Is it possible to improve the Whitney embedding theorem? - MathOverflow

Is it possible to improve the Whitney embedding theorem?

2011-03-06T08:38:17Z

Edited to fix the example, as per Zack's suggestion.

Edit 2: So it turns out that when I think 'manifold' I tend to assume the nicest possible object. As I believe is standard, I would like to assume that all manifolds are 2nd countable and Hausdorff. Furthermore, let's say that our manifolds are connected and closed.

The Whitney embedding theorem states that any smooth $n$-manifold may be smoothly embedded into $\mathbb{R}^{2n}$. If we consider embeddings into more general $k$-dimensional manifolds, is it possible find a '$n$-universal' manifold of dimension less than $2n$?

For example, a non-orientable 2-manifold cannot be embedded into $\mathbb{R}^3$, demonstrating the sharpness of the Whitney embedding theorem.

However, there are 3-manifolds into which we can embed any surface, such as $M = \mathbb{RP}^3 \sharp \mathbb{RP}^3$. Indeed, by the classification of surfaces we know that any surface may be decomposed as a connected sum of copies of $\mathbb{RP}^2$ and tori. In fact, by the monoid structure of closed surfaces under connected sums we may take this sum to have at most 2 copies of the projective plane. Now, embed 2 disjoint copies of the projective plane into $M$ and arbitrarily many copies of the torus. Taking the connected sum of these we see that any closed surface is embeddable into $M$.

Can we do something similar in higher dimensions?

Answer by Ryan Budney for Is it possible to improve the Whitney embedding theorem?

2011-03-06T20:56:49Z

Yes, the Whitney theorem can be improved in many cases. For example, C.T.C. Wall proved all 3-manifolds embed in $\mathbb R^5$.

Precisely what is the optimal minimal-dimensional Euclidean space that all $n$-manifolds embed in, I don't know what the answer to that is but Whitney's (strong) embedding theorem is only best-possible for countably-many $n$, not for all $n$. See Haefliger's work on embeddings -- I believe he noticed many cases where you can improve on Whitney.

The suggestion to improve Whitney's theorem that you're giving -- making the target not a Euclidean space but a manifold -- in a sense you're asking for something much weaker than Whitney's theorem. For example, given any $n$-manifold, you can take its Cartesian product with $S^1$. Take the connect sum of all manifolds obtained this way. It's a giant, non-compact $(n+1)$-manifold, and all $n$-manifolds embed in it. This isn't so interesting.

Answer by Bruno Martelli for Is it possible to improve the Whitney embedding theorem?

2011-03-06T21:16:26Z

I assume that all manifolds are closed and connected (embedding in $\mathbb R^n$ or in $S^n$ is the same), but not necessarily orientable.

Concerning dimension 3, a famous theorem of Wall states that every closed 3-manifold embeds in $S^5$.

It is not possible to improve this result. A theorem of Shiomi shows that there is no closed 4-manifold which contains every possible closed 3-manifold.

The question in dimension 4 seems more complicate. Since $4=2^2$, this is one of the dimensions where real projective plane $\mathbb R\mathbb P^4$ embeds in $S^8$ and not in $S^7$. Every orientable 4-manifold embeds topologically in $S^7$ by a theorem of Fang and Fuquan.

Answer by John Klein for Is it possible to improve the Whitney embedding theorem?

2011-03-06T23:12:18Z

If we restrict our interest to manifolds which are $k$-connected, then Wall proved that any $n$-manifold (closed) $M$ admits a locally flat PL embedding in $\Bbb R^{2n-k}$, thereby improving on Whitney by $k$ dimensions. If in addition we assume the metastable range condition $2k < n$, then we can even take the embedding to be smooth. The latter theorem was also known to Haefliger and Hirsch and is historically earlier.

One further thing worth mentioning: the Hirsch Conjecture says that a stably parallelizable $n$-manifold is supposed to embed in $\Bbb R^m$, where $m = \lceil (3/2)n\rceil $. The conjecture is still open. Partial results are known: for example it's true when the manifold is $[n/4]$-connected.