&S for two years and clearly forgot most of the contents... — Bombyx mori, Apr 21 '13 at 07:12

score 5 · Accepted Answer · answered Apr 21 '13 at 16:27

5

A proof of this can be found in Steenrod's "The Topology of Fiber Bundles": Theorem 24.10 (together with Corollary 27.3) implies that the clutching map $[T_6: S^4\rightarrow SO(5)]\in \pi_4(SO(5))$ of the tangent bundle of $S^5$ is of order two. Note that the tangent bundles of spheres are stably trivial and hence they are not detected by $K$-theory.

answered Apr 21 '13 at 16:27

user_11437

995

I see. I was in the impression that I can use the $H$-space structure and compute certain spectral sequences to solve the problem, but I was not familiar with the matter. You are certainly correct. – Bombyx mori Apr 21 '13 at 16:30

How can I show $\mathbb{TS^{5}}$ is not trivial?

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