Systoles and the generalized Geroch conjecture

Almost a year ago, I said here that I would write a sequence of posts on some simple facts and observations related to the systolic inequality but got distracted and didn’t manage to do much of that…

I was reminded last week as I heard professor Guth’s talk on systoles for the 4th time (Yes, the same talk! –in Toronto, Northwestern, India and here at the IAS). It’s interesting that I’m often thinking about different things each time I hear the same talk. This one is about the generalized Geroch conjecture.

Geroch conjecture: \mathbb{T}^n (the n-torus) does not admit a metric of positive scalar curvature.

The conjecture is proved by Schoen and Yau (1979).

Now, scalar curvature can be seen as a limit of volume of balls:

Definition: The scalar curvature of M at p is

\displaystyle \mbox{Sc}(p) = c_n \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^{n+2}}

where \mbox{Vol}_E is the Euclidian volume and c_n is a positive constant only depending on the dimension n.

Note that since our manifold does not have any cone points,

\displaystyle \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^n}

must vanish. Further more, the Riemannian structure on M forces the r^{n+1} term to vanish.

Since for this context we only care about is whether the scalar curvature is larger or smaller than 0, we can be even more simple-minded: M has positive scalar curvature at p all small enough balls around p has smaller volume than their Euclidean cousins (with a difference of order propositional to r^{n+2}). In light of this definition, we have:

Restatement of the Geroch conjecture: For all g on \mathbb{T}^n, there exists some point p s.t. \mbox{Sc}(p) \leq 0.

This is to say, small enough balls around some point $p$ are not small enough for it to have positive scalar curvature. What if instead we look at balls of a fixed radius instead of those infinitesimal balls? This naturally leads to

Generalized Geroch conjecture: For any (\mathbb{T}^n, g), for all r, there exists p s.t. \mbox{Vol}_g(B(p, r)) \geq \mbox{Vol}_E(B(\bar{0}, r)).

(For those r larger than the injectivity radius, we lift M to its universal cover so that all homotopically non-trival loops are ‘unfolded’)

Let’s take a look at the 2-torus to get a feel of the conjecture:

The flat torus, of course, has 0 r-scalar curvature at all points.

For the regular rotational torus, we take the ball around the saddle point of the gradient flow, the ball look like a saddle, as shown below.

To see that this has area larger than the analogous Euclidean ball, we can cut it along radial rays into thin triangles, each triangle can be ‘almost flattened’ to a Euclidean triangle, but we have a more triangles than in the Euclidean case.

What if we try to make the surface spherical for most of the area and having those negative scalar curvature points taking up a very small potion. One of my first attempts would be to connect a few spheres with cylinders:

We have a few parameters here: the number of balls n, the width of the connecting cylinders w, the length of the connecting cylinders l and the radius of each sphere R.

If cylinders are too long (longer than 2r), then we can just take the ball in the middle of the cylinder, the volume when lifted to universal cover would be equal to Euclidean.

If the width of cylinders are much smaller than r, then the ball around a point in the gluing line would have volume almost a full spherical ball plus a half Euclidean ball, which would obviously be larger than a full Euclidean ball.

Hence the more interesting case is to have very short, wide tubes and as a consequence, have many balls forming a loop. In this case, the ‘worst’ ball would be centered at the middle of the tube, it intersects the two spheres connected by the tube in something a bit larger than a spherical half-ball.

I haven’t figured out an estimate yet. i.e. can the advantage taken from the fact that spherical ball are smaller than Euclidean balls cancel out the ‘a bit larger than half’? I think that would be interesting to work out.

Finally, let’s say what does this has to do with systoles:

Theorem: Generalized Geroch conjecture \Rightarrow \mbox{Sys}(\mathbb{T}^n, g) \leq \frac{2}{\omega_n^{\frac{1}{n}}} \mbox{Vol}_g(\mathbb{T}^n)^{\frac{1}{n}} (which is the systolic inequality with a constant better than what we have so far)

Proof: Suppose not,

\mbox{Sys}(\mathbb{T}^n, g) > 2 (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}

Let r = (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}, by the generalized Geroch conjecture we have some B(p, r) larger than the Euclidean ball. i.e.

\mbox{Vol}_g(B(p, r))>\omega_n r^n = \omega_n \frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n} = \mbox{Vol}_g(\mathbb{T}^n)

Since the systole is at least 2r, hence B(p, r) cannot contain any homotopically non-trival loop i.e. it does not “warp around” and get unfolded when passing to the universal cover. Hence volume of a ball with radius r cannot be larger than the volume of the whole manifold. Contradiction

Proving the tameness conjecture

I have recently went through professor Gabai’s wonderful paper that gives a proof of the tameness conjecture. (This one is a simplified version of the argument given in Gabai and Calegari, where everything is done in the smooth category instead of PL). It’s been a quite exciting reading with many amazing ideas, hence I decided to write a summary from my childish viewpoint (as someone who knew nothing about the subject beforehand).

We say a manifold is tame if it an be embedded in a compact manifold s.t. the closure of the embedding is the whole compact manifold.

To motivate the concept, let’s look at surfaces: Any compact surface is, of course, tame. However, if we “shoot out” a few points of the surface to infinity, as the figure below, it become non-compact but still tame, as we can embed the infinite tube to a disk without a point.

Of course, we can also make a surface non-compact by shooting any closed subset to infinity (e.g. a Cantor set), but such construction will always result in a tame surface. (This can be realized using similar embeddings as above, we may embed the resulting surface into the original surface with image being the original surface subtract the closed set. If the closed set has interior, we further contract each interior components.)

On the other hand, any surface with infinite genus would be non-tame since if there is an embedding into a compact set, the image of ‘genesis’ would have limit points, which will force the compact space fail to be a manifold at that point.

Hence in spirit, being tame means that although the manifold may not be compact itself, but all topology happens in bounded regions (we can think of a complete embedding of the manifold into some \mathbb{R}^N so bounded make sense)

As usual, life gets more complicated for three-manifolds.

Tameness conjecture: Every complete hyperbolic 3-manifold with finitely generated fundamental group is tame.

A bubble chart for capturing the structure of the proof:

A few highlights of the proof: The key idea here is shrinkwrapping, very roughly speaking, to prove an geometrically infinite end is tame one needs to find a sequence of simplicial hyperbolic surfaces exiting at the end. Bonahon’s theorem gives us a sequence of closed geodesics exiting the end. By various pervious results, one is able to produce (topological) surfaces that are ‘in between’ those geodesics. Shrinkwrapping takes the given surface and shrinks it until it’s ‘tightly wrapped’ around the given sequence of geodesics. The fact that each of the curve the surface is wrapping around is a geodesic guarantees the resulting surface simplicial hyperbolic. (think of this as folding a piece of paper along a curve would effect its curvature, but alone a straight line would not; geodesics are like straight lines).

Once we have that, the remaining part would be showing the position of the surfaces are under control so that they would exit the end. Since simplicial hyperbolic surfaces has curvature \leq -1, by Gauss-Bonnet they have uniformly bounded area (given our surfaces also has bounded genus). By passing to a subsequence, we may choose the sequence of geodesics to be separated by some uniform constant, which will guarantee the wrapped surfaces are not too thin in the thick parts of the manifold, hence we have control over the diameter of the surface, from which we can conclude that the surfaces must exit the manifold.

Remark: Note that in general, unlike in two dimensions, a three manifold with finitely generated fundamental group does not need to be tame as the Whitehead manifold is homotopic to \mathbb{R}^3 (hence trivial fundamental group) but is not tame. On the other hand, if we have infinitely generated fundamental group, then the manifold can never be tame. The theorem says all examples of non-tame manifolds with finitely generated fundamental group does not admit hyperbolic structure.

Fibering the figure-8 knot complement over the circle

As I was making some false statements about how I think geometrically finite ends of a hyperbolic three manifold would look like, professor Gabai pointed out this super cool fact (proved by Cannon and Thurston, 2007) that the figure-eight knot complement admits a hyperbolic structure and fibers over the circle, but if we lift any fiber (which would be a surface) into the hyperbolic 3-space, the resulting surface would be an embedded topological disc with limit set being the whole limit 2-sphere (!) i.e. if we see \mathbb{H}^3 as a Euclidean open ball, then the boundary of such a disc is a Peano curve that covers the whole 2-sphere bounding \mathbb{H}^3.

I have read about the hyperbolic structure on the figure-8 knot complement in Thurston’s notes (4.3) (A similar construction can be found in my pervious post about hyperbolic structure on the Whitehead link complement), but I didn’t know the fibering over circle part, so I decided to figure out what this fibration would look like.

After playing with chicken wire and playdo for a few days, I am finally able to visualize the fibration. Here I want to point out a few simple points discovered in the process.

Start with the classical position of the figure-8 knot (two ends extends to infinity and meet at the point infinity in \mathbb{S}^3):

To find a fibration over the circle, we need to give a surface that spans the knot (such surface is called a Seifert surface) and a homotopy of the surface \varphi: S \times [0,1] \rightarrow \mathbb{S}^3 \backslash K which restricts to a bijection from S \times [0,1) to \mathbb{S}^3 \backslash K and \varphi(S\times\{1\}) = \varphi(S\times\{0\}).

For quite some time, I tried with the following surface:

Since it’s perfectly symmetric (via a rotation by \pi), we only need to produce a homotopy that sends S to the symmetric Seifert surface in the upper half plane. I was not able to find one. (I’m still curious if there is such homotopy, if so, then there are more than one way the knot complement can fiber)

It turns out that there are in fact non-homeomorphic (hence of course non-homotopic) Seifert surfaces spanning the knot, the one I end up using for the fibration is the following surface:

Or equivlently, we may connect the two ends at a finite point.

To see the boundary is indeed the figure-8 knot:

Note that this surface is not homeomorphic to the pervious one because this one is orientable and the pervious is not.

Now I’ll leave it as a brain exercise to see the homotopy. (well…this is largely because it takes forever to draw enough pictures for expressing that) A hint on how the it goes: think of the homotopy as a continuous family of disjoint Seifert surfaces that ‘swipes through’ the whole \mathbb{S}^3 \backslash K and returns to the initial one. As in the picture above, our surface is like a disc with two intertwined stripe handles on it, each handle is two twists in it. The major step is to see that one can ‘pass’ the disc through a double-twisted handle by making the interior of the old disc to become the interior of the new handle. i.e. we can homotope the bowl from under the strap to above the strap with a family of disjoint surfaces with same boundary.

In in figure-8 knot case, the disc would need to pass through both straps and return to itself.