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.

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