It's possible this question is trivial, in which case it will be answered quickly. In any case, I realized that it's a basic question the answer to which I should know but do not.

Everybody loves knots — one-dimensional compact manifolds mapped generically into three-dimensional compact manifolds — and it's natural to ask about "knots" in higher dimensions. Of course, the space of generic maps of a one-dimensional compact manifold into a four-dimensional compact manifold is connected, so there is no interesting "knotting". Instead, people usually think about "surface knots in 4d", which are usually defined as embedded compact 2-manifolds in a 4-manifold.

But surfaces can map into 4-space in much more interesting ways. In particular, whereas a generic map from a 1-manifold to a 3-manifold is an embedding, two generic surfaces in 4-d can be "stuck" on each other: the generic behavior is to have point intersections. So a richer theory than that of embedded surfaces in 4-space is one that allows for these point self-intersections — it would be the theory of connected components of the space of generic maps.

Still, though, thinking about these self-intersections is hard, and their existence is part of what makes 2-knot theory hard (for instance, it interferes with developing a good "Vassiliev" theory for 2-knots). If you really want to reproduce the fact that generic maps have no self intersections, you should move the ambient space one dimension higher.

Hence my question:

Can compact 2-manifold embedded into a compact 5-manifold be interestingly "knotted"? I.e. let $L$ be a compact 2-manifold and $M$ a compact 5-manifold; are there multiple connected components in the space of embeddings $L \hookrightarrow M$?

I expect the answer is "no", else I would have heard about it. But my intuition is sufficiently poor that I thought it best to ask.


5 Answers 5


If $M$ is a connected compact $2$-manifold, then it unknots in $\Bbb R^5$. More generally, $k$-connected $n$-manifolds embed in $\Bbb R^{2n-k}$, provided $k<\frac{n-2}2$, and unknot in $\Bbb R^{2n-k+1}$, provided $k<\frac{n-1}2$. This was proved around 1961 - by Roger Penrose, J.H.C. Whitehead, and Zeeman in the PL category; and by Haefliger in the smooth category. Later Zeeman and Irwin relaxed the metastable dimension restrictions in the PL result to codimension $\ge 3$ (see Zeeman's "Seminar on Combinatorial Topology").

On the other hand, the disjoint union of two $2$-spheres is a compact $2$-manifold. It definitely knots in $\Bbb R^5$ as detected by the linking number. That is the degree $\alpha$ of $S^2\times S^2\to S^4$, $(p,q)\mapsto \frac{f(p)-g(q)}{||f(p)-g(q)||}$, calling our link $f\sqcup g:S^2\sqcup S^2\to\Bbb R^5$. A nontrivial link is the Hopf link, whose components are the factors of the join $S^5=S^2*S^2$. Since $S^2$ unknots in $\Bbb R^5$, the exterior of one component is always homotopy equivalent to $S^2$, and the linking number is also the degree $\lambda$ of $p(S^2)\to S^5\setminus q(S^2)\simeq S^2$. [In different dimensions, where $\alpha$ and $\lambda$ are not numbers but homotopy classes of spheroids (more precisely $\alpha$ factors though a spheroid up to homotopy, upon killing the wedge), their relation is more interesting: $\alpha$ equals the suspension of $\lambda$ (up to a sign).]

By Haefliger's theorem (1963) that embeddings in the metastable range are classified by equivariant homotopy of two-point configuration spaces, the linking number for each pair of components is the only invariant of smoothly embedded $2$-manifolds in $\Bbb R^5$. This recovers the result that connected surfaces unknot in $\Bbb R^5$; and additionally implies that there is nothing new for $3$-component links. [In contrast, there are the Borromean rings of three $3$-spheres in $\Bbb R^6$, whose nontriviality is detected e.g. by a nonvanishing triple Massey product in the complement. Thinking of the usual Borromean rings in $\Bbb R^3$ as lying in the three coordinate planes, one can similarly do three copies of $S^1*S^1$ lying in the two-factor subproducts of $\Bbb R^2\times\Bbb R^2\times\Bbb R^2$.]

Also smooth, PL and topological knot theories coincide for smooth $n$-manifolds in smooth $m$-manifolds in the metastable range $m>\frac{3(n+1)}2$ (this includes $2$-manifolds in $\Bbb R^5$). In more detail, Haefliger's classification theorem implies that if two smooth embeddings in the metastable range are isotopic (=homotopic through topological embeddings, possibly wild) then they are smoothly isotopic. Weber's PL classification theorem (1967) implies additionally that every PL embedding of a smooth manifold in the metastable range is ambient isotopic to a smooth embedding. Also it follows from results of Edwards and Bryant that an arbitrary topological embedding in codimension $\ge 3$ is isotopic to a PL embedding, and, from results of Bryant-Seebeck, that a locally flat topological embedding in codimension $\ge 3$ is ambient isotopic to a PL embedding.


I will answer for embeddings of Sn in Sn+3.

No in the PL category by Zeeman.
No in the topological category by Stallings.
Yes in the differentiable category. See Levine and Haefliger.

EDIT: In the case $n=2$, which the question is about, the result is no in the differentiable category as well, as pointed out in other answers.

  • $\begingroup$ The result by Stallings is about locally nice topological embeddings? $\endgroup$ Jan 20, 2011 at 22:25
  • $\begingroup$ @Mariano- Of course. Otherwise there would be knotting in any codimension due to the existence of wild embeddings. $\endgroup$ Jan 20, 2011 at 22:32

I think the first reference in the literature for the result you want is Wen-Tsun Wu, On the isotopy of $C^r$-manifolds of dimension $n$ in Euclidean $(2n+1)$-space. Sci. Record (N.S.) 2 1958 271--275.

In it he proves any two $C^r$-embeddings $M^n \to \mathbb R^{2n+1}$ are isotopic, provided $n>1$. $M^n$ is any compact $n$-manifold (the title does not mention he also assumed connected, but it is). The techniques are now standard -- your two embeddings $f_0, f_1 : M \to \mathbb R^{2n+1}$ are homotopic $f_t : M \to \mathbb R^{2n+1}$ so you look at the "graph", $(x,t) \longmapsto (f_t(x), t)$ as a map $M \times [0,1] \to \mathbb R^{2n+!} \times [0,1]$ and then approximate by an immersion and consider appropriate usage of the Whitney trick.


An important reference concerning higher-dimensional knots is this paper of Zeeman of 1962. The paper considers the existence of knotted $k$-spheres inside $n$-spheres for general $n>k$. He proves that in the piecewise-linear setting a $k$-sphere in a $n$-sphere is always unknotted provided the codimension $n-k$ is strictly greater than 2, hence in particular there are no knotted 2-spheres in $S^5$ or $\mathbb R^5$ in the PL category.

As he says in the introduction of the paper, knotted spheres may exist however in the smooth category. The lowest-dimensional ones are some knotted $S^3$ inside $S^6$ found by Haefliger. However, Haefliger also shows that $S^k$ is always unknotted in $S^n$ when $n>3(k+1)/2$, and for $k=2$ we get $n>4.5$: hence 2-spheres unknot in $S^5$ or $\mathbb R^5$ also in the smooth category.

I don't know the proofs, but maybe one can prove that $S^2$ unknots in $S^5$ by decomposing $S^2$ as a 0- and 2-handle, thicken and extend this decomposition to the whole of $S^5$, and then simplify handles as in Smale's proof of Poincaré conjecture. By trading 1- and 4-handles with 3- and 2-handles we remain with 2- and 3-handles only, the core of each being attached to a (necessarily unknotted) circle in $S^4$ (the boundary of the 0-handle). The resulting decomposition of $S^5$ should be standard and hence the 2-sphere is unknotted (I guess).


I always thought that Haefliger's result applied in dimensions higher than 5, but in skimming the manuscript, it looks like 5 might be a critical case. Please read at page 404 and 405 carefully.

To address your original question, that surfaces in 4-space can still intersect and that they can be knotted may not effect the potential knottedness in dimension 5. On the other hand, it is easy to construct knottings of 3-manifolds in 5-space by spinning and twist-spinning.


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