Let $E$ be a Riemannian vector bundle over a manifold. Can $E$ be considered as a subbundle of trivial bundle $M\times \mathbb{R}^n$ for some $n$ such that the metric of each fiber is the restriction of Euclidean metric of $\mathbb{R}^n$?
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asked Dec 26, 2019 at 15:49
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1$\begingroup$ Should "trivial subbundle" be "subbundle of a trivial bundle"? $\endgroup$– Steven LandsburgCommented Dec 26, 2019 at 22:00
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5$\begingroup$ After applying Steven Landsburg's correction, this is a definition chase using the fact that there is a classifying map $f: M \to BO(k)$, where $BO(k)$ is the Grassmannian of $k$-planes in $\Bbb R^\infty$, and the fact that $f$ factors through some finite-dimensional Grassmannian $\text{Gr}(n,k)$. It is not analagous to Nash's theorem (which is much more difficult). $\endgroup$– mmeCommented Dec 26, 2019 at 23:42
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4$\begingroup$ Even more is true: if the vector bundle has a metric connection, then by a result of Narasimhan and Ramanan, there is a classifying map from $M$ to a sufficiently high-dimensional Grassmannian that classifies both metric and connection on $E$. This seems to me the correct "linearisation" of the Nash embedding theorem. $\endgroup$– Sebastian GoetteCommented Dec 27, 2019 at 14:04
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$\begingroup$ @StevenLandsburg sorry for my delay. Thanks for your correction. I revise it. $\endgroup$– Ali TaghaviCommented Dec 27, 2019 at 14:30
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