Suppose you have a closed $m$-dimensional manifold $M$, which embeds in $\mathbb{R}^{n+1}$ for some $n$. Can it have a closed submanifold $N$ (of dimension strictly smaller than $m$) which does *not* embed in $\mathbb{R}^n$? (By which I mean there is no embedding, not just that the restriction/projection of the first one doesn't work.)


I'm not sure how important the dimension of $N$ is; it seems like codimension 1 would be the easiest place to find an example, but I'm also interested in higher codimension examples. In fact, is there an upper limit to the codimension in which such examples can exist?


I would particularly like a codimension 1 example where both $M$ and $N$ are orientable. Or an example (orientable or not) for $n=2$ or $3$, but maybe there are obstructions in low dimensions...


I am primarily thinking about smooth manifolds, but examples in the topological category would also be interesting.



Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$
Here's another way to get examples, in codimension one and in low dimensions. There are lots of oriented closed 3-manifolds that don't embed in 4-space, for example any 3-manifold $M$ with $H\_1(M) \cong \mathbb{Z}\_{2n}$. But a theorem of Hirsch says that $M$ embeds in $\mathbb{R}^5$. By a standard transversality argument (like the one that produces Seifert surfaces for higher-dimensional knots) $M = \partial W$ for some oriented $W \subset \mathbb{R}^5$. Now $W$ has a trivial normal bundle, so $W \times I \subset \mathbb{R}^5$, and so $M \subset \partial(W \times I)$, which is just the double of $W$.


You can do a similar thing with an oriented closed $n$-manifold that doesn't embed in $n+1$-space but does embed in $n+2$ space. These aren't so easy to come by; for $n=4$ they were constructed by Tim Cochran (Inventiones Math. 77 (1984), 173--184). If you don't mind non-orientable examples, then you can do this with $n=2$. For a Klein bottle embeds in the quaternionic space form $S^3/Q8$, which in turn embeds in $\mathbb{R}^4$. (It is the boundary of a tubular neighborhood of an embedded $RP^2$.) But a Klein bottle cannot embed in $\mathbb{R}^3$.


