Normal bundle of Whitney embedding

Question

Let $X$ be a real $n$ dimensional manifold. One knows that it can be embedded into $\mathbb{R}^{2n}$ 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 $\mathbb{R}^{2n}$ 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 $\mathbb{R}^{2n}$, 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.

Answer 1

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 $n\le 3$ 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 $n\ge 4$, 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 $n\ge 4$, 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 $\mathbb Z\oplus H^{n-1}(X)/K$ 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.