For which values of $n$ does there exist an embedding of a smooth compact manifold $M\hookrightarrow R^n$ into $n$-dimensional Euclidean space such that the normal bundle to $M$ has no nonvanishing section?

If such an embedding does exist, can $M$ be taken to be orientable? What can be the dimension of $M$?

In other words, I'm asking for submanifolds $M$ of $R^n$ which cannot be "immediately pushed off of themselves" by the flow of a vector field. Conversely, a theorem of Hirsch states that if a nonvanishing normal field does exist then $M$ immerses in $R^{n-1}$.

If $n=4k$ for some integer $k$, then one can find non-orientable embedded $2k$-submanifolds $M$ in $R^{4k}$ with which have no such sections---the easiest (to me) way that I know of doing this is to ask for $M$ to be Lagrangian with respect to the standard symplectic structure, so that its normal bundle will be isomorphic to its tangent bundle and so one just needs $M$ to have nonzero Euler characteristic (since then the (integral) twisted Euler class in the local coefficient cohomology associated to $w_1$ will be nontrivial), and then examples can be constructed by Lagrangian surgery. On the other hand an orientable submanifold of $R^n$ of dimension $\frac{n}{2}$ will always have a nonvanishing normal vector field, since the only obstruction to constructing such is the Euler class of the normal bundle, which is the restriction of a cohomology class from $R^n$ and so vanishes.

If $\dim M>\frac{n}{2}$ then there are higher-order obstructions to the existence of a nonvanishing section of the normal bundle, which can be nontrivial in the orientable case. There are some examples of this described by Massey; for instance for $s\geq 1$, $CP^{2^{s}}$ embeds in $R^{4\cdot2^s-1}$, and there is no normal section because the secondary obstruction is nontrivial. I think (though haven't checked carefully) that a product of two of these $CP^{2^s}$ embeddings should also lack a nonvanishing normal vector field, due to product formulas for the obstruction classes. But these examples seem rather special, and I haven't been able to see a way of adapting the ideas underlying them to construct examples in arbitrary ambient dimension.

  • $\begingroup$ I've deleted my "answer". $\endgroup$ Nov 13, 2011 at 2:38
  • $\begingroup$ So, in orientable case, you have only $n=4⋅2^s−1$ and maybe $n=8⋅2^s−2$ (?) $\endgroup$ Nov 13, 2011 at 17:55
  • 2
    $\begingroup$ @Anton Petrunin: If I'm right about products then one could probably get $4(2^s+2^t)-2$, and Massey also gives a couple of isolated examples involving quaternionic or octonionic projective spaces in low dimensions. Massey's paper being 50 years old, there certainly may be other examples, but they seem to be hard to find. $\endgroup$
    – Mike Usher
    Nov 13, 2011 at 20:35

2 Answers 2


For non-orientable case the simplest example is: $RP^2$ can be embedded to $R^4$, but no such embedding admits a non-vanishing normal vector field.

The proof (sketch): This normal vector field could be made vertical by an isotopy, hence the embedding can be projected into $R^3$ as an immersion. There are 2 regular homotopy classes of such immersions: the Boy surface and its mirror image. Looking at the double points curve of any of these two immersions, we see that it represents a nontrivial double cover over $S^1$, hence it can not be made an embedding by adding a single function to $R^1$. Indeed, any function to $R^1$ on the double points set will have an odd number of such pair of points, that have the same image under the immersion and the function also takes the same value on them. This property -having an odd number of such pairs of points - is invariant under regular homotopy.


I gave an orientable example in my answer to this later question. In particular, any embedding $\mathbb{C}P^2\hookrightarrow \mathbb{R}^7$ will have this property.


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