Posts Tagged ‘counterexample’

Remarks from the Oxtoby Centennial Conference

November 2, 2010

A few weeks ago, I received this mysterious e-mail invitation to the ‘Oxtoby Centennial Conference’ in Philadelphia. I had no idea about how did they find me since I don’t seem to know any of the organizers, as someone who loves conference-going, of course I went. (Later I figured out it was due to Mike Hockman, thanks Mike~ ^^ ) The conference was fun! Here I want to sketch a few cool items I picked up in the past two days:

Definition:A Borel measure $\mu$ on $[0,1]^n$ is said to be an Oxtoby-Ulam measure (OU for shorthand) if it satisfies the following conditions:
i) $\mu([0,1]^n) = 1$
ii) $\mu$ is positive on open sets
iii) $\mu$ is non-atomic
iv) $\mu(\partial [0,1]^n) = 0$

Oxtoby-Ulam theorem:
Any Oxtoby-Ulam measure is the pull-back of the Lebesgue measure by some homeomorphism $\phi: [0,1]^n \rightarrow [0,1]^n$.

i.e. For any Borel set $A \subseteq [0,1]^n$, we have $\mu(A) = \lambda(\phi(A))$.

It’s surprising that I didn’t know this theorem before, one should note that the three conditions are clearly necessary: A homeo has to send open sets to open sets, points to points and boundary to boundary; we know that Lebesgue measure is positive on open sets, $0$ at points and $0$ on the boundary of the square, hence any pull-back of it must also has those properties.

Since I came across this at such a late time, my first reaction was: this is like Moser’s theorem in the continuous case! But much cooler! Because measures are a lot worse than differential forms: many weird stuff could happen in the continuous setting but not in the smooth setting.

For example, we can choose a countable dense set of smooth Jordan curves in the cube and assign each curve a positive measure (we are free to choose those values as long as they sum to $1$. Now we can define a measure supported on the union of curves and satisfies the three conditions. (the measure restricted to each curve is just a multiple of the length) Apply the theorem, we get a homeomorphism that sends each Jordan curve to a Jordan curve with positive $n$ dimensional measure and the $n$ dimensional measure of each curve is equal to our assigned value! (Back in undergrad days, it took me a whole weekend to come up with one positive measured Jordan curve, and this way you get a dense set of them, occupying a full measure set in the cube, for free! Oh, well…>.<)

Question: (posed by Albert Fathi, 1970)
Does the homeomorphism $\phi$ sending $\mu$ to Lebesgue measure depend continuously on $\mu$?

My first thought was to use smooth volume forms to approximate the measure, for smooth volume forms, Moser’s theorem gives diffeomosphisms depending continuously w.r.t. the form (I think this is true just due to the nature of the construction of the Moser diffeos) the question is how large is the closure of smooth forms in the space of OU-measures. So I raised a little discussion immediately after the talk, as pointed out by Tim Austin, under the weak topology on measures, this should be the whole space, with some extra points where the diffeos converge to something that’s not a homeo. Hence perhaps one can get the homeo depending weakly continuously on $\mu$.

Lifted surface flows:

Nelson Markley gave a talk about studying flows on surfaces by lifting them to the universal cover. i.e. Let $\phi_t$ be a flow on some orientable surface $S$, put the standard constant curveture metric on $S$ and lift the flow to $\bar{\phi}_t$ on the universal cover of $S$.

There is an early result:

Theorem: (Weil) Let $\phi_t$ be a flow on $\mathbb{T}^2$, $\bar{\phi}_t$ acts on the universal cover $\mathbb{R}^2$, then for any $p \in \mathbb{R}^2$, if $\displaystyle \lim_{t\rightarrow \infty} ||\bar{\phi}_t(p)|| = \infty$ then $\lim_{t\rightarrow \infty} \frac{\bar{\phi}_t(p)}{||\bar{\phi}_t(p)||}$ exists.

i.e. for lifted flows, if an orbit escapes to infinity, then it must escape along some direction. (No sprial-ish or wild oscillating behavior) This is due to the nature that the flow is the same on each unit square.

We can find its analogue for surfaces with genus larger than one:

Theorem: Let $\phi_t$ be a flow on $S$ with $g \geq 2$, $\bar{\phi}_t: \mathbb{D} \rightarrow \mathbb{D}$, then for any $p \in \mathbb{D}$, if $\displaystyle \lim_{t\rightarrow \infty} ||\bar{\phi}_t(p)|| = \infty$ then $\lim_{t\rightarrow \infty} \bar{\phi}_t(p)$ is a point on the boundary of $\mathbb{D}$.

Using those facts, they were able to prove results about the structure of $\omega$ limiting set of such orbits (those that escapes to infinity in the universal cover) using the geometric structure of the cover.

I was curious about what kind of orbits (or just non-self intersecting curves) would ‘escape’, so here’s some very simple observations: On the torus, this essentially means that the curve does not wind around back and forth infinitely often with compatible magnitudes, along both generators. i.e. the curve ‘eventually’ winds mainly in one direction along each generating circle. Very loosely speaking, if a somewhat similar thing is true for higher genus surfaces, i.e. the curve eventually winds around generators in one direction (and non-self intersecting), then it would not be able to have very complicated $\omega$ limiting set.

Measures on Cantor sets

In contrast to the Oxtoby-Ulam theorem, one could ask: Given two measures on the standard middle-third Cantor set, can we always find a self homeomorphism of the Cantor set, pushing one measure to the other?

Given there are so many homeomorphisms on the Cantor set, this sounds easy. But in fact it’s false! –There are countably many clopen subsets of the Cantor set (Note that all clopen subsets are FINITE union of triadic copies of Cantor sets, a countable union would necessarily have a limit point that’s not in the union), a homeo needs to send clopen sets to clopen sets, hence for there to exist a homeo the countably many values the measures take on clopen sets must agree.

So a class of ‘good measures’ on Cantor sets was defined in the talk and proved to be realizable by a pull back the standard Hausdorff measure via a homeo.

I was randomly thinking about this: Given a non-atomic measure $\mu$ on the Cantor set, when can it be realized as the restriction of the Lebesgue measure to an embedding of the Cantor set? After a little bit of thinking, this can always be done. (One can simple start with an interval, break it into two pieces according to the measure $\mu$ of the clopen sets before and after the largest gap, then slightly translate the two pieces so that there is a gap between them; iterate the process)

In any case, it’s been a fun weekend! ^^

Counterexamples to Isosystolic Inequality

June 20, 2009

This is a note on Mikhail Katz’s paper (1995) in which he constructed a sequence of Riemannian metrics $g_i$ on $S^n \times S^n$ s.t. $\lim_{i \rightarrow \infty} \mbox{Vol}_{2n}(S^n \times S^n, g_i) / (\mbox{\mbox{Sys}}_n(S^n \times S^n, g_i))^2 = 0$ for $n \geq 3$. Where $\mbox{Sys}_k(M)$ denotes the $k$-systole which is the infimum of volumes of $k$-dimensional integer cycles representing non-trivial homology classes. To find out more about systoles, here’s a nice 60-second introduction by Katz.

We are interested in whether there is a uniform lower bound for $\mbox{Vol}_{2n}(M) / (\mbox{Sys}_n(M))^2$ for $M$ being $S^n \times S^n$ equipped with any Riemann metric. For $n=1$, it is known that $\mbox{Vol}_2( \mathbb{T}, g)/(\mbox{Sys}_1(\mathbb{T})^2 \geq \sqrt{3}/2$. Hence the construction gave counterexamples for all $n \geq 3$. An counterexample for $n=2$ is constructed later using different techniques.

The construction breaks into three parts:

1) Construction a sequence of metrics $(g_i)$ on $S^1 \times S^n$ s.t. $\mbox{Vol}_{1+n}(S^1 \times S^n, g_i) / (\mbox{Sys}_1(S^1 \times S^n, g_i)\mbox{Sys}_n(S^1 \times S^n, g_i))$ approaches $0$ as $i \rightarrow \infty$.

2) Choose an appropriate metric $q$ on $S^{n-1}$ s.t. $M_i = S^1 \times S^n \times S^{n-1}$ equipped with the product metric $g_i \times q$ satisfy the property $\lim_{i \rightarrow \infty} \mbox{Vol}_{2n}(M_i) / (\mbox{Sys}_n(M_i))^2 = 0$

3) By surgery on $M_i = S^1 \times S^n \times S^{n-1}$ to obtain a sequence of metrics on $S^n \times S^n$, denote the resulting manifolds by $M_i'$, having the property that $\lim_{i \rightarrow \infty} \mbox{Vol}_{2n}(M_i') / (\mbox{Sys}_n(M_i'))^2 = 0$

The first two parts are done in previous notes (which are not published on this blog). Here I will talk about how is part 3) done given that we have constructed manifolds $M_i$ as in part 2).

Let $V_i = S^1 \times S^n$ equipped with metric $g_i$ as constructed in 1), $M_i$ be as constructed in 2).

Standard surgery: Let $B^{n-1} \subseteq S^{n-1}$ and let $U = S^1 \times B^{n-1}, U' = B^2 \times S^{n-2}$. $\partial B^2 = S^1, \partial B^{n-1} = S^{n-2}$. The resulting manifold from standard surgery along $S^1$ in $S^1 \times S^{n-1}$ is defined to be $C = S^1 \times S^{n-1} \setminus U \cup U'$ $= S^1 \times S^{n-1} \setminus S^1 \times B^{n-1} \cup B^2 \times S^{n-2}$ which is homeomorphic to $S^n$.

We perform the standard surgery on the $S^1 \times S^{n-1}$ component of $M_i$, denote the resulting manifold by $M_i'$. Hence $M_i' = S^n \times C = S^n \times S^n$ equipped with some metric.

Note that the metric depends on the surgery and so far we have only specified the surgery in the topological sense. Now we are going to construct the surgery taking the metric $g_i$ into account.

First we pick $B^{n-1} \subseteq S^{n-1}$ to be a small ball of radius $\varepsilon$, call it $B_\varepsilon^{n-1}$. Pick $B^2$ that fills $S^1$ to be a cylinder of length $L$ for some large $l$ with a cap $\Sigma$ on the top. i.e. $B_L^2 = S^1 \times [0,L] \cup \Sigma$ and $\partial \Sigma = S^1 \times \{1\}$. Hence the standard surgery can be performed with $U = S^1 \times B_\varepsilon^{n-1}$ and $U' = B_L^2 \times S_\varepsilon^{n-2}, \ S_\varepsilon^{n-2} = \partial B_\varepsilon^{n-1}$. The resulting manifold $M'_i (\varepsilon, L)$ is homeomorphic to $S^n \times S^n$ and has a metric on it that depends on $g_i, \varepsilon$ and $L$.

Let $H = B_L^2 \times S^n \times S_\varepsilon^{n-2}$ i.e. the part that’s glued in during the surgery, call it the ‘handle’.

The following properties hold:
i) For any fixed $L$, for $\varepsilon$ sufficiently small, $\mbox{Vol}(M'_i (\varepsilon, L)) \leq 2 \mbox{Vol}(M_i)$

Since $\mbox{Vol}(H) =\mbox{Vol}(B_L^2) \times \mbox{Vol}(S^n) \times \mbox{Vol}(S_\varepsilon^{n-2})$
$\mbox{Vol}(B_L^2) \sim L \times \mbox{Vol}(S^1), \ \mbox{Vol}(S_\varepsilon^{n-2}) \sim \varepsilon^{n-2}$
$\therefore \forall n \geq 3, n-2 > 0$ implies $\mbox{Vol}(H)$ can be made small by taking $\varepsilon$ small.

ii) The projection of $H$ to its $S^n$ factor is distance-decreasing.

iii) If we remove the the cap part $\Sigma \times S^n \times S_\varepsilon^{n-2}$ from $M'_i(\varepsilon, L)$ (infact from $H$), then the remaining part admits a distance-decreasing retraction to $M_i$.

i.e. project the long cylinder onto its base on $M_i$ which is $S^1 \times \{0\}$.

iv) Both ii) and iii) remain true if we fill in the last component of $H$ i.e. replace it with $B_L^2 \times S^n \times B_\varepsilon^{n-1}$ and get a $2n+1$-dimensional polyhedron $P$.

Since all we did in ii) and iii) is to project along the first and third component simultaneously or to project only the first component, filling in the third component does not effect the distance decreasing in both cases.

We wish to choose an appropriate sequence of $\varepsilon$ and $L$ so that $\lim_{i \rightarrow \infty} \mbox{Vol}(M'_i(\varepsilon_i, L_i))/(\mbox{Sys}_n(M'_i(\varepsilon_i, L_i))^2=0$.

In the next part we first fix any $i, \ \varepsilon_i$ and $L_i$ so that property i) from above holds and write $M'_i$ for $M'_i( \varepsilon_i, L_i)$.

We are first going to bound all cycles with a nonzero $[S^n]$ component and then consider the special case when the cycle is some power of $C$ and this will cover all possible non-trivial cycles.

Claim 1: $\forall n$-cycle $z \subseteq M'_i$ belonging to a class with nonzero $[S^n]$-component, we have $\mbox{Vol}(z) \geq 1/2 \ \mbox{Sys}_n(V_i)$.

Note that since $\mbox{Sys}_n(V_i) \geq \mbox{Sys}_n(M_i)$ and by part 2), $\lim_{i \rightarrow \infty} \mbox{Vol}_{2n}(M_i) / (\mbox{Sys}_n(M_i))^2 = 0$ and by property i), $\mbox{Vol}(M'_i) \leq 2 \ \mbox{Vol}(M_i)$. Let $\delta_i = \mbox{Vol}_{2n}(M_i) / (\mbox{Sys}_n(M_i))^2$, hence $\delta_i \rightarrow 0$. Therefore the bound in claim 1 would imply $\mbox{Vol}(M'_i)/(\mbox{Vol}(z))^2 \leq 2 \ \mbox{Vol}(M_i)/(1/2 \ \mbox{Sys}_n(M_i))^2$ $\leq 8 \delta_i \rightarrow 0$ which is what we wanted.

Proof:
a) If $z$ does not intersect $\Sigma \subseteq B_L^2 \times S^n \times S_\varepsilon^{n-1}$

In this case the cycle can be “pushed off” the handle to lie in $M_i$ without increasing the volume. i.e. we apply the retraction from proposition iii).

b) If $z \subseteq H$ then by proposition ii), $z$ projects to its $S^n$ component by a distance-decreasing map and $\mbox{Vol}_n(S^n) \geq \mbox{Sys}_n(V_i)$ by construction in part 2).

Now suppose $\exists z$ with $\mbox{Vol}(z) < 1/2 \ \mbox{Sys}_n(V_i)$.
Define $f: {M_i}' \rightarrow \mathbb{R}^+$ s.t. $f(p) = d(p, {M_i}' \setminus H)$.

Let $L_i \geq \mbox{Sys}_n(V_i)$, then by the coarea inequality, we have $\exists t \in (0, L)$ s.t. $\mbox{Vol}(z \cap f^{-1}(t)) < \mbox{Vol}(z) / L$ $< 1/2 \ \mbox{Sys}_n(V_i) / \mbox{Sys}_n(V_i) = 1/2$.

By our results in Gromov[83] and the previous paper of Larry Guth or Wenger’s paper, $\exists C(k)$ s.t. $\forall k$-cycle $c$ with $\mbox{Vol}(c) \leq 1$, $\mbox{FillVol}(c) \leq C(k) \ \mbox{Vol}(c)^(k+1)/k$. Hence $\mbox{Vol}(z \cap f^{-1}(t)) \leq 1/2 \Rightarrow \ \exists c_t \subseteq P$ with $\mbox{Vol}(c_t) \leq C(n-1) (\mbox{Vol}_{n-1}(z \cap f^{-1}(t)))^{n/(n-1)}$. By picking $L_i \geq 2^i \mbox{Sys}_n(V_i)$, we have $\mbox{Vol}(c_t) / \mbox{Sys}_n(V_i) \rightarrow 0$ as $i \rightarrow \infty$.

Recall that $f^{-1}(t) = S^1 \times S^n \times S_\varepsilon^{n-2}$; by construction $\mbox{Vol} (S^1) \geq 2$ and $\mbox{Vol}(S^n) \geq 2$.

Let $z_t = z \cap f^{-1}([0,t])$,

(1) If the cycle $z_t \cup c_t$ has non-trivial homology in $P$, then by proposition iv), the analog of proposition iii) for $P$ implies we may retract $z_t \cup c_t$ to $M_i$ without decreasing its volume. Then apply case a) to the cycle after retraction we obtain $\mbox{Vol}(z_t \cup c_t) \geq \mbox{Sys}_n(V_i)$.

$\therefore \mbox{Vol}(z) \geq \mbox{Vol}(z_t) \geq \mbox{Sys}_n(V_i) - \mbox{Vol}(c_t) \sim \mbox{Sys}_n(V_i)$

Contradicting the assumption that $\mbox{Vol}(z) \leq 1/2 \ \mbox{Sys}_n(V_i)$.

(2) If $z_t \cup c_t$ has trivial homology in $P$, then $z - z_t + c_t$ is a cycle with volume smaller than $z$ that’s contained entirely in $H$. By case b), $z - z_t + c_t$ projects to its $S^n$ factor by a distance decreasing map, and $\mbox{Vol}(S^n) \geq \mbox{Sys}_n(V_i)$. As above, $\mbox{Vol}(z) \geq \mbox{Sys}_n(V_i) - \mbox{Vol}(c_t) \sim \mbox{Sys}_n(V_i)$, contradiction.