Symplectic duality between complex domains
aa r X i v : . [ m a t h . S G ] M a r Symplectic duality between complex domains
Antonio J. Di Scala, Andrea Loi, Fabio Zuddas
Abstract
In this paper after extending the definition of symplectic duality (given in [3] for bounded symmetric domains )to arbitrary complex domains of C n centered at the origin we generalize some of the results proved in [3] and [4] tothose domains. Keywords : K¨ahler metric; bounded domain; symplectic duality; Hermitian symmetric space.
Subj.Class : 53D05, 58F06.
Let ( C H n , ω hyp ) be the n -dimensional complex hyperbolic space, namely the unit ball in C n equipped with the K¨ahlerform ω hyp = − i ∂ ¯ ∂ log(1 − n X j =1 | z j | ) (1)whose associated K¨ahler metric is the hyperbolic metric g hyp . It is well-known that ( C H n , ω hyp ) is globally sym-plectomorphic to ( C n , ω ) where ω = i P nj =1 dz j ∧ d ¯ z j is the standard symplectic form on C n = R n . An explicitdiffeomorphism Ψ hyp : C H n → C n satisfying Ψ ∗ hyp ω = ω hyp (2)is given by: Ψ hyp ( z ) = z p − | z | , (3)where z = ( z , . . . , z n ) and | z | = P nk =1 | z k | . A simple computation shows that the map Ψ hyp enjoys the followingadditional property: Ψ ∗ hyp ω F S = ω , (4)where we regard C n as the affine chart Z = 0 of the n -dimensional complex projective space C P n endowed withhomogeneous coordinates Z , . . . , Z n and ω F S = i ∂ ¯ ∂ log(1 + n X j =1 | z j | ) , z j = Z j Z is the restriction to C n ⊂ C P n of the Fubini-Study form of C P n .Properties (2) and (4) have been recently extended in [3] by the first two authors to all bounded symmetricdomains M ⊂ C n as expressed by the following theorem. Before stating it we recall that to each bounded symmetricdomain M ⊂ C n endowed with the hyperbolic form ω (which, in the irreducible case, is a suitable normalization ofthe Bergman form) one can associate its compact dual M ∗ equipped with the K¨ahler form ω ∗ which is given by thepull-back of the Fubini–Study form of C P N via the Borel–Weil embedding BW : M ∗ → C P N , i.e. BW ∗ ω F S = ω ∗ .Observe that M ∗ can be obtained by a suitable compactification of C n and the inclusion C n ⊂ M ∗ is often referredas the Borel embedding . Notice also that in the case M = C H n , M ∗ = C P n , the Borel embedding C n ⊂ C P n is theinclusion of the affine chart Z = 0 in C P n and the Borel–Weil embedding BW : C P n → C P n is the identity map of C P n . heorem 1.1 (Di Scala–Loi [3]) Let M ⊂ C n be a bounded symmetric domain endowed with the hyperbolic form ω .Then there exists a symplectic duality, namely a real analytic diffeomorphism Ψ : M → C n sending the origin to theorigin and such that: Ψ ∗ ω = ω, (5)Ψ ∗ ω ∗ = ω , (6) where ω is (the restriction to M of ) the flat K¨ahler form ω = i ∂ ¯ ∂ | z | = i n X j =1 dz j ∧ d ¯ z j on C n and where we are denoting by ω ∗ the restriction of ω ∗ to C n via the Borel embedding C n ⊂ M ∗ . Moreoverif T ⊂ M is a complex and totally geodesic submanifold of M of dimension k then Ψ( T ) = C k , i.e. the map Ψ takes complex and totally geodesic submanifolds through the origin of M to complex and totally geodesic submanifoldsthrough the origin of C n (the latter being equipped with the flat metric). In order to study to which extents the map Ψ is unique one needs to understand the set of real analytic maps B : M → M satisfying B ∗ ω = ω and B ∗ ω = ω . In [4], the set of these maps is called the bisymplectomorphismgroup of the bounded symmetric domain M and is denoted by B ( M ). The main result about this group is Theorem4 in [4]. In the case of C H n this theorem implies the following: Theorem 1.2 (Di Scala–Loi–Roos [4]) Let
Ψ : C H n → C n be a symplectic duality. Then Ψ( z ) = e ig ( z ) Ψ hyp ( z ) Az, (7) where g is an arbitrary smooth complex valued function on C H n depending only on | z | , A ∈ U ( n ) and Ψ hyp is givenby (2) above. The key ingredient in the proof of the previous theorems is that the dual K¨ahler form ω ∗ on C n can be obtainedby the hyperbolic form ω on M in the following way (see [2] and [10] for details). Since the K¨ahler form ω is realanalytic and M is contractible one can find a globally defined real analytic K¨ahler potential Φ : M → R for ω around the origin. The potential Φ can be expanded around the origin as a convergent power series of the variables z = ( z , . . . , z n ) and ¯ z = (¯ z , . . . , ¯ z n ), where z is the restriction to M of the Euclidean coordinates of C n . By thechange of variables ¯ z
7→ − ¯ z in this power series one gets a new power series which is convergent to a global definedand real valued function of C n , denoted by Φ( z, − ¯ z ). It turns out that Φ ∗ ( z, ¯ z ) = − Φ( z, − ¯ z ) is a strictly PSH functionof C n and, moreover, ω ∗ = i ∂ ¯ ∂ Φ ∗ ( z, ¯ z ).The aim of this paper is to address the problem of extending the previous procedure to an arbitrary n -dimensionalcomplex domain M ⊂ C n (open, bounded or unbounded connected subset of C n ) containing the origin 0 ∈ C n .Therefore, we assume that there exists a real analytic strictly PSH function Φ : M → R on M such that the functionΦ ∗ ( z, ¯ z ) = − Φ( z, − ¯ z ) is a real valued and strictly PSH function on an open domain M ∗ ⊂ C n containing the origin.The pair ( M ∗ , Φ ∗ ) is what we call in this paper a local dual of ( M, Φ). Notice that a local dual is not unique, indeedany neighbourhood of the origin contained in M ∗ is again a dual of ( M, Φ). Observe also that a dual does not existin general as shown by the following example.
Example 1.3
Consider the two potentials Φ hyp = − log(1 − | z | ) and Φ = Φ hyp + z + ¯ z for ω hyp on the unit disk C H ⊂ C . Then Φ does not admit a local dual. Indeed the function Φ ∗ = log(1 + | z | ) − z + ¯ z is not a real valuedfunction in any neighbourhood of the origin of C .Notice that the previous example also shows that the definition of local duality cannot be extended to the case ofK¨ahler forms. Indeed the same K¨ahler form can have two different potentials one admitting a (local) dual and theother not. Therefore when we speak of local dual of a K¨ahler form ω we always assume to have fixed a K¨ahlerpotential for it.Once we have defined a local dual ( M ∗ , Φ ∗ ) of ( M, Φ), we study the analogues of Theorem 1.1 and Theorem 1.2for the K¨ahler forms ω = i ∂ ¯ ∂ Φ and ω ∗ = i ∂ ¯ ∂ Φ ∗ . More precisely, we say that there exists a (local) λ -symplectic uality between ω and ω ∗ if there exist open neighbourhoods of the origin, say U ⊂ M and U ∗ ⊂ M ∗ , a positiveconstant λ and a diffeomorphism Ψ : U → U ∗ such thatΨ ∗ ω = λω, (8)Ψ ∗ λω ∗ = ω , (9)where ω is the flat K¨ahler form of C n . If λ = 1 we simply speak of symplectic duality instead of 1-symplectic duality.Therefore the existence of a λ -symplectic duality between ω and ω ∗ is equivalent to that of a local symplectic dualitybetween λω and λω ∗ (notice that we are not assuming Ψ(0) = 0).The presence of the constant λ in the previous equations is due to the fact that we want to include in our definitionalso those symplectic forms which do not admit a symplectic duality but for which there exists a λ -symplectic dualityas shown in the following simple example. Example 1.4
Let µ be a positive constant, µ = 1 and let Φ = µ Φ hyp , with Φ hyp as in the previous example. Thenthe dual of Φ is Φ ∗ = µ log(1 + | z | ) (defined on C ). Then it is not hard to see that there exists a λ -symplecticduality between ( C H , Φ hyp ) and ( C , Φ ∗ ) if and only if λµ = 1. Therefore, even if it does not exist a symplecticduality between ω = i ∂ ¯ ∂ Φ and ω ∗ = i ∂ ¯ ∂ Φ ∗ there exists a λ -symplectic duality (with λ = µ ) between them, given,for example, by the map (3) (with n = 1). Assumption
Throughout all this paper, to avoid triviality, we will assume that the form ω is not proportional to ω .This means it cannot exist any open subset of M and a real number c such that ω = cω on this open set. In factin this case ω ∗ = ω and the existence of a λ -symplectic duality is equivalent to a single equation Ψ ∗ ω = ω whichis easily solved by taking Ψ = Id. It is worth pointing out that by Darboux’s theorem each of the equations (8) and(9) can be separately solved (locally). With the assumption of non proportionality a λ -symplectic duality Ψ turnsout to be a simultaneous symplectomorphism with respect to different symplectic structures, namely λω and ω on U and ω and λω ∗ on U ∗ . This phenomenon could be of some interest from the physical point of view. Indeed,roughly speaking, it is telling us that the Darboux’s coordinates for λω are “the inverse” of those of λω ∗ . Moreoverthe existence of a λ -symplectic duality could give strong restrictions on the curvature of the K¨ahler metric ω (cf.Section 4 below).A very interesting case we consider in this paper is that of rotation invariant potentials, and, in particular,radial potentials, namely those Φ : M → R which depend only on | z | , . . . , | z n | and, in the radial case, on r = | z | + · · · + | z n | . Many interesting and important examples of K¨ahler forms on complex domains are rotationinvariant, since they often arise from solutions of ordinary differential equations on the variable r (cf. e.g. [1] and[11]). In the rotation invariant case it is easy to see that the local dual ( M ∗ , Φ ∗ ) of ( M, Φ) can be defined (namelyΦ ∗ is real valued and strictly PSH in a suitable neighborhood M ∗ of the origin) and Φ ∗ is rotation invariant.The main result of the present paper about λ -symplectic duality in the rotation invariant case is the followingtheorem which provides necessary and sufficient conditions for the existence of a special λ -symplectic duality solely interms of the potential Φ (see the beginning of next section for the definition of special map and for the terms involvedin the statement of the theorem). Theorem 1.5
Let M ⊂ C n be a complex domain containing the origin endowed with a rotation invariant K¨ahlerpotential Φ . Let Φ ∗ be the dual defined on M ∗ There exists a special λ -symplectic duality Ψ : U → U ∗ between ω = i ∂ ¯ ∂ Φ and ω ∗ = i ∂ ¯ ∂ Φ ∗ (where U ⊂ C n and U ∗ ⊂ C n are open subsets centered at the origin) if and only if thefollowing equations are satisfied: λ ∂ ˜Φ ∂x k ( x , . . . , x n ) · ∂ ˜Φ ∂x k − λ ∂ ˜Φ ∂x x , . . . , − λ ∂ ˜Φ ∂x n x n ! = 1 , k = 1 , . . . , n, (10) on an open neighbourhood of the origin of R n contained in ˜ M . Here ˜Φ (resp. ˜ M ) is the function (resp. the domain)associated to Φ (resp. M ) . Moreover Ψ is uniquely determined by Φ and it is rotation invariant. The authors believe it is an interesting and very challenging problem to classify all the λ -symplectic dualities Ψ inthe rotation invariant case without assuming that Ψ is special.In the radial case we have a complete classification of λ -symplectic dualities as expressed by the following theoremwhich can be considered a generalization of Theorem 1.2 above to all radial domains in C n centered at the origin. heorem 1.6 Let M ⊂ C n be a complex domain containing the origin endowed with a radial K¨ahler potential Φ .Let Ψ : U → U ∗ be a λ -symplectic duality between ω = i ∂ ¯ ∂ Φ and ω ∗ = i ∂ ¯ ∂ Φ ∗ . Then there exist an open subset V ⊂ U , containing the origin, a radial function g : V → R and a unitary n × n matrix A ∈ U ( n ) such that Ψ( z ) = e ig ( z ) ψ ( z ) A ( z ) , z ∈ V, (11) where ψ : V → R is the radial and real-analytic function on V given by ψ ( z ) = ( λf ′ ( x )) , x = | z | = | z | + · · · + | z n | (12) and where f : ˆ M → R is the function associated to Φ and ˆ M is the domain associated to M (see Section 3).Consequently there exists a λ -symplectic duality between ω and ω ∗ if and only if λ f ′ ( x ) f ′ ( − λxf ′ ( x )) = 1 , (13) on an open neighbourhood of the origin of R contained in ˆ M . The paper is organized as follows. The next two sections (Section 2 and Section 3) are dedicated to the proofs ofTheorem 1.5 and Theorem 1.6 respectively, In Section 4 we describe some applications and examples of our results.The paper ends with an appendix containing a technical lemma which is a key ingredient in the proof of our theorems.This lemma is indeed a simple corollary of the results developed in [11] for special symplectic maps. We have includedit here to make this paper self-contained as much as possible.
Let M ⊂ C n be a complex domain containing the origin and let Φ be a rotation invariant K¨ahler potential. Thismeans that there exists ˜Φ : ˜ M → R , defined on the open subset ˜ M ⊂ R n given by˜ M = { x = ( x , . . . , x n ) ∈ R n | x j = | z j | , z = ( z , . . . , z n ) ∈ M } (14)such that Φ( z ) = ˜Φ( x ). The function ˜Φ (resp. ˜ M ) will be called the function (resp. the domain) associated to Φ (resp. M ) . A real analytic map (not necessarily a diffeomorphism) Ψ : C → S : z = ( z , ..., z n ) (Ψ ( z ) , . . . , Ψ n ( z )) , between two complex domains C ⊆ C n and S ⊆ C n containing the origin is said to be special if Ψ j ( z ) = ψ j ( z ) z j , j =1 , . . . , n where ψ j , j = 1 , . . . , n , are real valued functions defined on C . We say that a a special map Ψ : C → S : z (Ψ ( z ) = ψ ( z ) z , . . . , Ψ n ( z ) = ψ n ( z ) z n ) is rotation invariant if there exist real valued functions ˜ ψ j : ˜ C R , whichwe call the functions associated to Ψ, such that ˜ ψ j ( x ) = ψ j ( z ) for x = ( x , . . . , x n ) ∈ ˜ C , x j = | z j | .We now prove Theorem 1.5. Proof of Theorem 1.5:
We start by proving the last part of the theorem, namely that a λ -symplectic duality which isspecial is necessarily rotation invariant. Actually we will show it for the special maps satisfying only the first equation(8) defining a λ -symplectic duality, namely Ψ ∗ ω = λω . We can assume λ = 1, namely Ψ ∗ ω = ω . In fact the proofextends easily to arbitrary λ . Notice that ω = i P nk,l =1 (cid:16) ∂ ˜Φ ∂x k ∂x l ¯ z l z k + ∂ ˜Φ ∂x k δ kl (cid:17) dz l ∧ d ¯ z k , where, with a slight abuseof notation, we are omitting the fact that the previous expression has to be evaluated at x = | z | , . . . , x n = | z n | .Hence equation Ψ ∗ ω = ω reads n X j =1 d Ψ j ∧ d ¯Ψ j = n X k,l =1 ∂ ˜Φ ∂x k ∂x l ¯ z l z k + ∂ ˜Φ ∂x k δ kl ! dz l ∧ d ¯ z k . (15)By comparing the (1 , , ,
2) components of the right-hand side and the left-hand side in this equality weget, for every k, m = 1 , . . . , n , n X j =1 ∂ Ψ j ∂z k ∂ ¯Ψ j ∂z m = n X j =1 ∂ Ψ j ∂z m ∂ ¯Ψ j ∂z k (16)and X j =1 (cid:20) ∂ Ψ j ∂z k ∂ ¯Ψ j ∂ ¯ z m − ∂ ¯Ψ j ∂z k ∂ Ψ j ∂ ¯ z m (cid:21) = ∂ ˜Φ ∂x k ∂x m ¯ z k z m + ∂ ˜Φ ∂x m δ km . (17)By inserting Ψ j = ψ j z j (and ¯Ψ j = ψ j ¯ z j ) into equations (16) and (17) we get respectively ∂ψ k ∂z m ψ k ¯ z k + n X j =1 ∂ψ j ∂z k ∂ψ j ∂z m | z j | = ∂ψ m ∂z k ψ m ¯ z m + n X j =1 ∂ψ j ∂z m ∂ψ j ∂z k | z j | (18) ∂ψ m ∂z k ψ m z m + ∂ψ k ∂ ¯ z m ψ k ¯ z k + ψ k δ km = ∂ ˜Φ ∂x k ∂x m ¯ z k z m + ∂ ˜Φ ∂x m δ km (19)which can be rewritten as 12 ∂ψ k ∂z m ¯ z k + n X j =1 ∂ψ j ∂z k ∂ψ j ∂z m | z j | = 12 ∂ψ m ∂z k ¯ z m + n X j =1 ∂ψ j ∂z m ∂ψ j ∂z k | z j | (20)and 12 ∂ψ m ∂z k z m + 12 ∂ψ k ∂ ¯ z m ¯ z k + ψ k δ km = ∂ ˜Φ ∂x k ∂x m ¯ z k z m + ∂ ˜Φ ∂x m δ km . (21)If we distinguish in equation (21) the cases m = k and m = k we get respectively Re (cid:18) ∂ψ k ∂z k z k (cid:19) = G k − ψ k k = 1 , . . . , n (22)12 ∂ψ m ∂z k z m + 12 ∂ψ k ∂ ¯ z m ¯ z k = ∂ ˜Φ ∂x k ∂x m ¯ z k z m (23)where G k = ∂ ˜Φ ∂x k | z k | + ∂ ˜Φ ∂x k is a rotation invariant function.Equation (20) implies that ∂ψ m ∂z k ¯ z m is symmetric in k, m . So if we multiply equation (23) by ¯ z m , assume z k = 0and divide by ¯ z k we can rewrite it as Re (cid:18) ∂ψ k ∂z m z m (cid:19) = H km ( k = m ) . (24)Up to changing the order of variables, we can assume k = 1. Let us set ψ = F . Equations (22) and (24) can bewritten then as Re (cid:18) ∂F∂z z (cid:19) = G − F (25) Re (cid:18) ∂F∂z m z m (cid:19) = H m ( m = 1) , (26)where we have set G = G and H m = H m . So we need to show that the real analytic function F is rotation invariant( F is real analytic since by definition a special map is real analytic). We will prove that ∂ i + ... + i n + j + ... + j n F∂z i . . . ∂z i n n ∂ ¯ z j . . . ∂ ¯ z j n n (0) = 0whenever ( i , . . . , i n ) = ( j , . . . , j n ). Let us assume first that i k = j k , where k = 1. Without loss of generality we canassume that i k > j k (otherwise we conjugate the derivative). Notice that equation (25) can be rewritten as F = G − ∂F∂z z − ∂F∂ ¯ z ¯ z . (27)Since G is rotation invariant, we get i k F∂z i k k = ∂ i k G∂x i k k ¯ z i k k − ∂ i k +1 F∂z ∂z i k k z − ∂ i k +1 F∂ ¯ z ∂z i k k ¯ z . (28)and then, since j k < i k ∂ i k + j k F∂ ¯ z j k k ∂z i k k = R ¯ z k − ∂ i k + j k +1 F∂z ∂ ¯ z j k k ∂z i k k z − ∂ i k + j k +1 F∂ ¯ z ∂ ¯ z j k k ∂z i k k ¯ z . (29)for some function R . By deriving equation (29) with respect to variables different from z , ¯ z , z k , ¯ z k , it is clear thatthe right-hand side writes as a sum of the kind A ¯ z k + Bz + C ¯ z and then vanishes when evaluated in z = 0. On theother hand, if we derive the equation with respect to z (the case ¯ z is analogous), then the right-hand side of (29)becomes ∂R∂z ¯ z k − ∂ i k + j k +1 F∂z ∂ ¯ z j k k ∂z i k k − ∂ i k + j k +2 F∂z ∂ ¯ z j k k ∂z i k k z − ∂ i k + j k +2 F∂z ∂ ¯ z ∂ ¯ z j k k ∂z i k k ¯ z . (30)so that equation rewrites as32 ∂ i k + j k +1 F∂z ∂ ¯ z j k k ∂z i k k = ∂R∂z ¯ z k − ∂ i k + j k +2 F∂z ∂ ¯ z kj k ∂z i k k z − ∂ i k + j k +2 F∂z ∂ ¯ z ∂ ¯ z kj k ∂z i k k ¯ z . (31)In general, deriving p times with respect to z and q times with respect to ¯ z the equation writes as follows c ∂ i k + j k + p + q F∂z p ∂ ¯ z q ∂ ¯ z j k k ∂z i k k = A ¯ z k + Bz + C ¯ z , (32)for some c > A, B, C . Then by deriving again this expression with respect to those variablesdifferent from z , ¯ z , z k , ¯ z k and evaluating in z = 0, it vanishes. In the case i > j , just derive equation (27) first i times with respect to z , j times with respect to ¯ z and apply arguments similar to the above in order to prove thatthe partial derivative vanishes at z = 0.Assume now that there exists a special λ -symplectic duality Ψ : U ⊂ M → U ∗ ⊂ M ∗ between ω and ω ∗ . Then,by what we showed Ψ is rotation invariant. By applying Lemma 5.1 in the Appendix at the end of the paper to C = U and S = U ∗ equipped first with the potentials α = λ Φ and β = | z | and then with the potentials α = | z | and β = λ Φ ∗ one gets that (8) and (9) are equivalent to the following equations on ˜ U (the open set associated to U ): ( ˜ ψ k = λ ∂ ˜Φ ∂x k , ˜ ψ k · λ ∂ ˜Φ ∗ ∂x k (cid:16) ˜ ψ x , . . . , ˜ ψ n x n (cid:17) = 1 , k = 1 , . . . , n. (33)Observe now that, by the very definition of duality, one has ˜Φ ∗ ( x ) = − ˜Φ( − x ) and so ∂ ˜Φ ∗ ∂x k ( x ) = ∂ ˜Φ ∂x k ( − x ). Thereforeequations (33) are equivalent to the following: ( ˜ ψ k = λ ∂ ˜Φ ∂x k , ˜ ψ k · λ ∂ ˜Φ ∂x k (cid:16) − ˜ ψ x , . . . , − ˜ ψ n x n (cid:17) = 1 , k = 1 , . . . , n. (34)By inserting the first equation of (34) into the second one we get that (10) is satisfied on ˜ U ⊂ ˜ M . Conversely,assume (10) holds true on a open neighbourhood of the origin, say ˜ W ⊂ ˜ M ⊂ R n . Since Φ is a rotation invariantK¨ahler potential we can assume, by shrinking ˜ W if necessary, that the function ∂ ˜Φ ∂x k is positive on ˜ W (cf. formula(15) above at z = 0). Hence we can define ˜ ψ k : ˜ W ⊂ R n → R , k = 1 , . . . , n by setting˜ ψ k ( x ) = ( λ ∂ ˜Φ ∂x k ( x )) , x ∈ ˜ W . (35)It follows by (10) that equations (34) (and hence equations (33)) are satisfied on ˜ W . Hence, again by Lemma 5.1, therotation invariant special map Ψ : W → M ∗ : z ( ψ ( z ) z , . . . , ψ n ( z ) z n ) defined by ψ j ( z ) = ˜ ψ j ( x ) (where W is theopen set whose associated set is ˜ W ) satisfies Ψ ∗ ω = λω and Ψ ∗ λω ∗ = ω . Since Ψ is a local diffeomorphism sending he origin to the origin it follows by the inverse function theorem that there exist open neighbourhoods of the origin U ⊂ W ⊂ M and U ∗ ⊂ M ∗ such that the restriction Ψ | U : U → U ∗ is a diffeomorphism and hence Ψ is a special λ -symplectic duality between ω and ω ∗ . Finally, notice that equation (35) shows that Ψ is uniquely determined bythe potential Φ. (cid:3) Let M ⊂ C n be a complex domain containing the origin. Let assume that Φ, the potential of the K¨ahler form ω is radialand real analytic. Therefore there exists a real analytic function f : ˆ M → R , defined on ˆ M = { x ∈ R | x = | z | , z ∈ M } such that Φ( z ) = f ( x ). The function f (resp. ˆ M ) will be called the function (resp. the domain) associated to Φ (resp. M ) . In what follows, due to the radiality of Φ, all the neighbourhoods of the origin involved can be taken tobe open balls centered at the origin (of a suitable radius).Before proving Theorem 1.6 we make a remark about it. Notice that the maps ϕ and ϕ from V to V given by ϕ ( z ) = A ( z ) , A ∈ U ( n ) and ϕ ( z ) = e ig ( z ) z where g is an arbitrary radial function on V satisfy ϕ ∗ ω = ϕ ∗ ω = ω and ϕ ∗ ω = ϕ ∗ ω = ω (the equalities regarding the map ϕ follow by the U ( n )-invariance of ω and ω while thoseregarding ϕ follow by straightforward computations). Hence Theorem 1.6 is telling us that, in the radial case, a λ -symplectic duality between ω and ω ∗ is uniquely determined, up to the composition with a unitary transformationand to the multiplication with a S -valued radial function, by the special λ -symplectic duality: z ψ ( z ) z, ψ ( z ) = ( λf ′ ( x )) , x = | z | . (36) Proof of the second part of Theorem 1.6:
We start proving the second part of the theorem (namely equation(12) and the fact that equation (13) is equivalent to the existence of a λ -symplectic duality). So assume that equation(13) is satisfied. Then, by Theorem 1.5 (cfr. formula (35)) the map Ψ given by (36) is (in a suitable neighbourhoodof the origin) a (special) λ -symplectic duality between ω and ω ∗ (this also proves (12)). Conversely, if Ψ : U → U ∗ isa λ -symplectic duality between ω and ω ∗ , then, by the first part of the theorem, it is of the form (11) in a suitableneighbourhood V ⊂ U of the origin. Therefore, by the previous remark it exists a special λ -symplectic dualitybetween ω and ω ∗ given by the map (36) and hence equation (13) holds true again by Theorem 1.5 (on a suitableneighbourhood of the origin of R ). (cid:3) Proof of the first part of Theorem 1.6:
The proof of the first part of the theorem, namely that a λ -symplectic duality can be written as (11) is quite involvedsince we are not assuming Ψ to be special. It is obtained by various steps. The first one deals with the complex onedimensional case. Step 1.
Let M ⊂ C be a -dimensional complex domain containing the origin. endowed with a radial K¨ahler potential Φ and let Ψ : U → U ∗ be a λ -symplectic duality between ω = i ∂ ¯ ∂ Φ and ω ∗ = i ∂ ¯ ∂ Φ ∗ . Then there exist radial functions g : U → R and ψ : U → R such that Ψ( z ) = e ig ( z ) ψ ( z ) z. (37) Moreover ψ is given by ψ ( z ) = ( λf ′ ( x )) , x = | z | . (38) Remark 3.1
Notice that in the one-dimensional case, in contrast to the general case, we are not forced to restrictto V ⊂ U in order to get (37). It should be possible to give an alternative proof of Theorem 1.6 when n ≥ U (this is obviously true if Ψ is assumed to bereal-analytic). Nevertheless for our purposes this is not really important since in this paper we are interested only onthe local behavior of a λ -symplectic duality. Proof:
Let us assume that U = D a (0), U ∗ = D a ∗ (0), where a and a ∗ are suitable real numbers. Let ( r, θ ) (resp.( ρ, η )) be polar coordinates on U (resp. on U ∗ ). Then we have ω = r dr ∧ dθ (39) = S ( r ) r dr ∧ dθ (40) ω ∗ = S ( − ρ ) ρ dρ ∧ dη (41)where we have set S ( x ) = ( xf ′ ) ′ . Notice that, by (40), S > ω is a K¨ahler form.Let Ψ be given in polar coordinates by Ψ( r, θ ) = ( ρ ( r, θ ) , η ( r, θ )). Then Ψ ∗ ω = λω writes ρ ( ρ r η θ − ρ θ η r ) dr ∧ dθ = λS ( r ) rdr ∧ dθ (42)and Ψ ∗ λω ∗ = ω writes λρS ( − ρ )( ρ r η θ − ρ θ η r ) dr ∧ dθ = rdr ∧ dθ. (43)Let us write these equalities as scalar equations as follows ρ ( ρ r η θ − ρ θ η r ) = λS ( r ) r, (44) λρS ( − ρ )( ρ r η θ − ρ θ η r ) = r. (45)Notice that ( ρ r η θ − ρ θ η r ) is the Jacobian determinant J Ψ of Ψ, and by (44) we have J Ψ > S > S ( − ρ ) S ( r ) = λ − . (46)If we derive this equation with respect to θ we get − S ′ ( − ρ ) ρρ θ S ( r ) = 0 . (47)Now, if ρ θ = 0 at some point, it does not vanish for r belonging to some open real interval. Then, since S > S ′ ≡ ω is proportional to ω , in contrast with ourassumption. We conclude that ρ θ = 0, i.e. ρ depends only on r .Moreover, (44) becomes ρρ r η θ = λS ( r ) r (48)Since J Ψ = ρ r η θ does not vanish, both ρ r and η θ are not zero, so this equation implies that ρ (0) = 0, that is Ψ(0) = 0.Now, if we divide (48) by ρρ r and integrate we get η = λS ( r ) rρρ r θ + c ( r ) (49)for some function c . Now, let us fix r and let us consider the map f : S r → S ρ ( r ) , e iθ e iη , induced by Ψ on thecircle centered at the origin and of radius r , where η is given by (49). On the one hand, the degree deg( f ) of thismap equals 1 because Ψ is an orientation-preserving diffeomorphism ( J Ψ > f ) = 12 π Z π dηdθ dθ = λS ( r ) rρρ r | r = r , (50)so that we get λS ( r ) rρρ r = 1 and thus η = θ + c ( r ). ThenΨ( re iθ ) = ρ ( r ) e iη = ρ ( r ) e iθ e ic ( r ) , which proves (38) for ψ ( z ) = ρ ( r ) /r and g ( z ) = c ( r ). Finally, formula (38) is exactly (12) (which we have alreadyproved in general) in the one-dimensional case. (cid:3) Before passing to the general case we pause to obtain additional results needed for the proof. Let Φ : M → R be aradial potential for ω , let ω ∗ be its dual sympletic form defined in M ∗ and let f : ˆ M → R be the function associatedto Φ. A simple computation shows that: = f ′′ ( | z | ) i ∂ | z | ∧ ∂ | z | ) + f ′ ( | z | ) ω (51) ω ∗ = − f ′′ ( −| z | ) i ∂ | z | ∧ ∂ | z | ) + f ′ ( −| z | ) ω . (52) Remark 3.2
Notice that the assumption that ω and ω are not proportional made at the beginning of the paper inthe radial case simply means that it cannot exists an open interval of R where f ′ is constant. In particular it cannotexist any constant c such that xf ′′ + f ′ = c in some open interval of R .Given a diffeomorphism Ψ : U → U ∗ between open subsets U, U ∗ ⊂ C n containing the origin we introduce theoperators B z , B ∗ z ∈ End ( T z U ) as follows: ω z ( · , · ) = ω ( B z · , · ) , ω ∗ z ( · , · ) = ω ( B ∗ z · , · ) . (53)We can compute explicitly both operators B z , B ∗ z by using equations (51) and (52). Namely, B z = f ′′ ( | z | ) z ⊙ ¯ z + f ′ ( | z | ) Id (54) B ∗ z = − f ′′ ( −| z | ) z ⊙ ¯ z + f ′ ( −| z | ) Id (55)where ( z ⊙ ¯ z )( v ) := h v, z i z = ( n X j =1 v j ¯ z j ) z, and where ω is the flat form, i.e. ω ( v, w ) = i ∂ ¯ ∂ | z | ( v, w ) = − Im ( h v, w i ) = − Im ( P j v j w j ) (so that g ( v, w ) = ω ( v, iw ), and h· , ·i = g − iω ) . Notice that both operators B z and B ∗ z satisfy B z ( C z ) = B ∗ z ( C z ) = C z . Define d Ψ sz : T Ψ( z ) U ∗ → T z U by the equation ω ( d Ψ z ( v ) , w ) = ω ( v, d Ψ sz ( w ))for all z ∈ U and for all v ∈ T z U and w ∈ T Ψ( z ) U ∗ .We can now translate the λ -symplectic duality conditions for Ψ : U → U ∗ in terms of the previous operators.Indeed, the equations of the symplectic duality give ω ( d Ψ z ( v ) , d Ψ z ( w )) = λω z ( v, w ) = λω ( B z v, w ) ,λω ∗ ( d Ψ z ( v ) , d Ψ z ( w )) = λω ( B ∗ Ψ( z ) d Ψ z ( v ) , d Ψ z ( w )) = ω ( v, w ) , for all v, w ∈ T z U. Then we get respectively: d Ψ sz ◦ d Ψ z = λB z . (56) d Ψ sz ◦ B ∗ Ψ( z ) ◦ d Ψ z = λ − Id (57)By inserting into (57) the explicit formula of B ∗ Ψ( z ) given by (55) we get: λ − Id = d Ψ sz ◦ ( − f ′′ ( −| Ψ( z ) | )Ψ( z ) ⊙ Ψ( z ) + f ′ ( −| Ψ( z ) | ) Id) ◦ d Ψ z == − f ′′ ( −| Ψ( z ) | ) d Ψ sz ◦ Ψ( z ) ⊙ Ψ( z ) ◦ d Ψ z + f ′ ( −| Ψ( z ) | ) d Ψ sz ◦ d Ψ z By (56), (54): λ − Id = − f ′′ ( −| Ψ( z ) | ) d Ψ sz ◦ Ψ( z ) ⊙ Ψ( z ) ◦ d Ψ z + f ′ ( −| Ψ( z ) | ) λB z == − f ′′ ( −| Ψ( z ) | ) d Ψ sz ◦ Ψ( z ) ⊙ Ψ( z ) ◦ d Ψ z + λf ′ ( −| Ψ( z ) | ) f ′′ ( | z | ) z ⊙ ¯ z ++ λf ′ ( −| Ψ( z ) | ) f ′ ( | z | ) Id =Finally, by the very defintion of ⊙ one gets: λ − Id = − f ′′ ( −| Ψ( z ) | ) d Ψ sz ( h d Ψ z ( · ) , Ψ( z ) i Ψ( z )) ++ λf ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) z ⊙ ¯ z + f ′ ( | z | ) Id (cid:1) . (58) e are now ready to continue the proof of the theorem.Let us come back to the general case. In all the following steps Ψ : U → U ∗ is a λ -symplectic duality betweenopen subsets of C n with n ≥ Step 2.
The map Ψ sends the origin to the origin, i.e., Ψ(0) = 0 . Consequently f ′ (0) = λ − and ω = ω ∗ = λ − ω at the origin ∈ U ⊂ C n . Proof:
Taking z = 0 in (56) and (57) and taking into account (54) and (55) one gets: d Ψ s ◦ d Ψ = λB = λf ′ (0) Id d Ψ s ◦ B ∗ Ψ(0) ◦ d Ψ = λ − Idwhich imply B ∗ Ψ(0) = λ − f ′ (0) Id . This together with (55) gives: B ∗ Ψ(0) = − f ′′ ( −| Ψ(0) | )Ψ(0) ⊙ Ψ(0) + f ′ ( −| Ψ(0) | ) Id = λ − f ′ (0) Id . (59)Assume now, by contradiction, that Ψ(0) = 0. Then the previous equation forces − f ′′ ( −| Ψ(0) | ) = 0 which,together with (52), implies that ω ∗ = cω at the point Ψ(0), where c = f ′ ( −| Ψ(0) | ). Since both forms ω , ω ∗ are U ( n )-invariant it follows that ω ∗ = cω at all points of the sphere centered at the origin of radius r = | Ψ(0) | . SinceΨ is a diffeomorphism there exists a non constant smooth curve γ : ( − ǫ, ǫ ) → U such that γ (0) = 0, δ = | γ ( ǫ ) | > | Ψ( γ ( t )) | = | Ψ( γ (0)) | = r , for all t ∈ ( − ǫ, ǫ ). We claim that ω and ω are proportional inside the ball D δ (0), i.e.the ball centered at the origin of radius δ . This will give the desired contradiction since we are assuming that ω and ω are not proportional (see Remark 3.2). In order to prove our claim let β = Ψ( γ ) be the image of γ under Ψ. Byconstruction, the curve β is contained in the sphere of radius r centered at the origin. It follows, from the previousdiscussion, that ω ∗ | β = c ω | β . This, together with the fact that Ψ is a λ -symplectic duality, implies (by restrictionto the curve γ ) that (Ψ ∗ ω ) | γ = λω | γ , (Ψ ∗ λω ∗ ) | γ = λc (Ψ ∗ ω ) | γ = ω | γ and so ω | γ = λ − c − ω | γ . Thus, since both forms ω, ω are U ( n )-invariant it follows that the above equalities holdfor all the points on the sphere centered at zero of radius | γ ( t ) | , for all t ∈ ( − ǫ, ǫ ). Now if t runs from 0 to ǫ theradius of these spheres runs from 0 to δ . So we get that ω and ω are proportional to each other on D δ (0), as weclaim. The last part of Step 2 is now straightforward. Indeed, since Ψ(0) = 0 by (59) we get ( f ′ (0)) = λ − and since f ′ (0) > ω is a K¨ahler form) it follows that f ′ (0) = λ − ,which again by (51) and (52) implies ω = ω ∗ = λ − ω at the origin. (cid:3) Step 3.
There exists an open subset W ⊂ U containing the origin and a nowhere dense subset S ⊂ W such that: h d Ψ z ( C z ) , Ψ( z ) i = C , ∀ z ∈ W \ S, (60) i.e., for each z ∈ W \ S and β ∈ C there exists α ∈ C such that h d Ψ z ( αz ) , Ψ( z ) i = β . Proof:
Let η = x + iy be a complex number. Then h d Ψ z ( ηz ) , Ψ( z ) i = xa ( z ) + yb ( z ) , where a ( z ) = h d Ψ z ( z ) , Ψ( z ) i , b ( z ) = h d Ψ z ( iz ) , Ψ( z ) i . To prove this step we need to find an open subset W ⊂ U containing the origin and a nowhere dense set S ⊂ W such that a ( z ) and b ( z ) are R -independent on W \ S . We first show that there exists an open subset W ⊂ U containingthe origin where b ( z ) = 0 for all z ∈ W \ { } . Indeed, assume, by contradiction, that such a set does not exist. Thenthere exists a sequence { z n } , z n ∈ U, z n = 0, with z n → n tends to infinity and such that b ( z n ) = 0 for all n . Set w n = z n | z n | and t n = | z n | . Then z n = t n w n , | w n | = 1 and t n →
0. Without loss of generality, since the unit sphere iscompact, we can assume that there exists ξ ∈ U, | ξ | = 1, such that w n → ξ . Therefore0 = b ( z n ) = h t n d Ψ t n w n ( iw n ) , Ψ( t n w n ) i , for all n . Dividing by t n and taking the limit as n → ∞ we get, h d Ψ ( iξ ) , d Ψ ( ξ ) i = 0 . n the other hand Im ( h d Ψ ( iξ ) , d Ψ ( ξ ) i ) = − ω ( d Ψ ( iξ ) , d Ψ ( ξ )) = − (Ψ ∗ ω ) ( iξ, ξ )= − ω ( iξ, ξ ) = Im ( h iξ, ξ ) i ) = | ξ | = 1 , which contradicts the previous equality. (The equality (Ψ ∗ ω ) = ω follows by (Ψ ∗ ω ) = λω and the fact that ω atthe origin equals λ − ω , by Step 2).Fix now an open set W containing the origin such that b ( z ) = 0 for all z ∈ W \ { } and let S be the set of pointsin W where the functions a and b are R -linearly dependent, i.e. S consists of those z ∈ W for which there exists areal number r ( z ) such that a ( z ) = r ( z ) b ( z ). Notice that 0 ∈ S . For each z ∈ S let X ( z ) be the vector at z defined by X ( z ) = (1 − ir ( z )) z. (61)Then it is immediate to see that h d Ψ z ( X ( z )) , Ψ( z ) i = 0 , ∀ z ∈ S. (62)The proof will be completed if we show that the interior of S is empty. Assume the contrary and let ˜ S be an opensubset contained in S . Then (61) gives rise to a smooth vector field X on ˜ S . By inserting X ( z ) in both sides ofequality (58), using (62) and ( z ⊙ ¯ z )( X ( z )) = h X ( z ) , z i z = | z | X ( z ), one gets: λ − X ( z ) = λf ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) z ⊙ ¯ z ( X ( z )) + f ′ ( | z | ) X ( z ) (cid:1) = λf ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) | z | + f ′ ( | z | ) (cid:1) X ( z )which implies λ − = f ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) | z | + f ′ ( | z | ) (cid:1) . Let now z ( t ) ⊂ ˜ S be an integral curve of the vector field X ( z ), where t is varying on an open interval, say I ⊂ R .Notice that (62) implies that ∂ | Ψ | ∂X ( z ) = 0 for all z ∈ ˜ S , and hence | Ψ( z ( t )) | is a constant, say d , for all t ∈ I . Byinserting z ( t ) in the above equality we then get: c = f ′′ ( | z ( t ) | ) | z ( t ) | + f ′ ( | z ( t ) | ) , t ∈ I, where c = ( λ f ′ ( − d )) − . On the other hand it follows by the very definition of X ( z ) that | z ( t ) | is not a constantfunction on I . Hence, when t is varying in I , x = | z ( t ) | is varying in a non-empty open interval of the real line. Inthis interval the function f satisfies the differential equation f ′′ ( x ) x + f ′ ( x ) = c contradicting our assumption (seeRemark 3.2). (cid:3) Step 4.
There exists an open subset V ⊂ U where the following condition is satisfied: given z ∈ V and β ∈ C onecan find a complex number δ (depending on β and z ) such that d Ψ z ( βz ) = δ Ψ( z ) . If this happens we will write d Ψ z ( C z ) = C Ψ( z ) , ∀ z ∈ V. (63) Proof:
Observe first that equation (63) is equivalent to d Ψ sz ( C Ψ( z )) = C z, ∀ z ∈ V, (64)i.e. for given z ∈ V and β ∈ C we can find δ ∈ C such that d Ψ sz ( β Ψ( z )) = δz . Indeed by (56) and by B z ( C z ) = C z one has d Ψ sz ( d Ψ z ( C z )) = λB z ( C z ) = C z and by applying ( d Ψ sz ) − on both sides we get (63).In order to prove (64) let β ∈ C and W and S as in Step 3. Then for z ∈ W \ S there exists α ∈ C (depending on β and z ) such that h d Ψ z ( αz ) , Ψ( z ) i = β . By inserting αz in both sides of formula (58) we obtain: λ − αz = − f ′′ ( −| Ψ( z ) | ) d Ψ sz ( β Ψ( z )) + λf ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) h αz, z i + αf ′ ( | z | ) (cid:1) z Hence f ′′ ( −| Ψ( z ) | ) d Ψ sz ( β Ψ( z )) = γz, (65) here γ = λf ′ ( −| Ψ( z ) | ) (cid:0) f ′′ ( | z | ) h αz, z i + αf ′ ( | z | ) (cid:1) − λ − α. Since f is real analytic and Ψ is a diffeomorphism f ′′ ( −| Ψ( z ) | ) can vanish only in a discrete number of points in W \ S . Let V ⊂ W be an open set around the originwhich does not contain any of these points. We want to prove the validity of (64) in the set V . This is obvious for z = 0 (since Ψ(0) = 0) and for all z ∈ V \ ( V ∩ S ) (this follows by (65)). So it remains to prove (64) for the pointsin V ∩ S \ { } . Let z ∈ V ∩ S , z = 0 and z n ∈ V \ ( V ∩ S ), z n = 0, be a sequence converging to z . Then, given β ∈ C there exists a sequence δ n of complex numbers such that d Ψ sz n ( β Ψ( z n )) = δ n z n (this follows again by (65)).By taking the limit as n → ∞ the left hand side of the previous equality converges and therefore the sequence δ n isforced to converge to a complex number, say δ , satisfying d Ψ sz ( β Ψ( z )) = δ z , and we are done. (cid:3) Step 5.
Let
L ⊂ C n be a complex line through the origin. Then there exists a complex line through the origin L ∗ such that Ψ( L ∩ U ) = L ∗ ∩ U ∗ . In particular d Ψ ∈ U ( n ) . Proof:
Let z ∈ L . By the U ( n )-invariance of ω , ω, ω ∗ we can assume Ψ( z ) ∈ L . Thus, we need to show thatΨ( L ∩ U ) = L ∩ U ∗ . Equivalently we have to show that for every ξ ∈ C n = R n orthogonal to L , i.e. g ( z, ξ ) = 0 forall z ∈ L , and for every smooth curve γ ( t ) ∈ L , such that γ (0) = z , one has g (Ψ( γ ( t )) , ξ ) = 0, in the interval ofdefintion of γ ( t ), say t ∈ ( − a, a ). Introduce the function φ ξ ( t ) = g (Ψ( γ ( t )) , ξ ). Then, by using Step 4, we get dφ ξ ( t ) dt = g ( d Ψ γ ( t ) ( γ ′ ( t )) , ξ ) = β ( t ) g (Ψ( γ ( t )) , ξ ) = β ( t ) φ ξ ( t ) , for some smooth function β ( t ) , t ∈ ( − a, a ). Then φ ξ verifies a first order ordinary differential equation. Since φ ξ (0)is zero then φ ξ ≡
0. Thus, Ψ( γ ( t )) ∈ L for all t , and this proves the first part of the step. In order to prove the lastassertion notice first that d Ψ is linear symplectomorphism from ( R n , ω ) to itself. Indeed, since Ψ is a symplecticduality one has d Ψ ∗ ω = λω | = λλ − ω (the last equality is due to the second part of Step 2). Moreover, by usingthe first part of the present step (namely the fact that Ψ sends complex lines through the origin to complex linesthrough the origin), a simple linear algebra argument yields d Ψ ( iv ) = ± i d Ψ ( v ), for all v ∈ C n . Since d Ψ preservesthe orientation d Ψ ( iv ) = i d Ψ ( v ), for all v ∈ C n , and hence d Ψ ∈ GL( n, C ) ∩ Symp( R n ) = U ( n ). (cid:3) Final step.
There exist an open V ⊂ U , a radial function h : V → C and A ∈ U ( n ) such that Ψ( z ) = h ( z ) Az.
Hence h ( z ) = e ig ( z ) ψ ( z ) where g and ψ are radial functions on V Proof:
By Step 5, Ψ restricted to a suitable open subset V ⊂ U sends complex lines through the origin (intersectedwith V ) to complex lines through the origin (intersected with Ψ( V )). Hence there exists a complex valued function h : V → C such that Ψ( z ) = h ( z ) d Ψ ( z ). Setting A = d Ψ ∈ U ( n ) it remains to prove that h is radial, i.e. it dependsonly on | z | . Since A ∗ ω = ω and A ∗ ω = ω we can assume that Ψ( z ) = h ( z ) z .We first show that | h ( z ) | is radial. Equivalently we will show that d ( | h | ) z ( v ) = 0 if v is a non-zero vectorperpendicular to z , i.e., g ( z, v ) = 0, for all z ∈ V, z = 0. Notice that this is true when v = iz , namely d ( | h | ) z ( iz ) = 0for all z ∈ V, z = 0. Actually a strongest condition is true, namely dh z ( iz ) = 0 for all z ∈ V \ { } . Indeed, if onerestricts Ψ to the complex line L ⊂ C n generated by z one gets a λ -symplectic duality between ( L ∩
V, ω | L∩ V ) and(Ψ( L ∩ V ) , ω ∗ | Ψ( L∩ V )) and the claim follows easily from the one-dimensional case (see Step 1 above). In order toprove our assertion for arbitrary v orthogonal to z we can then assume that v is perpendicular to span R { z , iz } . Thismeans that ω ( z, v ) = ω ( iz, v ) = 0. Using (51) and (52) we also get ω z ( z, v ) = ω z ( iz, v ) = 0. Hence, on the onehand, one gets (Ψ ∗ ω ) z ( iz, v ) = λω ( iz, v ) = 0. On the other hand,0 = (Ψ ∗ ω ) z ( iz, v ) = ω ( dh z ( iz ) z + h ( z ) iz , dh z ( v ) z + h ( z ) v ) == ω ( h ( z ) iz , dh z ( v ) z + h ( z ) v ) = ω ( h ( z ) iz , dh z ( v ) z ) + ω ( h ( z ) iz , h ( z ) v ) == ω ( h ( z ) iz , dh z ( v ) z ) = − Im ( h h ( z ) iz, dh z ( v ) z i ) = | z | Real ( h ( z ) dh z ( v )) == | z | h ( z ) dh z ( v ) + h ( z ) dh z ( v )) = | z | d ( | h | ) z ( v ) . It follows that d ( | h | ) z ( v ) = 0 and hence | h | just depends on | z | . e now show that h is radial. With the same considerations just made for | h | it is enough to show that dh z ( v ) = 0 for all z ∈ V \ { } and for all v perpendicular to span R { z, iz } . For such z and v one has, on the one hand,(Ψ ∗ ω ) z ( z, v ) = λω ( z, v ) = 0. On the other hand,0 = (Ψ ∗ ω ) z ( z, v ) = | z | i dh ∧ dh ) z ( z, v ) + i n X j =1 h ( z )¯ z j ( dz j ∧ d ¯ h ) z ( z, v )++ n X j =1 i z j h ( z )( dh ∧ d ¯ z j ) z ( z, v ) + | h | ω ( z, v ) == | z | i dh ∧ d ¯ h ) z ( z, v ) + | z | i h ( z ) d ¯ h z ( v ) − | z | i h ( z ) dh z ( v ) . Therefore 0 = ( dh ∧ d ¯ h ) z ( z, v ) + h ( z ) d ¯ h z ( v ) − h ( z ) dh z ( v ) = ( dh ∧ d ¯ h ) z ( z, v ) − h ( z ) dh z ( v ) , where the last equality is a consequence of the fact that we just prove that | h | is radial (and hence 0 = d ( | h | ) z ( v ) = h ( z ) d ¯ h z ( v ) + h ( z ) dh z ( v ) for all z ∈ V \ { } and for all v perpendicular to span R { z, iz } ).Multiplying both sides of the previous equality by | h ( z ) | we get:0 = | h ( z ) | dh z ( z ) dh z ( v ) − | h ( z ) | dh z ( v ) dh z ( z ) − | h ( z ) | h ( z ) dh z ( v ) == − h ( z ) dh z ( z ) h ( z ) dh z ( v ) − | h ( z ) | dh z ( v ) dh z ( z ) − | h ( z ) | h ( z ) dh z ( v ) == − h ( z ) dh z ( v ) (cid:16) h ( z ) dh z ( z ) + h ( z ) dh z ( z ) + 2 | h ( z ) | (cid:17) == − h ( z ) dh z ( v ) (cid:0) d ( | h | ) z ( z ) + 2 | h ( z ) | (cid:1) . Note that h ( z ) = 0 for z = 0 since the map Ψ : V → Ψ( V ) , z h ( z ) z is injective (it is a diffeomorphism). Hence,in order to show that h is a radial function it is enough to prove that d ( | h | ) z ( z ) + 2 | h ( z ) | cannot vanish on anyopen subset of V \ { } . Since | h | is radial we can restrict the problem to the real line R e , e = (1 , , . . . , σ ( t ) = | h ( te ) | and I = { t ∈ R | tσ ′ ( t ) + 2 σ ( t ) = 0 } the radiality of h will be guaranteed if I does not contain any open subset of the real line. Assume, by contradiction, that there exists such an open subset.Then in this set σ solves the differential equation tσ ′ ( t ) + 2 σ ( t ) = 0 and so σ ( t ) = ct ) for some real constant c . Thisis the desired contradiction since | h (0) | = σ (0) is a well defined real number. (cid:3) An immediate consequence of Theorem 1.6 is the following corollary which can be considered a generalization ofthe second part of Theorem 1.1.
Corollary 3.3
Let
Ψ : U → U ∗ be a λ -symplectic duality between two radial forms ω and ω ∗ . Then there exists anopen subset V ⊂ U where the restriction of Ψ takes complex and totally geodesic submanifolds through the origin of ( V, ω ) to complex and totally geodesic submanifolds through the origin of (Ψ( V ) , ω ) . Proof:
Let V ⊂ U ⊂ C n be an open subset containing the origin such that the restriction of Ψ to V is of theform (11), i.e. Ψ( z ) = e ig ( z ) ψ ( z ) A ( z ) , z ∈ V . Since ω is radial it is easy to see that a complex and totally geodesicsubmanifold of ( V, ω ) of (complex) dimension k is the intersection of V with a k -dimensional linear subspace of C n .Therefore Ψ( V ∩ T ) = T ∗ ∩ Ψ( V ) where T ∗ is the k -dimensional space of C n given by A ( T ). (cid:3) In this section we provide some examples and applications of our results. The first two subsections deal with Hartogsdomains and the Taub-NUT metric respectively which are important examples in the rotation invariant case. In thethird subsection, where we consider the radial case, we exhibit an example of radial K¨ahler form (different from thehyperbolic metric) for which there exists a λ -symplectic duality. In all this section given a rotation invariant (or evenradial) K¨ahler form ω = i ∂ ¯ ∂ Φ (on an open subset of C n containing the origin) we say that it admits a λ -symplecticduality if there exists a λ -symplectic duality Ψ : U → U ∗ between ω and ω ∗ = i ∂ ¯ ∂ Φ ∗ , where Φ ∗ is the (local) dualof Φ (defined on a suitable neighborhood M ∗ of the origin). .1 Hartogs domains Let x ∈ R + ∪ { + ∞} and let F : [0 , x ) → (0 , + ∞ ) be a decreasing real analytic function, on (0 , x ). The Hartogsdomain D F ⊂ C n associated to the function F is defined by D F = { ( z , z , ..., z n − ) ∈ C n | | z | < x , | z | + · · · + | z n − | < F ( | z | ) } . One can prove that, under the assumption − ( xF ′ ( x ) F ( x ) ) ′ > x ∈ [0 , x ), the natural (1 , D F given by ω F = i ∂∂ log 1 F ( | z | ) − | z | − · · · − | z n − | (66)is a K¨ahler form on D F . The previous equality is equivalent to the strongly pseudoconvexity of D F (see [5] for aproof and also [6], [8], [12] and [9] for other properties of these domains).Notice that, when x = 1 and F ( x ) = 1 − x , then the corresponding Hartogs domain is the n -dimensional unitball endowed with the hyperbolic form ω hyp . In this case, we have already observed in the introduction that ω hyp admits a special λ -symplectic duality. We now prove that in fact this is the only case among Hartogs domains,namely If ( D F , ω F ) admits a λ -symplectic duality then ( D F , ω F ) is holomorphically isometric to an open subset ofthe complex hyperbolic space. In order to prove our assertion notice first that the potential for the K¨ahler form ω F is rotation invariant and has ˜Φ( x , x , . . . , x n − ) = − log( F ( x ) − P n − j =1 x j ) as associated function. Therefore byTheorem 1.5 (cf. equations (10)) ( D F , ω F ) admits a λ -symplectic duality iff the following two equations are satisfiedon a neighbourhood of the origin of R n : λ F ′ ( x ) F ( x ) − P n − j =1 x j · F ′ ( − λ ∂ ˜Φ ∂x x ) F ( − λ ∂ ˜Φ ∂x x ) + P n − j =1 λ ∂ ˜Φ ∂x j x j = 1and 1 F ( x ) − P n − j =1 x j · F ( − λ ∂ ˜Φ ∂x x ) + P n − j =1 λ ∂ ˜Φ ∂x j x j = 1 . Substituting the second one into the first one we get λ F ′ ( x ) · F ′ λx F ′ ( x ) F ( x ) − P n − j =1 x j ! = 1 . If we fix x in this equation and let t = P n − j =1 x j take values in a small open interval contained in [0 , F ( x )), weget that F ′ ( x ) is constant on a sufficiently small interval, and hence F ′ is constant. So F ( x ) = c − c x for some c , c >
0, which implies that D F is holomorphically isometric to an open subset of the hyperbolic space C H n viathe map φ : D F → C H n , ( z , z , . . . , z n − ) (cid:18) z √ c /c , z √ c , . . . , z n − √ c (cid:19) . In [7] C. LeBrun constructed the following family of K¨ahler forms on C defined by ω m = i ∂ ¯ ∂ Φ m , whereΦ m ( U, V ) = U + V + m ( U + V ) , m ≥ U and V are implicitly defined by | z | = e m ( U − V ) U, | z | = e m ( V − U ) V. For m = 0 one gets the flat metric, while for m > C having the same volume form of the flat metric ω . Moreover, for m > C , ω m )is globally symplectomorphic to ( R , ω ) via a special symplectic map. e claim that there exists a special λ -symplectic duality Ψ for ( C , ω m ) if and only if m = 0. In order to proveour claim let x j = | z j | , j = 1 ,
2. By the inverse function theorem one easily gets ∂ ˜Φ m ∂x = (1 + 2 mV ) e m ( V − U ) , ∂ ˜Φ m ∂x = (1 + 2 mU ) e m ( U − V ) so that ∂ ˜Φ m ∂x x = U + 2 mx x and ∂ ˜Φ m ∂x x = V + 2 mx x . Equations (10) for x = 0 write respectively λ e − mU ( x , e − mU ( − λU ( x , , = 1 λ (1 + 2 mU ( x , e mU ( x , (1 + 2 mU ( − λU ( x , , e mU ( − λU ( x , , = 1 . By the first one we get U ( − λU ( x , ,
0) = 1 m log λ − U ( x , λ (1 + 2 mU ( x , λ − mU ( x , . The latter is a polynomial equation of degree 2 in 2 mU ( x , m = 0. Then, either this equation has not solution,and we are done, or it implies that U ( x ,
0) is constant in a neighbourhood of 0. But in this case, since x = e mU U ,also x must be constant, which is impossible. This proves our claim. Let M be an open neighbourhood of 0 in C n , endowed with a radial K¨ahler form ω = i ∂ ¯ ∂f , with f = f ( | z | ). In thiscase, we know that the existence of a radial invariant λ -symplectic duality is guaranteed by equation (13), namely λ f ′ ( x ) f ′ ( − λxf ′ ( x )) = 1 on a suitable neighbourhood of the origin of R .It is easy to see (in accordance with what we already knew) that f ( x ) = λ x (the flat metric), f ( x ) = − λ log(1 − x )(the hyperbolic metric) and f ( x ) = λ log(1 + x ) (the Fubini-Study metric) satisfy this equation. In order to see othersolutions different from these cases, notice that (13) can be rewritten as G ( G ( x )) = x (67)where G ( x ) = − λxf ′ ( x ). Thus if the graph of y = G ( x ) is symmetric with respect to the straight line y = x , then G ( x ) satisfies (67). Take for example G ( x ) = − √
22 + x + 12 q − √ x which is defined in a neighbourhood of 0, satisfies this condition (we obtained this function by simply rotating clockwisethe graph of the even function y = − x by an angle of π/ G is analytic in 0 and satisfies G (0) = 0, so that G ( x ) /x is also analytic. Then, by integrating f ′ ( x ) = − G ( x ) /λx , we get a function f ( x ) whichsatisfies the equation of symplectic duality and such that f ′ (0) = − G ′ (0) λ = λ >
0, so that it defines a K¨ahler metricin a sufficiently small neighbourhood of the origin. A simple calculation shows that the K¨ahler metric associated tothis potential f has not constant curvature and so this yields a K¨ahler metric which admits a λ -symplectic dualitybut which does not have constant curvature.Finally, an easy example of potential which admits a local dual but it does not admit a λ -symplectic duality isgiven by f ( x ) = x − x , x = | z | , in a suitable neighbourhood of the origin. The following lemma provides necessary and sufficient conditions for a given rotation invariant special map to besymplectic. emma 5.1 Let C ⊆ C n and S ⊆ C n be two complex domains containing the origin endowed with rotation invariantK¨ahler potentials α and β and corresponding K¨ahler forms ω α = i ∂ ¯ ∂α and ω β = i ∂ ¯ ∂β respectively. Then a rotationinvariant special map Ψ : C → S satisfies Ψ ∗ ( ω β ) = ω α if and only if ˜ ψ k ∂ ˜ β∂x k ( ˜ ψ x , . . . , ˜ ψ n x n ) = ∂ ˜ α∂x k , k = 1 , . . . , n, (68) where ˜ α : ˜ C ⊂ R n → R (resp. ˜ β : ˜ S ⊂ R n → R ) is the function associated to α (resp. β ) (see Section 2 for thedefinition of special maps). Proof:
Since ω β = i n X i,j =1 ∂ ˜ β∂x i ∂x j ¯ z j z i + ∂ ˜ β∂x i δ ij ! dz j ∧ d ¯ z i one gets Ψ ∗ ( ω β ) = i n X i,j =1 ∂ ˜ β∂x i ∂x j (Ψ)Ψ i ¯Ψ j + ∂ ˜ β∂x j (Ψ) δ ij ! d Ψ j ∧ d ¯Ψ i , where ∂ ˜ β∂x j (Ψ) = ∂ ˜ β∂x j ( ˜ ψ x , . . . , ˜ ψ n x n ) and ∂ ˜ β∂x i ∂x j (Ψ) = ∂ ˜ β∂x i ∂x j ( ˜ ψ x , . . . , ˜ ψ n x n ) . If one denotes by Ψ ∗ ( ω β ) = Ψ ∗ ( ω β ) (2 , + Ψ ∗ ( ω β ) (1 , + Ψ ∗ ( ω β ) (0 , the decomposition of Ψ ∗ ( ω β ) into addenda of type (2 , , (1 ,
1) and (0 ,
2) one has:Ψ ∗ ( ω β ) (2 , = i n X i,j,k,l =1 ∂ ˜ β∂x i ∂x j (Ψ)Ψ i ¯Ψ j + ∂ ˜ β∂x j (Ψ) δ ij ! ∂ Ψ j ∂z k ∂ ¯Ψ i ∂z l dz k ∧ dz l (69)Ψ ∗ ( ω β ) (1 , = i n X i,j,k,l =1 ∂ ˜ β∂x i ∂x j (Ψ)Ψ i ¯Ψ j + ∂ ˜ β∂x j (Ψ) δ ij ! (cid:18) ∂ Ψ j ∂z k ∂ ¯Ψ i ∂ ¯ z l − ∂ Ψ j ∂ ¯ z l ∂ ¯Ψ i ∂z k (cid:19) dz k ∧ d ¯ z l (70)Ψ ∗ ( ω β ) (0 , = i n X i,j,k,l =1 ∂ ˜ β∂x i ∂x j (Ψ)Ψ i ¯Ψ j + ∂ ˜ β∂x j (Ψ) δ ij ! ∂ Ψ j ∂ ¯ z k ∂ ¯Ψ i ∂ ¯ z l d ¯ z k ∧ d ¯ z l . (71)Since Ψ j ( z ) = ˜ ψ j ( | z | , ..., | z n | ) z j , one has: ∂ Ψ i ∂z k = ∂ ˜ ψ i ∂x k z i ¯ z k + ˜ ψ i δ ik , ∂ Ψ i ∂ ¯ z k = ∂ ˜ ψ i ∂x k z k z i (72)and ∂ ¯Ψ i ∂ ¯ z k = ∂ ˜ ψ i ∂x k z k ¯ z i + ˜ ψ i δ ik , ∂ ¯Ψ i ∂z k = ∂ ˜ ψ i ∂x k ¯ z k ¯ z i , (73)By inserting (72) and (73) into (69) and (70) after a long, but straightforward computation, one obtains:Ψ ∗ ( ω β ) (2 , = i n X k,l =1 A kl z k ¯ z l dz k ∧ dz l (74)and Ψ ∗ ( ω β ) (1 , = i n X k,l =1 " ( A kl + A lk ∂ ˜ β∂x k ∂x l (Ψ) ˜ ψ k ˜ ψ l )¯ z k z l + ∂ ˜ β∂x k (Ψ) δ kl ˜ ψ k dz k ∧ d ¯ z l , (75)where A kl = ∂ ˜ β∂x k (Ψ) ∂ ˜ ψ k ∂x l + ˜ ψ k n X j =1 ∂ ˜ β∂x j ∂x k (Ψ) ∂ ˜ ψ j ∂x l | z j | . (76) ow, we assume that Ψ ∗ ( ω β ) = ω α = i n X k,l =1 (cid:18) ∂ ˜ α∂x k ∂x l ¯ z k z l + ∂ ˜ α∂x l δ lk (cid:19) dz k ∧ d ¯ z l . Then the terms Ψ ∗ ( ω β ) (2 , and Ψ ∗ ( ω β ) (0 , are equal to zero. This is equivalent to the fact that (76) is symmetricin k, l .Hence, by setting Γ l = ˜ ψ l ∂ ˜ β∂x l (Ψ) , l = 1 , . . . , n (77)equation (75) becomesΨ ∗ ( ω β ) (1 , = i n X k,l =1 " ( A kl + ∂ ˜ β∂x k ∂x l (Ψ) ˜ ψ k ˜ ψ l )¯ z k z l + ∂ ˜ β∂x k (Ψ) δ kl ˜ ψ k dz k ∧ d ¯ z l == i n X k,l =1 (cid:18) ∂ Γ l ∂x k ¯ z k z l + Γ k δ kl (cid:19) dz k ∧ d ¯ z l . (78)So, Ψ ∗ ( ω β ) = ω α implies i n X k,l =1 (cid:18) ∂ Γ l ∂x k ¯ z k z l + Γ k δ lk (cid:19) dz k ∧ d ¯ z l = i n X k,l =1 (cid:18) ∂ ˜ α∂x k ∂x l ¯ z k z l + ∂ ˜ α∂x l δ kl (cid:19) dz k ∧ d ¯ z l . In this equality, we distinguish the cases l = k and l = k and get respectively ∂ Γ l ∂x k = ∂ ˜ α∂x k ∂x l ( k = l )and ∂ Γ k ∂x k x k + Γ k = ∂ ˜ α∂x k x k + ∂ ˜ α∂x k . By defining A k = Γ k − ∂ ˜ α∂x k , these equations become respectively ∂A k ∂x l = 0 ( l = k )and ∂A k ∂x k x k = − A k . The first equation implies that A k does not depend on x l and so by the second one we have A k = Γ k − ∂ ˜ α∂x k = c k x k , (79)for some constant c k ∈ R . Since the domains contains the origin this forces c k = 0 , ∀ k, and hence, by (77), we getΓ k = ˜ ψ k ∂ ˜ β∂x k (Ψ) = ∂ ˜ β∂x k ( ˜ ψ x , . . . , ˜ ψ n x n ) = ∂ ˜ α∂x k , k = 1 , . . . , n, namely (68).In order to prove the converse of Lemma 5.1, notice that by differentiating (68) with respect to l one gets: ∂ ˜ α∂x k ∂x l = A kl + ∂ ˜ β∂x k ∂x l ˜ ψ k ˜ ψ l with A kl given by (76). By ∂ ˜ α∂x k ∂x l = ∂ ˜ α∂x l ∂x k and ∂ ˜ β∂x k ∂x l ˜ ψ k ˜ ψ l = ∂ ˜ β∂x l ∂x k ˜ ψ l ˜ ψ k one gets A kl = A lk . Then, by (74), theaddenda of type (2,0) (and (0,2)) in Ψ ∗ ( ω β ) vanish. Moreover, by (76) and (78), it follows that Ψ ∗ ( ω β ) = ω α . (cid:3) eferences [1] F. Cuccu and A. Loi, Global symplectic coordinates on complex domains , J. Geom. and Phys. 56 (2006), 247-259.[2] A. J. Di Scala and A. Loi,
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E-mail: [email protected]
Universit`a di Cagliari,
E-mail: [email protected]
Universit`a di Cagliari,
E-mail: [email protected]@unica.it