MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
A nitpick: the second map in the example isn't well defined. $\mathbb{RP}^3$ doesn't contain two $\mathbb{RP}^2$s (the complement of one of them is $\mathbb{R}^3$). $\mathbb{RP}^3\sharp\mathbb{RP}^3$ should work, though. – Zack Mar 6 '11 at 9:42
This is an extremely silly comment; sorry about that. Let $S$ be a set of $n$-manifolds containing one manifold out of each diffeomorphism class. Then $U=\coprod_{M \in S} M$ is an $n$-manifold and each $n$-manifold embeds into $U$. Of course, $U$ is not second countable. – Johannes Ebert Mar 6 '11 at 10:25
@Johannes $U$ is not a $n$-manifold as it's not second countable. Definitions of "manifold" that I've seen all require second countability. – Kelly Davis Mar 6 '11 at 10:40
1. You should probably just sprinkle the word "connected" all over this question to keep it interesting. 2. We already have the "upper bound" of $\mathbb{R}^{2n}$ for $n$-manifolds, so I think the answer to whether we can do this sort of has to be yes. 3. Certainly we can compare $n$-universal manifolds if their dimensions are different, but a priori there might be two $n$-universal $N$-manifolds and neither covers the other. Or perhaps it's provable that one must cover the other; that'd be a neat result. – Aaron Mazel-Gee Mar 6 '11 at 17:35
I think that, in addition to assuming the domain manifolds to be connected we should also assume "closed," that is, compact without boundary. – John Klein Mar 6 '11 at 19:35

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.

share|cite|improve this answer
Dear Ryan, I vaguely remember a result about embedding in dimension 2n-b(n) where b(n) is the number of 1's in the binary expansion of n. But I dont remember if it was for the same problem. – Gil Kalai Mar 6 '11 at 21:42
@Ryan If I understand correctly, your answer is $(M_1 \times S^1) \# (M_2 \times S^1) \# \cdots$. This won't work. An infinite connected sum is not defined. For example, we know there exists an exotic $S^7$, notated as $S_+$ say, and it has an inverse $S_-$ such that $S^7 = S_- \# S_+$. If an infinite connected sum were defined, we would have $S_+ = S_+ \# (S_- \# S_+) \# (S_- \# S_+) \cdots$ $= (S_+ \# S_-) \# (S_+ \# S_-) \# \cdots$ $= S^7$ which is false as $S_+$ is exotic. – Kelly Davis Mar 6 '11 at 21:50
Kelly, I think there is a mis-communication over what is meant by "infinite connect-sum". Such arguments aren't a problem in the realm I'm working in because the manifolds are non-compact. If you want to make what I said more precise, the idea would be to take $\mathbb R^{n+1}$, and pick a countable discrete collection of disjoint points in $\mathbb R^{n+1}$, and use them as the "attaching points" for the connect sum operation. This doesn't cause any Mazur-swindle problems since the resulting manifold is non-compact. – Ryan Budney Mar 6 '11 at 21:54
@Gil : That's for immersions, not embeddings. It's the Immersion Conjecture, which was proven by Ralph Cohen. – Andy Putman Mar 6 '11 at 21:54
Hi Alessandro, there's many ways to argue this. One is that all smooth manifolds have triangulations, and there's only countably-many triangulations. Moreover, there's only finitely many smooth structures supported by a given triangulation. – Ryan Budney Mar 7 '11 at 15:38

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.

share|cite|improve this answer
This answers the question in dimensions 3. But I do not see how it does answers it in other dimensions. – John Klein Mar 6 '11 at 23:36
I agree. This isn't an answer to the full question. – Kelly Davis Mar 6 '11 at 23:41
Ok, my apologies, I am new to Mathoverflow. May I ask what is the proper etiquette for accepting answers? I assumed I was supposed to pick one, which was hard because all three are very informative and interesting, even if they didn't entirely answer my question (ie embeddable closed connected manifolds for arbitrary dimension). In fact, I did not expect to get a full answer to the question. – Ben McMillan Mar 7 '11 at 8:59
@Ben Thanks. Things were just getting interesting, and I didn't want the fun to stop early. – Kelly Davis Mar 7 '11 at 8:59

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.