## Posts Tagged ‘Larry Guth’

### The travelling salesman problem

March 19, 2012

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!)

### 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.

### A report of my Princeton generals exam

April 26, 2011

Well, some people might be wondering why I haven’t updated my blog since two weeks ago…Here’s the answer: I have been preparing for this generals exam — perhaps the last exam in my life.

For those who don’t know the game rules: The exam consists of 3 professors (proposed by the kid taking the exam, approved by the department), 5 topics (real, complex, algebra + 2 specialized topics chosen by the student). One of the committee member acts as the chair of the exam.

The exam consists of the three committee members sitting in the chair’s office, the student stands in front of the board. The professors ask questions ranging in those 5 topics for however long they want, the kid is in charge of explaining them on the board.

I was tortured for 4.5 hours (I guess it sets a new record?)
I have perhaps missed some questions in my recollection (it’s hard to remember all 4.5 hours of questions).

Conan Wu’s generals

Commitee: David Gabai (Chair), Larry Guth, John Mather

Topics: Metric Geometry, Dynamical Systems

Real analysis:

Mather: Construct a first category but full measure set.

(I gave the intersection of decreasing balls around the rationals)

Guth: $F:S^1 \rightarrow \mathbb{R}$ 1-Lipschitz, what can one say about its Fourier coefficients.

(Decreasing faster than $c*1/n$ via integration by parts)

Mather: Does integration by parts work for Lipschitz functions?

(Lip imply absolutely continuous then Lebesgue’s differentiation theorem)

Mather: If one only has bounded variation, what can we say?

($f(x) \geq f(0) + \int_0^x f'(t) dt$)

Mather: If $f:S^1 \rightarrow \mathbb{R}$ is smooth, what can you say about it’s Fourier coefficients?

(Prove it’s rapidly decreasing)

Mather: Given a smooth $g: S^1 \rightarrow \mathbb{R}$, given a $\alpha \in S^1$, when can you find a $f: S^1 \rightarrow \mathbb{R}$ such that
$g(\theta) = f(\theta+\alpha)-f(\theta)$ ?

(A necessary condition is the integral of $g$ needs to vanish,
I had to expand everything in Fourier coefficients, show that if $\hat{g}(n)$ is rapidly decreasing, compute the Diophantine set $\alpha$ should be in to guarantee $\hat{f}(n)$ being rapidly decreasing.

Gabai: Write down a smooth function from $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ with no critical points.

(I wrote $f(x,y) = x+y$) Draw its level curves (straight lines parallel to $x=-y$)

Gabai: Can you find a such function with the level curves form a different foliation from this one?

(I think he meant that two foliations are different if there is no homeo on $\mathbb{R}^2$ carrying one to the other,
After playing around with it for a while, I came up with an example where the level sets form a Reeb foliation, and that’s not same as the lines!)

We moved on to complex.

Complex analysis:

Guth: Given a holomorphic $f:\mathbb{D} \rightarrow \mathbb{D}$, if $f$ has $50$ $0$s inside the ball $B_{1/3}(\bar{0})$, what can you say about $f(0)$?

(with a bunch of hints/suggestions, I finally got $f(0) \leq (1/2)^{50}$ — construct polynomial vanishing at those roots, quotient and maximal modulus)

Guth: State maximal modulus principal.

Gabai: Define the Mobius group and how does it act on $\mathbb{H}$.

Gabai: What do the Mobius group preserve?

(Poincare metric)

Mather: Write down the Poincare metric, what’s the distance from $\bar{0}$ to $1$? (infinity)

(I don’t remember the exact distance form, so I tried to guess the denominator being $\sqrt{1-|z|}$, but then integrating from $0$ to $1$ does not “barely diverge”. Turns out it should be $(1-|z|^2)^2$.)

Gabai: Suppose I have a finite subgroup with the group of Mobius transformations acting on $\mathbb{D}$, show it has a global fixed point.

(I sketched an argument based on each element having finite order must have a unique fixed point in the interior of $\mathbb{D}$, if two element has different fixed points, then one can construct a sequence of elements where the fixed point tends to the boundary, so the group can’t be finite.)

I think that’s pretty much all for the complex.

Algebra:

Gabai: State Eisenstein’s criteria

(I stated it with rings and prime ideals, which leaded to a small discussion about for which rings it work)

Gabai: State Sylow’s theorem

(It’s strange that after stating Sylow, he didn’t ask me do anything such as classify finite groups of order xx)

Gabai: What’s a Galois extension? State the fundamental theorem of Galois theory.

(Again, no computing Galois group…)

Gabai: Given a finite abelian group, if it has at most $n$ elements of order divisible by $n$, prove it’s cyclic.

(classification of abelian groups, induction, each Sylow is cyclic)

Gabai: Prove multiplicative group of a finite field is cyclic.

(It’s embarrassing that I was actually stuck on this for a little bit before being reminded of using the previous question)

Gabai: What’s $SL_2(\mathbb{Z})$? What are all possible orders of elements of it?

(I said linear automorphisms on the torus. I thought it can only be $1,2,4,\infty$, but turns out there is elements of order $6$. Then I had to draw the torus as a hexagon and so on…)

Gabai: What’s $\pi_3(S^2)$?

($\mathbb{Z}$, via Hopf fibration)

Gabai: For any closed orientable $n$-manifold $M$, why is $H_{n-1}(M)$ torsion free?

(Poincare duality + universal coefficient)

We then moved on to special topics:

Metric Geometry:

Guth: What’s the systolic inequality?

(the term ‘aspherical’ comes up)

Gabai: What’s aspherical? What if the manifold is unbounded?

(I guessed it still works if the manifold is unbounded, Guth ‘seem to’ agree)

Guth: Sketch a proof of the systolic inequality for the n-torus.

(I sketched Gromov’s proof via filling radius)

Guth: Give an isoperimetric inequality for filling loops in the 3-manifold $S^2 \times \mathbb{R}$ where $S^2$ has the round unit sphere metric.

(My guess is for any 2-chain we should have

$\mbox{vol}_1(\partial c) \geq C \mbox{vol}_2(c)$

then I tried to prove that using some kind of random cone and grid-pushing argument, but soon realized the method only prove

$\mbox{vol}_1(\partial c) \geq C \sqrt{\mbox{vol}_2(c)}$.)

Guth: Given two loops of length $L_1, L_2$, the distance between the closest points on two loops is $\geq 1$, what’s the maximum linking number?

(it can be as large as $c L_1 L_2$)

Dynamical Systems:

Mather: Define Anosov diffeomorphisms.

Mather: Prove the definition is independent of the metric.

(Then he asked what properties does Anosov have, I should have said stable/unstable manifolds, and ergodic if it’s more than $C^{1+\varepsilon}$…or anything I’m familiar with, for some reason the first word I pulled out was structurally stable…well then it leaded to and immediate question)

Mather: Prove structural stability of Anosov diffeomorphisms.

(This is quite long, so I proposed to prove Anosov that’s Lipschitz close to the linear one in $\mathbb{R}^n$ is structurally stable. i.e. the Hartman-Grobman Theorem, using Moser’s method, some details still missing)

Mather: Define Anosov flow, what can you say about geodesic flow for negatively curved manifold?

(They are Anosov, I tried to draw a picture to showing the stable and unstable and finished with some help)

Mather: Define rotation number, what can you say if rotation numbers are irrational?

(They are semi-conjugate to a rotation with a map that perhaps collapse some intervals to points.)

Mather: When are they actually conjugate to the irrational rotation?

(I said when $f$ is $C^2$, $C^1$ is not enough. Actually $C^1$ with derivative having bounded variation suffice)

I do not know why, but at this point he wanted me to talk about the fixed point problem of non-separating plane continua (which I once mentioned in his class).

After that they decided to set me free~ So I wandered in the hallway for a few minutes and the three of them came out to shake my hand.

### Graph, girth and expanders

April 11, 2011

In the book “Elementary number theory, group theory and Ramanujan graphs“, Sarnak et. al. gave an elementary construction of expander graphs. We decided to go through the construction in the small seminar and I am recently assigned to give a talk about the girth estimate of such graphs.

Given graph (finite and undirected) $G$, we will denote the set of vertices by $V(G)$ and the set of edges $E(G) \subseteq V(G)^2$. The graph is assumed to be equipped with the standard metric where each edge has length $1$.

The Cheeger constant (or isoperimetric constant of a graph, see this pervious post) is defined to be:

$\displaystyle h(G) = \inf_{S\subseteq V_G} \frac{|\partial S|}{\min\{|S|, |S^c|\}}$

Here the notation $\partial S$ denote the set of edges connecting an element in $S$ to an element outside of $S$.

Note that this is indeed like our usual isoperimetric inequalities since it’s the smallest possible ratio between size of the boundary and size of the set it encloses. In other words, this calculates the most efficient way of using small boundary to enclose areas as large as possible.

It’s of interest to find graphs with large Cheeger constant (since small Cheeger constant is easy to make: take two large graphs and connect them with a single edge).

It’s also intuitive that as the number of edges going out from each vertice become large, the Cheeger constant will become large. Hence it make sense to restrict ourselves to graphs where there are exactly $k$ edges shearing each vertex, those are called $k$-regular graphs.

If you play around a little bit, you will find that it’s not easy to build large k-regular graphs with Cheeger constant larger than a fixed number, say, $1/10$.

Definition: A sequence of k-regular graphs $(G_i)$ where $|V_{G_i}| \rightarrow \infty$ is said to be an expander family if there exists constant $c>0$ where $h(G_i) \geq c$ for all $i$.

By random methods due to Erdos, we can prove that expander families exist. However an explicit construction is much harder.

Definition: The girth of $G$ is the smallest non-trivial cycle contained in $G$. (Doesn’t this smell like systole? :-P)

In the case of trees, since it does not contain any non-trivial cycle, define the girth to be infinity.

The book constructs for us, given pair $p, q$ of primes where $p$ is large (but fixed) and $q \geq p^8$, a graph $(p+1)$-regular graph$X^{p,q}$ with

$\displaystyle h(X^{p,q}) \geq \frac{1}{2}(p+1 - p^{5/6 + \varepsilon} - p^{1/6-\varepsilon})$

where $0 < \varepsilon < 1/6$.

Note that the bound is strictly positive and independent of $q$. Giving us for each $p$, $(X^{p,q})$ as q runs through primes larger than $p^8$ is a $(p+1)$-regular expander family.

In fact, this constructs for us an infinite family of expander families: a $(k+1)$-regular one for each prime $k$ and the uniform bound on Cheeger constant gets larger as $k$ becomes larger.

One of the crucial step in proving this is to bound the girth of the graph $X^{p,q}$, i.e. they showed that $g(X^{p,q}) \geq 2 \log_p(q)$ and if the quadratic reciprocity $(\frac{p}{q}) = -1$ then $g(X^{p,q}) \geq 4 \log_p(q) - \log_p(4)$. This is what I am going to do in this post.

Let $\mathbb H ( \mathbb Z)$ be the set of quaternions with $\mathbb Z$ coefficient, i.e.

$\mathbb H ( \mathbb Z) = \{ a+bi+cj+dk \ | \ a,b,c,d \in \mathbb Z \}$

Fix odd prime $p$, let

$\Lambda' = \{ \alpha \in \mathbb H(\mathbb Z) \ | \ \alpha \equiv 1 (mod 2) \}$

$\cup \ \{\alpha \in \mathbb H(\mathbb Z) \ | \ N(\alpha) = p^n \ \mbox{and} \ \alpha \equiv i+j+k (mod 2) \}$

where the norm $N$ on $\mathbb H(\mathbb Z)$ is the usual $N(a+bi+cj+dk) = a^2 + b^2 +c^2+d^2$.

$\Lambda'$ consists of points with only odd first coordinate or points lying on spheres of radius $\sqrt{p^n}$ and having only even first coordinate. One can easily check $\Lambda'$ is closed under multiplication.

Define equivalence relation $\sim$ on $\Lambda'$ by

$\alpha \sim \beta$ if there exists $m, n \in \mathbb{N}$ s.t. $p^m \alpha = \pm p^n \beta$.

Let $\Lambda = \Lambda' / \sim$, let $Q: \Lambda' \rightarrow \Lambda$ be the quotient map.

Since we know $\alpha_1 \sim \alpha_2, \beta_1 \sim \beta_2 \Rightarrow \alpha_1\beta_1 \sim \alpha_2\beta_2$, $\Lambda$ carries an induced multiplication with unit.

In elementary number theory, we know that the equation $a^2+b^2+c^2+d^2 = p$ has exactly $8(p+1)$ integer solutions. Hence the sphere of radius $p$ in $\mathbb H(\mathbb Z)$ contain $8(p+1)$ points.

In each $4$-tuple $(a,b,c,d)$ exactly one is of a different parity from the rest, depending on whether $p\equiv1$ or $3 (mod 4)$. Restricting to solutions where the first coordinate is non-negative, having different parity from the rest (in case the first coordinate is $0$, we pick one of the two solutions $\alpha, -\alpha$ to be canonical), this way we obtain exactly $p+1$ solutions.

Let $S'_p = \{ \alpha_1, \bar{\alpha_1}, \cdots, \alpha_k, \bar{\alpha_k}, \beta_1, \cdots, \beta_l \}$ be this set of $p+1$ points on the sphere. Note that the $\beta$s represent the solutions where the first coordinate is exactly $0$.

Check that $S'_p$ generates $\Lambda'$.

We have $\alpha_i \bar{\alpha_i} = p$ and $-\beta_j^2 = p$. By definition $S'_p \subseteq \Lambda'$ and $Q$ is injective on $S'_p$. Let $S_p = Q(S'_p)$.

Consider the Cayley graph $\mathcal G (\Lambda, S_p)$, this is a $(p+1)$-regular graph. Since $S_p$ generares $\Lambda$, $\mathcal G (\Lambda, S_p)$ is connected.

Claim: $\mathcal G (\Lambda, S_p)$ is a tree.

Suppose not, let $(v_0, v_1, \cdots, v_k=v_0)$ a non-trivial cycle. $k \geq 2$. Since $\mathcal G$ is a Cayley graph, we may assume $v_0 = e$.

Hence $v_1 = \gamma_1, \ v_2 = \gamma_1\gamma_2, \cdots, v_k = \gamma_1 \cdots \gamma_k$, for some $\gamma_1, \cdots, \gamma_k \in S_p$.

Since $v_{i-1} \neq v_{i+1}$ for all $1\leq i \leq k-1$, the word $\gamma_1, \cdots, \gamma_k$ cannot contain either $\alpha_i\bar{\alpha_i}$ or $\beta_i^2$, hence cannot be further reduced.

$\gamma_1, \cdots, \gamma_k = e$ in $\Lambda$ means for some $m, n$ we have

$\pm p^n \gamma_1, \cdots, \gamma_k = p^m$.

Since every word in $\Lambda'$ with norm $N(\alpha) = p^k$ must have a unique factorization $\alpha = \pm p^r w_m$ where $w_m$ is a reduces word of length $m$ in $S'_p$ and $2r+m = k$.

Now we reduce the group $\mathbb H (\mathbb Z)$ mod $q$:

$\pi_q: \mathbb H (\mathbb Z) \rightarrow \mathbb H (\mathbb{F}_q)$

One can check that $\pi_q(\Lambda') = \mathbb H (\mathbb{F}_q)^\times$.

Let $Z_q = \{ \alpha \in \mathbb H (\mathbb{F}_q)^\times \ | \ \alpha = \bar{\alpha} \}$, $Z_q < \mathbb H (\mathbb{F}_q)^\times$ is a central subgroup.

For $\alpha, beta \in \Lambda'$, $\alpha \sim \beta \Rightarrow \pi_q(\alpha)^{-1}\pi_q(\beta) \in Z_q$. Which means we have well defined homomorphism

$\Pi_q: \Lambda \rightarrow \mathbb H (\mathbb{F}_q)^\times / Z_p$.

Let $T_{p,q} = \Pi_q(S_p)$, if $q > 2\sqrt{q}$ we have $\Pi_q$ is injective on $S_p$ and hence $latex $| T_{p,q} | = p+1$. Now we are ready to define our expanding family: $X^{p,q} = \mathcal{G}( \Pi_q(\Lambda), T_{p,q})$. Since $S_p$ generates $\Lambda$, $T_{p,q}$ generates $\Pi_q(\Lambda)$. Hence $X^{p,q}$ is $(p+1)$-regular and connected. Theorem 1: $g(X^{p,q}) \geq 2 \log_p(q)$ Let $(e, v_1, \cdots, v_k=e)$ be a cycle in $X^{p,q}$, there is $t_1, \cdots, t_k \in T_{p,q}$ such that $v_i = t_1 t_2 \cdots t_k$ for $1 \leq i \leq k$. Let $\gamma_i = \Pi_q^{-1}(t_i), \ \gamma_i \in S_p$, $\alpha = a_0 + a_1 i+a_2 j +a_3 k = \gamma_1 \cdots \gamma_k \in \Lambda$. Note that from the above arguement we know $\alpha$ is a reduced word, hence $\alpha \neq e_{\Lambda}$. In particular, this implies $a_1, a_2, a_3$ cannot all be $0$. Also, since $\alpha$ is reduced, $\displaystyle N(\alpha) = \Pi_{i=1}^k N(\gamma_i) = p^k$. By Lemma, since $\Pi_q(\alpha) = 1$, $\alpha \in \mbox{ker}(\Pi_q)$ hence $q$ divide $a_1, a_2, a_3$, we conclude $N(\alpha) = a_0^2 + a_1^2 +a_2^2 +a_3^2 \geq q^2$ We deduce $p^k \geq q^2$ hence $k \geq \log_p(q)$ for all cycle. i.e. $g(X^{p,q}) \geq \log_p(q)$. Theorem 2: If $q$ does not divide $p$ and $p$ is not a square mod $q$ (i.e. $(\frac{p}{q}) = -1$), then $g(X^{p,q}) \geq 4 \log_p(q) - \log_p(4)$. For any cycle of length $k$ as above, we have $N(\alpha) = p^k \equiv a_0^2 (\mbox{mod} \ q)$, i.e. $(\frac{p^k}{q}) = 1$. Since $(\frac{p^k}{q}) = ((\frac{p}{q})^k$, we have $(-1)^k = 1, \ k$ is even. Let $k = 2l$. Note that $p^k \equiv a_0^2 (\mbox{mod} \ q^2)$, we also have $p^{2l} \equiv a_0^2 (\mbox{mod} \ q^2)$ Hence $p^l \equiv a_0 (\mbox{mod} \ q^2)$. Since $a_0^2 \leq p^{2l}$, $|a_0| \leq p^l$. If $2l < 4 \log_p(q) - \log_p(4)$ we will have $p^l < q^2/2$. Then $|p^l\pm a_0| \leq 2|p^l| < q^2$. But we know that $p^l \equiv a_0 (\mbox{mod} \ q^2)$, one of $p^l\pm a_0$ must be divisible by $q^2$, hence $0$. Conclude $p^l = \pm a_0$, $N(\alpha) = p^k = a_0^2$, hence $a_1=a_2=a_3=0$. Contradiction. ### The Carnot-Carathéodory metric March 22, 2011 I remember looking at Gromov’s notes “Carnot-Carathéodory spaces seen from within” a long time ago and didn’t get anywhere. Recently I encountered it again through professor Guth. This time, with his effort in explaining, I did get some ideas. This thing is indeed pretty cool~ So I decided to write an elementary introduction about it here. We will construct a such metric in $\mathbb{R}^3$. In general, if we have a Riemannian manifold, the Riemannian distance between two given points $p, q$ is defined as $\inf_\Gamma(p,q) \int_0^1||\gamma'(t)||dt$ where $\Gamma$ is the collection of all differentiable curves $\gamma$ connecting the two points. However, if we have a lower dimensional sub-bundle $E(M)$ of the tangent bundle (depending continuously on the base point). We may attempt to define the metric $d(p,q) = \inf_{\Gamma'} \int_0^1||\gamma'(t)||dt$ where $\Gamma'$ is the collection of curves connecting $p, q$ with $\gamma'(t) \in E(M)$ for all $t$. (i.e. we are only allowed to go along directions in the sub-bundle. Now if we attempt to do this in $\mathbb{R}^3$, the first thing we may try is let the sub-bundle be the say, $xy$-plane at all points. It’s easy to realize that now we are ‘stuck’ in the same height: any two points with different $z$ coordinate will have no curve connecting them (hence the distance is infinite). The resulting metric space is real number many discrete copies of $\mathbb{R}^2$. Of course that’s no longer homeomorphic to $\mathbb{R}^3$. Hence for the metric to be finite, we have to require accessibility of the sub-bundle: Any point is connected to any other point by a curve with derivatives in the $E(M)$. For the metric to be equivalent to our original Riemannian metric (meaning generate the same topology), we need $E(M)$ to be locally accessible: Any point less than $\delta$ away from the original point $p$ can be connected to $p$ by a curve of length $< \varepsilon$ going along $E(M)$. At the first glance the existence of a (non-trivial) such metric may not seem obvious. Let’s construct one on $\mathbb{R}^3$ that generates the same topology: To start, we first identify our $\mathbb{R}^3$ with the $3 \times 3$ real entry Heisenberg group $H^3$ (all $3 \times 3$ upper triangular matrices with “1”s on the diagonal). i.e. we have homeomorphism $h(x,y,z) \mapsto \left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right)$ Let $g$ be a left-invariant metric on $H_3$. In the Lie algebra $T_e(H_3)$ (tangent space of the identity element), the elements $X = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) , Y = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ and $Z = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ form a basis. At each point, we take the two dimensional sub-bundle $E(H_3)$ of the tangent bundle generated by infinitesimal left translations by $X, Y$. Since the metric $g$ is left invariant, we are free to restrict the metric to $E(M)$ i.e. we have $||X_p|| = ||Y_p|| = 1$ for each $p \in M$. The interesting thing about $H_3$ is that all points are accessible from the origin via curves everywhere tangent to $E(H_3)$. In other words, any points can be obtained by left translating any other point by multiples of elements $X$ and $Y$. The “unit grid” in $\mathbb{R}^3$ under this sub-Riemannian metric looks something like: Since we have $\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x+1 & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right)$, the original $x$-direction stay the same, i.e. a bunch of horizontal lines connecting the original $yz$ planes orthogonally. However, if we look at a translation by $Y$, we have $\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & x & z+x \\ 0 & 1 & y+1 \\ 0 & 0 & 1 \end{array} \right)$ i.e. a unit length $Y$-vector not only add a $1$ to the $y$-direction but also adds a height $x$ to $z$, hence the grid of unit $Y$ vectors in the above three $yz$ planes look like: We can now try to see the rough shape of balls by only allowing ourselves to go along the unit grid formed by $X$ and $Y$ lines constructed above. This corresponds to accessing all matrices with integer entry by words in $X$ and $Y$. The first question to ask is perhaps how to go from $(0,0,0)$ to $(0,0,1)$. –since going along the $z$ axis is disabled. Observe that going through the following loop works: We conclude that $d_C((0,0,0), (0,0,1)) \leq 4$ in fact up to a constant going along such loop gives the actual distance. At this point one might feel that going along $z$ axis in the C-C metric is always takes longer than the ordinary distance. Giving it a bit more thought, we will find this is NOT the case: Imagine what happens if we want to go from $(0,0,0)$ to $(0,0,10000)$? One way to do this is to go along $X$ for 100 steps, then along $Y$ for 100 steps (at this point each step in $Y$ will raise $100$ in $z$-coordinate, then $Y^{-100} X^{-100}$. This gives $d_C((0,0,0), (0,0,10000)) \leq 400$. To illustrate, let’s first see the loop from $(0,0,0)$ to $(0,0,4)$: The loop has length $8$. (A lot shorter compare to length $4$ for going $1$ unit in $z$-direction) i.e. for large $Z$, it’s much more efficient to travel in the C-C metric. $d_C( (0,0,0), (0,0,N^2)) = 4N$ In fact, we can see the ball of radius $N$ is roughly an rectangle with dimension $R \times R \times R^2$ (meaning bounded from both inside and outside with a constant factor). Hence the volume of balls grow like $R^4$. Balls are very “flat” when they are small and very “long” when they are large. ### Isoperimetric inequality on groups February 28, 2011 Back in high school, we have all learned and loved the isoperimetric inequality: “If one wants to enclose a region of area $\pi$ on the plane, one needs a rope of length at least $2 \pi$.” or equivalently, given $U \subseteq \mathbb{R}^2$ bounded open, we always have $\ell(\partial U) \geq 2 \sqrt{\pi} \sqrt{\mbox{Area}(U)}$ Of course this generalizes to $\mathbb{R}^n$: Theorem: Given any open set $U \subseteq \mathbb{R}^n$, we have $\mbox{vol}_{n-1}\partial (U) \geq n\cdot \omega_n^{1/n} \mbox{vol}_n(U)^{\frac{n-1}{n}}$ Here $\omega_n$ is the volume of the unit n-ball. Note that the inequality is sharp for balls in $\mathbb{R}^n$. One nature question to ask should be: for which other kind of spaces do we have such an inequality. i.e. when can we lower bound the volume of the boundary by in terms of the volume of the open set? If such inequality exists, how does the lower bound depend on the volume of the set? I recently learned to produce such bounds on groups: Let $G$ be a discrete group, fix a set of generators and let $C_G$ be its Cayley graph, equipped with the word metric $d$. Let $N(R) = |B_R(e)|$ be the cardinality of the ball of radius $R$ around the identity. For any set $U \subseteq G$, we define $\partial U = \{g \in G \ | \ d(g, U) = 1 \}$ i.e. points that’s not in $U$ but there is an edge in the Cayley graph connecting it to some point in $U$. Theorem:For any group with the property that $N(R) \geq c_n R^n$, then for any set $U \subseteq G$ with $|U| \leq \frac{1}{2}|G|$, $|\partial U| \geq c_n |U|^{\frac{n-1}{n}}$. i.e. If the volume of balls grow (w.r.t. radius) as fast as what we have in $\mathbb{R}^n$, then we can lower bound the size of boundary any open set in terms of its volume just like what we have in $\mathbb{R}^n$. Proof: We make use of the fact that right multiplications by elements of the group are isometries on Cayley graph. Let $R = (\frac{2}{c_n}|U|)^\frac{1}{n}$, so we have $|B_R(e)| \geq 2|U|$. For each element $g \in B_R(e)$, we look at how many elements of $U$ went outside of $U$ i.e. $| Ug \backslash U|$. (Here the idea being the size of the boundary can be lower bounded in terms of the length of the translation vector and the volume shifted outside the set. Hence we are interested in finding an element that’s not too far away from the identity but shifts a large volume of $U$ outside of $U$.) The average value of $|Ug \backslash U|$ as $g$ varies in the ball $B_R(e)$ is: $\displaystyle \frac{1}{|B_R(e)|} \sum_{g\in B_R(e)} |Ug \backslash U|$ The sum part is counting the elements of $U$ that’s translated outside $U$ by $g$ then sum over all $g \in B_R(e)$, this is same as first fixing any $u \in U$, count how many $g$ sends $u$ outside $U$, and sum over all $u \in U$ ( In fact this is just Fubini, where we exchange the order of two summations ). “how mant $g \in B_R(e)$ sends $u$ outside of $U$” is merely $|uB_R(e) \backslash U|$. Hence the average is $\displaystyle \frac{1}{|B_R(e)|} \sum_{u\in U}|uB_R(e) \backslash U|$. But we know $|B_R(e)| \geq 2 \cdot |U|$, hence $|uB_R(e) \backslash U|$ is at least $\frac{1}{2}|B_R(e)|$. Hence $\displaystyle \frac{1}{|B_R(e)|} \sum_{u\in U}|uB_R(e) \backslash U|$ $\geq \frac{1}{|B_R(e)|} \cdot |U| \cdot \frac{1}{2} |B_R(e)| = \frac{1}{2} |U|$ Now we can pick some $g \in B_R(e)$ where $|Ug \backslash U| \geq \frac{1}{2}|U|$ (at least as large as average). Since $g$ has norm at most $R$, we can find a word $g_1 g_2 \cdots g_k = g, \ k \leq R$. For any $ug \in (Ug \backslash U)$, since $u \in U$ and $ug \notin U$, there must be some $1\leq i\leq k$ s.t. $u(g_1\cdots g_{i-1}) \in U$ and $u(g_1\cdots g_i) \notin U$. Hence $u g_1 \cdots g_i \in \partial U$, $ug \in \partial U \cdot g_{i+1} \cdots g_k$. We deduce that $\displaystyle Ug \backslash U \subseteq \partial U \cup \partial U g_k \cup \cdots \cup \partial U g_2 \cdots g_k$ i.e. a union of $k$ copies of $\partial U$. Hence $R |\partial U| \geq k |\partial U| \geq |Ug \backslash U| \geq \frac{1}{2}|U|$ $|\partial U| \geq \frac{|U|}{2R}$ Since $|B_R(e)| \geq c_n R^n$, we have $R \leq c_n |B_R(e)|^{\frac{1}{n}}$. $R = (\frac{2}{c_n}|U|)^\frac{1}{n} = c_n |U|^\frac{1}{n}$, hence we have $|\partial U| \geq c_n |U|^\frac{n-1}{n}$ Establishes the claim. Remark: The same argument carries through as long as we have a lower bound on the volume of balls, not necessarily polynomial. For example, on expander graphs, the volume of balls grow exponentially: $B_R(e) \geq c \cdot e^R$, then trancing through the argument we get bound $|\partial U| \geq c \cdot \frac{|U|}{\log{|U|}}$ Which is a very strong isoperimetric inequality. However in fact the sharp bound for expanders is $|\partial U| \geq c \cdot |U|$. But to get that one needs to use more information of the expander than merely the volume of balls. On the same vein, we can also prove a version on Lie groups: Theorem:Let $G$ be a Lie group, $g$ be a left invariant metric on $G$. If $\mbox{vol}_n(B_R) \geq c_n R^n$ then for any open set $U$ with no more than half of the volume of $G$, $\mbox{vol}_{n-1}(\partial U) \geq c_n \mbox{vol}_n(U)^\frac{n-1}{n}$. Note that to satisfy the condition $\mbox{vol}_n(B_R) \geq c_n R^n$, our Lie group must be at least n-dimensional since if not the condition would fail for small balls. $n$ might be strictly larger than the dimension of the manifold depending on how ‘neigatively curved’ the manifold is in large scale. Sketch of proof: As above, take a ball twice the size of the set $U$ around the identity, say it’s $B_R(e)$. Now we consider all left translates of the set $U$ by element in $B_R(e)$. In average an element shifts at least half of $U$ outside of $U$. Pick element $g$ where $\mbox{vol}(gU \backslash U)$ is above average. Let $\gamma: [0, ||g||]$ be a unit speed geodesic connecting $e$ to $g$. Consider the union of left translates of $\partial U$ by elements in $\gamma([0, ||g||])$, this must contain all of $gU \backslash U$ since for any $gu \notin U$ the segment $\gamma([0,||g||]) \cdot u$ must cross the boundary of $U$, i.e. there is $c \in [0,||g||]$ where $\gamma(c) u \in \partial U$, hence $g\cdot u = \gamma(||g||) \cdot u = \gamma(||g||-c)\gamma(c) u \in \gamma(||g||-c) \cdot \partial U$ But since the geodesic has derivative $1$, the n-dimensional volume of the union of those translates is at most $\mbox{vol}_{n-1}(\partial U) \cdot ||g||$. We have $\mbox{vol}_{n-1}(\partial U) \cdot R \geq \mbox{vol}_n(gU \backslash U) \geq \mbox{vol}_n(U)/2$ Now since we have growth condition $2\mbox{vol}_n(U) = \mbox{vol}_n(B_R) \geq c_n R^n$ i.e. $R \leq c_n \mbox{vol}_n(U)^\frac{1}{n}$. Conbine the two inequalities we obtain $\mbox{vol}_{n-1}(\partial U) \geq c_n \mbox{vol}_n(U)^\frac{n-1}{n}$. Establishes the theorem. Remark: In general, this argument produces a lower bound on the size of the boundary in terms of the volume of the set as long as: 1. There is a way to ‘continuously translate’ the set by elements in the space. 2. We have a lower bound on the growth of balls in terms of radius. The key observation is that the translated set minus the original set is always contained in a ‘flattened cylinder’ of the boundary in direction of the translate, which then has volume controlled by the boundary size and the length of the translating element. Because of this, the constant is almost never strict as the difference (translate subtract original) can never be the whole cylinder (in case of a ball, this difference is a bit more than half of the cylinder). ### On Uryson widths February 21, 2011 This is a note on parts of Gromov’s paper ‘width and related invariants of Riemannian manifolds’ (1988). For a compact subset $C$ of $\mathbb{R}^n$, we define the k-codimensional width (or simply k-width) to be the smallest possible number $w$ where there exists a k-dimensional affine subspace $P_k \subseteq \mathbb{R}^n$ s.t. all points of $C$ is no more than $w$ away from $P_k$. i.e. $\displaystyle{W}_k(C) = \inf_{P_k \subseteq \mathbb{R}^n} \sup_{p\in C} \mbox{dist}(p, P_k)$ where $\mbox{dist}(p, P_k)$ is the length of the orthogonal segment from $p$ to $P_k$. It’s easy to see that, for any $C$, $\mathcal{W}_0(C) \geq \mathcal{W}_1(C) \geq \cdots \geq \mathcal{W}_n(C) = 0$. At the first glance it may seems that $\mathcal{W}_0(C) = \frac{\mbox{diam}(C)}{2}$. However it is not the case since for example the equilateral triangle of side length $1$ in $\mathbb{R}^2$ has diameter $1$ but 0-width $\frac{1}{\sqrt{3}}$. In fact, by a theorem of Jung, this is indeed the optimum case, i.e. we have: $\frac{1}{2}\mbox{diam}(C) \leq \mathcal{W}_0(C) \leq \sqrt{\frac{n}{2(n+1)}}\mbox{diam}(C)$ At this point one might wonder (at least I did), if we want to invent a notion that captures the ‘diameter’ after we ‘forget the longest k-dimensions’, a more direct way seem to be taking the smallest possible number $w'$ where there is an orthogonal projection of $C$ onto a $k$ dimensional subspace $P_k$ where any point $p \in P_k$ has pre-image with diameter $\leq w'$. i.e. $\displaystyle \widetilde{\mathcal{W}_k}(X) = \inf_{P_k \subseteq \mathbb{R}^n} \sup_{p \in P_k} \mbox{diam}(\pi^{-1}_{P_k}(p))$ Now we easily have $\mbox{diam}(C) = \widetilde{\mathcal{W}_0}(C) \geq \widetilde{\mathcal{W}_1}(C) \geq \cdots \geq \widetilde{\mathcal{W}_n}(C) = 0$. However, the disadvantage of this notion is, for example, there is no reason for a semicircle arc to have 1-width 0 but a three-quarters circular arc having positive 1-width. Since we are measuring how far is the set from being linear, taking convex hull should not make the set ‘wider’ $\widetilde{\mathcal{W}_k}$, unlike $\widetilde{\mathcal{W}_k}$ is not invariant under taking convex hulls. Note that for convex sets we do have $\frac{1}{2}\widetilde{\mathcal{W}_k}(C) \leq \mathcal{W}_k(C) \leq \sqrt{\frac{n-k}{2(n-k+1)}}\widetilde{\mathcal{W}_k}(C)$ $\mathcal{W}_k(C) = 0$ iff $C$ is contained in a $k$-plane. We now generalize this notion to general metric spaces: Definition: The Uryson k-width of a compact metric space $M$ is the smallest number $w$ where there exists $k$ dimensional topological space $X$ and a continuous map $\pi: M \rightarrow X$ where any point $x \in X$ has pre-image with diameter $\leq w$. i.e. $\displaystyle UW_k(M) = \inf \{ \ \sup_{x \in X} \mbox{diam}(\pi^{-1}(x)) \ |$ $\dim{X} = k, \pi:M \rightarrow X \ \mbox{is continuous} \}$ Note: Here dimension is the usual covering dimension for topological spaces: i.e. a topological space $X$ is $n$ dimensional if any finite cover of $X$ has a finite refinement s.t. no point of $X$ is contained in more than $n_1$ sets in the cover and $n$ is the smallest number with this property. For compact subsets $C$ of $\mathbb{R}^n$ with induced metric, we obviously we have $UW_k(C) \leq \widetilde{\mathcal{W}_k}(C)$ since the pair $(P_k, \pi_{p_k})$ is clearly among the pairs we are minimizing over. Speaking of topological dimensions, one of the classical results is the following: Lebesgue’s lemma: Let $M=[0,1]^n$ be the solid n-dimensional cube, then for any topological space $X$ with $\dim(X) and any continuous map $p: M \rightarrow X$, we have image of at least one pair of opposite $(n-1)$-faces intersect. Since the conclusion is purely topological, this applies equally well to rectangles. i.e. for $M = [0, L_1] \times [0, L_2] \times \cdots \times [0, L_n]$, $L_1 \geq L_2 \geq \cdots \geq L_n$, we have $UW_{n-1}(M) \geq L_n$; furthermore, $UW_k(M) \geq L_{k+1}$ for all $k$. (If the later statement does not hold, we write $M$ as $M_1 \times M_2$, $M_1$ being the product of the first $(k+1)$ coordinates. Now $UW_k(M) \geq UW_k(M_1) \geq L_{k+1}$). In light of the earlier post about minimax inequality, we should note that if we restrict $X$ to be a homeomorphic copy of $\mathbb{R}^k$ then the notion is the same as the minimax length of fibres. In particular as proved in the post the minimax length of the unit disc to $\mathbb{R}$ is 2. Exercise: Check that for the unit $2$-disk, $UW_1(D^2) = \sqrt{3}$, i.e. the optimum is obtained by contracting the disc onto a triod. Hence it can indeed be strictly smaller than merely taking $\mathbb{R}^k$ as the targeting space, even for simply connected sets. This gives a better measurement of ‘width’ in the sense that, for example, the $\varepsilon$ neighborhood of a tree will have $1-width$ about $2 \varepsilon$. ### Intergal geometry and the minimax inequality in a nutshell February 7, 2011 The goal for most of the posts in this blog has been to take out some very simple parts of certain papers/subjects and “blow them up” to a point where anybody (myself included) can understand. Ideally the simple parts should give some inspirations and ideas towards the more general subject. This one is on the same vein. This one is based on parts of professor Guth’s minimax paper. In an earlier post, we talked about the extremal length where one is able to bound the “largest possible minimum length” (i.e. the “maximum minimum length“) of a family of rectifiable curves under conformal transformation. When combined with the uniformization theorem in for surfaces, this becomes a powerful tool for understanding arbitrary Riemannian metrics (and for conformal classes of metrics in higher dimensions). However, in ‘real life’ we often find what we really want to bound is, instead, the “minimum maximum length” of a family of curves, for example: Question: Let $\mathbb{D} \subseteq \mathbb{R}^2$ be the unit disc. Given any family $\mathcal{F}$ of arcs with endpoints on $\partial \ \mathbb{D}$ and $\mathcal{F}$ foliates $\mathbb{D}$, then how short can the logest arc in $\mathcal{F}$ possibly be? In other words, let $\mathbb{F}$ be the collection of all possible such foliations $\mathcal{F}$ as above, what is $\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A)$? After playing around a little bit with those foliations, we should expect one of the fibres to be at least as long as the diameter ( i.e. no foliation has smaller maximum length leaf than foliating by straight lines ). Hence we should have $\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A) = 2$. This is indeed easy to prove: Proof: Consider the map $f: S^1 \rightarrow S^1$ where $S^1 = \partial \ \mathbb{D}$, $f$ switches the end-points of each arc in $\mathcal{F}$. It is easy to check that $f$ is a continuous, orientation reversing homeomorphism of the circle (conjugate to a reflection). Let $p, q$ be its fixed points, $L_1, L_2$ be the two arcs in $S^1$ connecting $p$ to $q$. Let $g: z \mapsto -z$ be the antipodal map on $S^1$. Suppose $p \neq g(q)$ then one of $L_1, L_2$ is longer than $\pi$, say it’s $L_1$. Then we have $f \circ g (L_1) \subseteq L_1$. Hence $f \circ g$ has a fixed point $m$ in $L_1$, i.e. $f(m) = -m$. There is a fibre $A$ in $\mathcal{F}$ with endpoints $m, -m$, the fibre must have length $\ell(A) \geq d(-m,m) = 2$. The remaining case is trivial: if $p = g(q)$ then both $L_1$ and $L_2$ gets mapped into themselves orientation-reversingly, hence fixed points still exists. Establishes the claim. Instead of the disc, we may look at circles that sweep out the sphere (hence to avoid the end-point complications): Theorem: Any one-parameter family of circles that foliates $S^2$ (except two points) must have the largest circle being longer than the equator. This is merely applying the same argument, i.e. one of the circles needs to contain a pair of antipodal points hence must be longer than the equator. In order for easier generalization to higher dimensions, with slight modifications, this can be formulated as: Theorem: For any $f: T^2 \rightarrow S^2$ having non-zero degree, there is $\theta \in S^1$ where $\ell(f(S^1 \times \{ \theta \})$ is larger than the equator. Hence in higher dimensions we can try to prove the same statement for largest image of a lever $k$-sphere under $f: S^k \times S^{n-k} \rightarrow S^n$. However before we do that I would like to highlight some intergal geometry machineries that are new to me but seemingly constantly used in proving those kinds of estimates. We shall get some idea of the method by showing: Theorem: Let $\mathbb{R}P^n$ be equipped with the round metric. $p^k \subseteq \mathbb{R}P^n$ be a ‘flat’ $k$-dimensional plane. Then any $k$-chain $z^k \subseteq \mathbb{R}P^n$ in the same $k$ dimensional homology class as $p^k$ must have volume at least as large as $p^k$. Proof: Let $Gr(\mathbb{R}P^n, n-k)$ be the set of all $(n-k)$-planes in $\mathbb{R}P^n$ (i.e. the Grassmannian). There is a standard way to associate a measure $\mu$ on $Gr(\mathbb{R}P^n, n-k)$: Let $\lambda$ be the Haar measure on $SO(n+1)$, fix some $Q \in Gr(\mathbb{R}P^n, n-k)$. Since $SO(n+1)$ acts on $\mathbb{R}P^n$, for open set $S \subseteq Gr(\mathbb{R}P^n, n-k)$, we set $\mu(S) = \lambda( \{ T \in SO(n+1) \ | \ T(Q) \in S \})$. –The measure of a collection of planes is the measure of linear transformations that takes the given plane to an element of the set. Now we are able to integrate over all $(n-k)$-planes! For almost all $Q \in Gr(\mathbb{R}P^n, n-k)$, since $P$ is $k$-plane, we have $| Q \cap P | = 1$. ( not $1$ only when they are ‘parallel’ ) Since $[z] = [p]$ in $H_k(\mathbb{R}P^n, \mathbb{Z}_2)$, for almost all $Q$, $z$ intersects $Q$ at least as much as $P$ does. We conclude that for almost all $Q, \ | z \cap Q | \geq 1$. Fact: There exists constant $C$ such that for any $k$-chain $\Sigma^k \in \mathbb{R}P^N$, $\mbox{Vol}_k(\Sigma^k) = \mathbb{E}(|\Sigma \cap Q |)$. The fact is obtained by diving the chain into fine cubes, observe that both volume and expectation are additive and translation invariant. Therefore we only need to show this for infinitesimal cubes (or balls) near $0$. We won’t work out the details here. Hence in our case, since for almost all $Q$ we have $| z \cap Q | \geq 1$, the expectation $\mathbb{E}(|z \cap Q |) \geq 1$. We therefore deduce $\mbox{Vol}_k(z) = \mathbb{E}(|z \cap Q |) \geq 1$. Establishes the theorem. Remark: I found this intergal geometry method used here being very handy: in the old days I always try to give lower bounds on volume of stuff by intersecting it with planes and then pretend the ‘stuff’ were orthogonal to the plane, which is the worst case in terms of having small volume. An example of such bound can be found in the knot distorsion post where in order to lower bound the length we look at its intersection number with a family of parallel planes and then integrate the intersection. This is like looking from one particular direction and record how many times did a curve go through each height, of course one would never get the exact length if we know the curve already. What if we are allowed to look from all directions? I always wondered if we know the intersection number with not only a set of parallel planes but planes in all directions, then are there anything we can do to better bound the volume? Here I found the perfect answer to my question: by integrating over the Grassmannian, we are able to get the exact volume from how much it intersect each plane! We get some systolic estimates as direct corollaries of the above theorem, for example: Corollary: $\mbox{Sys}_1(\mathbb{R}P^2) = \sqrt{\pi/2}$ where $\mathbb{R}P^2$ carries the round metric with total volume $1$. Back to our minimax problems, we state the higher dimensional version: Wish: For any $C^1$ map $f: S^k \times S^{n-k} \rightarrow S^n$ where $S^n$ carries the standard round metric, there exists some $\theta \in S^{n-k}$ with $\mbox{Vol}_k(f(S^k\times \{\theta\})) \geq \mbox{Vol}_k(E^k)$ where $E^k \subseteq S^n$ is the $k$-dimensional equator. But what we have is that there is a (small) positive constant $c(n,k)$ s.t. $\mbox{deg}(f) \neq 0$ implies $\displaystyle \sup_{\theta \in S^{n-k}} \mbox{Vol}_k(f(S^k \times \{\theta\})) \geq c(n,k) \mbox{Vol}_k(E^k)$ (shown by an inductive application of the isomperimetric inequality on $S^N$, which is obtained from applying intergal geometry methods) ### On length and volume November 29, 2010 About a year ago, I came up with an simple argument for the following simple theorem that appeared in a paper of professor Guth’s: Theorem: If $U$ is an open set in the plane with area $1$, then there is a continuous function $f$ from $U$ to the reals, so that each level set of $f$ has length at most $10$. Recently a question of somewhat similar spirit came up in a talk of his: Question: Let $\langle \mathbb{T}^2, g \rangle$ be a Riemannian metric on the torus with total volume $1$, does there always exist a function $f: \mathbb{T}^2 \rightarrow \mathbb{R}$ s.t. each level set of $f$ has length at most $10$? I have some rough thoughts about how might a similar argument on the torus look like, hence I guess it would be a good idea to review and (somewhat carefully) write down the original argument. Since our final goal now is to see how things work on a torus (or other manifolds), here I would only present the less tedious version where $U$ is bounded and all boundary components of $U$ are smooth Jordan curves. Here it goes: Proof: Note that if a projection of $U$ in any direction has length (one-dimensional measure) $\leq 10$, then by taking $f$ to be the projection in the orthogonal direction, all level sets are straight with length $\leq 10$ (see image below). Hence we can assume any $1$-dimensional projection of $U$ has length $\geq 10$. A typically bad set would ‘span’ a long range in all directions with small area, it can contain ‘holes’ and being not connected: Project $U$ onto $x$ and $y$-axis, by translating $U$, we assume $\inf \pi_x(U) = \inf \pi_y(U) = 0$. Look at the measure $1$ set $S$ in the middle of $\pi_y(U)$ (i.e. a measure 1 set $[a,b] \cap \pi_y(U)$ with the property $m_1(\pi_y(U) \cap [0,a]) = m_1(\pi_y(U) \cap [b, \infty]$) By Fubini, since the volume $\pi_y^{-1}(S)$ is at most $1$, there must be a point $p\in S$ with $m_1(\pi_y^{-1}(p))\leq 1$: Since the boundary of $U$ is smooth, we may find a very small neighborhood $B_\delta(p) \subseteq \mathbb{R}$ where for each $q \in B_\delta(p), m_1(\pi_y^{-1}(q) \leq 1+\epsilon$. (we will call this pink region a ‘neck’ of the set for it has small width and is roughly in the middle) Now we define a $\varphi_1: U \rightarrow \mathbb{R}^2$ that straches the neck to fit in a long thin tube (note that in general $\pi_y(U)$ may not be connected, but everything is still well-defined and the argument does go through.) and then bend the neck to make the top chunk vertically disjoint from the bottom chunk. We can take $\varphi$ so that $\varphi^{-1}$ sends the vertical foliation of $\varphi(U)$ to the following foliation in $U$ (note that here we drew the neck wider for easier viewing, in fact the horizontal lines are VERY dense in the neck). If the $y$-projection of the top or bottom chunk is larger than $2$, we repeat the above process t the chunks. i.e. Finding a neck in the middle measure $1$ set in the chunk, starch the neck and shift the top chunk, this process is guaranteed to terminate in at most $m_1(\pi_y(U))$ steps. The final $\varphi$ sends $U$ to something like: Where each chunk has $y$-width $L$ between $1$ and $2$. Define $f = \pi_x \circ \phi$. Claim: For any $c \in \mathbb{R}, m_1(f^{-1}(c)) \leq 5$. The vertical line $x=c$ intersects $\varphi(U)$ in at most one chunk and two necks, taking $\varphi^{-1}$ of the intersection, this is a PL curve $C$ with one vertical segment and two horizontal segment in $U$: The total length of $f^{-1}(c) = C \cap U$ is less than $2+2\delta$ (length of $U$ on the vertical segment) $+ 2 \times (1+\epsilon)$ (length of $U$ on each horizontal segment). Pick $\epsilon, \delta$ both less than $1/4$, we conclude $m_1(f^{-1}(c)) < 5$. Establishes the theorem. Remark:More generally,any open set of volume $V$ has such function with fibers having length $\leq 5 \sqrt{V}$. T he argument generalizes by looking at the middle set length $\sqrt{V}$ set of each chunk. Moving to the torus Now let’s look at the problem on $\langle \mathbb{T}^2, g \rangle$, by the uniformization theorem we have a flat torus $T^2 = \mathbb{R}^2/\Gamma$ where $\Gamma$ is a lattice, $\mbox{vol}(T^2) = 1$ and a function $h: T^2 \rightarrow \mathbb{R}^{+}$ s.t. $\langle T^2, h g_0 \rangle$ is isometric to $\langle \mathbb{T}^2, g \rangle$. $g_0$ is the flat metric. Hence we only need to find a map on $T^2$ with short fibers. Note that $\int_{T^2} h^2 d V_{g_0} = 1$ and the length of the curve $\gamma$ from $p$ to $q$ in $\langle T^2, h \dot{g_0} \rangle$ is $\int_I h |\gamma'(t)| dt$. Consider $T^2$ as the parallelogram given by $\Gamma$ with sides identified. w.l.o.g. assume one side is parallel to the $x$-axis. Let $L$ be a linear transformation preserving the horizontal foliation and sends the parallelogram to a rectangle. Let $F$ be a piece-wise isometry that “folds” the rectangle: (note that $F$ is four-to-one except for on the edges and the two medians) Since all corresponding edges are identified,$lates F$is continuous not only on the rectangle but on the rectangular torus. Now we consider $F \circ L$, pre-image of typical horizontal and vertical lines in the small rectangle are union of two parallel loops: Note that vertical loops might be very long in the flat $T^2$ due to the shear while the horizontal is always the width. (to be continued) ### Systoles and the generalized Geroch conjecture October 25, 2010 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