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Let $X$ be a real $n$ dimensional manifold. One knows that it can be embedded into $mathbbR^2n$ by the Whitney embedding theorem. The normal bundle for such an embedding will be a rank $n$ real vector bundle. I would be interested in understanding the connection between this embedding, its normal bundle and the rank $n$ bundles on $X$.

1. How does the normal bundle change under the isotopy of this embedding?




2. Is it possible to obtain all the isomorphism classes of rank $n$ vector bundles by such isotopies?




3. If the answer for 2. is negative, is it still possible to find an embedding of $X$ into $mathbbR^2n$ for each vector bundle of rank $n$, such that the corresponding normal bundle is isomorphic to it?

I was thinking that the answer should have something to do with the self-intersection of $X$ in the vector bundles and its self-intersection in $mathbbR^2n$, but don't really see how to use that.

Edit: I have realized that it shouldn't be possible to obtain a vector bundle with a non-zero self intersection of the zero section like this. However, I would still like to know if it works for all the other cases.

Edit 2: Based on Mike's comment, I have realized I have missed something very obvious. I am just thinking whether one still gets all the rank $n$ vector bundles which complete the tangent bundle to a trivial $2n$ bundle.

This is to shed some light on Part 3 which asks for a classification of normal vector bundles of a smooth $n$-dimensional submanifold $X$ of $mathbb R^2n$.

The normal bundle $nu$ to $X$ is stably isomorphic to the negative of $TX$, and in particular the Pontryagin and Stiefel-Whitney classes of $nu$ are completely determined by those of $X$. Also if $X$ is orientable, then the Euler class of $nu$ vanishes (contactibility of $mathbb R^2n$ allows to push the embedding off itself). On the other hand, if $X$ is non-orientable, the twisted
Euler class (i.e., the first obstruction to the existence of a nowhere zero section) of $nu$ can be nonzero and is an important invariant (see below).

For simplicity let us assume that $X$ is closed. Let us also ignore the cases $nle 3$ where if $X$ is orientable one can use the classification of oriented bundles Dold-Whitney in [Classification of Oriented Sphere Bundles Over A 4-Complex], and presumably with some work one deal with the nonorientable case.

Instead of classifying normal bundles to $X$ let us describe isotopy classes of embeddings of $X$ into $mathbb R^2n$. Of course, isotopic embeddings have isomorphic normal bundles. One quick statement is a theorem of Haefliger-Hirsch [On the existence and classification of differentiable embeddings.
Topology 2 1963 129–135] that

If $X$ is simply-connected and $nge 4$, then there is only one isotopy class of embeddings from $X$ into $mathbb R^2n$, and hence, only one normal bundle.

More generally,
if $X$ is orientable and $nge 4$, then the set of isotopy classes of smooth embeddings of $X$ into $mathbb R^2n$ is bijective to $H_1(X)$ if $n$ is odd, and to $H_1(X;mathbb Z_2)$ if $n$ is even.

If $X$ is non-orientable and $n$ is even and $ge 4$, there is a more complicated classification by Kitada in
[Classification of embeddings of a non-orientable manifold.
J. Fac. Sci. Univ. Tokyo Sect. IA Math. 18 (1971/72), 435–442]. The conclusion is that the set of isotopy classes is bijective to $mathbb Zoplus H^n-1(X)/K$ where $H$ is a certain subgroup of $H^n-1(X)$. The $mathbb Z$-factor corresponds to the twisted Euler class of $nu$, and in particular, if $H^n-1(X)=0$, then $nu$ is completely determined by the twisted Euler class.

I don't know what happens if $X$ is non-orientable and $n$ is odd.