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 is an open set in the plane with area , then there is a continuous function from to the reals, so that each level set of has length at most .
Recently a question of somewhat similar spirit came up in a talk of his:
Question: Let be a Riemannian metric on the torus with total volume , does there always exist a function s.t. each level set of has length at most ?
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 is bounded and all boundary components of are smooth Jordan curves. Here it goes:
Proof: Note that if a projection of in any direction has length (one-dimensional measure) , then by taking to be the projection in the orthogonal direction, all level sets are straight with length (see image below).
Hence we can assume any -dimensional projection of has length . A typically bad set would ‘span’ a long range in all directions with small area, it can contain ‘holes’ and being not connected:
Project onto and -axis, by translating , we assume . Look at the measure set in the middle of (i.e. a measure 1 set with the property )
By Fubini, since the volume is at most , there must be a point with :
Since the boundary of is smooth, we may find a very small neighborhood where for each . (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 that straches the neck to fit in a long thin tube (note that in general 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 so that sends the vertical foliation of to the following foliation in (note that here we drew the neck wider for easier viewing, in fact the horizontal lines are VERY dense in the neck).
If the -projection of the top or bottom chunk is larger than , we repeat the above process t the chunks. i.e. Finding a neck in the middle measure set in the chunk, starch the neck and shift the top chunk, this process is guaranteed to terminate in at most steps. The final sends to something like:
Where each chunk has -width between and .
Claim: For any .
The vertical line intersects in at most one chunk and two necks, taking of the intersection, this is a PL curve with one vertical segment and two horizontal segment in :
The total length of is less than (length of on the vertical segment) (length of on each horizontal segment). Pick both less than , we conclude .
Establishes the theorem.
Remark:More generally,any open set of volume has such function with fibers having length . T he argument generalizes by looking at the middle set length set of each chunk.
Moving to the torus
Now let’s look at the problem on , by the uniformization theorem we have a flat torus where is a lattice, and a function s.t. is isometric to . is the flat metric. Hence we only need to find a map on with short fibers.
and the length of the curve from to in is
Consider as the parallelogram given by with sides identified. w.l.o.g. assume one side is parallel to the -axis. Let be a linear transformation preserving the horizontal foliation and sends the parallelogram to a rectangle.
Let be a piece-wise isometry that “folds” the rectangle:
(note that 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 , 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 due to the shear while the horizontal is always the width.
(to be continued)