# Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$

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.

• What category are you talking about? Smooth? Topological? Etc? – Igor Rivin Nov 8 '16 at 11:32
• Presumably you want $M$ and $N$ to be closed, otherwise you get trivial examples by taking $M=\mathbb{R}^{n+1}$. – Mark Grant Nov 8 '16 at 11:50

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 $\mathbb{R}^3$.

If I have understood the table at

http://www.lehigh.edu/~dmd1/immtable

correctly, then $\mathbb{RP}^{10}$ embeds into $\mathbb{R}^{17}$. But by

Mahowald, Mark On the embeddability of the real projective spaces. Proc. Amer. Math. Soc. 13 1962 763–764.

$\mathbb{RP}^9$ does not embed into $\mathbb{R}^{16}$.

If you are interested in an example in codimension 2 examples (which also happens to involve two orientable manifolds), according to the table on page 2 in the survey

Davis, Donald M. Embeddings of real projective spaces, Bol. Soc. Mat. Mex. 4 (1998) 115-122.

$\mathbb{RP}^{39}$ embeds in $\mathbb{R}^{71}$, but $\mathbb{RP}^{37}$ does not embed in $\mathbb{R}^{70}$.

The first result is credited to:

Rees, Elmer, Embeddings of real projective spaces, Topology 10 (1971) 309-312.

and the second to

Adem, José, Gitler, Samuel, and Mahowald, Mark E. Embedding and immersion of projective spaces, Bol. Soc. Mat. Mex. 10 (1965) 84-88.

Here is a simple example, using $2$-manifolds and $3$-manifolds.

A Klein bottle does not embed in $\mathbb R^3$. But it is a submanifold of a 3-manifold that embeds in $\mathbb R^4$. For the $3$-manifold take the unit normal bundle of the standard embedding of the Klein bottle in $\mathbb R^4$. This bundle has a section, so it contains the Klein bottle. This bundle's Euler class is zero.

• This is Danny's Klein bottle example, with $S^3 / Q8$ removed from the discussion. $S^3 / Q8$ is the unit normal bundle of any embedding of $\mathbb RP^2$ into $\mathbb R^4$ so they're very similar examples, in spirit. – Ryan Budney Nov 8 '16 at 21:11
• The unit normal bundle of the Klein bottle in $\mathbb R^4$ is non-orientable. Are you talking about the twisted Euler class? – Igor Belegradek Nov 8 '16 at 22:21
• Yes, I'm using "Euler class" in the sense of Whitney, i.e. with twisted coefficients. My impression is this is standard usage among 4-manifold theorists. – Ryan Budney Nov 8 '16 at 22:31