# The sponge problem

Here’s a cute open problem professor Guth told me about yesterday (also appeared in [Guth 2007], although that version is slightly stronger and more general:

Problem: Does there exist $c>0$ s.t. $\forall U \subseteq \mathbb{R}^2$ with $m(U) < c$, we have $\exists f: U \rightarrow \mathbb{R}^2$ that’s bi-lipschitz with constant 2 and $f(U) \subseteq B_1(0)$?

Here bi-lipschitz with constant $k$ is defined in the infinitesimal sense, i.e. $\forall p \in U, 1/k < |Df(p)| < k$.

i.e. given any open set of a very small measure, say $1/100$, does there always exist a map that locally does not deform the metric too much ( i.e. bi-lipschitz with constant $2$ ) and “folds” the set into the unit ball.

My initial thought are to approximate the set with some kind of skeleton (perhaps a 1-dimensional set that has $U$ contained in its $\delta$ neighborhood) and fold the skeleton instead. Not sure exactly what to do yet…maybe Whitney’s construction? Or construction similar to creating the nerve in Čech homology? Obviously there is a smaller lipschitz constant allowed when we pass to the skeleton, but that might not be an issue since we can pick the measure of our set arbitrarily small.

By the way, personally I believe the answer is affective…There got to be a way to “fold stuff in” when we don’t have much stuff to start with, right?

# Counterexamples to Isosystolic Inequality

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.

# Generic Accessibility (part 1) – Andy Hammerlindl

Pugh-Shub Conjecture: Generic measure preserving partially hyperbolic diffeomorphism is ergodic. [PS (2000)]

The accessibility approach breaks this into two conjectures (and both are open):

Conjecture A: Generic partially hyperbolic diffeomorphism (measure preserving or not) is accessible.

Conjecture B: All measure preserving accessible partially hyperbolic diffeomorphisms are ergodic.

Note that if both conjecture A and conjecture B are true, then the Pugh-Shub Conjecture is true, but the failure of either won’t imply the conjecture being wrong.

Here we discuss the recent result of RHRHU (2008) which proves Pugh-Shub conjecture in the case where $\dim (E^c)=1$ by using accessibility.

We are going to focus on the proof of Conjecture A, here’s a sketch of proof of Conjecture B when $\dim (E^c)=1$ is assumed:

Theorem B: Let $M$ be compact manifold, $f: M \rightarrow M$ be a measure preserving accessible partially hyperbolic diffeomorphism, $\dim (E^c)=1$, then $f$ is ergodic.

Proof: Let $\phi: M \rightarrow \mathbb{R}$ be $f$ invariant

Let $A = \phi^{-1}(( - \infty, c])$, if $m(A)>0$ then $\exists p \in A$ s.t. $p$ is a density point of $A$.

At this point there are some technical details in the paper which we are going to skip, but the main idea is to the fact that $\dim (E^c)=1$ (or in this case even the weaker hypothesis center brunching would work) to prove that in our case $y \in M$ is a density point iff $y$ is a “leaf density point” in both its center-stable and center-unstable leaves. Hence by accessibility from $p$ to $y$, we can “push” the point $p$ along the us-path that joins $p$ to $y$ and induce that $y$ is a “leaf density point” in $A$ hence a density point in $A$.

$\therefore$ all points $y \in M$ are density points of $A$ hence $m(A)=1$.

Note that here if we replace accessibility by essential accessibility, we still get $m(A)=1$.

Hence $\forall c \in \mathbb{R}$, either $m(\phi^{-1}(( - \infty, c]))=0$ or $m(\phi^{-1}(( - \infty, c])) = 1$

$\therefore \ \phi$ is essentially constant. $\therefore f$ is ergodic.

This establishes theorem B.

Let $PH^r(M)$ be the set of measure preserving diffeomorphisms on $M$ that are of class $C^r$

Theorem A: Accessibility is open dense in the space of diffeomorphisms in $PH^r(M)$ with $\dim(E^c) = 1$.

For any $x \in M$, let $AC(x)$ denote the set of points that’s accessible from $x$

Let $\mathcal{D} = \{ f \in {PH}^r (M) | \forall \ x \in Per(f), AC(x)$ is open $\}$

Fact: $\mathcal{D} \subseteq PH^r(M)$ with $\dim(E^c) = 1$ is $G_\delta$ and $\mathcal{D} = \mathcal{A} \sqcup \mathcal{B}$
where $\mathcal{A} = \{ \ f \ | \ f$ is accessible $\}$ and
$\mathcal{B} = \{ \ f \ | \ per(f) = \phi$ and $E^u \oplus E^s$ is integrable $\}$

Note that this actually requires some rather technical work which was done in the paper, here we skip the proof of this.

Let $U(f) = \{ \ x \in M \ | \ AC(x)$ is open $\}$

It’s easy to see that $U(f)$ is automatically open hence $V(f) = M \setminus U(f)$ is compact.

Proposition: Let $x \in M$, the following are equivalent:

1) $AC(x)$ has non-empty interior

2) $AC(x)$ is open

3) $AC(x) \cap \mathcal{W}^c_{loc}(x)$ has non-empty interior in $\mathcal{W}^c_{loc}(x)$

Proof: 1) $\Rightarrow$ 2) $\Rightarrow$ 3) $\Rightarrow$ 1)

Mainly by drawing pictures and standard topology.

Unweaving lemma: $\forall x \in Per(f), \ \exists g \in$latex PH^r(M)$with $\dim(E^c) = 1$ s.t. the $C^r$ distance between $f$ and $g$ is arbitrarily small, $x \in Per(g)$ and $AC_g(x)$ is open. The proof is left to the second part of the talk… # LaTeX?$latex \LaTeX\$!

Heard that we can use $\LaTeX$ here…but how?

$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$

is this working?