Posts Tagged ‘Riemannian manifold’

Longest shortest geodesic on a 2-sphere

October 31, 2011

This is a little note about constructing a Riemannian 2-sphere which has longer shortest geodesic than the round 2-sphere of same area.

—–  Background Story  —–

So there has been this thing called ‘mathematical conversations’ at the IAS, which involves someone present a topic that’s elementary enough to be accessible to mathematicians in all fields and yet can be expanded in different directions and lead into interesting interdisciplinary discussions.

Nancy Hingston gave one of those conversations about simple geodesics on the two-sphere one night and I was (thanks to Maria Trnkova who dragged me in) able to attend.

So she talked about some fascinating history of proving the existence of closed geodesics and later simple closed geodesics on generic Riemannian two-spheres.

Something about this talk obviously touched my ‘systolic nerve’, so when the discussion session came up I asked whether we have bounds on ‘length of longest possible shortest closed geodesic on a sphere with unit area’. The question seem to have generated some interest in the audience and resulted in a back-and-forth discussion (some of which I had no clue what they were talking about). So the conclusion was at least nobody knows such a result on top of their head and perhaps optimum is obtained by the round sphere.

—–  End of Story —-

A couple of post-docs caught me afterwards (Unfortunately I didn’t get their names down, if you happen to know who they are, tell me~) and suggested that suspending a smooth triangular region and smoothen the corners might have longer shortest geodesic than the round sphere:

The evidence being the fact that on the plane a rounded corner triangular contour has larger ‘width’ than the disc of same area. (note such thing can be made to have same width in all directions)

Well that’s pretty nice, so I went home and did a little high-school computations. The difficulty about the pillow is that the shortest geodesic isn’t necessarily the one that goes through the ‘tip’ and ‘mid-point of the base’, something else might be shorter. I have no idea how to argue that.

A suspicious short geodesic:

So I ended up going with something much simpler, namely gluing together two identical copies of the flat equilateral triangles. This can be made to a Riemannian metric by smoothing the edge and corners a little bit:

Okay, now the situation is super simple~ I want to prove that this ‘sphere’ (let’s call it $S$ from now on) has shortest geodesic longer than the round sphere ($\mathbb{S}^2$)!

Of course we suppose both $S$ and $\mathbb{S}^2$ has area 1.

Claim: The shortest geodesic on $S$ has length $\sqrt[4]{12}$ (which is length of the one through the tip and mid-point of the opposite edge.)

Proof: The shortest closed geodesic passing through the corner is the one described above, since any other such geodesics must contain two symmetric segments from the corner to the bottom edge on the two triangles, those two segments alone is longer than the one orthogonal to the edge.

That middle one has length $2h$ where

$A(\Delta) = 1/2 = h^2/\sqrt{3}$

i.e. $h = \sqrt[4]{3} / \sqrt{2}, \ \ell = 2h = \sqrt[4]{12}$

The good thing about working with flat triangles is that now I know what the closed geodesics are~

First we observe any closed geodesic not passing through the corner is a periodical billiard path in the triangular table with even period.

So let’s ‘unfold’ the triangles on the plane. Such periodic orbits correspond to connecting two corresponding points on a pair of identified parallel edges and the segment between them intersecting an even number of tiles.

W.L.O.G we assume the first point in on edge $e$. Since we are interested in orbits having shortest length, let’s take neighborhood of radius $\sqrt[4]{12} + \epsilon$ around our edge $e$: (all edges with arrows are identified copies of $e$)

There are only 6 parallel copies of $e$ in the neighborhood:

Note that no matter what point $p$ on $e$ we start with, the distance from $p$ to another copy of it on any of the six edges is EQUAL to $\sqrt[4]{12}$. (easy to see since one can slide the segments to begin and end on vertices.)

Hence we conclude there are no shorter periodic billiard paths, i.e. the shortest closed geodesic on $S$ has length $\sqrt[4]{12}$.

Note it’s curious that there are a huge amount of closed geodesics of that particular length, most of them are not even simple! However it seems that after we smoothen $S$ to a Riemannian metric, the non-simple ones all become a little longer than that simple one through the corner. I wonder if it’s possible that on a Riemannian sphere the shortest closed geodesic is a non-simple one.

Anyways, now let’s return to $\mathbb{S}^2$~ So the surface area is $1$ hence the radius is $r= \sqrt{1/4\pi} = \frac{1}{2\sqrt{\pi}}$

Any closed geodesics is a multiple of a great circle, hence the shortest geodesic has length $2 \pi r = \sqrt{\pi}$, which is just slightly shorter than $\sqrt[4]{12} \approx \sqrt{3.4}$.

Now the natural question arises: if the round sphere is not optimum, then what is the optimum?

At this point I looked into the literature a little bit, turns out this problem is quite well-studied and there is a conjecture by Christopher Croke that the optimum is exactly $\sqrt[4]{12}$. (Of course this optimum is achieved by our singular triangle metric hence after smoothing it would be $< \sqrt[4]{12}$.

There is even some recent progress made by Alex Nabutovsky and Regina Rotman from (our!) University of Toronto! See this and this. In particular, one of the things they proved was that the shortest geodesic on a unit area sphere cannot be longer than $8$, which I believe is the best known bound to date. (i.e. there is still some room to $\sqrt[4]{12}$.)

Random remark: The essential difference between this and the systolic questions is that the sphere is simply connected. So the usual starting point, namely ‘lift to universal cover’ for attacking systolic questions does not work. There is also the essential difference where, for example, the question I addressed above regarding whether the shortest geodesic is simple would not exist in systolic situation since we can always split the curve into two pieces and tighten them up, at least one would still be homotopically non-trivial. In conclusion since this question sees no topology but only the geometry of the metric, I find it interesting in its own way.

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)

Systoles and the generalized Geroch conjecture

October 25, 2010

Almost a year ago, I said here that I would write a sequence of posts on some simple facts and observations related to the systolic inequality but got distracted and didn’t manage to do much of that…

I was reminded last week as I heard professor Guth’s talk on systoles for the 4th time (Yes, the same talk! –in Toronto, Northwestern, India and here at the IAS). It’s interesting that I’m often thinking about different things each time I hear the same talk. This one is about the generalized Geroch conjecture.

Geroch conjecture: $\mathbb{T}^n$ (the $n$-torus) does not admit a metric of positive scalar curvature.

The conjecture is proved by Schoen and Yau (1979).

Now, scalar curvature can be seen as a limit of volume of balls:

Definition: The scalar curvature of $M$ at $p$ is

$\displaystyle \mbox{Sc}(p) = c_n \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^{n+2}}$

where $\mbox{Vol}_E$ is the Euclidian volume and $c_n$ is a positive constant only depending on the dimension $n$.

Note that since our manifold does not have any cone points,

$\displaystyle \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^n}$

must vanish. Further more, the Riemannian structure on $M$ forces the $r^{n+1}$ term to vanish.

Since for this context we only care about is whether the scalar curvature is larger or smaller than $0$, we can be even more simple-minded: $M$ has positive scalar curvature at $p$ all small enough balls around $p$ has smaller volume than their Euclidean cousins (with a difference of order propositional to $r^{n+2}$). In light of this definition, we have:

Restatement of the Geroch conjecture: For all $g$ on $\mathbb{T}^n$, there exists some point $p$ s.t. $\mbox{Sc}(p) \leq 0$.

This is to say, small enough balls around some point $p$ are not small enough for it to have positive scalar curvature. What if instead we look at balls of a fixed radius instead of those infinitesimal balls? This naturally leads to

Generalized Geroch conjecture: For any $(\mathbb{T}^n, g)$, for all $r$, there exists $p$ s.t. $\mbox{Vol}_g(B(p, r)) \geq \mbox{Vol}_E(B(\bar{0}, r))$.

(For those $r$ larger than the injectivity radius, we lift $M$ to its universal cover so that all homotopically non-trival loops are ‘unfolded’)

Let’s take a look at the $2$-torus to get a feel of the conjecture:

The flat torus, of course, has $0$ $r$-scalar curvature at all points.

For the regular rotational torus, we take the ball around the saddle point of the gradient flow, the ball look like a saddle, as shown below.

To see that this has area larger than the analogous Euclidean ball, we can cut it along radial rays into thin triangles, each triangle can be ‘almost flattened’ to a Euclidean triangle, but we have a more triangles than in the Euclidean case.

What if we try to make the surface spherical for most of the area and having those negative scalar curvature points taking up a very small potion. One of my first attempts would be to connect a few spheres with cylinders:

We have a few parameters here: the number of balls $n$, the width of the connecting cylinders $w$, the length of the connecting cylinders $l$ and the radius of each sphere $R$.

If cylinders are too long (longer than $2r$), then we can just take the ball in the middle of the cylinder, the volume when lifted to universal cover would be equal to Euclidean.

If the width of cylinders are much smaller than $r$, then the ball around a point in the gluing line would have volume almost a full spherical ball plus a half Euclidean ball, which would obviously be larger than a full Euclidean ball.

Hence the more interesting case is to have very short, wide tubes and as a consequence, have many balls forming a loop. In this case, the ‘worst’ ball would be centered at the middle of the tube, it intersects the two spheres connected by the tube in something a bit larger than a spherical half-ball.

I haven’t figured out an estimate yet. i.e. can the advantage taken from the fact that spherical ball are smaller than Euclidean balls cancel out the ‘a bit larger than half’? I think that would be interesting to work out.

Finally, let’s say what does this has to do with systoles:

Theorem: Generalized Geroch conjecture $\Rightarrow$ $\mbox{Sys}(\mathbb{T}^n, g) \leq \frac{2}{\omega_n^{\frac{1}{n}}} \mbox{Vol}_g(\mathbb{T}^n)^{\frac{1}{n}}$ (which is the systolic inequality with a constant better than what we have so far)

Proof: Suppose not,

$\mbox{Sys}(\mathbb{T}^n, g) > 2 (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}$

Let $r = (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}$, by the generalized Geroch conjecture we have some $B(p, r)$ larger than the Euclidean ball. i.e.

$\mbox{Vol}_g(B(p, r))>\omega_n r^n = \omega_n \frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n} = \mbox{Vol}_g(\mathbb{T}^n)$

Since the systole is at least $2r$, hence $B(p, r)$ cannot contain any homotopically non-trival loop i.e. it does not “warp around” and get unfolded when passing to the universal cover. Hence volume of a ball with radius $r$ cannot be larger than the volume of the whole manifold. Contradiction

Systolic inequality on the 2-torus

March 1, 2010

Starting last summer with professor Guth, I’ve been interested in the systolic inequality for Riemannian manifolds. As a starting point of a sequence of short posts I plan to write on little observations I had related to the subject, here I’ll talk about the baby case where we find the lower bound of the systole on the $2$-torus in terms of the area of the torus.

Given a Riemannian manifold $(M, g)$ where $g$ is the Riemannian metric.

Definition: The systole of $M$ is the length of smallest homotopically nontrivial loop in $M$.

We are interested in bounding the systole in terms of the $n$-th root of the volume of the manifold ( where $n$ is the dimension of $M$ ).

Note that the systole is only defined when our manifold has non-trivial fundamental group. I wish to remark that for the case of n-torus, having an inequality of the form $(\mbox{Sys}(\mathbb{T}^n))^n \leq C \cdot \mbox{Vol}(\mathbb{T}^n)$ is intuitive as we can see in the case of an embedded $2$-torus in $\mathbb{R}^3$, we may deform the metric (hence the embedding) to make a non-contactable loop as small as we want while keep the volume constant, however when we attempt to make the smallest such loop large when not changing the volume, we can see that we will run into trouble. Hence it’s expected that there is an upper bound for the length of the smallest loop.

Since if only one loop in some homotopy class achieves that minimal length, we should be able to enlarge it and contract some other loops in that class to enlarge the systole and keep the volume constant. Hence it’s tempting to assume that all loops in the same class are of the same length. In the $2$-torus case, such thing is the flat torus. Since any flat torus has systole proportional to $(\mbox{Vol}(\mathbb{T}^2))^{\frac{1}{2}}$, we have reasons to expect the optimal case fall inside this family. i.e.

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq C \cdot \mbox{Vol}(\mathbb{T}^2)$.

This is indeed the case. The result was given in an early unpublished result by Loewner.

Let’s first optimize in the class of flat torus:

My first guess was that $C$ cannot be made less than $1$ i.e. the torus $\mathbb{R}^2/ \mathbb{Z}^2$ is the optimum case. However, this is not true. Let’s be more careful:

$\mathbb{T}^2 = \mathbb{R}^2 / (0,c)\mathbb{Z} \times (a, b)\mathbb{Z}$

Since by scaling does not change ratio between $(\mbox{Vol}(\mathbb{T}^2))$ and $(\mbox{Sys}(\mathbb{T}^2))^2$, we may normalize and let $c=1$

Let $\alpha, \beta$ be generators of the fundamental group of $\mathbb{T}^2$ length of all geodesic loops in class $[\alpha], \ [\beta]$ are the side lengths of the parallelepiped i.e. $1$ and $||(a,b)||$. W.L.O.G we suppose $a, b > 0$. $\alpha \beta^{-1}$ has length $||((1-a),b)||$ and all geodesics in other classes are at least twice as long as one of the above three.

Hence the systole is maximized when those three are equal, we get $a=1/2, b=\sqrt{3}/2$. The systole in this case is $1$ and the volume is $\sqrt{3}/2$. Hence for any flat torus, we have

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq \frac{2}{\sqrt{3}} \cdot \mbox{Vol}(\mathbb{T}^2)$.

Theorem (Loewner): This bound holds for any metric $g$ on $\mathbb{T}^2$.

Proof: We will show this by reducing the case to flat metric.

$g$ induced an almost complex structure on $\mathbb{T}^2$, on surfaces, any almost complex structure is integrable. Hence there exists $f:\mathbb{T}^2 \rightarrow \mathbb{R}^+$ and $g= f \dot g_0$ where $(\mathbb{T}^2, g_0)$ is a Riemann surface.

By uniformization theorem, $(\mathbb{T}^2, g_0)$ is the quotient of $\mathbb{C}$ by a discrete lattice. i.e. $(\mathbb{T}^2, g_0)$ is a flat torus $\mathbb{R}^2 / (0,c)\mathbb{Z} \times (a, b)\mathbb{Z}$.

By scaling of the torus, we may assume the volume of the manifold is $1$ i.e.

$\displaystyle \int_{\mathbb{T}^2} f \ dV_{g_0} = 1 = \mbox{Vol}(\mathbb{T}^2, g_0)$

Any nontrivial homotopy class of loops on $(\mathbb{T}^2, g)$ can be represented by a straight loop on the flat torus. The length of such a loop in $(\mathbb{T}^2, g)$ is merely integration of $f$ along the segment.

Here we have a family of loops in the homotopy class that is straight, by taking a segment of appropriate length orthogonal to the loops, we have the one-parameter family of parallel loops foliate the torus. Hence integrating over the segment of the length of the loops gives us the total volume of the torus. By Fubini, we have at least one loop is longer than volume of the torus over length of the segment we integrated on, which is the length of the straight loop in the flat torus.

Therefore the systole of $(\mathbb{T}^2, g)$ is smaller than the minimum length of straight loops which is smaller than that of the flat torus. While the volume are the same. Hence it suffice to optimize the ratio in the class of flat tori. Establishes the theorem.

Combining the pervious statement, we get

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq \frac{2}{\sqrt{3}} \cdot \mbox{Vol}(\mathbb{T}^2)$

for any metric on the torus.