The gap between near commutativity and almost commutativity in symplectic category

On any symplectic manifold of dimension greater than 2, we construct a pair of smooth functions, such that on the one hand, the uniform norm of their Poisson bracket equals to 1, but on the other hand, this pair cannot be reasonably approximated (in the uniform norm) by a pair of Poisson commuting smooth functions. This comes in contrast with the dimension 2 case, where by a partial case of a result of Zapolsky, an opposite statement holds.


Introduction and results
During the last ten years, function theory on symplectic manifolds has attracted a great deal of attention [B, BEP, CV, EP-1, EP-2, EP-3, EPR, EPZ, P, Z-1, Z-2].The C 0 -rigidity of the Poisson bracket [EP-2] (cf.[B]) is one of the achievements of this theory, and it states that on a closed symplectic manifold (M, ω), the uniform norm of the Poisson bracket of a pair of smooth functions on M is a lower semi-continuous functional, when we consider the uniform (or the C 0 ) topology on C ∞ (M) × C ∞ (M).Informally speaking, this means that one cannot significantly reduce the C 0 norm of the Poisson bracket of two smooth functions by an arbitrarily small C 0 perturbation.The C 0 -rigidity of the Poisson bracket holds also when the symplectic manifold (M, ω) is open, if we restrict to smooth compactly supported functions.In this context it is natural to ask, how strongly one should perturb a given pair of smooth functions in order to significantly reduce the C 0 norm of their Poisson bracket.The following question was asked privately by Polterovich in 2009 (later it also appeared in [Z-2]): Question 1.1.Let (M, ω) be a closed symplectic manifold.Does there exist a constant C > 0, such that for any pair of smooth functions f, g : M → R satisfying {f, g} = 1, there exists a pair of smooth functions F, G : M → R, such that F −f , G−g C and such that {F, G} = 0 on M?
The 2-dimensional case of this question was answered affirmatively by Zapolsky, and in fact, it appeared as a particular case of a more general statement which applies to functions on a manifold with a volume form [Z-2].
The main result of this note is Theorem 1.2.Let (M, ω) be a symplectic manifold of dimension 2n > 2. Then for any 0 < p < q there exists a pair of smooth compactly supported functions f, g : M → R, such that f = g = 1, {f, g} = q, and such that for any s ∈ [0, q] we have Here ρ f,g (s) is the profile function [BEP], which is defined as Now let (M, ω) be a symplectic manifold of dimension 2n > 2. Note that for a pair f, g : M → R of smooth compactly supported functions, the value ρ f,g (0) is precisely the uniform distance from the pair f, g of functions to the set of pairs F, G : M → R of smooth compactly supported Poisson commuting functions.Hence, if we take any C > 0, and set q = 1/(64C 2 ), then Theorem 1.2 implies the existence of a pair f, g : M → R of smooth compactly supported functions, such that {f, g} = q = 1/(64C 2 ) and ρ f,g (0) = 1/2, and hence for the functions f 1 = 8Cf , g 1 = 8Cg we first of all get {f 1 , g 1 } = 1, and moreover we get ρ f 1 ,g 1 (0) = 4C, which implies that there does not exist a pair F, G : M → R of smooth compactly supported Poisson commuting functions, such that F − f 1 , G − g 1 C. Thus, we obtain Corollary 1.3.The answer to Question 1.1 is negative in dimension > 2.
The proof of Theorem 1.2 relies on a certain computation of the pb 4 invariant, which is carried out similarly as an analogous computation in the proof of Proposition 1.21 in [BEP].Some part of the proof of Theorem 1.2 is reminiscent of the proof of Theorem 1.4 (ii) in [BEP].
Remark 1.4.Question 1.1 has an analogue in the setting of Hermitian matrices equipped with a commutator (see [PeS, H]).

Proofs
Let us first remind the definition of the pb 4 invariant which was initially introduced in [BEP] and which participates in the proof of Theorem 1.2.
where the infimum is taken over the set of all pairs of smooth compactly supported functions F, G : and G 1 on Y 1 .Note that such set of pairs of functions is non-empty whenever The following result was proved in [BEP] (in [BEP] this is Proposition 1.21): Proposition 2.2.Let (M, ω) be a symplectic surface of area B > 0. Consider a curvilinear quadrilateral Π ⊂ M of area A > 0 with sides denoted in the cyclic order by a 1 , a 2 , a 3 , a 4 -that is Π is a topological disc bounded by the union of four smooth embedded curves a 1 , a 2 , a 3 , a 4 connecting four distinct points in M in the cyclic order as listed here and (transversally) intersecting each other only at their common endpoints.Let L be an exact section of T * S 1 .Assume that M = S 2 and that 2A B. Then in the symplectic manifold M × T * S 1 (with the split symplectic form) we have The proof of Proposition 2.2, which is presented in [BEP], is divided into two parts.The first part proves the inequality pb 4 (a 1 × L, a 3 × L, a 2 × L, a 4 × L) 1/A by providing a concrete example of a pair of functions F, G as in Definition 2.1.The second part proves the inequality pb 4 (a 1 × L, a 3 × L, a 2 × L, a 4 × L) 1/A, and it uses the Gromov's theory of pseudo-holomorphic curves.Now we turn to the proofs of our results.
Then given any open set U ⊆ R 2n and any A > 0, there exists a smooth symplectic embedding (B 2 (A), σ std ) → (U, ω std ), where B 2 (A) ⊂ R 2 is the disc of area A centred at the origin, and σ std is the standard area form on R 2 (and on B 2 (A)).
Proof.It is enough to construct a smooth embedding of B 2 (π) into B 2n (2π) (where B 2n (2π) is the 2n-dimensional ball of capacity 2π, or equivalently, radius √ 2, centred at the origin), such that its image is a symplectic curve having arbitrarily large symplectic area.An example of such an embedding is the map u : B 2 (π) → C n given by u(z) = (z k , z, 0, 0, ..., 0), where k ∈ N is large enough.
Lemma 2.4.Assume that we have a curvilinear quadrilateral of area A > 0 on the plane R 2 , with sides a 1 , a 2 , a 3 , a 4 (written in the cyclic order), and let L ⊂ T * T n−1 be an exact section.Then in the symplectic manifold R 2 × T * T n−1 (with the split symplectic form) we have Proof.The proof goes similarly as the proof of Proposition 2.2 (which is Proposition 1.21 in [BEP]).
Consider functions f1 , g1 , as in Claim 2.5.Choose 0 < ρ 1 < ρ < ρ 2 such that πρ 2 2 < ǫ ′ , and choose a smooth function h1 : B 2 (ǫ ′ ) → R such that its support lies inside the g 1 (z, w 1 , ..., w n−1 ) = g1 (z) h1 (w 1 )... h1 (w n−1 ), for (z, w 1 , ..., w n−1 ) ∈ W , and then define f, g : for any (z, w 1 , ..., w n−1 ) ∈ W , and hence {f, g} = q.Also we have that f, g 0 on M, and that f = g = 1.Let us show that the constructed functions f, g satisfy (1.1).For showing the upper bound, choose a smooth compactly supported function h : M → [0, 1] such that h = 1 on the union of supports supp(f ) ∪ supp(g), and define functions F, G : M → R by F = ( 1 2 − s 2q + s q f )h, G = g (the upper bound in (1.1) is essentially the statement of Theorem 1.4 (ii), inequality (8) in [BEP]).Let us show the lower bound, for a given value of s.Since the profile function ρ f,g is non-negative and non-increasing, it is sufficient to prove the lower bound for s ∈ (0, p).Take s ∈ (0, p) and denote t = 1 2 − 1 2p s > 0. Now assume that we have a pair of smooth compactly supported functions ⊂ W , then the supports of F ′ , G ′ lie inside φ(W ′ ), so in particular the supports of φ * F ′ , φ * G ′ lie inside W ′ , and moreover we have that φ ) n−1 is the (n − 1)-dimensional torus), given as a product of maps, such that at the factor B 2 (2l 2 + 2ǫ) we have the standard embedding into R 2 (given by the identity map), and such that at each factor D 2 ρ 1 ,ρ 2 , the circle S 1 ρ is mapped onto the zero section of T * S 1 .Denote the push-forwards which are a priori defined on ψ(W ′ ), and extend them by 0 to obtain functions on the whole R 2 × T * T n−1 .We have F 2 = 0 on a 1 × L 0 , F 2 1 − (2 + δ)α on a 3 × L 0 , G 2 = 0 on a 2 × L 0 , and G 2 1 − (2 + δ)β on a 4 × L 0 , where L 0 ⊂ T * T n−1 is the zero section.Therefore in view of Lemma 2.4, we get Since we can choose δ > 0 to be arbitrarily small, we in fact get {F, G} (1 − 2t)p = s.
Thus we have shown that for any F, G ∈ C ∞ c (M) with F − f + G − g t we have {F, G} s.This immediately implies ρ f,g (s) t = 1 2 − 1 2p s.