11

I've been thinking a little bit about smooth embeddings recently. In particular, I was wondering:

Do the $3$-manifolds $\mathbb{S}^2 \times \mathbb{S}^1$ and $\mathbb{RP}^2 \times \mathbb{R}$ embed smoothly into $\mathbb{R}^4$?

As with most of the questions I ask on this site, I ask entirely out of curiosity, and largely because I have no idea how to even begin.

Further question: If either of the above has a positive answer, could we (more generally) embed $\Sigma^2 \times \mathbb{S}^1$ or $M^2 \times \mathbb{R}$ into $\mathbb{R}^4$, where $\Sigma^2$, $M^2$ are compact (orientable, non-orientable, resp.) smooth surfaces.

Note: I'm aware that the $3$-manifolds $\Sigma^2 \times \mathbb{S}^1$ can be immersed into $\mathbb{R}^4$ and embedded into $\mathbb{R}^5$.

Jesse Madnick
  • 31,524

2 Answers2

13

If $\Sigma^2$ is a compact orientable surface, then $\Sigma^2\times\mathbb{S}^1$ can be smoothly embedded in $\mathbb{R}^4$. The trick is simple: first embed $\Sigma^2$ into $\mathbb{R}^4$, and let $T$ be a tubular neighborhood of the surface. Then the closure of $T$ is isomorphic to $\Sigma^2\times\mathbb{D^2}$, and the boundary of $T$ is isomorphic to $\Sigma^2\times\mathbb{S}^1$. Note that $\Sigma^2\times\mathbb{R}$ also embeds in $\mathbb{R}^4$, since it is a submanifold of $\Sigma^2\times\mathbb{S}^1$.

No closed, non-orientable 3-manifold can be embedded in $\mathbb{R}^4$, for the same reason that no closed, non-orientable surface can be embedded in $\mathbb{R}^3$. (If there were such an embedding, then the "inward-pointing" normals could be used to define an orientation of the manifold.) In particular, if $M^2$ is a compact, non-orientable surface, then $M^2\times\mathbb{S}^1$ cannot be embedded in $\mathbb{R}^4$.

The case $M^2\times\mathbb{R}$ for $M^2$ a non-orientable surface is covered briefly in this paper (near the top of the fourth page). In particular:

  • If $M^2$ has odd Euler characteristic (such as the projective plane), then $M^2\times\mathbb{R}$ cannot embed in $\mathbb{R}^4$. (This is related to, but does not follow from this question.) The paper linked to above gives the following argument for this:

    Although $M^2$ is not orientable, the normal Euler class $\bar{e}(M^2)\in\mathbb{Z}$ of an embedding $M^2\subset\mathbb{R}^4$ is well-defined and $\bar{e}(M^2) = 2\,\chi(M^2)\;\mathrm{mod}\;4$ [several citations given]. Hence the normal bundle of an embedding $M^2\subset\mathbb{R}^4$ has no cross-sections.

  • If $M^2$ has even Euler characteristic, then $M^2\times\mathbb{R}$ can be smoothly embedded in $\mathbb{R}^4$. This should be clear for the Klein bottle—the standard immersion of the Klein bottle $K$ in $\mathbb{R}^3$ can be "thickened" to give an immersion of $K\times\mathbb{R}$ into $\mathbb{R}^3$, which can then be modified slightly to give an embedding into $\mathbb{R}^4$. The same argument works for the other surfaces with even Euler characteristic, since each of them is simply a Klein bottle with handles attached.

Community
  • 1
Jim Belk
  • 49,278
  • 2
    At the end of the first bullet point, I assume you mean $\mathbb{R}^4$, not $S^4$. I appreciate, by the way, that you discussed the case of $M^2 \times \mathbb{S}^1$ although I did not explicitly ask for it. As you probably guessed, I was in fact thinking more generally about $3$-manifolds that split as products of $1$- and $2$-manifolds :-) – Jesse Madnick Jun 26 '13 at 09:00
  • @JesseMadnick Actually, I don't think my argument there was correct. The answer to the question I linked to only applies in the case that the submanifold is closed. I have replaced this with a quote of the more technical argument given in the paper. – Jim Belk Jun 26 '13 at 16:05
  • 3
    The case of when a 2-manifold cross an interval embeds into $\mathbb R^4$ was first handled (and completely answered) by Hassler Whitney in his paper: "On the topology of differentiable manifolds" Lectures in topology, U.Michigan Press, 1941. – Ryan Budney Jun 28 '13 at 03:06
  • In your first paragraph, are you claiming that any embedding of $\Sigma$ into $\mathbb{R}^4$ extends to an embedding of $\Sigma \times S^1$? Or, more specifically, are you claiming that the normal bundle of $\Sigma$ in $\mathbb{R}^4$ is independent of embedding? (I agree that if we embed $\Sigma$ into $\mathbb{R}^3 $ in the usual way and then embed $\mathbb{R}^3$ into $\mathbb{R^4}$ in the usual way, that the normal bundle of $\Sigma is trivial - and that this is enough to answer the OP's question. It is just not clear to me what happens for an arbitrary embedding.) – Jason DeVito - on hiatus Oct 21 '17 at 19:42
  • @JasonDeVito The normal bundle to a compact, orientable $n$-manifold embedded in $\mathbb{R}^{n+1}$ or $\mathbb{R}^{n+2}$ is always trivial. See W. Massey, On the Normal Bundle of a Sphere Imbedded in Euclidean Space. Proceedings of the American Mathematical Society 10.6 (1959), 959-964. See also this MathOverflow answer. – Jim Belk Oct 22 '17 at 03:09
  • @Jim: Well, I learned something today. Thanks! – Jason DeVito - on hiatus Oct 23 '17 at 12:13
5

Any product of spheres embeds in codimension $1$. That's a nice exercise with the Tubular Neighborhood Theorem. I'm not sure yet about the second.

Ted Shifrin
  • 115,160