A symplectic embedding of the cube with minimal sections and a question by Schlenk

I prove that the open unit cube can be symplectically embedded into a longer polydisc in such a way that the area of each section satisfies a sharp bound and the complement of each section is path-connected. This answers a variant of a question by F. Schlenk.


The main result
Let n ≥ 2. By q 1 , p 1 , . . . , q n , p n we denote the standard coordinates in R 2n , and we equip R 2n with the standard symplectic form ω 0 := n i=1 dq i ∧dp i . 1 We denote by B m r resp. B m r the open resp. closed ball in R m of radius r around 0. M. Gromov's famous nonsqueezing theorem [Gro85,Corollary,p. 310] implies that B 2n r does not symplectically embed into the closed unit symplectic cylinder B 2 1 × R 2n−2 if r > 1. In [Sch03] F. Schlenk investigated how flexible symplectic embeddings are in the case r ≤ 1. More precisely, for every z ∈ R 2n−2 we define ι z : R 2 → R 2n , ι z (y) := (y, z).
Answering a question by D. McDuff [McD98], in [Sch03, Theorem 1.1] Schlenk proved that for every a > 0 there exists a symplectic embedding ϕ of B 2n 1 into B 2 1 × R 2n−2 , such that for every z ∈ R 2n−2 the section ι −1 z ϕ B 2n 1 has area 2 at most a. Schlenk's lifting method [Sch05, Section 8.4] also shows that for every positive integer k and every a > 1 k there exists a symplectic embedding of the open cube (0, 1) 2n into the open polydisc (0, 1) 2n−1 ×(0, k), whose sections have area at most a. The main result of the present article answers the following two questions: Question 1. Is this statement true with the integer k replaced by a general real number c ≥ 1?
Question 2. Can the bound a on the areas of the sections be made sharp, i.e., equal to 1 c ? 3 I also answer a variant of the following question by Schlenk. For every bounded subset S of R m we define the bounded hull of S to be the union of S and all bounded connected components of R m \ S.
The main result of this article is the following.
This theorem answers Questions 1 and 2 affirmatively. It also provides a negative answer to Schlenk's Question 3 with the word "closure" dropped. It even implies that there exists a symplectic embedding for which the bounded hull of each section has arbitrarily small area: Corollary 5. For every n ≥ 2 and a > 0 there exists a symplectic embedding ψ : B 2n 1 → B 2 1 × R 2n−2 , such that the bounded hull of each section of ψ B 2n 1 has area at most a.
(For a proof see p. 8.) This corollary is optimal in the sense that its statement becomes false if we replace B 2n 1 and B 2 1 by the closed balls B In particular the bounded hull of this section equals B 2 1 , which has area π.
Remark. Let ϕ be as in the statement of Theorem 4. Then each section of the image of ϕ equals its own bounded hull. Hence ϕ is a sharp counterexample to a variant of Question 3 concerning embeddings of cubes. Figure 1. The green arrow depicts the Lagrangian shear p → P := p 1 , cp 1 + p 2 , and the orange arrow the induced shear in the q-plane. The black arrows depict the wrapping maps. The magenta line segment is a P 2 -section of the image of the square under the composed map in the p-plane, where P 2 ∈ R/(cZ). The violet set depicts a Q 2 -section of the image of the open square under the composed map in the q-plane, where Q 2 ∈ R/Z.

}
The bracket } indicates that the product of these two sets is given by the red ribbon on the blue cylinder. The image of this ribbon under some area-preserving map is a section the image of the desired symplectic embedding ϕ. It has area equal to 1 c .
In the case n = 2 the idea of proof of Theorem 4 is to consider the linear symplectic map Ψ : (q, p) → (Q, P ), induced by the Lagrangian shear p → P := p 1 , cp 1 + p 2 . The P 2 -sections of the image of the square (0, 1) 2 under this shear have length at most 1 c . Hence the area of each section of Ψ (0, 1) 4 is at most 1 c . To make the image of Ψ fit in the polydisc (0, 1) 3 × (0, c), we wrap its upper part (in P 2 -direction) back to the lower part, by passing to the quotient R/cZ. We also wrap the Q 1 -coordinate. See Figure 1. Finally, we compose the resulting map with the product of two area preserving embeddings of finite cylinders into rectangles. This yields a symplectic embedding with the desired properties.
Remarks (method of proof, related result, terminology).
• This construction is similar to L. Traynor's symplectic wrapping construction, which she used e.g. to show that certain polydiscs embed into certain cubes, see [Tra95] and [Sch05, Chapter 7]. One difference is that I wrap coordinates of mixed type (Q and P ), whereas Traynor wraps coordinates of pure type. • Schlenk proved a nonsharp result regarding the areas of the bounded hulls of the sections. More precisely, his folding method [Sch05, Section 8.3] can be used to prove that for every n ≥ 2, positive integer k, and ℓ ∈ (0, 1) there exists a symplectic embedding ϕ : (0, ℓ) 2n → (0, 1) 2n−1 × (0, k), such that the bounded hull of every section of ϕ (0, ℓ) 2n has area at most 1 k . Theorem 4 improves this in the following ways: -It treats the critical case ℓ = 1.
-It makes the area estimate sharp.
-It holds for any real number c ≥ 1, not only for an integer c = k. -The proof of Theorem 4 is easier than the folding method.

Proofs of the main result and of Corollary 5
In the proofs of Theorem 4 and Corollary 5 we will use the following lemma.
Lemma 7 (squaring the disc and the cylinder). We denote r := π − 1 2 . (i) There exists a homeomorphism  The idea of proof of this lemma is explained by Figure 2. In the proof of Lemma 7 we will use the following.
Remark 8 (straightening corners). We denote by Σ the square [0, 1] 2 without the corners. Let r > 0 and S be a subset of the circle of radius r consisting of four points. There exists homeomorphism θ : [0, 1] 2 → B 2 r that restricts to a diffeomorphism from Σ onto B 2 r \ S, such that (θ|Σ) * ω 0 extends to a nonvanishing smooth 2-form on B 2 r . To see this, observe that the map is a homeomorphism that restricts to a diffeomorphism from [0, The desired map θ can be constructed from four copies of θ (one for each corner), using charts for B Proof of Lemma 7. To prove (i), we define r := π − 1 2 and choose a map θ as in Remark 8. We define Hence the hypotheses of Proposition 9 are satisfied. We choose a diffeomorphism ϕ as in the statement of this proposition. The map has the required properties.
We prove (ii). There exists a symplectomorphism We choose a symplectomorphism ξ of [0, 1] 2 that equals the identity in a neighbourhood of the boundary and maps κ(0) to y 0 . 5 We obtain such a map as the Hamiltonian flow of a suitable function on (0, 1) 2 with compact support. The map has the required properties. This proves (ii) and completes the proof of Lemma 7.
To prove property (ii), consider the continuous path The point y(0) lies on the boundary of the square [0, 1] 2 . It follows from (2) that the path y lies inside the complement of ι −1 z ϕ (0, 1) 4 in R 2 . Every point outside (0, 1) 2 can be connected to y(0) through a continuous path outside of (0, 1) 2 . Every point in the complement of ι −1 z ϕ (0, 1) 4 in (0, 1) 2 can be connected to a point on the path y through a path in this complement. This follows from (4) and the facts Figure 3. This proves (ii).
Hence ϕ has the desired properties. This proves Theorem 4 in the case n = 2. For n ≥ 3 we take the product of ϕ with the identity map.
In the proof of Corollary 5 we will use the following.
Remark 10 (monotonicity). The bounded hull is monotone in the sense that if A ⊆ B ⊆ R m are bounded sets then the bounded hull of A is contained in the bounded hull of B.
Proof of Corollary 5. We define r := π − 1 2 . By a rescaling argument it suffices to show that for every a ∈ (0, 1] there exists a symplectic embedding ψ : B 2n r → B 2 r × R 2n−2 , such that the bounded hull of each section of ψ(B 2n r ) has area at most a. To prove this statement, we choose ϕ is as in the conclusion of Theorem 4 with c := 1 a . We choose a map κ as in Lemma 7(i). The map ψ := (κ −1 × id) • ϕ • κ × · · · × κ : B 2n r → B 2 r × R 2n−2 is a symplectic embedding. Let z ∈ R 2n−2 . Property (ii) in Theorem 4 implies that the complement of V := κ −1 ι −1 z ϕ (0, 1) 2n in R 2 is path-connected. Hence V equals its bounded hull. The section ι −1 z ψ(B 2n r ) is contained in V . Using Remark 10, it follows that the bounded hull of this section is also contained in V . Using Theorem 4(i) and that κ is area-preserving, it follows that this bounded hull has area at most 1 c = a. Hence ψ has the desired properties. This proves Corollary 5.

Acknowledgments
I would like to thank Felix Schlenk for an interesting discussion and for proof-reading the first version of this article.