Yesterday and today Marty Scharlemann gave two talks on the Schoenflies Conjecture, one of the great open problems in low dimensional topology. These talks were very clear and inspiring, and I thought it would be useful to summarize what Marty said in a blog post, just for my own benefit.

The story starts with the following classical theorem, usually called the Jordan curve theorem, or Jordan-Schoenflies theorem:

**Theorem (Jordan-Schoenflies):** Let P be a simple closed curve in the plane. Then its complement has a unique bounded component, whose closure is homeomorphic to the disk in such a way that P becomes the boundary of the disk.

In order to make the relationship between the two complementary components more symmetric, one could express this theorem by saying that a simple closed curve P in the 2-sphere separates the 2-sphere into two components X and Y, each of which has closure homeomorphic to a disk with P as the boundary.

Based on this simple but powerful fact in dimension 2, Schoenflies asked: is it true for every n that every n-sphere P in the (n+1)-sphere splits the (n+1)-sphere into two standard (n+1)-balls?

For n=2 (i.e. 2-spheres in the 3-sphere) Alexander showed in 1924 that the answer is *no*: there is an embedding of the 2-sphere in the 3-sphere for which a complementary region is not homeomorphic to a ball (in fact, it it not even simply-connected). This counterexample is the well-known *Alexander’s horned sphere*, illustrated in the figure below:

For the example indicated in the figure, the “outside” region is not homeomorphic to a ball, and in fact its fundamental group is infinite. Interestingly enough, Alexander duality implies that the complementary regions have the *homology* of a ball, and the fundamental group, though infinite, is therefore perfect (i.e. every element can be expressed as a product of commutators).

Alexander’s sphere has a Cantor set of “wild” points where the sphere is not *locally flat*; i.e. where there is no neighborhood U in which the 2-sphere sits in the 3-sphere locally like a flat plane in 3-space. So Schoenflies question was modified to ask about locally flat n-spheres in the (n+1)-sphere. Perhaps surprisingly, the answer to this modified question turns out to be *yes:*

**Theorem (M. Brown 1960):** Every locally flat n-sphere in the (n+1)-sphere bounds a standard (n+1)-ball.

Brown’s argument depends on a certain remarkably simple infinite construction, introduced by Barry Mazur, and called the *Mazur swindle*. Morally, the argument is as follows. If some locally flat sphere were not standard, it would exhibit the (n+1)-sphere S as the connect sum of two manifolds X and Y, neither of which were themselves (n+1)-spheres; i.e. X#Y=S. But then we can form an infinite connect sum X#Y#X#Y#X#Y# . . . which is still homeomorphic to S. On the other hand, since Y#X=S we can bracket this infinite sum as X#(Y#X#Y#X# . . .)=X#S=X, so X=S contrary to hypothesis.

Because of the infinite nature of this construction, the resulting manifolds are only shown to be *topologically* standard, and not *smoothly* standard, even if P is smooth. So it is natural to wonder whether every smooth n-sphere in the (n+1)-sphere bounds a smooth (n+1)-ball. This is a question where the dimension is very important. For n=2, this is a classical theorem of Alexander:

**Theorem (Alexander 1924):** Every smooth 2-sphere in the 3-sphere bounds a smoothly standard 3-ball.

This is proved by a kind of Morse theory argument. We let P be the 2-sphere in question, and we look at its intersection with a foliation of the 3-sphere (minus the north/south poles, and assume by general position that all but finitely many planes are transverse to P, and at the exceptional level sets we have a standard Morse critical point – a local minimum, a local maximum, or a saddle. At a non-critical level, the intersection of the plane with P is a compact smooth 1-manifold, and hence a collection of circles. By the Jordan-Schoenflies theorem, some innermost circle bounds a disk, and one can cut along this disk to produce two simpler spheres which, by induction, bound balls. Thinking about how these balls are glued together along the disk we cut along proves the theorem. The base step of the induction involves looking at pieces with two critical points, which are analyzed directly. qed

In high enough dimensions too, the question is known to have a positive answer:

**Theorem (S. Smale 1960):** For n at least 4, every smooth n-sphere in the (n+1)-sphere bounds a smoothly standard (n+1)-ball.

This follows (at least for n>4) from Smale’s **h-cobordism Theorem**, which says that if W is a smooth cobordism between two simply-connected manifolds U and V which are both deformation retracts of W, then W is a smooth product UxI, and therefore U and V are diffeomorphic. A smooth n-sphere in the (n+1)-sphere is cobordant to a tiny standard sphere around a point, and therefore the region between them is a smooth product, and when capped off with a tiny ball around a point, is a smooth ball.

The last remaining case is n=3; this is the

**Schoenflies Conjecture:** Every smooth 3-sphere P in the 4-sphere bounds a smoothly standard 4-ball.

As a technical point: of course, we want P to bound a smoothly standard 4-ball on *both* sides. But it turns out that if one side is smoothly standard, the other side is too, since (for example), we could shrink one side down (by a smooth isotopy) to a very small, round ball in a small coordinate patch where a Riemannian metric looks almost flat, and recognize its complement as a standard smooth ball once it is small enough.

OK, let’s get started! It is natural to try to reproduce Alexander’s argument one dimension lower, and consider the intersections of P with a foliation of the 4-sphere minus the north/south poles by 3-spheres of constant “latitude”. We can put P into general position, so that the height function defining these level sets is Morse, and put the critical points on distinct levels in increasing index; a technical improvement due to Kearton-Lickorish says that we can arrange for all handles to be horizontal (ie contained in a level sphere), and for all collars (between handles) to be vertical.

By Alexander duality, P divides the 4-sphere into two submanifolds X and Y (Marty had the clever mnemonic that one should think of these as the Xterior and Ynterior), each with the homology of a 4-ball (actually, by Brown, we even know that they are homeomorphic to 4-balls, but perhaps not diffeomorphic). As we build up P by a handle decomposition, we can also imagine that we are building up X and Y at the same time. The effect on X and Y of attaching a handle to P depends on which “side” of P the handle is added (in its level 3-sphere); one has the

**Rising Water Principle:** adding a 3-dimensional i-handle to P on the Y side has no effect on Y, but adds a 4-dimensional i-handle to X (and vice-versa).

This is perhaps a bit counter-intuitive, unless one thinks of a “4-d printer”, building up X and Y as we go. During the collar regions between critical levels, the printer adds layer after layer to the “top” of X and the “top” of Y, building them higher, but not changing their diffeomorphism type. Adding an i-handle to the Y side has the effect of putting a “cap” on the top of some subset of Y; above this level, the printer lays down material on Y only above the part in the complement of this “cap”. From this point of view it is clear that the topology of Y is not changing – we are just adding a product collar on the top of some subset of the top face. But on the X side we are adding a new “bridge” running over the i-handle, which is unsupported on lower levels.

This is illustrated schematically (and one dimension lower) in the figure above. The Xterior is in red, and the Ynterior in blue. At some level, a (2-dimensional) 1-handle is added on the Ynterior side (the green square in the second figure). Above this level, the effect on the Xterior is to add a (3-dimensional) 1-handle, while the effect on the Ynterior is nothing.

There are also two kinds of “duality” to think about: the core of an “ascending” i handle in P can be “turned upside down” to be the cocore of a “descending” 3-i handle in P. But an i handle in P corresponds to an i handle in X or Y (depending on whether it is on the Y or the X side), so when it is turned upside-down in corresponds to a descending 4-i handle in X or Y.

Marty gave a nice example to illustrate these ideas. Suppose P can be built in such a way that all the (3-dimensional) 0 and 1 handles are attached on the X side. If we turn this picture upside down, a 0 handle on the X side becomes a 3 handle on the Y side, and a 1 handle on the X side becomes a 2 handle on the Y. side. So turning the picture upside down, Y is built without any (4-dimensional) 2 or 3 handles; i.e. it is made just from 0 and 1 handles. But this means Y is diffeomorphic to a thickened neighborhood of a graph, and since it is homeomorphic to a 4-ball (by Brown’s theorem), it is diffeomorphic to a thickened neighborhood of a *tree*, and hence is standard.

One of the first observations to make is that if we cut P along a surface H above all the 0 and 1 handles, and below all the 2 and 3 handles, then the two sides of H in P are actually handlebodies, and H is a Heegaard surface. Every Heegaard splitting of the 3-sphere is standard (by an old theorem of Haken), so this is quite reassuring. The genus of H is called the *genus** of the embedding*. An embedding P is said to be a *Heegaard embedding* if *every* (nonsingular) level set is a Heegaard surface (not just the ones between the 1 and the 2 handles). A recent preprint of Agol-Freedman shows that every embedding can be isotoped to a Heegaard embedding, possibly at the cost of raising the genus dramatically.

It is natural to try to get some insight into the Schoenflies conjecture by restricting attention to a specific (low) genus. Marty Scharlemann famously proved the conjecture for genus at most 2; his paper appeared in the journal *Topology* in 1984. Something that Marty emphasized is the (*a priori* unexpected) fact that (hard) 3-manifold topology can be used to get insight in the Schoenflies conjecture, at least in the low genus case. For example, suppose P is a smooth 3-sphere in the 4-sphere and (with increasing height function) all 0 and 1 handles are attached on the X side. It follow that X can be built using only 2 and 3 handles. Turning the handle decomposition of X upside down, we see that X can be built using only 1 and 2 handles. If there is *only one* 1 handle and canceling 2 handle, then after attaching the 1 handle X is a circle times 3-ball, with boundary a circle times 2-sphere, and then the result of attaching a 2-handle is to do 0-framed surgery on a knot in the boundary circle times 2-sphere in such a way as to obtain the 3-sphere. Turning the handle decomposition of this 3-sphere upside down, we can say conversely that a circle times 2-sphere is obtained by 0 frame surgery a knot K (the co-core of the 2 handle in X) in the 3-sphere. Now, the famous Property R conjecture, proved in 1987 by Gabai, says that if 0-framed surgery on a knot K in the 3-sphere gives rise to a circle times 2-sphere, then K was the unknot. This shows that X is standard, and therefore Y too, and therefore P.

In general, knowing that X is built only from 1 and 2 handles is *not* known to be sufficient to show that X is standard. In the particular context of this example, one can get around this by studying the handle decomposition of Y: if we turn the original Morse function upside down, all 2 and 3 handles of P are attached on the Y side, so Y is built only from 0 and 1 handles. Any 4-manifold built from 0 and 1 handles is a smooth thickening of a graph; if it is contractible, the graph is a tree, and the 4-manifold (i.e. Y) is the smooth 4-ball. So in this particular case, we find a shortcut to the proof, bypassing the need for property R in this case.

But the idea of using 3-manifold topology to tackle Schoenflies is too good to pass up, and in fact, a certain purely 3-dimensional generalization of Property R would imply the Schoenflies conjecture. We explain how.

We have a smooth 3-sphere P in the 4-sphere, and to show it is standard it suffices to show that the two sides X and Y are standard 4-balls. In fact, just showing that *one* of them is standard implies that the other is, and that P is standard. Suppose that we somehow have some completely different smooth 3-sphere P’ in the 4-sphere, with sides X’ and Y’, and suppose we know that X and X’ are diffeomorphic (but *a priori* we don’t know anything about the relationship of Y and Y’). If we could show that X’ was standard, then of course X would be standard, and therefore also Y, and P. How might we find such a 3-sphere P’? Remember, the handles attached on the X side do not affect the topology of X. So if we build up P’ with the same abstract handle decomposition as P, attaching the handles on the Y side in the “same” way, but the handles on the X side in a possibly different way, we will construct X’ and Y’ for which we know that X and X’ are diffeomorphic, without immediately knowing anything about Y and Y’.

This new 3-sphere P’, with the same exterior as P, is called a *reimbedding*. Marty showed that for a genus 2 splitting, reimbedding can always make one side (say Y’) a handlebody. Just as above, a handlebody is always standard, so Y’ is standard, and therefore so is X’, and therefore X, and therefore Y, and therefore P.

It is worth remarking that reimbedding circumvents one natural drawback in a naive approach on the Schoenflies conjecture. Suppose one wanted to show directly that any smooth 3-sphere P was standard, by performing some canonical sequence of simplifying moves on P, ultimately obtaining a standard round 3-sphere. For instance, one could hope to find a flow which gradually straightened out the kinks, making P flatter and flatter until it could be recognized. The existence of such a flow would prove more than just Schoenflies: it would prove not just that the space of smooth (oriented) embeddings of the 3-sphere in the 4-sphere is path-connected (which is another reformulation of Schoenflies) but that its homotopy type was that of SO(4), the space of embeddings of round 3-spheres in the round 4-sphere. By contrast, reimbedding just jumps magically from one point in the space of embeddings to another, and if the Schoenflies conjecture were true, one would know that the two points were joined by a path, but without having to choose an explicit path from one to the other.

Let’s return to Schoenflies. Our original morse function has handles in increasing order, so we can always arrange to find some level 3-sphere with the 0 and 1 handles below, and the 2 and 3 handles above. This 3-sphere splits X into two sides, which are both 4-dimensional handlebodies. Suppose one further knew that the intersection of X with this level 3-sphere was itself a 3-dimensional handlebody. Then X could be represented as a *Heegaard union*. This implies (by a direct argument) that X admits a handle decomposition with only 1 and 2 handles. Is such an X a smooth 4-ball? By looking at the boundary of X in the dual handle decomposition, we see this is equivalent to a 3-dimensional question:

**Generalized Property R Conjecture: **if surgery on an n-component link L in the 3-sphere gives a connect sum of n circle times 2-spheres, can L be transformed into the unlink by handle slides?

It is known that this conjecture would imply Schoenflies for embeddings P with a single minimum or maximum (this follows from a recent result of Agol-Freedman; for details see their preprint) . Unfortunately, it seems likely that this conjecture is *false*: Gompf produced an example of a genus 4 splitting of the 3-sphere in the 4-sphere which gives the fundamental group of X the following presentation of the trivial group:

Generalized Property R would imply that this presentation could be reduced to the trivial presentation of the trivial group by a sequence of *Andrews-Curtis* moves; i.e. Nielsen transformations and conjugation of relators. The Andrews-Curtis Conjecture says that every balanced presentation of the trivial group can be reduced to a trivial presentation (with the same number of generators and relations) by Andrews-Curtis moves, but this conjecture is widely believed to be false, and the presentation above is widely considered to be a premier candidate counterexample.

One can try to weaken this Generalized Property R conjecture by allowing extra kinds of moves, for instance stabilization, corresponding to adding canceling 1-2 handle pairs or canceling 2-3 handle pairs at the level of X. Are these Weak Generalized Property R Conjectures true or false? Let’s find out!

(Update 10/18/13: made a couple of corrections due to Marty Scharlemann)

## 6 comments

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October 20, 2013 at 10:27 pm

Ian AgolMike Freedman pointed out to me an observation about the connection between 4D smooth Poincare and Schoenflies.

If the 4D smooth Poincare conjecture is true, then the Schoenflies conjecture is also true. This follows because if one has a 3-sphere embedded in the 4-sphere which doesn’t bound a standard 4-ball, then cutting along it produces an punctured exotic sphere on either side, since capping off with a ball cannot produce a standard sphere by the observation you made.

On the other hand, if the 4D smooth Poincare conjecture is false, then the set of oriented exotic 4-spheres forms a monoid under connect sum. Then there is an invertible element of this monoid if and only if the Schoenflies conjecture is true. One direction is the essentially the observation above: if the Schoenflies is false, then a non-standard 3-sphere in the 4-sphere exhibits the 4-sphere as a connect sum of inverse exotic 4-spheres. On the other hand, if an exotic 4-sphere is invertible, then puncturing it along a ball embeds in the standard 4-sphere, so is a standard ball by Schoenflies. But this implies that it is glued from two standard balls, so by the Smale conjecture (proved by Hatcher), the manifold is standard.

October 21, 2013 at 5:45 am

Danny CalegariHi Ian – this is a lovely observation; thanks for sharing it!

July 1, 2014 at 8:42 am

Roger AlperinIn the presentation for the trivial group that you give, I think you want to assume n>1; the case n=1 can be reduced to the trivial group by A-C moves.

July 1, 2014 at 9:03 am

Danny CalegariHi Roger! You’re quite right – thanks for pointing that out. Explicitly, substituting the second relation in the first reduces the presentation to which reduces to and substituting the first into the second reduces it to the trivial presentation .

In fact, maybe n needs to be at least 4 to get to examples which are not known to be Andrews-Curtis trivial (but I am probably out of date on the state of the art).

July 1, 2014 at 10:06 am

Roger AlperinI’m enjoying your blog. I found it mentioned on facebook! I like your writing style.

July 1, 2014 at 10:08 am

Danny CalegariIf it’s on Facebook, I must be famous! Thanks very much for your compliment. :) How is life in the Bay Area these days?