What is the total space of the tangent bundle of $S^4$?

Rank-4 vector bundles over $S^4$ are classified by $\pi_3(SO(4))=\mathbb{Z}\times \mathbb{Z}$, so it seems there are many possibilities. Also, it cannot be the trivial bundle, by the Poincare-Hopf theorem.


I'm not sure what description you're looking for. It has Euler class $e=2$ (because $\chi(S^4) = 2)$ and first Pontryagin class $p_1 = 0$ (because $S^4$ is stably parallelizable, all stable characteristic classes vanish.)

In Milnor's On manifolds homeomorphic to the 7-sphere, one finds the following description of an isomorphicm $\pi_3(SO(4))\cong \mathbb{Z}\oplus\mathbb{Z}$:

To the integers $(i,j)\in \mathbb{Z}\oplus\mathbb{Z}\cong \pi_3(SO(4))$, we associate the $S^3$ bundle over $S^4$ with clutching function $f_{i,j}:S^3\rightarrow SO(4)$ given by $f_{i,j}(x)(v) = x^i v x^j$, where we interpreter $x,v\in \mathbb{R}^4\cong \mathbb{H}$.

Milnor proves that $p_1$ is given by $\pm 2(i-j)$ and $e = i+j$.

Since $T^1 S^4$ has $p_1 = 0$ and $e=2$, it follows that, under Milnor's isomorphism $\pi_3(SO(4))\cong \mathbb{Z}\oplus\mathbb{Z}$, that $T^1 S^4$ corresponds to $(1,1)$.


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