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)

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