## Archive for October, 2011

### Longest shortest geodesic on a 2-sphere

October 31, 2011

This is a little note about constructing a Riemannian 2-sphere which has longer shortest geodesic than the round 2-sphere of same area.

—–  Background Story  —–

So there has been this thing called ‘mathematical conversations’ at the IAS, which involves someone present a topic that’s elementary enough to be accessible to mathematicians in all fields and yet can be expanded in different directions and lead into interesting interdisciplinary discussions.

Nancy Hingston gave one of those conversations about simple geodesics on the two-sphere one night and I was (thanks to Maria Trnkova who dragged me in) able to attend.

So she talked about some fascinating history of proving the existence of closed geodesics and later simple closed geodesics on generic Riemannian two-spheres.

Something about this talk obviously touched my ‘systolic nerve’, so when the discussion session came up I asked whether we have bounds on ‘length of longest possible shortest closed geodesic on a sphere with unit area’. The question seem to have generated some interest in the audience and resulted in a back-and-forth discussion (some of which I had no clue what they were talking about). So the conclusion was at least nobody knows such a result on top of their head and perhaps optimum is obtained by the round sphere.

—–  End of Story —-

A couple of post-docs caught me afterwards (Unfortunately I didn’t get their names down, if you happen to know who they are, tell me~) and suggested that suspending a smooth triangular region and smoothen the corners might have longer shortest geodesic than the round sphere:

The evidence being the fact that on the plane a rounded corner triangular contour has larger ‘width’ than the disc of same area. (note such thing can be made to have same width in all directions)

Well that’s pretty nice, so I went home and did a little high-school computations. The difficulty about the pillow is that the shortest geodesic isn’t necessarily the one that goes through the ‘tip’ and ‘mid-point of the base’, something else might be shorter. I have no idea how to argue that.

A suspicious short geodesic:

So I ended up going with something much simpler, namely gluing together two identical copies of the flat equilateral triangles. This can be made to a Riemannian metric by smoothing the edge and corners a little bit:

Okay, now the situation is super simple~ I want to prove that this ‘sphere’ (let’s call it $S$ from now on) has shortest geodesic longer than the round sphere ($\mathbb{S}^2$)!

Of course we suppose both $S$ and $\mathbb{S}^2$ has area 1.

Claim: The shortest geodesic on $S$ has length $\sqrt[4]{12}$ (which is length of the one through the tip and mid-point of the opposite edge.)

Proof: The shortest closed geodesic passing through the corner is the one described above, since any other such geodesics must contain two symmetric segments from the corner to the bottom edge on the two triangles, those two segments alone is longer than the one orthogonal to the edge.

That middle one has length $2h$ where

$A(\Delta) = 1/2 = h^2/\sqrt{3}$

i.e. $h = \sqrt[4]{3} / \sqrt{2}, \ \ell = 2h = \sqrt[4]{12}$

The good thing about working with flat triangles is that now I know what the closed geodesics are~

First we observe any closed geodesic not passing through the corner is a periodical billiard path in the triangular table with even period.

So let’s ‘unfold’ the triangles on the plane. Such periodic orbits correspond to connecting two corresponding points on a pair of identified parallel edges and the segment between them intersecting an even number of tiles.

W.L.O.G we assume the first point in on edge $e$. Since we are interested in orbits having shortest length, let’s take neighborhood of radius $\sqrt[4]{12} + \epsilon$ around our edge $e$: (all edges with arrows are identified copies of $e$)

There are only 6 parallel copies of $e$ in the neighborhood:

Note that no matter what point $p$ on $e$ we start with, the distance from $p$ to another copy of it on any of the six edges is EQUAL to $\sqrt[4]{12}$. (easy to see since one can slide the segments to begin and end on vertices.)

Hence we conclude there are no shorter periodic billiard paths, i.e. the shortest closed geodesic on $S$ has length $\sqrt[4]{12}$.

Note it’s curious that there are a huge amount of closed geodesics of that particular length, most of them are not even simple! However it seems that after we smoothen $S$ to a Riemannian metric, the non-simple ones all become a little longer than that simple one through the corner. I wonder if it’s possible that on a Riemannian sphere the shortest closed geodesic is a non-simple one.

Anyways, now let’s return to $\mathbb{S}^2$~ So the surface area is $1$ hence the radius is $r= \sqrt{1/4\pi} = \frac{1}{2\sqrt{\pi}}$

Any closed geodesics is a multiple of a great circle, hence the shortest geodesic has length $2 \pi r = \sqrt{\pi}$, which is just slightly shorter than $\sqrt[4]{12} \approx \sqrt{3.4}$.

Now the natural question arises: if the round sphere is not optimum, then what is the optimum?

At this point I looked into the literature a little bit, turns out this problem is quite well-studied and there is a conjecture by Christopher Croke that the optimum is exactly $\sqrt[4]{12}$. (Of course this optimum is achieved by our singular triangle metric hence after smoothing it would be $< \sqrt[4]{12}$.

There is even some recent progress made by Alex Nabutovsky and Regina Rotman from (our!) University of Toronto! See this and this. In particular, one of the things they proved was that the shortest geodesic on a unit area sphere cannot be longer than $8$, which I believe is the best known bound to date. (i.e. there is still some room to $\sqrt[4]{12}$.)

Random remark: The essential difference between this and the systolic questions is that the sphere is simply connected. So the usual starting point, namely ‘lift to universal cover’ for attacking systolic questions does not work. There is also the essential difference where, for example, the question I addressed above regarding whether the shortest geodesic is simple would not exist in systolic situation since we can always split the curve into two pieces and tighten them up, at least one would still be homotopically non-trivial. In conclusion since this question sees no topology but only the geometry of the metric, I find it interesting in its own way.

### Graph of groups in relation to 3-manifolds

October 24, 2011

(some images might appear soon)

Somehow I decided to wake up at 6:30 a.m. every Thursday to attend Bruce Kleiner‘s 9:30 course in NYU this semester. So far it’s been fun~

I learned about this thing called graph of groups. If you have been reading posts on this blog regarding any geometric group theory stuff (especially those posts related to Kleiner), then warning: this ‘graph’ has nothing to do with the Cayley graph. It’s not much about geometry but a rather ‘category-theoretical’ thing. Well, at this point you may think that you hate those algebra prople and is ready to leave…just don’t do that yet, because I hated them too, and now I finally got a tiny bit of understanding and appreciation on what those abstract non-sense was all about! :-P

So how do we connect cool 3-manifold stuff (incompressible surfaces, loops, embedded discs Heegaard splittings etc.) to groups?

Well, one handy thing is of course the Dehn’s lemma:

Theorem: For 3-manifold $M$ with boundary, if the inclusion map $i: \pi_1(\partial(M)) \rightarrow \pi_1(M)$ is not injective, then there exists a simple non-trivial loop in $\partial M$ bounding an embedded disc in $M$.

Note: Dehn’s theorem was proved by Papakyriakopoulos, I talked about it in this pervious post, although not exactly stated in this form, we can see that Dehn’s lemma follows easily from the loop theorem.

This means we can say things about the 3-manifold by only looking solely at maps between groups!

That’s cool, but sometimes we find groups and just one map between two groups are not enough, and that’s when graph of groups comes in:

Definition: A graph of groups is a graph with vertice set $V$, edge set $E$, to each vertex $v$ we associate a group $G_v$ and to each edge $e$ (say connecting $v_1, v_2$) we also associate a group $G_e$, together with a pair of injective homomorphisms $f_1: G_e \rightarrow G_{v_1}$, $f_2: G_e \rightarrow G_{v_2}$.

In our context, we should think of this as gluing together a bunch of spaces and take the fundamental group of those spaces, along with their pairwise intersections, as our vertice and edge groups. Just note that we need to have injections from the edge group to vertice groups. For simplicity one may first restrict oneself to the case where all edge groups are trivial (say spaces glued along contractible spaces).

There is something called the fundamental group of a graph of groups which is essentially the fundamental group of the resulting space after you glued spaces according to the given graph of groups. Note that the injection associated to edges takes into account how gluing of different pairs interact with each other (that is to say, for example, on a homotopy level it knows about triple intersections, etc.)

Let’s look at an application in this paper of Kleiner and Kapovich which I also talked about in an earlier post. Continue from that pervious post, now we know that

Theorem: Any hyperbolic group (plus obvious conditions, namely torsion free and does not split over a finite cyclic group) with 1-dimensional boundary has $\partial_\infty G$ homeomorphic to $\mathbb{S}^1$, the Sierpinski carpet or the Menger curve.

When $\partial_\infty G = \mathbb{S}^1$, my wonderful advisor Dave Gabai proved that $G$ would act discrete and cocompactly on $\mathbb{H}^2$ by isometries. (i.e. it’s almost the fundamental group of some hyperbolic surface except for possible finite order elements which make the action not properly discontinuous.)

Now the next step is of course figuring out when does groups act on $\mathbb{H}^3$, we have:

Cannon’s conjecture: If hyperbolic group $G$ has boundary $\mathbb{S}^2$, then $G$ acts discretely and cocompactly on $\mathbb{H}^3$ by isometries.

This conjecture was also mentioned another pervious post. Turns our we do not know much about groups with $\mathbb{S}^2$ boundary. However, using graph of groups, they were able to show:

Theorem: If Cannon’s conjecture is true, then those hyperbolic groups with Sierpinski carpet boundary are fundamental groups of hyperbolic 3-manifolds with totally geodesic boundary.

i.e. the idea is to ‘extend’ the group with Sierpinski carpet boundary to a group having sphere boundary. Of course as sets we can embed the carpet into a sphere and start to ‘reflect it along the boundary of the ‘holes’, continue the process and eventually the union of all copies of the carpets is the entire $\mathbb{S}^2$. The problem is how to ‘reflect’ a group?

First, since the boundary is homeomorphic to the carpet, there are countably many well-defined ‘boundary circles’, the group $G$ acts on the set of boundary circles. They showed this action has only finitely many different orbits. (those orbits of boundary circles will eventually correspond to those totally geodesic boundary components of our resulting 3-manifold). We pick one boundary circle from each orbit and denote their stabilizers $H_1, \cdots, H_k$ each $H_i < G$.

Define a graph of groups $\mathcal{G}$ with two vertices both labeled $G$, with $k$ edges, all going from one vertex to the other. Let the edge groups be $H_1, \cdots, H_k$.

Now we can start to 'unfold' the graph： Let $X_G$ be a 2-complex associated to a set of generators and relations for $G$ and $X_i$ be 2-complexes associated to $H_i$. The inclusion map induces cellular maps $h_i: X_i \rightarrow X_G$. Hence we have

$\displaystyle h: \sqcup_{i=1}^n X_i \rightarrow X_G$

Let $X$ be the mapping cylinder of $h$. i.e. $X$ has boundary components $\sqcup_{i=1}^n X_i$ and $X_G$.

Let $DX$ be the complex obtained by gluing together two copies of $X$ along $\sqcup_{i=1}^n X_i$, take it’s universal cover $\widetilde{DX}$. Now the fundemental group $\hat{G}$ of $DX$ is, in some sense, the group obtained by doubling $G$ along each $H_i$. i.e. $\hat{G}$ is the fundamental group of the graph graph of groups $\mathcal{G}$.

Now by studying the 1-skeleton of the complex $\widetilde{DX}$, one is able to conclude that $\hat{G}$ is Gromov hyperbolic with $\mathbb{S}^2$ boundary, as expected.

Hence from groups with Sierpinski carpet boundary we are able to produce groups with sphere boundary. Now if Cannon’s conjecture is true, $\hat{G}$ is fundamental group of some hyperbolic 3-manifold, together with Gabai’s result that $H_i$ are fundamental groups of hyperbolic surfaces, we would have that $G$ is the fundamental group of a hyperbolic 3-manifold with $n$ totally geodesic boundary components.

Well, since now we don’t have Cannon’s conjecture, there is still something we can conclude:

Definition: A n-dimensional Poincare duality group is a group which has group cohomology satisfying n-dimensional Poincare duality.

Those should be thought of as fundamental groups of manifolds in the level of homology. Well, I know nothing about group cohomologies, luckily we have:

Theorem: (Bestvina-Mess)

$\Gamma$ is a n-dimensional Poincare duality group iff it’s torsion free and $\partial \Gamma$ has integral Cech cohomology of $\mathbb{S}^{n-1}$

Great! In our case $\partial \hat{G}$ IS the sphere! So it’s a 3-dimensional Poincare duality group~ Now we have a splitting of $\hat{G}$ over a bunch of 2-dimensional Poincare duality groups (namely $H_i$) it follows that $(G; H_1, \cdots, H_n)$ is a Poincare duality pair.

It is not known whether all such pairs can be realized as fundamental groups of 3-manifolds with boundary. If so, then by Thurston’s geometrization we can also obtain what we derived assuming Cannon’s conjecture.

### A Bosuk-Ulam-kind theorem for simplexes

October 17, 2011

This little note came out of a lunch discussion with NYU grad student Alfredo Hubard earlier this week. I think the problem-solving process was quite amusing hence worth shearing.

Back in kindergarten, we all learned this theorem called ‘at any given time, there are two opposite places on Earth having exactly the same temperature and air pressure’. Yes that’s the Bosuk-Ulam theorem. I remember at some point I came across a much less famous theorem in some kind of discrete/combinatorial geometry, saying:

Theorem: Any map from a n-dimensional simplex to $\mathbb{R}^{n-1}$ must have a pair of intersecting opposite faces.

Note: each k-dimensional face of the n-dimensional simplex has a unique, well-defined $(n-1)-k$ dimensional opposite face, as shown:

Some examples of maps from the 3-simplex to $\mathbb{R}^2$:

i.e. in general they can be quite a mess. I think it can be proved by Thurston’s simplex straightening argument. (haven’t checked carefully)

To me this is like Bosuk-Ulam except for instead of considering a large amount of antipodal pairs, we consider only finitely many such pairs. Hence a discrete analoge.

However, one should note that although they are of the same nature, neither follows from the other.

So somehow this theorem came up during the lunch and Alfredo mentioned to me that professor Guth wondered whether the theorem can be proved for mappings from the simplex to lower dimensional simplicial complexes (instead of $\mathbb{R}^n$). i.e.

Question: Given $f: \Delta^{n+1} \rightarrow S$ where $\Delta^{n+1}$ denote the $(n+1)$-dimensional simplex and $S$ is a $n$-dimensional simplicical complex. Then must there be a pair of opposite faces with intersecting image?

So we started to throw out random ideas.

First of all, although only boundary faces of the simplex has non-empty opposite faces (hence can possibly be intersecting pair), it is important that $f: \Delta^n \rightarrow S$ is defined on the solid simplex. (i.e. if one just map the boundary, then we may let $S$ be topologically a sphere and make the map a homeomorphism!) So the moral is, we kind of need to ‘claps’ the simplex to a simply connected lower dimensional thing first, then map it to our $S$, hence $S$ having non-trivial topology won’t be of much help. Looks almost like the $\mathbb{R}^n$ case, doesn’t it?

Perhaps the image on $S$ can be complicated and has non-trivial topology, but this is merely ‘wrapping’ a contractible, $n$-dimensional thing around. But wrapping around can only cause more overlapping hence making the faces intersecting more, not less.

The above line of thoughts give an immediate proof in the case when $S$ is a surface and $n=2$: Lift the map to the universal cover and apply the theorem for $\mathbb{R}^2$. (It’s a little more tricky when the surface is $\mathbb{S}^2$ but you can work it out~ the map restricted to $\partial \Delta$ must be of even degree) Note this won’t generalize to higher dimensions (even for manifolds) since universal covers are no longer that similar to $\mathbb{R}^n$.

So what’s the main difference between complexes and manifolds? well, one can have more than two $n$ dimensional faces attached to the same $n-1$ dimensional face. I decided to first think of whether Bosuk-Ulam is still true if we further assume that the map extends to the solid ball (as seen above, it is true in the surface case).

After trying to generalize that ‘lifting’ proof for a while, we realized the Bosuk-Ulam does not work for simplicial complexes in all dimensions! For very simple reason, i.e. if we map the $n-sphere$ to $\mathbb{R}^n$ by projecting down, the pre-image of the center point is the only pair of antipodal points that’s mapped together. Hence if instead of $\mathbb{R}^n$ we have three $n$-simplicies attached along a single $(n-1)$-simplex (think of it as a piece of $\mathbb{R}^n$ with a vertical simplex attached in the middle):

Now we can still project to that piece of $\mathbb{R}^n$, with the image of antipodal point lying on the middle $(n-1)$ simplex, all we need to do is to separate this pair while not creating new pairs! But this is easy, just ‘drug’ the upper sheet into the third $n$-simplex a little bit:

(the region outside of the red neibourhood is unchanged) It’s easy to see that no new antipodal pair is moved together.

So now we turned back to the simplex and realized it’s even easier to argue: project in orthogonal direction to a $n$-face, originally the only intersecting pair of opposite faces was the vertex and that $n$-face. Now if we lift the vertex int the third sheet, nothing can intersect~

OK~ perhaps not a useful answer but problem solved!

So far I do not know the answer to the following:

Questions:

1. Is this intersecting faces theorem true for mapping $n$-simplexes to $n$-dimensional manifolds?

2. Is the Bosuk-Ulam true if we consider maps from spheres to $n$-dimensional manifolds which extends to the ball?

The later might be well-known. But so far I can only find a theorem by Conner and Floyd, stating that any map from $S^n$ to a lower dimensional manifold must have a pair of antipodal points mapped together.

### Stable isoperimetric inequality

October 10, 2011

Eric Carlen from Rutgers gave a colloquium last week in which he bought up some curious questions and facts regarding the ‘stability’ of standard geometric inequalities such as the isoperimetric and Brunn-Minkowski inequality. To prevent myself from forgetting it, I’m dropping a short note on this matter here. Interestingly I was unable to locate any reference to this nor did I take any notes, hence this post is completely based on my recollection of a lunch five days ago.

–Many thanks to Marco Barchies, serval very high-quality references are located now. It turns out that starting with Fusco-Maggi-Pratelli ’06  which contains a full proof of the sharp bound, there has been a collective progress on shorter/different proofs and variations of the theorem made. See comments below!

As we all know, for sets in $\mathbb{R}^n$, the isoperimetric inequality is sharp only when the set is a round ball. Now what if it’s ‘almost sharp’? Do we always have to have a set that’s ‘close’ to a round sphere? What’s the appropriate sense of ‘closeness’ to use?

One might first attempt to use the Hausdorff distance:

$D(S, B_1(\bar{0})) = \inf_{t \in \mathbb{R}^n}\{D_H(S+t, B_1(\bar{0}))$.

However, we can easily see that, in dimension $3$ or higher, a ball of radius slightly small than $1$ with a long and thin finger sticking out would have volume $1$, surface volume $\varepsilon$ larger than that of the unit ball, but huge Hausdorff distance:

In the plane, however it’s a classical theorem that any region $S$ of area $\pi$ and perimeter $m_1(\partial S) \leq 2\pi +\varepsilon$ as $D(S, B_1(\bar{0})) \leq f(\varepsilon)$ where $f(\varepsilon) \rightarrow 0$ as $\varepsilon \rightarrow 0$ (well, that $f$ is because I forgot the exact bound, but should be linear in $\varepsilon$).

So what distance should we consider in higher dimensions? Turns out the nature thing is the $L^1$ norm:

$D(S, B_1(\bar{0})) = \inf_{t\in \mathbb{R}^n} \mbox{vol}((S+t)\Delta B_1(\bar{0}))$

where $\Delta$ is the symmetric difference.

First we can see that this clearly solves our problem with the thin finger:

To simplify notation, let’s normalize our set $S \subseteq \mathbb{R}^n$ to have volume 1. Let $B_n$ denote the ball with n-dimensional volume 1 in $\mathbb{R}^n$ (note: not the unit ball). $p_n = \mbox{vol}_{n-1}(\partial B_n)$ be the ($n-1$ dimensional) measure of the boundary of $B_n$.

Now we have a relation $D(S, B_n)^2 \leq C_n (\mbox{vol}_{n-1}(\partial S) - p_n)$

As said in the talk (and I can’t find any source to verify), there was a result in the 90’s that $D(S, B_n)^4 \leq C_n (\mbox{vol}_{n-1}(\partial S) - p_n)$ and the square one is fairly recent. The sharp constant $C_n$ is still unknown (not that I care much about the actual constant).

At the first glance I find the square quite curious (I thought it should depend on the dimension maybe like $n/(n-1)$ or something, since we are comparing some n-dimensional volume with (n-1) dimensional volume), let’s see why we should expect square here:

Take the simplest case, if we have a n-dimensional unit cube $C_n$, how does the left and right hand side change when we perturbe it to a rectangle with small eccentricity?

As we can see, $D(R_\varepsilon, C_n)$ is roughly $p_n \varepsilon$. The new boundary consists of two faces with measure $1+\varepsilon$, two faces of measure $1-\varepsilon$ and $2 \times (n-2)$ faces with volume $1-\epsilon^2$. Hence the linear term cancels out and we are left with a change in the order of $\varepsilon^2$! (well, actually to keep the volume 1, we need to have $1-\varepsilon/(1+\varepsilon)$ instead of $1-\varepsilon$, but it would still give $\varepsilon^2$)

It’s not hard to see that ellipses with small eccentricity behaves like rectangles.

Hence the square here is actually sharp. One can check that no matter how many of the $n$ side-length you perturbe, as long as the volume stay the same (up to $O(\varepsilon^2)$) the linear term of the change in boundary measure always cancels out.

There is an analoge of this stability theorem for the Brunn-Minkowski inequality, i.e. Given two sets of volume $V_1, V_2$, if the sum set has volume only a little bit larger than that of two round balls with those volumes, are the sets $L^1$ close to round balls? I believe it’s said this is only known under some restrictions on the sets (such as convex), which is strange to me since non-convex sets would only make the inequality worse (meaning the sum set has larger volume), don’t they?

I just can’t think of what could possibly go wrong for non-convex sets…(Really hope to find some reference on that!)

Anyways, speaking of sum sets, the following question recently caught my imagination (pointed out to me by Percy Wong, thank him~ and I shall quote him ‘this might sound like probability, but it’s really geometry!’):

Given a set $T \subseteq l^2$ (or $\mathbb{R}^n$), we define two quantities:

$G(T)=\mathbb{E}(\sup_{p \in T} \Sigma p_i g_i)$ and

$B(T) = \mathbb{E}(\sup_{p \in T} \Sigma p_i \epsilon_i)$

where $\mathbb{E}$ is the expected value, $\{g_i\}$ are independent random variables with a standard normal distribution (mean 0, variance 1) and $\{\epsilon_i\}$ are independent Bernoulli random variables.

Question: Given any $T$, can we always find $T'$ such that

$T \subseteq T' + B_{l^1}(\bar{0}, c B(T))$ and

$G(T') \leq c B(T)$

To find out more about the question, see Chapter 4 of this book. By the way, I should mention that there is a \$5000 prize for this :-P