# The travelling salesman problem

This is one of those items I should have written about long ago: I first heard about it over a lunch chat with professor Guth; then I was in not one, but two different talks on it, both by Peter Jones; and now, finally, after it appeared in this algorithms lecture by Sanjeev Arora I happen to be in, I decided to actually write the post. Anyways, it seem to live everywhere around my world, hence it’s probably a good idea for me to look more into it.

Has everyone experienced those annoy salesman who keeps knocking on you and your neighbors’ doors? One of their wonderful properties is that they won’t stop before they have reached every single household in the area. When you think about it, in fact this is not so straight foreword to do; i.e. one might need to travel a long way to make sure each house is reached.

Problem: Given $N$ points in $\mathbb{R}^2$, what’s the shortest path that goes through each point?

Since this started as an computational complexity problem (although in fact I learned the analysts’s version first), I will mainly focus on the CS version.

Trivial observations:

In total there are about $N!$ paths, hence the naive approach of computing all their lengths and find the minimum takes more than $N! \sim (N/e)^N$ time. (which is a long time)

This can be easily improved by a standard divide and concur:

Let $V \subseteq \mathbb{R}^2$ be our set of points. For each subset $S \subseteq V$, for each $p_1, p_2 \in S$, let $F(S, p_1, p_2) =$ length of shortest path from $p_1$ to $p_2$ going through all points in $S$.

Now the value of this $F$ can be computed recursively:

$\forall |S| = 2$, $F(S, p_1, p_2) = d(p_1, p_2)$;

Otherwise $F(S, p_1, p_2) =$

$\min \{ F(S \backslash \{p_2\}, p_1, q) + d(q, p_2) \ | \ q \in S, q \neq p_1, p_2 \}$

What we need is minimum of $F(V, p_1, p_2)$ where $p_1, p_2 \in V$. There are $2^N$ subsets of $V$, for any subset there are $\leq N$ choices for each of $p_1, p_2$. $F(S, p_1, p_2)$ is a minimum of $N$ numbers, hence $O(N)$ time (as was mentioned in this pervious post for selection algorithm). Summing that up, the running time is $2^N\times N^2\times N \sim O(2^N)$, slightly better than the most naive way.

Can we make it polynomial time? No. It’s well known that this problem is NP-hard, this is explained well in the wikipedia page for the problem.

Well, what can we do now? Thanks to Arora (2003), we can do an approximate version in polynomial time. I will try to point out a few interesting ideas from that paper. The process involved in this reminded me of the earlier post on nonlinear Dvoretzky problem (it’s a little embracing that I didn’t realize Sanjeev Arora was one of the co-authors of the Dvoretzky paper until I checked back on that post today! >.< ) it turns out they have this whole program about ‘softening’ classic problems and produce approximate versions.

Approximate version: Given $N$ points in $\mathbb{R}^2$, $\forall \varepsilon > 0$, find a path $\gamma$ through each point such that length $l(\gamma) < (1+\varepsilon)l(\mbox{Opt})$.

Of course we shall expect the running time $T$ to be a function of $\varepsilon$ and $N$, as $\varepsilon \rightarrow 0$ it shall blow up (to at least exponential in $N$, in fact as we shall see below, it will blow up to infinity).

The above is what I would hope is proved to be polynomial. In reality, what Arora did was one step more relaxed, namely a polynomial time randomized approximate algorithm. i.e. Given $V$ and $\varepsilon$, the algorithm produces a path $\gamma$ such that $E(l(\gamma)-l(\mbox{Opt}) < \varepsilon)$. In particular this means more than half the time the route is within $(1+\varepsilon)$ to the optimum.

Theorem (Arora ’03): $T(N, \varepsilon) \sim O(N^{1/\varepsilon})$ for the randomized approximate algorithm.

Later in that paper he improved the bound to $O(N \varepsilon^{C/\varepsilon}+N\log{N})$, which remains the best know bound to date.

Selected highlights of proof:

One of the great features in the approximating world is that, we don’t care if there are a million points that’s extremely close together — we can simply merge them to one point!

More precisely, since we are allowing a multiplicative error of $\varepsilon$, we also have trivial bound $l(\mbox{Opt}) > \mbox{ diam}(V)$, Hence the length can be increased by at least $\varepsilon \mbox{diam}(V)$ which means if we move each point by a distance no more than $\varepsilon \mbox{diam}(V) / (4N)$ and produce a path $\gamma'$ connecting the new points with $l(\gamma')< (1+\varepsilon/2)l(\mbox{Opt})$, then we can simply get our desired $\gamma$ from $\gamma'$, as shown:

i.e. the problem is “pixelated”: we may bound $V$ in a square box with side length $\mbox{diam}(V)$, divide each side into $8N/\varepsilon$ equal pieces and assume all points are in the center of the gird cell it lies in (for convenience later in the proof we will assume $8N/\varepsilon = 2^k$ is a power of $2$, rescale the structure so that each cell has side length $1$. Now the side length of the box is $8N/\varepsilon = 2^k$):

Now we do this so-called quadtree construction to separate the points (reminds me of Whitney’s original proof of his extension theorem, or the diatic squares proof of open sets being countable) i.e. bound $V$ in a square box and keep dividing squares into four smaller ones until no cell contains more than one point.

In our case, we need to randomize the quad tree: First we bound $V$ in a box that’s 4 times as large as our grid box (i.e. of side length $2^{k+1}$), shift the larger box by a random vector $(-i/2^k,-j/2^k)$ and then apply the quad tree construction to the larger box:

At this point you may wonder (at least I did) why do we need to pass to a larger square and randomize? From what I can see, doing this is to get

Fact: Now when we pick a grid line at random, the probability of it being an $i$th level dividing line is $2^i/2^k = 2^{i-k}$.

Keep this in mind.

Note that each site point is now uniquely defined as an intersection of no more than $k$ nesting squares, hence the total number of squares (in all levels) in this quad tree cannot exceed $N \times k \sim N \times \log{N/\varepsilon}$.

Moving on, the idea for the next step is to perturb any path to a path that cross the sides of the square at some specified finite set of possible “crossing points”. Let $m$ be the unique number such that $2^m \in [(\log N)/\varepsilon, 2 (\log N)/ \varepsilon ]$ (will see this is the best $m$ to choose). Divide sides of each square in our quad tree into $2^m$ equal segments:

Note: When two squares of different sizes meet, since the number of equally squares points is a power of $2$, the portals of the larger square are also portals of the smaller one.

With some simple topology (! finally something within my comfort zone :-P) we may assume the shortest portal-respecting path crosses each portal at most twice:

In each square, we run through all possible crossing portals and evaluate the shortest possible path that passes through all sites inside the square and enters and exists at the specified nodes. There are $(2^{4 \times 2^m})^2 \sim ($side length$)^2 \sim (N/\varepsilon)^2$ possible entering-exiting configurations, each taking polynomial time in $N$ (in fact $\sim N^{O(1/\varepsilon)}$ time) to figure out the minimum.

Once all subsquares has their all paths evaluated, we may move to the one-level larger square and spend another $\log(N/\varepsilon) \times (N/\varepsilon)^2$ operations. In total we have

$N \times \log{N/\varepsilon} \times (N/\varepsilon)^2 \times N^{O(1/\varepsilon)}$

$\sim N^{O(1/\varepsilon)}$

which is indeed polynomial in $N/\varepsilon$ many operations.

The randomization comes in because the route produced by the above polynomial time algorithm is not always approximately the optimum path; it turns out that sometimes it can be a lot longer.

Expectation of the difference between our random portal respecting minimum path $\mbox{OPT}_p$ and the actual minimum $\mbox{OPT}$ is bounded simply by the fact that minimum path cannot cross the grid lines more that $\mbox{OPT}$ times. At each crossing, the edge it crosses is at level $i$ with probability $2^{i-k}$. to perturb a level $i$ intersection to a portal respecting one requires adding an extra length of no more than $2 \times 2^{k-i}/2^m \sim 2^{k+1-i}/(\log N / \varepsilon)$:

$\displaystyle \mathbb{E}_{a,b}(\mbox{OPT}_p - \mbox{OPT})$

$\leq \mbox{OPT} \times \sum_{i=1}^k 2^{i-k} \times 2^{k+1-i} / (\log N / \varepsilon)$

$= \mbox{OPT} \times 2 \varepsilon / \log N < \varepsilon \mbox{OPT}$

P.S. You may find images for this post being a little different from pervious ones, that’s because I recently got myself a new iPad and all images above are done using iDraw, still getting used to it, so far it’s quite pleasant!

Bonus: I also started to paint on iPad~

–Firestone library, Princeton. (Beautiful spring weather to sit outside and paint from life!)

# Stable isoperimetric inequality

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 # Ultrametrics and the nonlinear Dvoretzky problem Hi guys~ The school year here at Princeton is finally (gradually) starting. So I’m back to this blog :-P In this past week before anything has started, Assaf Naor came here and gave a couple rather interesting talks, which I decided to make a note about. For the complete content of the talk please refer to this paper by Naor and Mendel. As usual, I make no attempt on covering what was written on the paper nor what was presented in the talk, just some random pieces and bits which I found interesting. A long time ago, Grothendieck conjectured what is now known as Dvoretzky’s theorem: Theorem: For any $D>1$, for any $k \in \mathbb{N}$, there exists $N$ depending only on $k, D$ such that any normed vector space of dimension $N$ or larger has a k-dimensional linear subspace that embeds (via a linear transformation) with distorsion at most $D$ into the Hilbert space. i.e. this says in the case where $D$ is close to $1$ (let’s write $D = 1+\varepsilon$) for any given k, *any* norm on a vector space of sufficiently high dimension will have a $k$ dimensional subspace that looks almost Eculidean (meaning unit ball is round up to a multiple $1+\varepsilon$). Well, that’s pretty cool, but linear. As we all know, general metric spaces can be much worse than normed vector spaces. So here comes a nonlinear version of the above, posted by Terrence Tao: Nonlinear Dvoretzky problem: Given $\kappa>0, D>1$, does there exist a $\alpha$ so that every metric space of Hausdorff dimension $\geq \alpha$ has a subset $S$ of Hausdorff dimension $\kappa$ that embeds into the Hilbert space with distorsion $D$? Indeed, we should expect this to be a lot more general and much harder than the linear version, since we know literally nothing about the space except for having large Hausdorff dimension and as we know metric spaces can be pretty bizarre. That why we should find the this new result of theirs amazing: Theorem 1 (Mendel-Naor): There is a constant $C$ such that for any $\lambda \in (0,1)$, every compact metric space with dimension $\alpha$ has a closed subset $S$ of dimension at least $(1-\lambda) \alpha$ that admits an embedding into Hilbert space with distorsion $C/ \lambda$. Note that, in the original linear problem, $N = N(k, D)$ of course blows up to infinity whenever $k \rightarrow \infty$ or $D \rightarrow 1$. (whenever we need a huge dimensional space with fixed ‘flatness’ or a fixed dimension but ‘super-flat’ subspace). That is, when we are looking for subspaces inside a random space with fixed (large) dimension $N$, the larger dimension we need, the less flat (more distorsion) we can expect it to be. i.e. $k \rightarrow N$ necessarily forces $D \rightarrow \infty$ and $D \rightarrow 1$ forces $k \rightarrow 0$. We can see that this nonlinear theorem is great when we need large dimensional subspaces (when $\lambda$ is close to $0$), but not so great when we want small distorsion (it does not apply when distorsion is smaller than $C$, and blows up when it gets close to $C$). In the original statement of the problem this gives not only that a large $\alpha$ exists, but $\alpha(\kappa, D) \leq \frac{D}{D-C} \kappa$! In fact, when we are allowing relatively large distorsion (compare to constant $C$) this theorem is sharp: Theorem 2 (Mendel-Naor): There exists constant $C'$ and a family of metric spaces $\{ X_\alpha \ | \ \alpha \in \mathbb{R}^+ \}$ such that for any small $\varepsilon$, no subset of $X_\alpha$ with dimension $\geq (1-\varepsilon)\alpha$ embeds into Hilbert space with distorsion $\leq C'/\varepsilon$. This construction makes use of our all-time favorite: expander graphs! (see pervious post for definitions) So what if $D$ is small? Turns out, unlike in the linear case when $D<2$, there is no $\alpha(\kappa, D)$, in the paper they also produced spaces $X_\alpha$ for each $\alpha$ where the only subset that embeds with distorsion < $2$ are of Hausdorff dimension 0! In his words, the proof of theorem 1 (i.e. the paper) takes five hours to explain, hence I will not do that. But among many neat things appeared in the proof, instead of embedding into the Hilbert space directly they embedded it into an ultrametric space with distorsion $D$, and then make use of the fact that any such space sits in the Hilbert space isometrically. Definition: An ultrametric on space $X$ is a metric with the additional property that for any three points $x, y, z \in X$, we have the so-called strong triangle inequality, i.e. $d(x, y) \leq \max \{ d(x, z), d(y,z) \}$ If one pause and think a bit, this is a quite weird condition: for example, all triangles in an ultrametric space are isosceles with the two same length edge both longer than the third edge; every point is the center of the ball, etc. Exercise: Come up with an example of an ultrametric that’s not the discrete metric, on an infinite space. Not so easy, hum? My guess: the first one you came up with is the space of all words in 2-element alphabet $\Sigma_2$, with the dictionary metric (two points are distance $2^{-n}$ apart where $n$ is the first digit they differ), right? In fact in some sense all ultrametrics look like that. (i.e. they are represented by an infinite tree with weights assigned to vertices, in the above case a 2-regular tree with equal weights on same level) Topologically our $\Sigma_2$ is a Cantor set and sits isometrically in $l_2$. It’s not hard to see in fact all such space embeds isometrically in $l_2$. I just want to say that this operation of first construct an ultrametric space according to our given metric space, embed, and then embed the ultrametric metric space into the Hilbert space somehow reminds me of when we want to construct a measure satisfying a certain property, we first construct it on a Cantor set, then divide up the space and use the values on parts of the Cantor set for parts of the space…On this vein the whole problem gets translated to producing a tree that gives the ultrametric and work with the tree (by no means simple, though). Finally, let’s see a cute statement which can be proved by applying this result: Urbanski’s conjecture: Any compact metric space of Hausdorff dimension > $n$ can be mapped surjectively [thanks to Tushar Das for pointing out the typo] onto the unit ball $B_1^n$ by a Lipschitz map. (note strictly larger is important for our theorem to apply…since we need to subtract an $\varepsilon$) and hence another cute statement that’s still not known: Question: Does any compact metric space with positive Hausdorff $n$-dimensional measure maps Lipschitzly onto $B_1^n$? # A remark on a mini-course by Kleiner in Sullivan’s 70th birthday I spent the last week on Long Island for Dennis Sullivan’s birthday conference. The conference is hosted in the brand new Simons center where great food is served everyday in the cafe (I think life-wise it’s a wonderful choice for doing a post-doc). Anyways, aside from getting to know this super-cool person named Dennis, the talks there were interesting~ There are many things I found so exciting and can’t help to not say a few words about, however due to my laziness, I can only select one item to give a little stupid remark on: So Bruce Kleiner gave a 3-lecture mini-course on boundaries of Gromov hyperbolic spaces (see this related post on a piece of his pervious work in the subject) Cannon’s conjecture: Any Gromov hyperbolic group with $\partial_\infty G \approx \mathbb{S}^2$ acts discretely and cocompactly by isometries on $\mathbb{H}^3$. As we all know, in the theory of Gromov hyperbolic spaces, we have the basic theorem that says if a groups acts on a space discretely and cocompactly by isometries, then the group (equipped with any word metric on its Cayley graph) is quasi-isometric to the space it acts on. Since I borrowed professor Sullivan as an excuse for writing this post, let’s also state a partial converse of this theorem (which is more in the line of Cannon’s conjecture): Theorem: (Sullivan, Gromov, Cannon-Swenson) For $G$ finitely generated, if $G$ is quasi-isometric to $\mathbb{H}^n$ for some $n \geq 3$, then $G$ acts on $\mathbb{H}^n$ discretely cocompactly by isometries. This essentially says that due to the strong symmetries and hyperbolicity of $\mathbb{H}^n$, in this case quasi-isometry is enough to guarantee an action. (Such thing is of course not true in general, for example any finite group is quasi-isometric to any compact metric space, there’s no way such action exists.) In some sense being quasi-isometric is a much stronger condition once the spaces has large growth at infinity. In light of the above two theorems we know that Cannon’s conjecture is equivalent to saying that any hyperbolic group with boundary $\mathbb{S}^2$ is quasi-isometric to $\mathbb{H}^3$. At first glance this seems striking since knowing only the topology of the boundary and the fact that it’s hyperbolic, we need to conclude what the whole group looks like geometrically. However, the pervious post on one dimensional boundaries perhaps gives us some hint on the boundary can’t be anything we want. In fact it’s rather rigid due to the large symmetries of our hyperbolic group structure. Having Cannon’s conjecture as a Holy Grail, they developed tools that give raise to some very elegant and inspring proofs of the conjecture in various special cases. For example: Definition: A metric space $M$, is said to be Alfors $\alpha$-regular where $\alpha$ is its Hausdorff dimension, if there exists constant $C$ s.t. for any ball $B(p, R)$ with $R \leq \mbox{Diam}(M)$, we have: $C^{-1}R^\alpha \leq \mu(B(p,R)) \leq C R^\alpha$ This is saying it’s of Hausdorff dimension $\alpha$ in a very strong sense. (i.e. the Hausdorff $\alpha$ measure behaves exactly like the regular Eculidean measure everywhere and in all scales). For two disjoint continua $C_1, C_2$ in $M$, let $\Gamma(C_1, C_2)$ denote the set of rectifiable curves connecting $C_1$ to $C_2$. For any density function $\rho: M \rightarrow \mathbb{R}^+$, we define the $\rho$-distance between $C_1, C_2$ to be $\displaystyle \mbox{dist}_\rho(C_1, C_2) = \inf_{\gamma \in \Gamma(C_1, C_2)} \int_\gamma \rho$. Definition: The $\alpha$-modulus between $C_1, C_2$ is $\mbox{Mod}_\alpha(C_1, C_2) = \inf \{ \int_M \rho^\alpha \ | \ \mbox{dist}_\rho(C_1, C_2) \geq 1 \}$, OK…I know this is a lot of seemingly random definitions to digest, let’s pause a little bit: Given two continua in our favorite $\mathbb{R}^n$, new we are of course Hausdorff dimension $n$, what’s the $n$-modulus between them? This is equivalent to asking for a density function for scaling the metric so that the total n-dimensional volume of $\mathbb{R}^n$ is as small as possible but yet the length of any curve connecting $C_1, \ C_2$ is larger than $1$. So intuitively we want to put large density between the sets whenever they are close together. Since we are integrating the $n$-th power for volume (suppose $n>1$, since our set is path connected it’s dimension is at least 1), we would want the density as ‘spread out’ as possible while keeping the arc-length property. Hence one observation is this modulus depends on the pair of closest points and the diameter of the sets. The relative distance between $C_1, C_2$ is $\displaystyle \Delta (C_1, C_2) = \frac{\inf \{ d(p_1, p_2) \ | \ p_1 \in C_1, \ p_2 \in C_2 \} }{ \min \{ \mbox{Diam}(C_1), \mbox{Diam}(C_2) \} }$ We say $M$ is $\alpha$-Loewner if the $\alpha$ modulus between any two continua is controlled above and below by their relative distance, i.e. there exists increasing functions $\phi, \psi: [0, \infty) \rightarrow [0, \infty)$ s.t. for all $C_1, C_2$, $\phi(\Delta(C_1, C_2)) \leq \mbox{Mod}_\alpha(C_1, C_2) \leq \psi(\Delta(C_1, C_2))$ Those spaces are, in some sense, regular with respect to it’s metric and measure. Theorem: If $\partial_\infty G$ is Alfors 2-regular and 2-Loewner, homeomorphic to $\mathbb{S}^2$, then $G$ acts discrete cocompactly on $\mathbb{H}^3$ by isometries. Most of the material appeared in the talk can be found in their paper. There are many other talks I found very interesting, especially that of Kenneth Bromberg, Mario Bonk and Peter Jones. Unfortunately I had to miss Curt McMullen, Yair Minski and Shishikura… # Stabilization of Heegaard splittings In the last lecture of a course on Heegaard splittings, professor Gabai sketched an example due to Hass-Thompson-Thurston of two genus $g$ Heegaard splittings of a $3$-manifold that requires at least $g$ stabilization to make them equivalent. The argument is, in my opinion, very metric-geometric. The connection is so striking (to me) so that I feel necessary to give a brief sketch of it here. (Side note: This has been a wonderful class! Although I constantly ask stupid questions and appear to be confused from time to time. But in fact it has been very interesting! I should have talked more about it on this blog…Oh well~) The following note is mostly based on professor Gabai’s lecture, I looked up some details in the original paper ( Hass-Thompson-Thurston ’09 ). Recall: (well, I understand that I have not talked about Heegaard splittings and stabilizations here before, hence I’ll *try to* give a one minute definition) A Heegaard splitting of a 3-manifold $M$ is a decomposition of the manifold as a union of two handlebodies intersecting at the boundary surface. The genus of the Heegaard splitting is the genus of the boundary surface. All smooth closed 3-manifolds has Heegaard splitting due the mere existence of a triangulation ( by thicken the 1-skeleton of the triangulation one gets a handlebody perhaps of huge genus, it’s easy exercise to see its complement is also a handlebody). However it is of interest to find what’s the minimal genus of a Heegaard splitting of a given manifold. Two Heegaard splittings are said to be equivlent if there is an isotopy of the manifold sending one splitting to the other (with boundary gluing map commuting, of course). A stabilization of a Heegaard splitting $(H_1, H_2, S)$ is a surgery on $S$ that adds genus (i.e. cut out two discs in $S$ and glue in a handle). Stabilization will increase the genus of the splitting by $1$) Let $M$ be any closed hyperbolic $3$-manifold that fibres over the circle. (i.e. $M$ is $F_g \times [0,1]$ with the two ends identified by some diffeomorphism $f: F_g \rightarrow F_g$, $g\geq 2$): Let $M'_k$ be the $k$ fold cover of $M$ along $S^1$ (i.e. glue together$k$copies of $F_g \times I$ all via the map $f$: Let $M_k$ be the manifold obtained by cut open $M'_k$ along$F_g\$ and glue in two handlebodies $H_1, H_2$ at the ends:

Since $M$ is hyperbolic, $M'_k$ is hyperbolic. In fact, for any $\varepsilon > 0$ we can choose a large enough $k$ so that $M_k$ can be equipped with a metric having curvature bounded between $1-\varepsilon$ and $1+\varepsilon$ everywhere.

( I’m obviously no in this, however, intuitively it’s believable because once the hyperbolic part $M'_k$ is super large, one should be able to make the metric in $M'_k$ slightly less hyperbolic to make room for fitting in an almost hyperbolic metric at the ends $H_1, H_2$). For details on this please refer to the original paper. :-P

Now there comes our Heegaard splittings of $M_k$!

Let $k = 2n$, let $H_L$ be the union of handlebody $H_1$ together with the first $n$ copies of $M$, $H_R$ be $H_2$ with the last $n$ copies of $M$. $H_L, H_R$ are genus $g$ handlebodies shearing a common surface $S$ in the ‘middle’ of $M_k$:

Claim: The Heegaard splitting $\mathcal{H}_1 = H_L \cup H_R$ and $\mathcal{H}_2 = H_L \cup H_R$ cannot be made equivalent by less than $g$ stabilizations.

In other words, first of all one can not isotope this splitting upside down. Furthermore, adding handles make it easier to turn the new higher genus splitting upside down, but in this particular case we cannot get away with adding anything less than $g$ many handles.

Okay, not comes the punchline: How would one possible prove such thing? Well, as one might have imagined, why did we want to make this manifold close to hyperbolic? Yes, minimal surfaces!

Let’s see…Suppose we have a common stabilization of genus $2g-1$. That would mean that we can sweep through the manifold by a surface of genus (at most) $2g-1$, with $1$-skeletons at time $0, 1$.

Now comes what professor Gabai calls the ‘harmonic magic’: there is a theorem similar to that of Pitts-Rubinstein

Ingredient #1: (roughly thm 6.1 from the paper) For manifolds with curvature close to $-1$ everywhere, for any given genus $g$ Heegaard splitting $\mathcal{H}$, one can isotope the sweep-out so that each surface in the sweep-out having area $< 5 \pi (g-1)$.

I do not know exactly how is this proved. The idea is perhaps try to shrink each surface to a ‘minimal surface’, perhaps creating some singularities harmless in the process.

The ides of the whole arguement is that if we can isotope the Heegaard splittings, we can isotope the whole sweep-out while making the time-$t$ sweep-out harmonic for each $t$. In particular, at each time there is (at least) a surface in the sweep-out family that divides the volume of $M'_n$ in half. Furthermore, the time 1 half-volume-surface is roughly same as the time 0 surface with two sides switched.

We shall see that the surfaces does not have enough genus or volume to do that. (As we can see, there is a family of genus $2g$ surface, all having volume less than some constant independent of $n$ that does this; Also if we have ni restriction on area, then even a genus $g$ surface can be turned.)

Ingredient #2: For any constant $K$, there is $n$ large enough so no surfaces of genus $ and area $ inside the middle fibred manifold with boundary $M'_n$ can divide the volume of $M'_n$ in half.

The prove of this is partially based on our all-time favorite: the isoperimetric inequality:

Each Riemannian metric $\lambda$ on a closed surface has a linear isoperimetric inequality for 1-chains bounding 2-chains, i.e. any homologically trivial 1-chain $c$ bounds a $2$ chain $z$ where

$\mbox{Vol}_2(z) \leq K_\lambda \mbox{Vol}_1(c)$.

Fitting things together:

Suppose there is (as described above) a family of genus $2g-1$ surfaces, each dividing the volume of $M_{2n}$ in half and flips the two sides of the surface as time goes from $0$ to $1$.

By ingredient #1, since the family is by construction surfaces from (different) sweep-outs by ‘minimal surfaces’, we have $\mbox{Vol}_2(S_t) < 5 \pi (2g-2)$ for all $t$.

Now if we take the two component separated by $S_t$ and intersect them with the left-most $n$ copies of $M$ in $M'_{2n}$ (call it $M'_L$), at some $t$, $S_t$ must also divide the volume of $M'_L$ in half.

Since $S_t$ divides both $M'_2n$ and $M'_L$ in half, it must do so also in $M'_R$.

But $S_t$ is of genus $2g-1$! So one of $S_t \cap M'_L$ and $S_t \cap M'_R$ has genus $< g$! (say it's $M'_L$)

Apply ingredient #2, take $K = 5 \pi (2g-2)$, there is $n$ large enough so that $S_t \cap M'_L$, which has area less than $K$ and genus less than $g$, cannot possibly divide $M'_L$ in half.