Let $M$ be a path connected smooth manifold and $E$ be a vectorbundle over $M$ of rank at least two. My question is: Under which conditions is the space of global nonvanishing sections path connected?(in a suitable topology)

2$\begingroup$ The case of trivial bundles reduces to studying the set of homotopy classes of maps from the base to the sphere fiber, which up to basepoint issues is the cohomotopy group, see en.wikipedia.org/wiki/Cohomotopy_group. $\endgroup$– Igor BelegradekCommented Jan 30, 2019 at 19:33

2$\begingroup$ If the rank of the bundle is two more than the dimension of the manifold this’ll hold. $\endgroup$– Dylan WilsonCommented Jan 31, 2019 at 3:20
1 Answer
Let $S^k\to SE\to M$ be the unit sphere bundle of $E$ (I am assuming your vector bundle has rank $k+1$). Nowhere vanishing sections of $E$ are the same thing as sections of $SE$. The obstructions to finding a path of sections between two given sections $M\to SE$ will lie in $H^{i}(M;\pi_iS^k)$. Therefore, if $k\geq d+1$ ($d$=dimension of $M$), all the obstructions vanish. The extension provides a desired path.
Note that the condition on the dimension is necessary. For example, let $E\to S^3$ be the trivial rank $3$ (or 4) vector bundle over the 3sphere. The space of nowhere vanishing sections is the space of maps from $S^3$ to $S^2$ (or $S^3$), which has infinitely many components.