## Posts Tagged ‘volume’

### Stable isoperimetric inequality

October 10, 2011

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

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

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

One might first attempt to use the Hausdorff distance:

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

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

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

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

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

where $\Delta$ is the symmetric difference.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

To find out more about the question, see Chapter 4 of this book. By the way, I should mention that there is a $5000 prize for this :-P Advertisements ### 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. ### 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) ### C^1 vs. C^1 volume preserving January 23, 2011 One of the things I’ve always been interested in is, for a given compact set say in $\mathbb{R}^n$, what maps defined on the set into $\mathbb{R}^n$ can be extended to a volume preserving map (of certain regularity) on a larger set (for example, some open set containing the original set). The analogues extension question without requiring the extended map to be volume preserving is answered by the famous Whitney’s extension theorem. It gives a beautiful necessary and sufficient condition on when the map has $C^r$ extension – See this pervious post for more details. A simple case of this type of question was discussed in my earlier Moser’s theorem post: Question: Given a diffeomorphism on the circle, when can we extend it to a volume preserving diffeomorphism on the disc? In the post, we showed that any $C^r$ diffeomorphism on the circle can be extended to a $C^{r-1}$ volume preserving diffeomorphism on the disc. Some time later Amie Wilkinson pointed out to me that, by using generating function methods, in fact one can avoid losing derivative and extend it to a $C^r$ volume preserving. Anyways, so we know the answer for the circle, what about for sets that looks very different from the circle? Is it true that whenever we can $C^r$ extend the map, we can also so it volume-preserving? (Of course we need to rule out trivial case such as the map is already not volume-preserving on the original set or the map sends, say a larger circle to a smaller circle.) Question: Is it true that for any compact set $K \subseteq \mathbb{R}^n$ with connected complement, for any function $f: K \rightarrow \mathbb{R}^n$ satisfying the Whitney condition with all candidate derivatives having determent $1$, one can always extend $f$ to a volume preserving $F: \mathbb{R}^n \rightarrow \mathbb{R}^n$. Note: requiring the set to have connected complement is to avoid the ‘larger circle to small circle’ case and if some candidate derivative does not have determent $1$ then the extended map cannot possibly be volume preserving near the point. After thinking about this for a little bit, we (me, Charles and Amie) came up with the following simple example where the map can only be $C^1$ extended but not $C^1$ volume preserving. Example: Let $K \subset \mathbb{R}^2$ be the countable union of segments: $K = \{0, 1, 1/2, 1/3, \cdots \} \times [0,1]$ As shown below: Define $f: K \rightarrow K$ be the map that sends the vertical segment above $1/n$ to the vertical segment above $1/(n+1)$, preserves the $y$-coordinate and fixes the segment $\{0\} \times [0,1]$: Claim: $f$ can be extended to a $C^1$ map $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Proof: Define $g: \mathbb{R} \rightarrow \mathbb{R}$ s.t. 1) $g$ is the identity on $\mathbb{R}^{\leq 0}$ 2) $g(x) = x-1/2$ for $x>1$ 3) $g: 1/n \mapsto 1/(n+1)$ 4) $g$ is increasing and differentiable on each $[1/n, 1/(n-1)]$ with derivative no less than $(1-1/n)(n^2-n)/(n^2+n)$ and the one sided derivative at the endpoints being $1$. It’s easy to check such $g$ exists and is continuous: Since $\lim_{n \rightarrow \infty} (1-1/n)(n^2-n)/(n^2+n) = 1$, we deduce $g$ is continuously differentiable with derivative $1$ at $0$. Let $F = g \otimes \mbox{id}$, $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a $C^1$ extension of $f$. Establishes the claim. Hence the pair $(K, f)$ satisfies the Whitney condition for extending to $C^1$ map. Furthermore, since the $F$ as above has derivative being the identity matrix at all points of $K$, the determent of candidate derivatives are uniformly $1$. In other words, this example satisfies all conditions in the question. Claim: $f$ cannot be extended to a $C^1$ volume preserving diffeomorphism of the plane. Proof: The idea here is to look at rectangles with sides on the set $K$, if $F$ preserves area, they have to go to regions enclosing the same area as the original rectangles, then apply the isoperimetric inequality to deduce that image of some edges of the rectangle would need to be very long, hence at some point on the edge the derivative of $F$ would need to be large. Suppose such extension $F$ exists, consider rectangle $R_n = [1/n, 1/(n-1)] \times [0,1]$. We have $m_2(R_n) = 1/(n^2-n)$ $m_2(R_n) - m_2(R_{n+1})$ $=1/(n^2-n)-1/(n^2+n)=2/(n^3-n)$ Hence in order for $F(R_n)$ to have the same area as $R_n$, the image of the two segments $s_{n,0} = [1/n, 1/(n-1)] \times \{ 0\}$ and $s_{n,1}= [1/n, 1/(n-1)] \times \{ 1\}$ would need to enclose an area of $2/(n^3-n) \sim n^{-3}$ outside of the rectangle $R_{n+1}$. By isoparametric inequality, the sum of the length of the two curves must be at least $\sim n^{-3/2}$, while the length of the original segments is $2/(n^2-n) \sim n^{-2}$. Hence somewhere on the segments $F$ needs to have derivative having norm at least $\ell(F(s_{n,0} \cup s_{n,1}) / \ell(s_{n,0} \cup s_{n,1})$ $\sim n^{-3/2}/n^{-2} = n^{1/2}$ We deduce that there exists a sequence of points $(p_n)$ converging to either $(0,0)$ or $(0,1)$ where $|| F'(p_n) || \sim n^{1/2} \rightarrow \infty$. Hence $F$ cannot be $C^1$ at the limit point of $(p_n)$. Remark: In fact we have showed the stronger statement that no volume preserving Lipschitz extension could exist and gave an upper bound $1/2$ on the best possible Holder exponent. From this we know the answer to the above question is negative, i.e. not all $C^1$ extendable map can me extended in a volume preserving fashion. It would be very interesting to give criteria on what map on which sets can be extended. By applying same methods we are also able to produce an example where the set $K$ is a Cantor set on the plane. ### Extremal length and conformal geometry December 6, 2010 There has been a couple of interesting talks recently here at Princeton. Somehow the term ‘extremal length’ came up in all of them. Due to my vast ignorance, I knew nothing about this before, but it sounded cool (and even somewhat systolic); hence I looked a little bit into that and would like to say a few words about it here. One can find a rigorous exposition on extremal length in the book Quasiconformal mappings in the plane. Let $\Omega$ be a simply connected Jordan domain in $\mathbb{C}$. $f: \Omega \rightarrow \mathbb{R}^+$ is a conformal factor on $\Omega$. Recall from my last post, $f$ is a Lebesgue measurable function inducing a metric on $\Omega$ where $\mbox{Vol}_f(U) = \int_U f^2 dx dy$ and for any $\gamma: I \rightarrow \Omega$ ($I \subseteq \mathbb{R}$ is an interval) with $||\gamma'(t)|| = \bar{1}$, we have the length of $\gamma$: $l_f(\gamma) = \int_I f dt$. Call this metric $g_f$ on $\Omega$ and denote metric space $(\Omega, g_f)$. Given any set $\Gamma$ of rectifiable curves in $U$ (possibly with endpoints on $\partial U$), each comes with a unit speed parametrization. Consider the “$f$-width” of the set $\Gamma$: $\displaystyle w_f(\Gamma) = \inf_{\gamma \in \Gamma} l_f(\gamma)$. Let $\mathcal{F}$ be the set of conformal factors $f$ with $L^2$ norm $1$ (i.e. having the total volume of $\Omega$ normalized to $1$). Definition: The extremal length of $\Gamma$ is given by $\mbox{EL}(\Gamma) = \displaystyle \sup_{f \in \mathcal{F}} w_f(\Gamma)^2$ Remark: In fact I think it would be more natural to just use $w_f(\Gamma)$ instead of $w_f(\Gamma)^2$ since it’s called a “length”…but since the standard notion is to sup over all $f$, not necessarily normalized, and having the $f$-width squared divide by the volume of $\Omega$, I can’t use conflicting notation. One should note that in our case it’s just the square of sup of width. Definition:The metric $(\Omega, g_f)$ where this extremal is achieved is called an extremal metric for the family $\Gamma$. The most important fact about extremal length (also what makes it an interesting quantity to study) is that it’s a conformal invariant: Theorem: Given $h: \Omega' \rightarrow \Omega$ bi-holomorphic, then for any set of normalized curves $\Gamma$ in $\Omega$, we can define $\Gamma' = \{ h^{-1}\circ \gamma \ | \ \gamma \in \Gamma \}$ after renormalizing curves in $\Gamma'$ we have: $\mbox{EL}(\Gamma) = \mbox{EL}(\Gamma')$ Sketch of a proof: (For simplicity we assume all curves in $\Gamma'$ are rectifiable, which is not always the case i.e. for bad maps $h$ the length might blow up when the curve approach $\partial \Omega'$ this case should be treated with more care) This is indeed not hard to see, first we note that for any $f: \Omega \rightarrow \mathbb{R}^+$ we can define $f' : \Omega' \rightarrow \mathbb{R}^+$ by having $f^\ast (z) = |h'(z)| (f \circ h) (z)$ It’s easy to see that $\mbox{Vol}_{f^\ast}(\Omega') = \mbox{Vol}_{f}(\Omega)$ (merely change of variables). In the same way, $l_{f^\ast}(h^{-1}\circ \gamma) = l_f(\gamma)$ for any rectifiable curve. Hence we have $w_{f^\ast}(\Gamma') = w_f(\Gamma)$. On the other hand, we know that $\varphi: f \mapsto f^\ast$ is a bijection from $\mathcal{F}_\Omega$ to $\mathcal{F}_{\Omega'}$, deducing $\mbox{EL}(\Gamma) = \displaystyle ( \sup_{f \in \mathcal{F}} w_f(\Gamma))^2 = \displaystyle ( \sup_{f' \in \mathcal{F}'} w_{f'}(\Gamma'))^2 = \mbox{EL}(\Gamma')$ Establishes the claim. One might wonder how on earth should this be applied, i.e. what kind of $\Gamma$ are useful to consider. Here we emphasis on the simple case where $\Omega$ is a rectangle (Of course I would first look at this case because of the unresolved issues from the last post :-P ): Theorem: Let $R = (0,w) \times (0, 1/w)$, $\Gamma$ be the set of all curves starting at a point in the left edge $\{0\} \times [0, 1/w]$, ending on $\{1\} \times [0, 1/w]$ with finite length. Then $\mbox{EL}(\Gamma) = w^2$ and the Euclidean metric $f = \bar{1}$ is an extremal metric. Sketch of the proof: It suffice to show that any metric $g_f$ with $\mbox{Vol}_f(R) = 1$ has at least one horizontal line segment $\gamma_y = [0,w] \times \{y\}$ with $l_f(\gamma_y) \leq w$. (Because if so, $w_f(\Gamma) \leq w$ and we know $w_{\bar{1}}(\Gamma) = w$ for the Euclidean length) The average length of $\gamma_y$ over $y$ is $w \int_0^{1/w} l_f(\gamma_y) dy$ $= w \int_0^{1/w} (\int_0^w f(t, y) dt) dy = w \int_R f$ By Cauchy-Schwartz this is less than $w (\int_R f^2)^{1/2} |R|^{1/2} = w$ Since the shortest curve cannot be longer than the average curve, we have $w_f(\Gamma) \leq w$. Hence $\mbox{EL}(\Gamma) = \displaystyle \sup_{f \in \mathcal{F}}w_f(\Gamma)^2 = w^2$ Note it’s almost the same argument as in the proof of systolic inequality on the 2-torus. Corollary: Rectangles with different eccentricity are not conformally equivalent (i.e. one cannot find a bi-homomorphic map between them sending each edge to an edge). Remark: I was not aware of this a few days ago and somehow had the silly thought that there are conformal maps between any pair of rectangles while discussing with Guangbo >.< then tried to see what would those maps look like and was of course not able to do so. (there are obviously Riemann maps between the rectangles, but they don't send conners to conners, i.e. can't be extended to a conformal map on the closed rectangle). An add-on: While I came across a paper of Odes Schramm, applying the techniques of extremal length, the following theorem seemed really cool. Let $G = (V, E)$ be a finite planar graph with vertex set $V$ and edges $E \subseteq V^2$. For each vertex $v$ we assign a simply connected domain $D_v$. Theorem: We can scale and translate each $D_v$ to $D'_v$ so that $\{ D_v \ | \ v \in V \}$ form a packing (i.e. are disjoint) and the contact graph of $D'_v$ is $G$. (i.e. $\overline{D'_{v_1}} \cap \overline{D'_{v_2}} \neq \phi$ iff $(v_1, v_2) \in E$. Note: This is vastly stronger than producing a circle packing with prescribed structure. ### 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)