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In the proof of unoriented cobordism ring being isomorphic to homotopy group of Thom spectra, one considers a large enough dimensional Euclidean space where a given manifold has all the embeddings isotopic. This is needed to show the Pontrjagin Thom collapse map does not depend on the embedding.

Here is my question: Why does one need isotopy of embeddings? Given a Euclidean space any embedding will have normal bundle isomorphic to a trivial bundle modulo the tangent bundle of the given manifold. It does not depend on embedding in that particular Euclidean space. Any help will be appreciated.

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Well, why would "a trivial bundle modulo the tangent bundle" be uniquely defined? That depends on a choice of an embedding of the tangent bundle in a trivial bundle, and there may be multiple different such embeddings (even into trivial bundles of the same rank) which give quotients which are not isomorphic.

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  • $\begingroup$ I was exactly thinking this. But I am unable to find an explicit example. It will be helpful if you provide one. Thank you. $\endgroup$
    – Paladin
    Nov 29, 2019 at 17:04
  • $\begingroup$ I guess an easy example can be given by using some stably trivial bundle. $\endgroup$
    – Paladin
    Nov 29, 2019 at 17:13
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    $\begingroup$ Right, stably trivial bundles that are not trivial show that you can't uniquely "subtract" vector bundles in general. I don't know off the top of my head of an example specifically with normal bundles though. $\endgroup$ Nov 29, 2019 at 17:19
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Levine calculates normal bundles of embedded spheres in his paper "A Classification of Differentiable Knots", notably they are sometimes nontrivial. Since the canonical embedding of spheres into Euclidean space always has trivial normal bundle, these give examples of manifolds with embeddings into the same Euclidean space that have nonisomorphic normal bundles.

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  • $\begingroup$ Thank you for the reference. $\endgroup$
    – Paladin
    Aug 2, 2021 at 9:07

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