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Mar 11, 2022 at 1:24 comment added Michael Albanese @inkievoyd: No. For every $n$ there could be an obstruction. Of course, for $n < r$ the groups $H^n(X; \pi_{n-1}(S^{r-1}))$ are trivial. If the bundle is orientable (which is the case if it is stably trivial), then the obstruction in $H^r(X; \pi_{r-1}(S^{r-1})) = H^r(X; \mathbb{Z})$ is the Euler class.
Mar 11, 2022 at 0:01 comment added inkievoyd In the first paragraph, should the $n$'s be $r$'s?
Mar 10, 2022 at 1:48 vote accept inkievoyd
Mar 10, 2022 at 1:34 history edited Michael Albanese CC BY-SA 4.0
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Mar 10, 2022 at 1:09 history edited Michael Albanese CC BY-SA 4.0
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Mar 10, 2022 at 0:01 history answered Michael Albanese CC BY-SA 4.0