On C^1 closing lemma

May 25, 2010

Let f: M \rightarrow M be a diffeomorphism. A point p is non-wandering if for all neighborhood U of p, there is increasing sequence (n_k) \subseteq \mathbb{N} where U \cap f^{n_k}(U) \neq \phi. We write p \in \mathcal{NW}(f).

Closing lemma: For any diffeomorphism f: M \rightarrow M, for any p \in \mathcal{NW}(f). For all \varepsilon>0 there exists diffeomorphism g s.t. ||f-g||_{C^1} < \varepsilon and g^N(p) = p for some N \in \mathbb{N}.

Suppose p \in \mathcal{NW}(f), \overline{\mathcal{O}(p)} is compact, then for any \varepsilon>0, there exists x_0 \in B(p, \varepsilon), k \in \mathbb{N} s.t. f^k(x) \in B(p, \varepsilon).

First we apply a selection process to pick an appropriate almost-orbit for the closing. Set x_i = f^i(x_0), \ 0 \leq i \leq k.

If there exists 0 < j < k where

\min \{ d(x_0, x_j), d(x_j, x_k) \} < \sqrt{\frac{2}{3}}d(x_0, x_k)

then we replace the origional finite sequence by (x_0, x_1, \cdots, x_j) or (x_j, \cdots, x_k). Iterate the above process. since the sequence is at least one term shorter after each shortening, the process stops in finite time. We obtain final sequence (p_0, \cdots, p_n) s.t. for all 0 < i < n,

\min \{ d(p_0, p_i), d(p_i, p_n) \} \geq \sqrt{\frac{2}{3}}d(p_0, p_n).

Since the process is applied at most k times, x_0, x_k \in B(p, \varepsilon), after the first shortening, d(p, x_{i_1}) \leq \max \{d(p, x_0), d(p, x_k) \} + \sqrt{\frac{2}{3}}d(x_0, x_k) \leq \varepsilon +  2 \sqrt{\frac{2}{3}} \varepsilon.

i.e. both initial and final term of the sequence is at most (\frac{1}{2}+ \sqrt{\frac{2}{3}}) 2 \varepsilon. Along the same line, we have, at the i-th shortening, the distance between the initial and final sequence and p is at most (\frac{1}{2} + \sqrt{\frac{2}{3}} + (\sqrt{\frac{2}{3}})^2 + \cdots (\sqrt{\frac{2}{3}})^i) 2 \varepsilon. Hence for the final sequence p_0, p_n \in B(p, 1+2 \sqrt{\frac{2}{3}}/(1-\sqrt{\frac{2}{3}}) \varepsilon) \subseteq B(p, 10 \varepsilon).

There is a rectangle R \subseteq M where p_0, p_n \in \sqrt{\frac{3}{4}}R
(i.e. shrunk R by a factor of \sqrt{\frac{3}{4}} w.r.t. the center) and for all 0 < i < n, \ p_i \notin R.

Next, we perturb f in R i.e. find h: M \rightarrow M with ||h||_{C^1} < \delta and h|_{M \backslash R} = id. Hence ||h \circ f - f ||_{C^1} < \delta.

Suppose R = I_1 \times I_2; L_1, L_2 are the lengths of I_1, I_2, L_1 < L_2.
By main value theorem, for all x \in M, \ d(x, h(x)) < \delta L_1.
On the other hand, since p_0 \in \sqrt{\frac{3}{4}}R, it's at least \frac{1}{2}(1-\sqrt{\frac{3}{4}})L_1 away from the boundary of R. i.e. there exists bump function h satisfying the above condition and d(p_0, h(p_0)) > \frac{\delta}{8}(1-\sqrt{\frac{3}{4}})L_1.

Hence in order to move a point by a distance L_1, we need about 1/ \delta such bump functions, to move a distance L_2, we need about \frac{L_2}{\delta L_1} bumps.

For simplicity, we now suppose M is a surface. By starting with an \varepsilon (and hence R) very small, we have for all 0 \leq i \leq N+M, \ f^i(R) is contained in a small neighbourhood of p_i. Hence on f^i(B), f^i is C^1 close to the linear map p_i + Df^i(p_0)(x-p_0). Hence mod some details we may reduce to the case where f is linear in a neighborhood of \mathcal{O}(p_0).

By choosing appropiate coordinate system in R, we can have f preserving the horizontal and vertical foliations and the horizontal vectors eventually grow more rapidly than the vertical vectors.

It turns out to be possible to choose R to be long and thin such that for all i \leq 40 / \delta, f^i(R) has height greater than width. (note that M = \lfloor 40/ \delta \rfloor bumps will be able to move the point by a distance equal to the width of the original rectangle R. Since horizontal vectors eventually grow more rapidly than the vertical vectors, there exists N s.t. for all N \leq i \leq N+M, f^i(R) has width greater than its height.
For small enough \epsilon, the boxes f^i(R) are disjoint for 0 \leq i \leq N+40/ \delta. Construct h to be identity outside of

\displaystyle \bigsqcup_{i=0}^M f^i(R) \sqcup \bigsqcup_{i=N}^{N + M} f^i(R)

For the first M boxes, we let h preserve the horizontal foliation and move along the width so that g = h \circ f has the property that g^M(p_n) lies on the same vertical fiber as f^M(p_0).

On the boxes f^{N+i}(R), \ 0 \leq i \leq M, we let h pushes along the vertical direction so that

g^{N+M}(p_n) = f^{N+M}(p_0)

Since iterates of the rectangle are disjoint, for N+M \leq i \leq n, \ h(p_i) = p_i, g(p_i) = f(p_i).

Hence g^n(p_n) = g^{n-(N+M)} \circ g^{N+M}(p_n) = g^{n-(N+M)} f^{N+M}(p_0) = g^{n-(N+M)} (p_{N+M}) = p_n.

Therefore we have obtained a periodic point p_n of g.

Since p_n \in B(p, 10 \varepsilon), we may further perturb g to move p_n to p. This takes care of the linear case on surfaces.

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