Can we smoothly embed $\mathbb{S}^2 \times \mathbb{S}^1$ or $\mathbb{RP}^2 \times \mathbb{R}$ in $\mathbb{R}^4$?

Question

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 $\mathbb{R}^5$.

Answer 1

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.

Answer 2

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.