Posts Tagged ‘Larry Guth’

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 The result was given in an early unpublished result by Loewner.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.

The sponge problem

June 26, 2009

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

June 20, 2009

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.