Normal bundle of Whitney embedding

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.

• (1) It is unchanged up to isomorphism. (2) No. All normal bundles obtained by an embedding into $\Bbb R^N$ for any $N$ have the property that $TM \oplus \nu$ is trivial. That is, $\nu$ is a "stable additive inverse" of $TM$. This completely determines its Stiefel-Whitney classes from those of $TM$ and thus is hugely constrained. – Mike Miller Jul 27 '19 at 19:53
• Your remaining question is interesting. This can be done for immersions by the Smale-Hirsch theorem (and embeddings one dimension up). Using the assumption that $e(\nu) = 0$ (essentially your condition on self-intersection), can one cancel the double points a la the Whitney trick? I don't know. – Mike Miller Jul 27 '19 at 22:25
• May be this post 9and its related-connected post) would be related to the question: mathoverflow.net/questions/237384/… – Ali Taghavi Jul 28 '19 at 10:54
• To follow up on what Mike wrote: the Smale-Hirsch h-principle shows that any open parallelizable $m$-manifold immerses into $\mathbb R^m$, and the corrsponding statement for embeddings is not true (I think). One just has to find a counterexample that is the total space of a $\mathbb R^n$ vector bundle over a smooth $n$-manifold. – Igor Belegradek Jul 28 '19 at 14:18

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.

• Just a comment: if $n$ is odd, then $X$ embeds in $\mathbb{R}^{2n-1}$. But presumably there can be embeddings in $\mathbb{R}^{2n}$ for which the normal bundle is not stabilized. mathscinet.ams.org/mathscinet-getitem?mr=149494 – Ian Agol Jul 30 '19 at 16:38