I’ve been looking through the proceedings of the Heegaard splittings conference at the Technion in 2005, in particular at a list of open problems Cameron Gordon compiled based on the talks there.  Lots of good problems are discussed, but I wanted to mention a couple that were suggested by Yair Minsky.  Recall that given a Heegaard surface in a 3-manifold, for each handlebody in the complement one can consider the set of vertices in the curve complex for the surface corresponding to loops that bound disks in the handlebody. This is called a handlebody set and every Heegaard splitting determines two such sets.

The self-homeomorphisms of a handlebody act on the curve complex in a way that preserves the corresponding handlebody set.  For the two handlebodies in a Heegaard splitting, the intersection of their mapping class groups (as subgroups of the mapping class group of the surface) is precisely the mapping class group of the Heegaard splitting (the group of automorphisms of the ambient manifold that take each handlebody onto itself.)  In addition to the intersection, one can also consider the subgroup of the mapping class group for the surface generated by the mapping class groups of the handlebodies.  Yair asks whether this group is a free product with amalgamation of the mapping class groups of the handlebodies, amalgamated along their intersection.  It seems like a reasonable problem.

Second, Yair suggests looking at the set of all loops in the curve complex for the Heegaard surface that are homotopy trivial in the ambient manifold.  The loops in the handlebody sets are all homotopy trivial, and in fact are all unknots.  Loops outside the handlebody set may also be homotopy trivial, whether or not they’re unknotted.  For example, in a Heegaard splitting of the 3-sphere, every loop in the curve complex is homotopy trivial.  Yair asks when the set of homotopy trivial loops is equal to the image of the handlebody sets under the action of the group generated by the mapping class groups of the handlebodies.  My guess is that this is not true except in a few simple cases, though it seems more reasonable that the first set is always contained in the second.  Given a 3-manifold with a homogeneous metric, one might also ask what the set of loops that are isotopic to geodesics looks like.  This is probably a much harder problem.

Back in May I pointed out a paper on spacial graphs [1] that answered a question I had asked earlier: whether it is possible to embed a theta graph in the 3-sphere so that each of the three knots that result from removing edges is a trefoil. (They actually prove a more general property for a larger family of graphs.)  Akira Yasuhara, one of the authors of the paper has just sent me a link to a picture of a theta graph containing three trefoils.   It’s surprisingly simple.  Thanks Akira.

[1] Realization of knots and links in a spatial graph. (English summary) Topology Appl. 112 (2001), no. 1, 87–109.

Recall that every 2-sphere in the 3-sphere cuts it into two balls.  A knot K in the 3-sphere is in bridge position with respect to this sphere if it intersects each ball in a collection of boundary parallel arcs.  One of the nice things about bridge position is that in the double branched cover over the knot, the bridge sphere lifts to a Heegaard surface whose genus is one less than the number of arcs (i.e. bridges) in each ball.  Thus the Heegaard genus of the double branched cover is at most the bridge number of the knot minus one.

Of course, not every Heegaard splitting for the double branched cover comes from such a construction, so it’s possible that the Heegaard genus is much lower than the bound given by the bridge number.  For example, Scott Taylor has pointed out that a Heegaard surface for the knot complement also lifts to a Heegaard surface for the double branched cover.  The genus of the lifted surface is one less than twice the original genus.  Thus if the Heegaard genus of the knot complement is much smaller than the bridge number, the lift of the bridge surface will not be minimal genus.  This is the case, for example, with torus knots; they all have Heegaard genus two, but their bridge numbers can be arbitrarily large.  (One can apply a similar argument to higher genus bridge surfaces, in which case the examples of Moriah, Minsky and Schleimer [1] or my examples with Thompson [2] for genus one, can be used.)

Of course, just because these lifted surfaces are not minimal genus does not mean they’re stabilized/reducible.  For example, the double branched cover of a torus knot is a Seifert fibered space.  Some Seifert fibered spaces have strongly irreducible, non-minimal genus Heegaard splittings called horizontal splittings.  It’s possible that the bridge surface lifts to an irreducible horizontal surface.  (The way the bridge surface intersects the fibering of the knot complement makes this plausible, though I haven’t checked this carefully.)  Thus one can still ask when the non-minimal Heegaard surfaces that come from lifting a minimal bridge surface to the double branched cover are in fact reducible.

Math blog God Plays Dice has pointed out a site called the Open Problem Garden, a Wiki that lists open problems.  As I’m writing this, it has just three topology problems, all of which are of the set-theoretic variety.  Graph theory, on the other hand, has over a hundred.  I think this could be very useful if enough people update it and check it.  You can also get an RSS feed of the different pages (I just signed up for topology.)

Though not quite as exciting as a possible proof of the Riemann hypothesis, a paper on realizing the mapping class group of a surface as a subgroup of the group of self-homeomorphisms [1] caught my eye a couple of days ago. Recall that the set of all homeomorphisms from a surface S to itself form a group that I’ll call Aut(S) or the automorphism group. The mapping class group of S, which I’ll write Mod(S), is the quotient of this group by isotopies of the surface. The quotient construction implies a homomorphism from Aut(S) onto Mod(S). The paper above proves that for a surface of genus at least two, there is no reverse homomorphism from Mod(S) into Aut(S) such that composing the maps produces the identity on Mod(S).

I had though this was already known, but apparently it was only known for genus at least 5 (or 3 if you replace Aut(S) with the group Diff(S) of diffeomorphisms.) The introduction to the paper lists the previously known results. You can ask a similar question for subgroups of Mod(S) as well, i.e. whether there’s a map from the subgroup into Aut(S) or Diff(S) that composes to the identity on that subgroup. For infinite cyclic groups, the answer is an almost immediate yes (just pick any representative for a generator). For finite subgroups, the answer is a much harder to prove yes; this is the Nielsen Realization Theorem (which was proved by Steve Kerckhoff, not by Nielsen). I wonder which infinite, non-cyclic subgroups of Mod(S) can be realized as groups of homeomorphisms?

The proof in the paper examines a certain relation that is discussed in Farb and Margalit’s primer on mapping class groups: Given a separating loop in a surface, you can write a Dehn twist around that loop as a composition of Dehn twists along loops in the interior of one of the complementary components or along loops in the other complementary component. Since both give you a Dehn twist along the same loop, these give you a relation. The authors then use the machinery defined in [2] (where the “no” answer for genus at least 5 is proved) to show that such a relation cannot exist in Aut(S).

[2] V. Markovic, Realization of the mapping class group by homeomorphisms. Inventiones Mathematicae
168 , no. 3, 523–566 (2007)

It appears that someone has just posted a 40 page preprint that claims to prove the Riemann hypothesis. I don’t know anything about the author or the research program that this came out of, so I can’t judge how likely it is to be correct. Does anyone else know?

I thought I’d follow up on the construction beginning Commensurability and make a post rather than a comment. Gonzalez-Acuna and Whitten actually show that if a knot complement (i.e. the complement of a knot in S^3) finitely covers another knot complement, then the covering is cyclic [1]. This was done algebraically, and now in the post-Perelman world it gives a characterization of knots in S^3 with lens space surgeries through the construction Jesse described. Genevieve Walsh pointed this out to me last summer.
One may find the prospect of using this characterization to approach the Berge Conjecture tantalizing.

Instead of performing a p/1-lens space surgery on a knot followed by taking the p-fold S^3 cover of the resulting lens space to get another knot in S^3, one can do it the other way around. Take a cover then do surgery. More specifically, take the p-fold cyclic branched cover of the knot, then do 1-surgery to obtain S^3 again.

Through Heegaard Floer homology, we now know that such knots must be fibered (see Ni and the references therein). Consequentially we can describe the monodromy of the covering knot in S^3. If a knot with p/1-lens space surgery has monodromy , then the knot with covering complement is a fibered knot of the same genus with monodromy (where, is a Dehn twist along the boundary).

Dunno if this approach will really help elucidate a resolution to the Berge conjecture, but it does present the following problem. Given a knot in S^3, you can either (a) take a p-fold cyclic branched cover and then do 1-surgery on the result to obtain manifold A or (b) do p/1-surgery and then take a p-fold cover (a smallest covering in which the knot lifts to a null homologous one perhaps?) to obtain manifold B. To what extent do the manifolds A and B differ?

[1] F. Gonzalez-Acuna and W. C. Whitten. Imbeddings of knot groups in knot groups. Geometry and topology (Athens, Ga., 1985), 147–156, Lecture Notes in Pure and Appl. Math., 105, Dekker, New York, 1987

Here’s an interesting construction that I recently encountered: It’s possible to find a knot in the 3-sphere whose complement has a finite cover that is also a 3-sphere knot complement. Let K be a knot with a Dehn surgery producing a lens space (for example a Berge knot). The lens space is finitely covered by the 3-sphere and the image of K lifted to the 3-sphere is a new knot K’ . The complement of K’ finitely covers the complement of K. It turns out this is the only way to build a knot complement covered by a knot complement, which is proved in [1]. It also shows up in Reid and Walsh’s paper on commensurability classes of 2-bridge knots [2]. (For the record, it was grad. student Neil Hoffman of UT Austin who told me about this construction.)

Two compact 3-manifolds are called commensurable if one has a finite cover that is homeomorphic to a finite cover of the other. Two groups are called commensurable if one has a finite index subgroup that is isomorphic to a finite index subgroup of the other. I don’t know which definition came first, but thanks to some basic algebraic topology, the definitions are more or less equivalent: A cover of a topological space is uniquely determined by (the conjugacy class of) a subgroup of the fundamental group. Thus if two 3-manifolds are commensurable then their fundamental groups are commensurable. Conversely, a closed (or cusped) hyperbolic 3-manifold is determined by its fundamental group. Thus two hyperbolic 3-manifolds have commensurable fundamental groups if and only if they’re commensurable. Note that commensurable groups are quasi-isometric, so these ideas are related to coarse geometry as well.

Walsh and Reid prove their result by showing that in the commensurability class for a knot complement, there is a unique minimal element (i.e. it is covered by every 3-manifold in the class) that is the quotient of the hyperbolic plane by the normalizer in Isom(H^3) of the group of isomitries that produce the knot complement. (The knot complement is a regular cover of this manifold, so they need to show that the knot complement doesn’t have any hidden symmetries, i.e. doesn’t irregularly cover any smaller 3-manifold.) They then show that this minimal element of the commensurability class covers exactly one knot complement.

[1] F. Gonzalez-Acuna and W. C. Whitten, Imbeddings of three-manifold groups,
Mem. Amer. Math. Soc. 474 (1992).

A knot in R^3 is in bridge position if the horizontal plane (y = 0, say) cuts it into two sets of arcs, each of which can be isotoped into the plane (though both can’t be isotoped into the plane at the same time). This is roughly equivalent to the condition that in the restriction of the height function f(x,y,z) = y to the knot, every maximum is above every minimum. This definition suggests a connection between bridge positions and Heegaard splittings since a Heegaard splitting can be defined as a level set of a Morse function in a 3-manifold in which every index i critical point is below every index (i + 1) critical point. (The Heegaard surface sits between the index 2 and index 3 critical points.) In this definition, we don’t care if the Morse function is induced by an embedding of the 3-manifold into a higher dimensional space. But what if we did?

Scharlemann [1], [2] has used an idea along these lines in his work on the Schoenflies conjecture (that every 3-sphere smoothly embedded in R^4 bounds a smooth ball), showing, in particular, how to interpret the embedding into R^4 as a picture in R^3. Scott Taylor [3], [4] has studied the picture in R^3 in more detail. Of course, Scharlemann focussed entirely on embeddings of the 3-sphere. It might be interesting to apply these sorts of methods to Heegaard splittings of other 3-manifolds. (Note: The link from [1] was originally to the wrong paper, but it’s fixed now.)

For example, given a 3-manifold that can be embedded in R^4, can it be embedded so that the restriction of the height function is a Morse function inducing a minimal genus Heegaard splitting? Can every embedding be isotoped so the height function induces a minimal genus Heegaard splitting? For 3-manifolds that can’t be embedded in R^4, one can ask the question for the smallest R^n that it can be embedded in.

My guess is that the answer is no, but I don’t know how one might prove it. (I’d start with Scharlemann’s methods for the 3-sphere in R^4.) Note that if a 3-manifold can be embedded in R^n then it can be embedded in R^(n+1) in a way that can be isotoped to induce any Heegaard splitting. It might also be interesting to try to characterize which gluing maps produce Heegaard splittings that come from embeddings in R^4, R^5, etc.

[2] M. Scharlemann, Smooth spheres in R4 with four critical points are standard, Invent. Math. 79 (1985) 125–141.

It appears that some crafty biologists have figured out a way to trick unsuspecting internet users into helping them find minimal energy embeddings of complex proteins in R^3 (i.e. protein folding). The game fold it allows players to manipulate proteins and scores them based on how efficient an embedding they can find. The best solutions are then recorded by the main site. (There’s also a nice tutorial for players who don’t know any biology.)

Recall that a protein is a chain of amino acids linked by single-bonded carbon atoms that allow the joints to rotate. Different angles will determine different distances between atoms in the protein, so different embeddings have different amounts of potential energy. In nature, the protein will twist along the joints to take on an embedding that minimizes the energy. Scientists have figured out how to read off the sequence of amino acids in a protein, but figuring out the lowest energy embedding is not so easy, since there are infinitely many possible configurations.

There have been computer simulations of protein folding for a number of years now, but this only solves part of the problem - all the computer simulations may just just be finding local minima and missing the actual solutions. Fold it takes advantage of some of the things the human mind still does better than a computer. (I think this is called crowd sourcing.) I wonder if there are any math problems that could benefit from a similar campaign. One could probably get ropelength estimates this way, but they would still just be estimates rather than a proof.