Skip to main content
3 events
when toggle format what by license comment
Aug 17, 2020 at 21:32 comment added Qiaochu Yuan Said another way, the space of Riemannian metrics on a given vector bundle is convex, hence contractible. So the natural forgetful map from Riemannian vector bundles to vector bundles has contractible fibers.
Aug 17, 2020 at 20:19 comment added mme Two Riemannian vector bundles are (1)-equivalent iff they are isomorphic as vector bundles, so you can get nothing new from this. One way of thinking of this is the fact that a Riemannian vector bundle is classified by a map to $BO(n)$, while a vector bundle is classified by a map to $BGL(n)$. Because the inclusion $O(n) \to GL(n)$ is a homotopy equivalence the induced map on classifying spaces is too.
Aug 17, 2020 at 18:11 history asked Ali Taghavi CC BY-SA 4.0