# Oriented vector bundle with odd-dimensional fibers

Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere?

Let us consider some vector bundles over $$S^4$$. Since $$S^4$$ is simply connected all vector bundles are oriented and rank $$k$$ vector bundles over $$S^4$$ are classified by $$\pi_{3}(SO(k))$$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $$\pi_3(SO(1))$$ and $$\pi_3(SO(2)=\pi_3(S^1)$$ are trivial groups, so there do not exist non-trivial rank $$1$$ and rank $$2$$ bundles over $$S^4$$. What about rank $$3$$? Well $$SO(3)\cong \mathbb RP^3$$ and the long exact sequence of homotopy groups of the fiber bundle $$\mathbb{Z}_2\rightarrow S^3\rightarrow \mathbb{RP}^3$$ shows that $$\pi_3(SO(3))\cong\pi_3(\mathbb{RP}^3)\cong \mathbb{Z}$$. Hence there are $$\mathbb{Z}$$ different rank $$3$$ vector bundles over $$S^4$$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $$2$$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $$2$$ bundles over $$S^4$$.
• Moreover an important example (often denoted $\Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer. – BS. Jul 22 at 10:28
• @fredy: The map that sends each fiber $v$ to $-v$ is orientation reversing if the vector space is odd dimensional. Hence $e(E)=-e(E)$, so $2e(E)=0$, which in the de Rham theory implies that $e(E)=0$ (but not when the Euler class is defined with integral coefficients). – Thomas Rot Aug 6 at 9:42