Archive for November 29th, 2010

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)