The tangent bundle of the 2-sphere is nontrivial

Let $S^2$ be the 2-sphere. Its tangent bundle $\pi:TS^2\to S^2$ is a smooth real vector bundle of rank $2$ .

The bundle $TS^2$ is not isomorphic (as a rank-2 real vector bundle) to the trivial rank-2 bundle $S^2\times \mathbb R^2$ .

A standard proof uses the “hairy ball” phenomenon: every continuous tangent vector field on $S^2$ has a zero. Since a global nowhere-zero section would trivialize a rank-1 subbundle and (together with a second independent section) produce a global frame, this obstructs triviality.

Equivalent characterizations

Equivalently:

$TS^2$ does not admit a global frame of smooth sections, and in particular
$TS^2$ admits no nowhere-vanishing smooth vector field on $S^2$ .

Examples

Local triviality is easy: remove a point. On $S^2\setminus\{p\}\cong \mathbb R^2$ , the tangent bundle is trivial: $T(S^2\setminus\{p\})\cong (S^2\setminus\{p\})\times\mathbb R^2$ . The obstruction is global, not local.

Contrast with the circle. The circle $S^1$ admits a nowhere-vanishing tangent vector field (rotation), so $TS^1$ is trivial. This highlights that the failure for $S^2$ is not automatic for spheres.

Frame bundle reflection. Nontriviality of $TS^2$ implies that the frame bundle $\mathrm{Fr}(TS^2)$ is not globally a product $S^2\times \mathrm{GL}(2,\mathbb R)$ , even though it is locally trivial by definition.