aa r X i v : . [ m a t h . S G ] A p r A note on symmetrical symplectic capacities
Kun Shi and Guangcun Lu ∗ April 28, 2020
Abstract
For a convex domain in the standard Euclidean symplectic space which is invariant undera linear anti-symplectic involution τ we show that its Ekeland-Hofer-Zehnder capacity is equalto the τ -symmetrical symplectic capacity of it. Contents
Recall that a real symplectic manifold is a triple (
M, ω, τ ) consisting of a symplectic manifold(
M, ω ) and an anti-symplectic involution τ on ( M, ω ), i.e. τ ∗ ω = − ω and τ = id M . Acanonical example is the standard linear symplectic space ( R n , ω ) with ω = P ni =1 dx i ∧ dy i and with respect to the canonical involution τ given by τ ( x, y ) = ( x, − y ) . Let τ be a linear anti-symplectic involution on ( R n , ω ), and let the convex boundeddomain D ⊂ R n be τ -invariant, i.e., τ D = D . A nonconstant absolutely continuous loop u from [0 , T ] (for some T >
0) to the boundary S of D is called a generalized τ -brake closedcharacteristic on S if it satisfies˙ u ( t ) ∈ J N S ( u ( t )) a.e. on [0 , T ] and u ( T − t ) = τ ( u ( t )) ∀ t ∈ [0 , T ] . Rongrong Jin and the second named author proved in [4] that the symmetrical Ekeland-Hofer-Zehnder symplectic capacity of (
D, ω , τ ) is given by c EHZ ,τ ( D, ω ) = min { A ( u ) > | u is a τ -brake closed characteristic on ∂D } . where A ( u ) is the action of u defined by A ( u ) = 12 Z T h− J ˙ u, u i dt (1.1)with J = (cid:18) − I n I n (cid:19) . Our following result relates c EHZ ,τ ( D, ω ) to the classical Ekeland-Hofer-Zehnder symplectic capacity c EHZ ( K, ω ) of D . ∗ Corresponding authorPartially supported by the NNSF 11271044 of China.2010
Mathematics Subject Classification. heorem 1.1. For any linear anti-symplectic involution τ on ( R n , ω ) and any τ -invariantconvex domain D ⊂ R n there holds c EHZ ( D, ω ) = c EHZ ,τ ( D, ω ) . Clearly, we can assume that D is bounded. By [4, (1.18), (1.22)], c EHZ ,τ ( D, ω ) = c EHZ ,τ (Ψ D, ω ),where Ψ is any symplectic matrix Ψ of 2 n such that Ψ τ = τ Ψ (which always exist). Since c EHZ (Ψ D, ω ) = c EHZ ( D, ω ), we only need to prove the case where τ = τ .Let E ′ = { u ∈ W , ([0 , , R n ) | u (1) = u (0) , A ( u ) = 1 } and E = { u ∈ E ′ | Z u ( t ) dt = 0 } , F ′ = { u ∈ E ′ | u (1 − t ) = τ u ( t ) ∀ t ∈ [0 , } , F = { u ∈ E | u (1 − t ) = τ u ( t ) ∀ t ∈ [0 , } . For an invariant convex bounded domain D ⊂ R n there holds c EHZ ( D, ω ) = min u ∈E Z H ∗ D ( − J ˙ u ) (2.1)([2, 3, 8]). If D is also τ -invariant it was proved in [4, Theorem 1.3] that c EHZ ,τ ( D, ω ) = min u ∈F Z H ∗ D ( − J ˙ u ) . (2.2)It’s worth noting that c EHZ ( D, ω ) = min u ∈E ′ Z H ∗ D ( − J ˙ u ) (2.3)(cf. [6, § u ∈E ′ R H ∗ D ( − J ˙ u ) ≤ min u ∈E R H ∗ D ( − J ˙ u ) because E ⊂ E ′ . For any u ∈ E ′ let u ( t ) = u ( t ) − Z u ( s ) ds, ∀ t ∈ [0 , . Then u ∈ E because A ( u ) = 12 Z h− J ˙ u , u i dt = 12 Z h− J ˙ u, u i dt = 12 Z h− J ˙ u, u i dt = A ( u ) = 1 . Note that R H ∗ D ( − J ˙ u ) = R H ∗ D ( − J ˙ u ). From (2.1) we deduceinf u ∈E ′ Z H ∗ D ( − J ˙ u ) ≥ inf u ∈E Z H ∗ D ( − J ˙ u ) = min u ∈E Z H ∗ D ( − J ˙ u )and thus (2.3).It was claimed in [5, Exercise 12.1.8] (resp. [4, Proposition 1.2]) that any symplecticcapacity (resp. the symmetrical symplectic capacity) on R n with its standard symplecticstructure is continuous on compact convex subsets (resp. the symmetrical compact convexsubsets) of R n with respect to the Hausdorff metric. Moreover, every convex body can beapproximated by convex polytopes in the Hausdorff metric by [7, Theorem 1.8.13]. Hence fromnow on we may assume that D is a convex polytope K . roposition 2.1 ([6, Proposition 3.6]) . Let K ⊂ R n be a convex polytope containing theorigin and with (2 n − -dimensional facets { F i } F K i =1 . Let { n i } K F i =1 be the normal vectors tothe n − -dimensional facets of K , and let p i = J∂g K | F i = h i Jn i . Recall that h i := h K ( n i ) ,where h K ( x ) = sup {h y, x i| y ∈ K } . Let c > be a constant and let u ∈ W , ([0 , , R n ) be aloop that satisfies that for almost every t , there is a non-empty face of K , F j ∩ · · · ∩ F j l = ∅ ,with ˙ u ( t ) ∈ c · conv { p j , · · · , p j l } . Then Z H ∗ K ( − J ˙ u ( t )) dt = c . Let K ⊂ R n be a convex polytope as above. Suppose that it is also τ -invariant below.Then F K is even and we can write the normals to the (2 n − K as { n , · · · , n F K / , − τ n , · · · , − τ n F K / } .By (2.1) and (2.2) it is obvious that c EHZ ( K, ω ) c HZ ,τ ( K, ω ) since F ⊂ E .In order to prove that c EHZ ( K, ω ) = c EHZ ,τ ( K, ω ), by (2.2) we only need to prove thatthere exists a v ∈ F to satisfy Z H ∗ K ( − J ˙ v ) = c EHZ ( K, ω ) . (2.4)By the proof of [6, Theorem 1.5] we may get a u ∈ E ′ such that Z H ∗ K ( − J ˙ u ) = c EHZ ( K, ω ) (2.5)and ˙ u ( t ) ∈ d · { p , · · · , p F K / , − τ p , · · · , − τ p F K / } , a.e.. with d = p c EHZ ( K, ω ). Now let us use this u to construct a v ∈ F satisfying (2.4).Let L := Fix( τ ) = { ( x, y ) ∈ R n | y = 0 } . We claim that there exist t ∈ [0 , ) such that u ( t ) − u ( + t ) ∈ L . In fact, if u (0) − u (1 / ∈ L , we have done. If u (0) − u (1 / / ∈ L ,denote θ ( t ) by the angle between u ( t ) − u ( t + ) and L . Because u (0) − u (1 / / ∈ L , we have θ (0) / ∈ { , π } . Thus when t varies from 0 to , θ varies continuously from θ (0) to θ (0) − π or θ (0) + π . Therefore, there exist t ∈ (0 , ) such that θ ( t ) = 0 or θ ( t ) = π , that is u ( t ) − u ( + t ) ∈ L .Define u = u − u ( t ). Then u ∈ E ′ and Z H ∗ K ( − J ˙ u ) = Z H ∗ K ( − J ˙ u ) = c EHZ ( K, ω )by (2.5). Note that u ( t ) , u ( + t ) ∈ L . Define u ∗ by u ∗ ( t ) = ( u ( t + t ) , t ∈ [0 , − t ) ,u ( t + t − , t ∈ [1 − t , . Then u ∗ ∈ E ′ , and satisfies Z H ∗ K ( − J ˙ u ∗ ) = Z H ∗ K ( − J ˙ u ) = c EHZ ( K, ω ) = d ,u ∗ (0) = u ∗ (1) = u ( t ) , u ∗ ( 12 ) = u ( 12 + t ) , ˙ u ∗ ( t ) ∈ d · { p , · · · , p F K / , − τ p , · · · , − τ p F K / } , a.e.. (2.6)Hence u ∗ (0) , u ∗ ( ) ∈ L . ince A ( u ∗ ) = A ( u ) = A ( u ) = 1, we have either12 Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt >
12 or 12 Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt > . (2.7)Firstly, we assume that the former holds. Since u ∗ ( ) ∈ L , u ′ ( t ) = u ∗ ( t ) , t ∈ [0 ,
12 ) ,τ u ∗ (1 − t ) , t ∈ [ 12 ,
1] (2.8)defines an element in W , ([0 , , R n ) satisfying u ′ (1) = u ′ (0). Put v ( t ) = u ′ ( t ) − Z u ′ ( s ) ds. Since Z u ′ ( s ) ds = Z u ∗ ( s ) ds + τ Z u ∗ ( s ) ds ∈ L , we deduce v (1 − t ) = u ′ (1 − t ) − Z u ′ ( t ) dt = τ u ′ ( t ) − τ Z u ′ ( s ) ds = τ v ( t ) . Moreover, a straightforward computation yields A ( v ) = 12 Z h− J ˙ u ′ ( t ) , u ′ ( t ) − Z u ′ ( s ) ds i dt = 12 Z h− J ˙ u ′ ( t ) , u ′ ( t ) i dt = 12 Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt + 12 Z h Jτ ˙ u ∗ (1 − t ) , τ u ∗ (1 − t ) i dt = 12 Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt + 12 Z h− τ J ˙ u ∗ (1 − t ) , τ u ∗ (1 − t ) i dt = 12 Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt + 12 Z h− τ J ˙ u ∗ (1 − t ) , τ u ∗ (1 − t ) i dt = Z h− J ˙ u ∗ ( t ) , u ∗ ( t ) i dt > . In order to prove that A ( v ) = 1, let w = v/ p A ( v ). Then w ∈ F and satisfies˙ w ( t ) ∈ d/ p A ( v ) · { p , · · · , p F K / , − τ p , · · · , − τ p F K / } a . e . because ˙ v ( t ) = ˙ u ′ ( t ), (2.6) and (2.8). By Proposition 2.1, we get Z H ∗ K ( − J ˙ w ) = d /A ( v ) d = Z H ∗ K ( − J ˙ u ) . Moreover Z H ∗ K ( − J ˙ w ) ≥ min γ ∈F Z H ∗ D ( − J ˙ γ ) ≥ min γ ∈E ′ Z H ∗ D ( − J ˙ γ ) = d . Hence A ( v ) = 1. n summary, we have proved that v ∈ F and Z H ∗ K ( − J ˙ v ) = d = Z H ∗ K ( − J ˙ u ) = c EHZ ( K, ω ) . (2.4) is proved.Next, suppose that the second inequality in (2.7) holds true. We only need to change thedefinition of u ′ in (2.7) as follows: u ′ ( t ) = τ u ∗ (1 − t ) , t ∈ [0 ,
12 ) ,u ∗ ( t ) , t ∈ [ 12 , , and then repeat the above proof to get (2.4) again. References [1] A. Artstein-Avidan and Y. Ostrover, Bounds for Minkowski billiard trajectories in convexbodies,
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