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Assume that $M$ is a $k$ dimensional manifold which is embedded in $\mathbb{R}^n$. We define the map $\phi_{M,n}: M \to G(k,n)$ with $\phi_{M,n} (x)= T_x M$, the tangent space to $M$ at point $x\in M$.

For a given manifold, is there an embedding of $M$ in some $\mathbb{R}^n$ such that $\phi_{M,n}$ would be an immersion?

We can pull back the tangent bundle of the grassmanian via $\phi$ to a $k(n-k)$ dimensional bundle over $M$. Do the characterstic classes of this bundle depend on a particular embedding of $M$ in $\mathbb{R}^n$?

In particular is there a non trivial knot in space for which the correspondind plane bundle over $S^1$, is a trivial bundle?

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    $\begingroup$ The tangent bundle of $G(k,n)$ at $L$ is canonically isomorphic to $Hom(L,\mathbb{R}^n/L)$. So the pullback of the tangent bundle is isomorphic to $Hom(TM,\mathbb{R}^n/TM)$. The bundle does(I think so) depend on the embedding but if n is big enough it doesn't. The characteristic classes don't depend on n, they are algebraic combination of classes of $TM$, and $TM^\bot$. Concerning your first question, I think that the derivative of the map $\phi_{M,n}$ is Porteous's intrinsic derivative (see for example wisdom.weizmann.ac.il/~dnovikov/Manifolds5775/… page 149) $\endgroup$
    – Omar
    Commented Sep 24, 2017 at 15:37
  • $\begingroup$ @Omar Thank you for your very helpfull comment and usefull information. $\endgroup$
    – Ali Taghavi
    Commented Sep 24, 2017 at 21:20
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    $\begingroup$ @Omar The characteristic classes of the tangent bundle $TM$ and the (stable) normal bundle $TM^\perp$ do not depend on $n$. Because their sum is trivial, those of $TM$ can be computed from those of $TM^\perp$ and vice versa. But assume that $TM\oplus TM^\perp\cong\underline{\mathbb R^n}$ is an unstable realisation of the normal bundle. Then $\operatorname{Hom}(TM,\mathbb R^{n+1}/TM)\cong\operatorname{Hom}(TM,TM^\perp)\oplus T^*M$, which has different characteristic classes. The upshot is: the characteristic classes in question depend only on those of $TM$ and on $n$. $\endgroup$
    – Sebastian Goette
    Commented Sep 26, 2017 at 10:39
  • $\begingroup$ @SebastianGoette thank you for your very helpfull comment $\endgroup$
    – Ali Taghavi
    Commented Sep 27, 2017 at 13:54

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