Assume that we have a vector bundle $E$ over $S^n$.



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> Is there a continuous family of invertible linear maps $T\_x:E\_x \to E\_{-x}$?
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Here continuity has the obvious meaning as soon as we have trivialization for the bundle around $x$ and $-x$


There is an obvious affirmative answer for the non orientable line bundle over $S^1$, for the tangent bundle of $S^n$ and for the $2$ dimensional real vector bundle $E \to S^2 \simeq \mathbb{C}P^1$ where $E$ is the tautological complex line bundle over $\mathbb{C}P^1$.


If the answer is yes, what about if we consider $E$ as a Riemannian bundle and we require that every $T\_x$ be an isometric linear map?



Invertible (isometric) sections of certain hom bundles over sphere
There is no such family of *complex* linear maps for the tautological bundle on $\mathbb CP^1$. The antipodal map is orientation reversing, hence the first Chern class changes its sign. The same argument should work for the positive complex spinor bundle on any even-dimensional sphere.


The same argument also works for a real vector bundle on $S^{4k}$ with non-vanishing $k$-th Pontryagin class. Probably again the positive real spinor bundle is an example.


On the other hand, the answer to the second question is "yes" because $O(r)$ or $U(r)$ is a deformation retract of $GL(r,\mathbb R)$ or $GL(r,\mathbb C)$, respectively.


