Download - Classifying triples of Lagrangians in a Hermitian vector space

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Topology and its Applications 144 (2004) 1–27

www.elsevier.com/locate/topo

Classifying triples of Lagrangians in a Hermitianvector space

Elisha Falbel∗, Jean-Pierre Marco, Florent Schaffhauser

Université Pierre et Marie Curie, Analyse Algébrique, Institut de Mathématiques, 4 place Jussieu,175 rue du Chevaleret, Paris 75013, France

Received 24 March 2003; accepted 3 August 2003

Abstract

The purpose of this paper is to study the diagonal action of the unitary groupU(n) on triplesof Lagrangian subspaces ofC

n. The notion of angle of Lagrangian subspaces is presentedand we show how pairs(L1,L2) of Lagrangian subspaces are classified by the eigenvalues ounitary mapσL2 ◦ σL1 obtained by composing the Lagrangian involutions relative toL1 andL2.We then study the case of triples of Lagrangian subspaces ofC2 and give a complete descriptionthe orbit space. Our main result is Theorem 3.5. As an application of the methods presentewe give a way of computing the inertia index of a triple(L1,L2,L3) of Lagrangian subspacesCn from the measures of the angles(Lj ,Lk), and relate the classification of Lagrangian trip

of C2 to the classification of two-dimensional unitary representations of the fundamentalπ1(S

2\{s1, s2, s3}). 2004 Elsevier B.V. All rights reserved.

MSC:53D12; 22E

Keywords:Lagrangian; Hermitian; Inertia index; Unitary representation

1. Introduction

Let (V ,ω) be a 2n-dimensional real symplectic vector space. Acomplex structureonV is an automorphismJ of V such thatJ 2 = −Id. The complex structureJ is said to beadapted toω if the bilinear formg(u, v) := ω(u,J v) is a Euclidean scalar product onV .Given a complex structureJ , the vector spaceV can be endowed with a complex vect

* Corresponding author.E-mail address:[email protected] (E. Falbel).

0166-8641/$ – see front matter 2004 Elsevier B.V. All rights reserved.
doi:10.1016/j.topol.2003.08.027

2 E. Falbel et al. / Topology and its Applications 144 (2004) 1–27

space structure by settingi . v = Jv for all v ∈ V . If J is adapted toω, the formh = g − iω

p

lce

pertiesroup

ult

rbitWeces ofsional

nian

r

then is a Hermitian scalar product onV endowed with its complex structure and the maJis both orthogonal and symplectic. LetU(V ) be the unitary group ofV relative toh, O(V )

the orthogonal group ofV relative tog andSp(V ) the symplectic group ofV relative toω.Then by definition ofh, we haveU(V ) = O(V ) ∩ Sp(V ).

A real subspaceL of V is said to beLagrangianif its orthogonal complementL⊥ω

with respect toω is L itself. Equivalently,L is Lagrangian if and only if its orthogonacomplement with respect tog is L⊥g = JL. In particular, given a Lagrangian subspaL of V , we have theg-orthogonal decompositionV = L ⊕ JL. Denoting byL(V ) theset of all Lagrangian subspaces ofV (the Lagrangian Grassmannianof V ), the abovedecomposition enables us to associate to every LagrangianL anR-linear map

σL :V → V,

x + Jy �→ x − Jy

called theLagrangian involution associated toL (see also [8,7]). Observe thatσL is anti-holomorphic: σL ◦ J = −J ◦ σL and that the map

L(V ) → GL(V ),

L �→ σL

is continuous.The purpose of this paper is to use these Lagrangian involutions, some pro

of which are stated in Section 2.4, to study the diagonal action of the unitary gU(V ) on L(V ) × L(V ) × L(V ). We shall first recall the unitary classification resfor pairs of Lagrangian subspaces ofV , then describe the diagonal action ofU(V ) onL(V ) × L(V ) × L(V ), and give a complete classification and description of the ospace in the case whereV is of complex dimension 2 (Theorems 3.1, 3.5 and 3.6).then deduce a way to compute the inertia index of a triple of Lagrangian subspaV (Theorem 4.4). Finally, we relate these results to the classification of two-dimenunitary representations of the fundamental groupπ1(S

2\{s1, s2, s3}) (Corollary 4.10).By choosing a unitary basis ofV , we may identifyV with the Hermitian vector

space(Cn,h(z, z′) = ∑nk=1 zkz

′k) and denoteL(V ) by L(n), U(V ) by U(n), O(V ) by

O(2n) and Sp(V ) by Sp(n). We setg = Reh and ω = − Imh and denote byJ theR-endomorphism ofCn R2n corresponding to multiplication byi ∈ C in Cn, and wehave indeedg = ω(. , J.).

2. Pairs of Lagrangian subspaces

The unitary groupU(n) acts smoothly and transitively on the Lagrangian GrassmanL(n). Fixing a LagrangianL in L(n), its stabilizer can be identified toO(n), andL(n)

therefore is a compact homogeneous space diffeomorphic toU(n)/O(n). We shall here beconcerned with the diagonal action ofU(n) onL(n) ×L(n). Observe that requiringψ(L)

to be Lagrangian whenL is Lagrangian andψ ∈ O(2n) is equivalent to requiring thatψbe unitary (sinceL ⊕ JL = C

n, ag-orthogonal basisB for L overR is a unitary basis fo

E. Falbel et al. / Topology and its Applications 144 (2004) 1–27 3

Cn overC, and ifL is Lagrangian andψ orthogonal,ψ(B) then also is a unitary basis, sos

: to

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ns

thatψ is a unitary map). Equivalently, the orbit of a pair(L1,L2) of Lagrangian subspaceunder the diagonal action ofU(n) is the intersection withL(n) × L(n) of the orbit of(L1,L2) under the diagonal action ofO(2n). The orbit[L1,L2] of the pair(L1,L2) underthe diagonal action ofU(n) may therefore be called theLagrangian angleformed byL1 andL2. In the following, we shall simply speak of the angle(L1,L2) to designatethe orbit[L1,L2]. We now wish to find complete numerical invariants for this actioneach angle(L1,L2) we shall associate ameasure, denoted by meas(L1,L2), in a way thattwo pairs(L1,L2) and(L′

1,L′2) lie in the same orbit of the action ofU(n) if and only if

meas(L1,L2) = meas(L′1,L

′2). This can be done in three equivalent ways, which we s

describe and compare.

2.1. Projective properties of Lagrangian subspaces ofCn

A real subspaceW of Cn is said to betotally real if h(u, v) ∈ R for all u,v ∈ W . A realsubspaceL of V therefore is Lagrangian if and only if it is totally real and of maximadimension with respect to this property. Letp be the projectionp :Cn\{0} → CPn−1 onthe(n − 1)-dimensional complex projective space, and for any real subspaceW of Cn, letp(W) be the image ofW\{0}.

WhenL is a Lagrangian subspace ofCn, recall that we denote byσL the only anti-holomorphic involution ofCn leavingL pointwise fixed (called the Lagrangian involutioassociated toL). The mapσL being anti-holomorphic, it induces a map

σL :CPn−1 → CPn−1,

[z] �→ [σL(z)

].

If we endowCPn−1 with the Fubini–Study metric,σL becomes an isometry, andp(L) isthe fixed point set of that isometry. Therefore, for any LagrangianL of C

n, the subspacl = p(L) of CPn−1, called aprojective Lagrangian, is a totally geodesic embeddesubmanifold ofCPn−1. More generally, every totally real subspaceW of Cn is sent byp to a closed embedded submanifold ofCPn−1 which is diffeomorphic toRP(W) (see [11,p. 73]). These projective properties can be used to prove the first diagonalization(Theorem 2.1), as shown in [11]. They will also be important to us in the study of projeLagrangians ofCP1.

2.2. First diagonalization lemma and unitary classification of Lagrangian pairs

We state here the results obtained by Nicas in [11]. Let(L1,L2) be a pair of Lagrangiasubspaces ofCn and letB1 = (u1, . . . , un) andB2 = (v1, . . . , vn) be orthonormal basefor L1 andL2, respectively. LetA be then × n complex matrix with coefficientsAij =h(vj , ui).

Definition 1 (Souriau matrix,[11]). The matrixAAt , whereAt is the transpose ofA, iscalled theSouriau matrixof the pair(L1,L2) with respect to the basesB1 andB2.


The matrixAAt is both unitary and symmetric. As shown in [11, p. 73], two Souriauion.

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matrices of the pair(L1,L2) are conjugate. We can therefore make the following definit

Definition 2 [11]. The characteristic polynomialof the pair (L1,L2), denoted byP(L1,L2), is the characteristic polynomial of any Souriau matrix of the pair(L1,L2).

Theorem 2.1 (First diagonalization lemma, [11]).Let (L1,L2) be a pair of Lagrangiansubspaces ofCn. Then there exists an orthonormal basis(u1, . . . , un) for L1 and unitcomplex numberseiλ1, . . . , eiλn such that(eiλ1u1, . . . , e

iλnun) is an orthonormal basisfor L2. Furthermore, the squaresei2λ1, . . . , ei2λn of these numbers are the roots of tcharacteristic polynomial of the pair(L1,L2), counted with their multiplicities.

Theorem 2.2 (Unitary classification of Lagrangian pairs ofCn, [11]). Let (L1,L2) and(L′

1,L′2) be two pairs of Lagrangian subspaces ofCn. A necessary and sufficient conditi

for the existence of a unitary mapψ ∈ U(n) such thatψ(L1) = L′1 andψ(L2) = L′

2 isthat the characteristic polynomialsP(L1,L2) andP(L′

1,L′2) be equal.

2.3. Second diagonalization lemma

It is possible to express the result of the first diagonalization lemma in terms of umaps sendingL1 to L2, in a way that generalizes the situation of real lines inC.

Proposition 2.3 (Second diagonalization lemma).Given two Lagrangian subspaces ofL1andL2 of Cn, there exists a unique unitary mapϕ12 ∈ U(n) sendingL1 to L2 and verifyingthe following diagonalization conditions:

(i) the eigenvalues ofϕ12 are unit complex numberseiλ1, . . . , eiλn verifying π > λ1 �· · · � λn � 0;

(ii) there exists an orthonormal basis(u1, . . . , un) for L1 such thatuk is an eigenvectofor ϕ12 (with eigenvalueeiλk ).

Proof. The existence is a consequence of the firstdiagonalization lemma. As for unicityobserve that two such unitary maps have the same eigenspaces and the same correeigenvalues, and therefore are equal.�

It is also possible to give a direct proof of this result, which then proves thediagonalization lemma without making use of projective geometry (see [2] or [Observe that condition (i) is essential for the unicity part: for any LagrangianL, the twomapsJ and−J are both unitary, they both sendL to JL = (−J )L and verify condition (ii)for any orthonormal basis ofL, butJ is the only one of these two maps whose eigenvaare located in the upper half of the unit circle ofC.

Observe that the Souriau matrix of the pair(L1,L2) with respect to the base(u1, . . . , un) and(eiλ1u1, . . . , e

iλnun) is the diagonal matrix diag(ei2λ1, . . . , ei2λn). There-fore, the roots of the characteristic polynomialP(L1,L2) are the squares of the eigenvaluof ϕ12.


At last, observe that if(L1,L2) and (L′ ,L′ ) are located in the same orbit of the

tion

the

e

.r any

he

f

c

l.

1 2diagonal action ofU(n) on L(n) × L(n), then the two associated unitary mapsϕ12 andϕ′

12 are conjugate inU(n). Indeed, ifψ(L1) = L′1 andψ(L2) = L′

2 with ψ ∈ U(n), thenψ ◦ ϕ12 ◦ ψ−1 sendsL′

1 to L′2 and verifies the conditions of the second diagonaliza

lemma, hence by unicityψ ◦ ϕ12 ◦ ψ−1 = ϕ′12. The unitary mapsϕ12 will be very useful

in the study of the diagonal action ofU(2) on triples of Lagrangian subspaces ofC2 (seeSection 3).

2.4. Lagrangian involutions

We give here the properties of Lagrangian involutions that we shall need infollowing. Finite groups generated by such involutions are studied in [6].

Proposition 2.4. LetL ∈ L(n) be a Lagrangian subspace ofCn. Then:

(i) There exists a unique anti-holomorphic mapσL whose fixed point set is exactlyL.(ii) If L′ is a Lagrangian subspace such thatσL = σL′ , thenL = L′: there is a one-to-on

correspondence between Lagrangian subspaces and Lagrangian involutions.(iii) σL is anti-unitary:for all z, z′ ∈ Cn, h(σL(z), σL(z′)) = h(z, z′).

Observe then that the composite mapσL2 ◦σL1 of two Lagrangian involutions is unitaryAs a direct consequence of the definition of a Lagrangian involution, we have, foLagrangianL and any unitary mapψ , σψ(L) = ψ ◦ σL ◦ ψ−1 (sinceψ ◦ σL ◦ ψ−1 isanti-holomorphic and leavesψ(L) pointwise fixed). The following result establishes trelation between Lagrangian involutions and angles of Lagrangian subspaces.

Proposition 2.5. Let L1 andL2 be two Lagrangian subspaces ofCn. The eigenvalues oσL2 ◦ σL1 are the roots of the characteristic polynomialP(L1,L2) of the pair(L1,L2),with the same multiplicity. Equivalently, sinceP(L1,L2) is monic, it is the characteristipolynomial of the holomorphic mapσL2 ◦ σL1.

Proof. By the first diagonalization lemma, there exists an orthonormal basis(u1, . . . , un)

for L1 and unit complex numbersα1, . . . , αn such that(α1u1, . . . , αnun) is an orthonormabasis forL2 andα2

1, . . . , α2n are the roots ofP(L1,L2), counted with their multiplicities

Let ψ be the unitary morphism sendinguk to αkuk for k = 1, . . . , n. Thenα21, . . . , α2

n arethe eigenvalues ofψ2, counted with their multiplicities, and it is therefore sufficient toprove thatσL2 ◦ σL1 = ψ2.

The mapψ ◦ σL1 ◦ ψ−1 is anti-holomorphic and leavesL2 pointwise fixed, henceσL2 = ψ ◦ σL1 ◦ ψ−1. Furthermore, for allj = 1, . . . , n, we haveσL1 ◦ ψ−1(uj ) =σL1(

1αj

uj ) = αjσL1(uj ) = ψ(uj ) = ψ ◦ σL1(uj ). �In particular, settingψ = ϕ12 in the above proof, we obtain the following corollary.

Corollary 2.6. Let ϕ12 be the only unitary map sendingL1 to L2 and verifying theconditions of Proposition2.3. Thenϕ2

12 = σL2 ◦ σL1.


2.5. Measure of a Lagrangian angle

ition.

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Keeping the classification theorem in mind, we are then led to the following defin

Definition 3 (Measure of a Lagrangian angle). Let L1 andL2 be two Lagrangians ofCn

and letei2λ1, . . . , ei2λn be the eigenvalues ofσL2 ◦ σL1, counted with their multiplicitiesThe symmetric groupSn acts onS1 × · · · × S1 by permuting the elements of then-tuplesof unit complex numbers, and we denote by[ei2λ1, . . . , ei2λn] the equivalence class o(ei2λ1, . . . , ei2λn) ∈ S1 × · · · × S1, and call it themeasureof the angle formed byL1 andL2: meas(L1,L2) = [ei2λ1, . . . , ei2λn] ∈ (S1 × · · · × S1)/Sn.

As σψ(L) = ψ ◦ σL ◦ ψ−1 for any unitary mapψ ∈ U(n), we have meas(ψ(L1),

ψ(L2)) = meas(L1,L2), so this notion is well defined. This definition of a measure dnot extend the usual one (in the casen = 1, we obtainei2λ, whereλ ∈ [0,π[ is the usualmeasure). It will nonetheless prove to be relevant.

Observe that, sinceσL1 ◦ σL2 = (σL2 ◦ σL1)−1, if ei2λ is an eigenvalue ofσL2 ◦ σL1

then e−i2λ is an eigenvalue ofσL1 ◦ σL2. As a consequence, if(ei2λ1, . . . , ei2λn) is arepresentative of meas(L1,L2) then (ei2ξ1, . . . , ei2ξn), whereξk = π − λk mod π , is arepresentative of meas(L2,L1). In particular, meas(L1,L2) = meas(L′

1,L′2) if and only if


′1).

In the following, we shall identifyS1 × · · · × S1 with the n-torus Tn = R

n/πZn,

to which it is homeomorphic. The measure of the angle(L1,L2) will be denoted bymeas(L1,L2) = [ei2λ1, . . . , ei2λn] ∈ Tn/Sn, where(ei2λ1, . . . , ei2λn) is a representative omeas(L1,L2) verifying π > λ1 � · · · � λn � 0. In view of Proposition 2.5 above, we manow reformulate the classification theorem for Lagrangian pairs in the following way

Proposition 2.7. Given two pairs of Lagrangian subspaces(L1,L2) and (L′1,L

′2) of

Lagrangian subspaces ofCn, a necessary and sufficient condition for the existenceunitary mapψ ∈ U(n) such thatψ(L1) = L′

1 and ψ(L2) = L′2 is that meas(L1,L2) =

meas(L′1,L

′2). Equivalently, the map

χ :(L(n) ×L(n)

)/U(n) → T

n/Sn,

[L1,L2] �→ meas(L1,L2)

is one-to-one.

The mapχ is in fact a bijection: given[ei2λ1, . . . , ei2λn] ∈ Tn/Sn, consider anyLagrangianL1 ∈ L(n), (u1, . . . , un) an orthonormal basis forL1 and letL2 be the realsubspace ofCn generated by(eiλ1u1, . . . , e

iλnun). Since(eiλ1u1, . . . , eiλnun) is a unitary

basis ofCn overC, L2 is Lagrangian and meas(L1,L2) = [ei2λ1, . . . , ei2λn].

Corollary 2.8. The angle space(L(n)×L(n))/U(n), endowed with the quotient topologis homeomorphic to the quotient spaceT

n/Sn, both being Hausdorff and compact.

As a final remark, observe that the corresponding symplectic problem admits a sanswer: a necessary and sufficient condition for the existence of a symplectic maψ ∈


Sp(n) such thatψ(L1) = L′ andψ(L2) = L′ is that dim(L1 ∩ L2) = dim(L′ ∩ L′ ); the

ssify

.lt

cee

onar

1 2 1 2measure of the symplectic angle formed by two Lagrangian subspaces ofCn simply is thedimension of their intersection.

2.5.1. Orthogonal decomposition ofL1 associated tomeas(L1,L2)

The presentation given here follows that of [11]. This notion will enable us to clatriples of Lagrangian subspaces ofC2 (Theorem 3.1).

Let (L1,L2) be a pair of Lagrangian subspaces ofCn, and let (α21, . . . , α2

n) be arepresentative of meas(L1,L2) ∈ Tn/Sn. The unit complex numbersα2

1, . . . , α2n then are

the roots of the characteristic polynomialP(L1,L2) of the pair(L1,L2) (Proposition 2.5)Let α2

j1, . . . , α2

jmbe the distinct roots ofP(L1,L2). For k = 1, . . . ,m, define the rea

subspaceWk of L1 by Wk = {u ∈ L1 | αjku ∈ L2}. Observe thatWk is independenof the choice of the square root ofα2

jk, and thatW1 ⊕ · · · ⊕ Wm is independent, up

to permutation of the subspaces, of the choice of the representative(α21, . . . , α2

n) ofmeas(L1,L2) ∈ Tn/Sn.

Proposition 2.9 [11]. L1 decomposes as an orthogonal direct sum: L1 = W1 ⊕ · · · ⊕ Wm,the dimension ofWk being the multiplicity ofα2

jkas a root ofP(L1,L2).

Observe thatL2 then also decomposes as an orthogonal direct sum:L2 = αj1W1⊕· · ·⊕αjmWm. Furthermore, by considering the representative(ei2λ1, . . . , ei2λn) of meas(L1,L2),whereeiλ1, . . . , eiλn are the eigenvalues of the unitary mapϕ12, we see that the subspaWk of L1 is the intersection withL1 of the eigenspace ofϕ12 with respect to the eigenvalueiλjk . Given a Lagrangian triple(L1,L2,L3), the unitary mapsϕ12 andϕ13 therefore havethe same eigenspaces if and only if the orthogonal decompositions ofL1 associated tomeas(L1,L2) and meas(L1,L3) are the same (see Definition 5).

3. Triples of Lagrangian subspaces

3.1. First classification result for triples of Lagrangian subspaces ofC2

The following remark is valid for anyn. If (L1,L2,L3) and (L′1,L

′2,L

′3) are two

triples of Lagrangian subspaces ofCn that lie in the same orbit of the diagonal actiof U(n) on L(n) × L(n) × L(n), it follows from Section 2 that we have in particulmeas(L1,L2) = meas(L′

1,L′2) and meas(L1,L3) = meas(L′

1,L′3). Let L1 = W1 ⊕ · · · ⊕

Wm be the orthogonal decomposition ofL1 associated to meas(L1,L2) and letL1 = Z1 ⊕· · · ⊕ Zp be the orthogonal decomposition ofL1 associated to meas(L1,L3). DefineL′

1 =W ′

1 ⊕ · · · ⊕ W ′m andL′

1 = Z′1 ⊕ · · · ⊕ Z′

p similarly. Since meas(L1,L2) = meas(L′1,L

′2)

and meas(L1,L3) = meas(L′1,L

′3), the respective numbers of factorsm and p in the

above decompositions are indeed pairwise the same, and furthermore dimWk = dimW ′k

for k = 1, . . . ,m and dimZl = dimZ′l for l = 1, . . . , p. More specifically, if the unitary

mapψ ∈ U(n) sendsLj to L′j for j = 1,2,3, thenψ(Wk) = W ′

k for k = 1, . . . ,m andψ(Zl) = Z′

l for l = 1, . . . , p, as follows from the definition ofWk and Zl . Sinceψ is


unitary, we even haveψ(Wk ⊕ JWk) = W ′ ⊕ JW ′ for all k andψ(Zl ⊕ JZl) = Z′ ⊕ JZ′

cationn

de

at

esreal

nt

the

r

le

e

k k l l

for all l.Whenn = 2, the above remark admits an easy converse, which gives a first classifi

result for triples of Lagrangians ofC2. We shall use the following notations: givetwo triples (L1,L2,L3) and (L′

1,L′2,L

′3) of Lagrangian subspaces ofC2, let ϕ12 be

the only unitary map sendingL1 to L2 and verifying the conditions of the secondiagonalization lemma (Theorem 2.3), and let(eiλ12, eiµ12) be its eigenvalues, wherπ > λ12 � µ12 � 0, and defineϕ13, ϕ

′12, ϕ

′13 and(eiλ13, eiµ13), (eiλ′

12, eiµ′12), (eiλ′

13, eiµ′13)

similarly. As a preliminary remark to the statement of the classification result, observe thwhen bothϕ12 andϕ13 have two distinct eigenvalues, respectively denoted by(eiλ12, eiµ12)

and by(eiλ13, eiµ13), whereπ > λ12 > µ12 � 0 andπ > λ13 > µ13 � 0, thenW1 = {u ∈L1 | eiλ12u ∈ L2} and Z1 = {u ∈ L1 | eiλ13u ∈ L3} are one-dimensional real subspacof the Euclidean spaceL1, and therefore form a (non-oriented) angle measured by anumberθ ∈ [0, π

2 ], that will be denoted by meas(W1,Z1). A real numberθ ′ may be definedsimilarly in L′

1, sinceW ′1 areZ′

1 are also one-dimensional.

Theorem 3.1 (Unitary classification of Lagrangian triples ofC2). Given two triples(L1,L2,L3) and (L′

1,L′2,L

′3) of Lagrangian subspaces ofC2, a necessary and sufficie

condition for the existence of a unitary mapψ ∈ U(n) such thatψ(L1) = L′1,ψ(L2) = L′

2andψ(L3) = L′

3 is that either

(A) λ12 �= µ12, λ13 �= µ13 and

(λ12,µ12) = (λ′12,µ

′12),

(λ13,µ13) = (λ′13,µ

′13),

θ = θ ′,whereθ = meas(W1,Z1) ∈ [0, π

2 ] andθ ′ = meas(W ′1,Z

′1) are defined as above, or

(B) λ12 = µ12 or λ13 = µ13 and

{(λ12,µ12) = (λ′

12,µ′12),

(λ13,µ13) = (λ′13,µ

′13).

Observe that, in each case, the condition(λjk,µjk) = (λ′jk,µ

′jk) is equivalent to the

condition meas(Lj ,Lk) = meas(L′j ,L

′k).

Proof. Suppose that such aψ ∈ U(2) exists. Then, as we have seen earlier, meas(L1,

L2) = meas(L′1,L

′2) and meas(L1,L3) = meas(L′

1,L′3). Furthermore,ψ(W1) = W ′

1 andψ(Z1) = Z′

1, so that ifϕ12 andϕ13 both have distinct eigenvalues (that is, we are insituation of (A) above), we haveθ = θ ′ sinceψ|L1 :L1 → L′

1 is an orthogonal map.Conversely, suppose first that conditions (A) are fulfilled. Letw1 ∈ L1 be a generato

of W1 and letw′1 ∈ L′

1 be a generator ofW ′1. By choosingw2 in L1 orthogonal tow1

andw′2 in L′

1 orthogonal tow′1, we may define an orthogonal mapν :L1 → L′

1 sendingW1 to W ′

1 (and thereforeW2 = W⊥1 to W ′

2 = (W ′1)

⊥). Then the measure of the ang(W ′

1, ν(Z1)) = (ν(W1), ν(Z1)) is θ = θ ′, so there exists an orthogonal mapξ ∈ O(L′1)

such thatξ ◦ ν(W1) = W ′1 andξ ◦ ν(Z1) = Z′

1. The subspaceL1 being Lagrangian, thorthogonal mapξ ◦ ν can be extendedC-linearly to a unitary transformationψ ∈ U(2) ofC2 = L1 ⊕ JL1 sendingL1 to L′

1 by construction. ButL2 = eiλ12W1 ⊕ eiµ12W2 andL3 =


eiλ13Z1 ⊕ eiµ13Z2 (corollary of Proposition 2.9, henceψ(L2) = eiλ12W ′ ⊕ eiµ12W ′ = L′

e

in apairsalentt goingive

l

ld

1 2 2andψ(L3) = eiλ13Z′

1 ⊕ eiµ13Z′2 = L′

3.If now the conditions (B) are fulfilled, then for instanceL2 = eiλL1 and the result is a

consequence of the classification of pairs.�Observe that, given real numbers(λ12,µ12, λ13,µ13, θ) as in (A), it is always possibl

to find a triple (L1,L2,L3) such that meas(L1,L2) = [ei2λ12, ei2µ12], meas(L1,L3) =[ei2λ13, ei2µ13] and meas(W1,Z1) = θ . Indeed, letL1 be any Lagrangian ofC2 and let(u1, u2) be an orthonormal basis forL1, let d1 = Ru1, d2 = Ru2, and letd be the image

of d1 by the rotation of the euclidean spaceL1 with matrix( cosθ −sinθ

sinθ cosθ

)in the basis

(u1, u2), and setL2 = eiλ12d1⊕eiµ12d2 andL3 = eiλ13d ⊕eiµ13d⊥. Given numbers(λ12 =µ12 = λ,λ13,µ13) as in (B), we only need to setL2 = eiλL1 andL3 = eiλ13d1 ⊕ eiµ13d2.

Thus, the orbits of the diagonal action ofU(2) onL(2) ×L(2) ×L(2) are genericallycharacterized by the five invariantsλ12,µ12, λ13,µ13 andθ .

3.2. Geometric study of projective Lagrangians ofCP1

The aim of this section is to study the space(L(2) ×L(2) × L(2))/U(2) of the orbitsof the diagonal action ofU(2) on triples of Lagrangians subspaces ofC2, and morespecifically to describe it in terms of the map

ρ :(L(2) ×L(2) ×L(2)

)/U(2) → T

2/S2 × T2/S2 × T

2/S2,

[L1,L2,L3] �→ (meas(L1,L2),meas(L2,L3),meas(L3,L1)

)which will enable us to state the classification result for Lagrangian triplesway (Theorem 3.6) that is similar to the corresponding result for Lagrangian(Theorem 2.7). We shall see in Section 3.3 that this way of doing things is equivto considering orthogonal decompositions of one of the three subspaces. We are firsto describe the image of the mapρ and then prove that it is one-to-one. This will also ga topological description of the orbit space.Our main tool to characterize the image ofρ

will be the study of projective Lagrangians ofCP1.

3.2.1. Configurations of projective Lagrangians ofCP1

In the following, we shall constantly identify the complex projective lineCP1, endowedwith the Fubini–Study metric, with the Euclidean sphereS2 ⊂ R3 endowed with its usuastructure of oriented Riemannian manifold. We will denote byp the projection

p :C2\{0} → CP1,

z = (z1, z2) �→ p(z) = [z] = [z1, z2].As seen in 2.1, the image of a Lagrangian subspace ofC2 is a totally geodesic submanifoof CP1 S2 that is diffeomorphic toRP1 S1. Thereforel = p(L) is a great circleof S2, and the isometryσL of CP1, induced by the Lagrangian involutionσL, acts onS2 as the reflexion with respect to the plane ofR3 containing the great circlel = p(L).Recall that the unitary groupU(2) acts transitively on the Lagrangian GrassmannianL(2).


The action ofU(2) on CP1 is the same as the action of the special unitary groupSU(2),

g

r

ons ofived thenign for

isrd

.

t

which acts onS2 by the 2-sheeted universal covering maph : SU(2) → SO(3). The mapL ∈ L(2) �→ l = p(L) ⊂ CP1 S2 is equivariant for these actions. For anyϕ ∈ GL(2,C),we shall denote byϕ the induced map ofCP1 S2 into itself: ϕ . [z] = [ϕ(z)]. If ϕ ∈ U(2)

then ϕ acts onS2 as an element ofSO(3): indeedϕ = ei δ2 ψ , whereeiδ = detϕ and

ψ ∈ SU(2), and thenϕ = ψ in Aut(CP1), the action onS2 being obtained by considerinh(ψ), which we shall from now on simply denote byψ .

In the following, let (L1,L2,L3) be a triple of Lagrangian subspaces ofC2 andlet (l1, l2, l3) be the triple of corresponding great circles ofS2: lj = p(Lj ) for j =1,2,3. As above, we denote byϕjk the only unitary map sendingLj to Lk andverifying the conditions of the second diagonalization lemma. Let(eiλjk , eiµjk ) be itseigenvalues, whereπ > λjk � µjk � 0, and let(ujk, vjk) be an orthonormal basis foLj formed by eigenvectors ofϕjk: ϕjk(ujk) = eiλjkujk andϕ(vjk) = eiµjk vjk . Recall that(ei2λjk , ei2µjk ) then is a representative of meas(Lj ,Lk) ∈ T2/S2. We denote byL0 theLagrangian subspaceL0 = {(x, y) ∈ C2 | x, y ∈ R} of C2. We denote its projection onCP1

by l0 = p(L0).We are now going to relate the angles of projective Lagrangians ofCP1 S2 with the

Lagrangian angles defined in Section 2. Furthermore, in order to study configuratiprojective Lagrangians ofCP1, we are going to define a notion of sign of a projectLagrangian triple. To do so, we shall first define such a notion in a generic case anextend it to the remaining cases. At last, we shall see that there is also a notion of sLagrangian triples ofCn and that in the casen = 2, the triples(L1,L2,L3) and(l1, l2, l3)

have same sign.

Proposition 3.2 (Projection of a Lagrangian pair).Let (L1,L2) be a pair of Lagrangiansubspaces ofC2 and let(eiλ12, eiµ12) be the eigenvalues ofϕ12. Thenl1 = l2 if and onlyif λ12 = µ12. Furthermore, ifλ12 �= µ12, thenl2 is the image ofl1 by the(direct) rotationof angleα12 = λ12 − µ12 ∈]0,π[ around the point[v12] ∈ CP1 S2 ⊂ R3 ([v12] = Cv12being the complex eigenline ofϕ12 associated to the eigenvalueeiµ12 of lowest argument).

Proof. If λ12 = µ12 = λ thenL2 = eiλL1 and thereforel2 = l1 in CP1.If now λ12 �= µ12, suppose first thatL1 = L0 and that(u12, v12) is the standard bas

of C2. ThenL2 is the image ofL1 by the unitary map whose matrix in the standa

basis ofC2 is( eiλ12 0

0 eiµ12

)so thatL2 = {(eiλ12x, eiµ12y) | x, y ∈ R} andl2 = p(L2) =

{[eiλ12x, eiµ12y] | x, y ∈ R}. Therefore, in the chart[z1, z2] �→ z1/z2 of CP1 containing[v12] = [0,1], l2 is sent diffeomorphically onto the real line{ei(λ12−µ12) x

y| x, y ∈ R, y �=

0} = ei(λ12−µ12)d0 of the planeC R2, whered0 is the image ofl0 = l1 in this same chartThus,l2 and l1 intersect ata12 = [u12] andb12 = [v12], andl2 is the image ofl1 by therotation of angleα12 = λ12 − µ12 ∈]0,π[ around the pointb12 = v12, which means thathe oriented angle formed byl1 andl2 at b12 is of measureα12 = λ12 − µ12. Note that theoriented angle ata12 is of measureπ − α12 ∈]0,π[, since in the chart[z1, z2] �→ z2/z1, l2is diffeomorphic to the real lineei(µ12−λ12)d0 = ei(π−(λ12−µ12))d0.

If now (u12, v12) is not the standard basis ofC2, consider the unitary mapψ ∈ U(2)

sending the standard basis(e, f ) of C2 to (u12, v12). Then L0 = ψ−1(L1), and let


of the

ts

ctive

ianglein

e

le ones

tons

Fig. 1. Two projective Lagrangians ofCP1.

L = ψ−1(L2). Then[v12] = ψ . [f ], l2 = ψ(l) andl1 = ψ(l0). Since then meas(L0,L) =meas(L1,L2), we deduce from the above paragraph thatl = p(L) is the image ofl0 bythe rotation of angle theα12 around the point[f ]. Sinceψ ∈ SO(3), the oriented anglebetweenl1 andl2 at b12 = [v12] ∈ l1 ∩ l2 therefore also has measureα12. �

Observe that this proof also provides an elementary way to see whyL0, and thereforeevery Lagrangian subspace ofC2, projects to a great circle ofS2 CP1. We shall state aconverse to the above resultlater (see Proposition 3.4).

Note that the preceding result gives a complete description of the relative positionprojective Lagrangiansl1 andl2 only by means of the unitary mapϕ12. In particular, therotation described above is no other that the mapϕ12 of CP1 S2 into itself: l2 = ϕ12(l1).The axis of this rotation is the real line ofR3 generated by any of the antipodal poina12 = [u12] or b12 = [v12] of S2, (u12, v12) being a unitary basis ofC2 into which thematrix ofϕ12 is diagonal. See Fig. 1.

We are now going to describe all the possible configurations of the projeLagrangiansl1, l2 and l3 of CP1 S2 satisfying the following condition for(j, k) =(1,2), (2,3), (3,1): if lj �= lk thenlk is the image oflj by the direct rotationϕjk of S2 ⊂ R3

of angleαjk ∈]0,π[ around a specified pointbjk ∈ lj ∩ lk .

First case: l1, l2 andl3 are pairwise distinct.(a) Suppose first that the three pointsb12, b23, b31 are linearly independent inR3 (that

is, l1, l2, l3 do not have a common diameter). We may then consider the spherical tr(b12, b23, b31), whose sides[b12, b23], [b23, b31], [b31, b12] are, respectively, containedthe geodesicsl2, l3, l1. Sincelk is the image oflj by a direct rotation aroundbjk, the onlypossible configurations are the in ones Fig. 2.

On each sphere, we represent the anglesαjk around the pointbjk and we shall continuto do so in the following. We call the first trianglenegativeand the second trianglepositive.Let us explain this terminology and prove that these cases are indeed the only possibwhen thebjk are pairwise distinct.

Let ϕ = ϕ31 ◦ ϕ23 ◦ ϕ12 ∈ U(2). Thenϕ(L1) = L1 and thereforeϕ(l1) = l1. There areonly two possible cases: eitherϕ preserves a given orientation onl1, or it reverses thaorientation. Butϕ = ϕ31 ◦ ϕ23 ◦ ϕ12 is the map obtained by composing the three rotatiϕjk around thebjk ’s. Whenϕ reverses the orientation ofl1, which we will call thenegative


fs

same

l cell

ing

Fig. 2. Triples of projective Lagrangians ofCP1 in general position.

Fig. 3. The tetrahedron(∆).

case, then(α12, α23, α31) are the angles of the spherical triangle(b12, b23, b31). Whenϕ

preserves the orientation ofl1, which we will call thepositive case, then the angles othe triangle(b12, b23, b31) areβjk, whereβjk = π − αjk ∈]0,π[. Observe that this givea series of necessary conditions for the existence of a triple(L1,L2,L3) of Lagrangiansubspaces ofC2 projecting onto a triple(l1, l2, l3) of great circles ofS2 that do not havea common diameter. Namely, for instance, if the triangle(b12, b23, b31) has anglesαjk

(negative case), we necessarily have

(∆)

α12, α23, α31 ∈]0,π[,α12 + α23 + α31 > π,

α12 + π > α23 + α31,

α23 + π > α31 + α12,

α31 + π > α12 + α23,

since(α12, α23, α31) are the angles of a spherical triangle. In the positive case, theconditions apply to(β12, β23, β31). In the following we shall write(α12, α23, α31) ∈ ∆

to say that(α12, α23, α31) verify this set of conditions.∆ is an open subset ofR3 andits closure∆ in R3 is a tetrahedron, and is therefore endowed with a 3-dimensionacomplex structure (see Fig. 3).

(b) Suppose now thatb12, b23 and b31 are not linearly independent. Thenl1, l2, l3have a common diameter and we either haveb12 = b23 = b31 or, for instance,b12 = b23and b31 �= b12. Sincel1, l2 . l3 are still supposed to be pairwise distinct and satisfylk = ϕjk(lj ), the only possible configurations are the ones shown in Fig. 4.


ave the

e

Fig. 4. Exceptional triples of pairwisedistinct projective Lagrangians ofCP1.

These 4 cases correspond to degenerate spherical triangles, so that we hfollowing respective necessary conditions:

α12, α23, α31 ∈]0,π[α12 + α23 + α31 = π

α12 + π > α23 + α31α23 + π > α31 + α12α31 + π > α12 + α23

β12, β23, β31 ∈]0,π[β12 + β23 + β31 = π

β12 + π > β23 + β31β23 + π > β31 + β12β31 + π > β12 + β23

β12, β23, β31 ∈]0,π[β12 + β23 + β31 > π

β12 + π > β23 + β31β23 + π > β31 + β12β31 + π = β12 + β23

α12, α23, α31 ∈]0,π[α12 + α23 + α31 > π

α12 + π > α23 + α31α23 + π > α31 + α12α31 + π = α12 + α23

This means that either theαjk or theβjk, depending on the negativity or positivity of thtriple (l1, l2, l3), are located in an open 2-cell of the 3-dimensional complex∆ (see Fig. 3).The remaining 2-cells are obtained whenb23 = b31 andb12 �= b23, and whenb31 = b12 andb23 �= b31.

Second case: l1, l2 andl3 are not pairwise distinct.


rnd

es

Fig. 5. Triples of non-pairwise distinct projective Lagrangians ofCP1.

(a) Suppose first, for instance, thatl1 = l2 andl3 �= l1. Sincel1 = l2, we may consideeither thatα12 = 0 or thatα12 = π and that it is the angle of a direct rotation aroub23 ∈ l2 ∩ l3 = l1 ∩ l3, so that the notion of negative and positive triples is still valid. Thenthe only possible configurations ofl1, l2, l3 are the ones shown in Fig. 5.

Those configurations correspond to open 1-cells of∆ (see Fig. 3):

α12 = 0 α23, α31 ∈]0,π[α12 + α23 + α31 = π

α12 + π = α23 + α31α23 + π > α31 + α12α31 + π > α12 + α23

β12 = π β23, β31 ∈]0,π[β12 + β23 + β31 > π

β12 + π > β23 + β31β23 + π = β31 + β12β31 + π = β12 + β23

β12 = 0 β23, β31 ∈]0,π[β12 + β23 + β31 = π

β12 + π = β23 + β31β23 + π > β31 + β12β31 + π > β12 + β23

α12 = π α23, α31 ∈]0,π[α12 + α23 + α31 > π

α12 + π > α23 + α31α23 + π = α31 + α12α31 + π = α12 + α23

The remaining 1-cells are obtained whenl2 = l3 andl1 �= l2 and whenl3 = l1 andl2 �= l3.(b) Suppose at last thatl1 = l2 = l3. The notion of negative and positive tripl

remains valid by considering either thatαjk = 0 or that αjk = π , and that thebjk ’s


all are a sameb chosen arbitrarily inl1 = l2 = l3. Then the possible configurations

ionsthee

n

itive.only

arysll give

y

veates

le

on S2 correspond to the 0-cells of∆; that is, in the negative case,(α12, α23, α31) =(π,0,0), (0,π,0), (0,0,π) or (π,π,π), and in the positive case:(β12, β23, β31) =(π,0,0), (0,π,0), (0,0,π) or (π,π,π). Observe that in the cases where the three rotatϕjk occur around a same pointbjk or around two diametrically opposed points, thennegative case corresponds to(α12 + α23 + α31) ≡ π (mod 2π), and the positive cascorresponds to(β12+β23+β31) ≡ π (mod 2π), that is to(α12+α23+α31) ≡ 0 (mod 2π).Also note that ifl1, l2, l3 are pairwise distinct great circles ofS2 that do not have a commodiameter, the pairwise intersectionslj ∩ lk determine 6 points onS2, which in turn giverise to 8 spherical triangles, four of which are negative, the other four being posTwo triangles with a common edge have opposite sign, whereas two triangles witha common vertex have same sign.

From the study above, we deduce that a Lagrangian triple(L1,L2,L3) projects ona triple (l1, l2, l3) of great circles ofS2, that is either positive with(α12, α23, α31) ∈∆ or negative with(β12, β23, β31) ∈ ∆. In particular, these conditions are necessconditions for([ei2λ12, ei2µ12], [ei2λ23, ei2µ23], [ei2λ31, ei2µ31]) to be the triple of measureof a Lagrangian triple. Before showing that these conditions are sufficient, we shaanother way of determining if a triple(l1, l2, l3) is negative or positive.

Proposition 3.3. Let (L1,L2,L3) be a triple of Lagrangian subspaces ofC2, and setϕ = ϕ31◦ϕ23◦ϕ12. Writedetϕ = eiδ, whereδ = (λ12+µ12)+ (λ23+µ23)+ (λ31+µ31).Then δ ≡ 0 (mod π), and (l1, l2, l3) is negative ifδ ≡ π (mod 2π) and positive ifδ ≡ 0 (mod 2π).

Observe that whenδ ≡ 0 (mod 2π) we have detϕ = 1, so that we might also sathat the triple(L1,L2,L3) of Lagrangian subspaces ofC2 is positive. Similarly, whenδ ≡ π (mod 2π), detϕ = −1 and(L1,L2,L3) will then be said to be negative. The aboproposition then says that the triples(L1,L2,L3) and(l1, l2, l3) have same sign. Note ththe notion of sign of a Lagrangian triple(L1,L2,L3) is also valid for Lagrangian subspacof Cn.

Proof. Suppose first thatL1 = L0 and that(u12, v12) is the standard basis ofC2. Write

ϕjk = eiλjk+µjk

2 ψjk whereψjk ∈ SU(2) andei(λjk+µjk) = detϕjk . Setψ = ψ31◦ψ23◦ψ12,

so thatϕ = ei δ2 ψ , whereδ = ∑

j,k(λjk + µjk). Note thatϕjk = ψjk and ϕ = ψ . In

particular,ψ(l0) = l0. But the matrix ofψ in the standard basis ofC2 is of the form

A = ( s −t

t s

)wheres, t ∈ C and verify|s|2+|t|2 = 1. Sincel0 = {[x, y] ∈ CP1, x, y ∈ R},

ψ(l0) = l0 if and only if A = ( a −b

b a

)or A = ( ia ib

ib −ia

)wherea, b ∈ R and verify

a2 + b2 = 1.In the first caseψ(L0) = L0, so thatL0 = ϕ(L0) = ei δ

2 , and sinceL0 is totally realwe haveδ

2 ≡ 0 (modπ), that isδ ≡ 0 (mod 2π). In the second caseψ(L0) = i . L0, so

thatL0 = ϕ(L0) = ei δ2 i . L0 and thereforeδ2 ≡ π

2 (modπ), that isδ ≡ π (mod 2π). Now

recall thatϕ(l0) = (l0). WhenA = ( a −b

b a

), a given orientation onl0 is preserved byψ

(since, in the chart[z1, z2] �→ z1z2

, the mapx ∈ R �→ ax−bbx+a

is increasing), so that the trip


(l0, l2, l3) is positive. WhenA = ( ia ib ), a given orientation onl0 is reversed byψ

le

p

tede

sle

een

es

ib −ia

(since, in the chart[z1, z2] �→ z1z2

, the mapx ∈ R �→ ax+bbx−a

is decreasing), so that the trip(l0, l2, l3) is negative.

Suppose now that(u12, v12) is not the standard basis ofC2, and define the unitary maν ∈ U(2) sending the standard basis to(u12, v12). Let L′

2 = ν−1(L2), L′3 = ν−1(L3), l′2 =

p(L′2) andl′3 = p(L3). Thenν−1 ◦ϕ ◦ ν sendsL0 to L0 and det(ν−1 ◦ϕ ◦ ν) = detϕ = eiδ.

From the study above, the triple(l0, l′2, l′3) is positive if and only ifδ ≡ 0 (mod 2π), andnegative if and only ifδ ≡ π (mod 2π). But sincel1 = ν(l0), l2 = ν(l′2) andl3 = ν(l′3) withν ∈ SO(3), the triples(l1, l2, l3) and(l′0, l′2, l′3) have same sign.�3.2.2. Second classification result for triples of Lagrangian subspaces ofC2

As a converse to Proposition 3.2, it is possible, given two distinct great circlesl1 �= l2 ofS2 CP1, to describe the measure of the angle(L1,L2) between two Lagrangians ofC2

that project respectively tol1 andl2. Recall that two distinct great circlesl1 �= l2 intersectalong two antipodal pointsa, b, and thatα ∈]0,π[ is said to be the measure of the orienangle betweenl1 andl2 atb ∈ l1 ∩ l2 if l2 is the image ofl1 by the (direct) rotation of anglα aroundb.

Proposition 3.4 (Lifting lemma). Let l1 �= l2 be two distinct projective Lagrangianof CP1 S2, let b ∈ l1 ∩ l2 and let α ∈]0,π[ be the measure of the oriented ang(l1, l2) at b. Then, givenλ and µ such thatπ > λ > µ � 0, and given a LagrangiansubspaceL1 ∈ p−1(l1), there exists a unique Lagrangian subspaceL2 ∈ p−1(l2) such thatmeas(L1,L2) = [ei2λ, ei2µ].

Proof. Let v ∈ L1 such thatp(v) = b. We may choosev such that‖v‖ = h(v, v) = 1. Letthenu ∈ L1 such that(u, v) is an orthonormal basis forL1. SinceL1 is Lagrangian,(u, v)

is a unitary basis forC2. Let ψ be the unitary transformation ofC2 whose matrix in the

basis(u, v) is( eiλ 0

0 eiµ

), and letL = ψ(L1). ThenL is Lagrangian and meas(L1,L) =

[ei2λ, ei2µ]. Therefore, by Proposition 3.2,l = p(L) is a great circle ofS2, distinct of l1sinceλ �= µ, that intersectsl1 at p(v) = b and the measure of the oriented angle betwl1 andl atb is λ − µ = α, so thatl = l2.

As for unicity, if L′ ∈ p−1(l2), then, again by Proposition 3.2, we know thatL′ = eiθ . L,whereθ ∈]0,π[. The unitary mapeiθ . ψ then sendsL1 to L′, and its matrix in the

unitary basis(u, v), which is an orthonormal basis forL1 is( ei(θ+λ) 0

0 ei(θ+µ)

)so that

meas(L1,L′) = [ei2((θ+λ) mod π), ei((θ+µ) modπ)], with π > (θ + λ) mod π > (θ + µ)

modπ � 0. Since meas(L1,L) = meas(L1,L′), we have in particular(θ + λ) modπ = λ,

henceθ modπ = 0 (and soθ = 0) andL′ = eiθ . L = L. �The next theorem is our main result: it completely describes the image of the mapρ and

lays the ground for the second classification theorem for triples of Lagrangian subspacof C2.


Theorem 3.5. Given a triple of measures([ei2λ12, ei2µ12], [ei2λ23, ei2µ23], [ei2λ31, ei2µ31])

ds

s

at

ts

tion

e

f

tier.ce

a

satisfying the conditionsπ � λjk � µjk � 0, setαjk = λjk − µjk ∈ [0,π], βjk = π −αjk ∈ [0,π] and δ = (λ12 + µ12) + (λ23 + µ23) + (λ31 + µ31). Then, a necessary ansufficient condition for the existence of a triple(L1,L2,L3) of Lagrangian subspaceof C2 such thatmeas(L1,L2) = [ei2λ12, ei2µ12], meas(L2,L3) = [ei2λ23, ei2µ23], andmeas(L3,L1) = [ei2λ31, ei2µ31] is that

δ ≡ π (mod 2π) and (α12, α23, α31) ∈ ∆ (negative case),

or

δ ≡ 0 (mod 2π) and (β12, β23, β31) ∈ ∆ (positive case).

(Here we allowλjk = π so that we may haveαjk = π andβjk = 0.)

Proof. The study made in Section 3.2.1 shows that these conditions are necessary.Conversely, suppose first thatδ ≡ π (mod 2π) and that(α12, α23, α31) lie in the open

set ∆. Then there exists a negative triple(l1, l2, l3) of pairwise distinct great circleof S2 such thatlk is the image oflj by the direct rotation of angleαjk around acertain pointbjk ∈ lj ∩ lk for (j, k) = (1,2), (2,3), (3,1), and we may suppose thl1 = l0. Let L1 = L0. Then, by Proposition 3.4, there exists a unique LagrangianL2 ∈p−1(l2) such that meas(L1,L2) = [ei2λ12, ei2µ12]. Again by Proposition 3.4, there exisa unique LagrangianL3 ∈ p−1(l3) such that meas(L2,L3) = [ei2λ23, ei2µ23], and a uniqueLagrangianL4 ∈ p−1(l1) such that meas(L3,L4) = [ei2λ31, ei2µ31]. Let ϕ34 be the uniqueunitary map sendingL3 to L4 and satisfying the conditions of the second diagonaliza

Lemma 2.3, and letϕ = ϕ34◦ϕ23◦ϕ12. Thenϕ(L1) = L4 and detϕ = eiδ. Writeϕ = ei δ2 ψ ,

whereψ ∈ SU(2). Thenψ(l1) = l1, and since(l1, l2, l3) is negative, we have, from thstudy made in Section 3.2.1, thatψ(L1) = i . L1, hence, asδ ≡ π (mod 2π), we have

L4 = ϕ(L1) = ei δ2 i . L1 = L1.

Suppose now that(α12, α23, α31) ∈ ∂∆. If (α12, α23, α31) lay in an open 2-cell o∆, there exists a negative triple(l1, l2, l3) of pairwise distinct great circles ofS2 suchthat lk is the image oflj by the direct rotation of angleαjk around a certain poinbjk ∈ lj ∩ lk for (j, k) = (1,2), (2,3), (3,1), and we can therefore conclude as earlIf now (α12, α23, α31) lay in an open 1-cell of∆, there exists a negative triple, for instanof the form(l1, l2 = l1, l3 �= l2), such thatl3 is the image ofl2 by the rotation of angleα23aroundb23 ∈ l2 ∩ l3 and such thatl1 is the image ofl3 by the rotation of angleα31 aroundb31 ∈ l3 ∩ l1. Sincel2 = l1, α12 is either 0 orπ , and by settingb12 = b23 (or b12 = b31),we have thatl2 is the image ofl1 by the rotation of angleα12 aroundb12 ∈ l1 ∩ l2 (seeFig. 5). LetL1 = L0. If α12 = 0, thenλ12 = µ12 and we setL2 = eiλ12 . L1. If α12 = π ,thenλ12 = π andµ12 = 0, and we setL2 = L1. In both casesL2 ∈ p−1(l2) = p−1(l1)

and meas(L1,L2) = [ei2λ12, ei2µ12]. Sincel1 = l2 �= l3, there exists, by Proposition 3.4,unique LagrangianL3 ∈ p−1(l3) such that meas(L2,L3) = [ei2λ23, ei2µ23], and a uniqueLagrangianL4 ∈ p−1(l1) such that meas(L3,L4) = [ei2λ31, ei2µ31]. As earlier, since the

triple (l1, l2, l3) is negative, we haveL4 = ei δ2 i . L1 = L1.

At last, if (α12, α23, α31) is a 0-cell of∆, that is, if(α12, α23, α31) = (π,0,0), (0,π,0),(0,0,π) or (π,π,π), thenL1 = L2 = L3 = L0 meet the required conditions.


If now, δ ≡ 0 (mod 2π), the condition(β12, β23, β31) ∈ ∆ implies the existence of a.

es

d

re

n

ined

s

eults

positive triple(l1, l2, l3) of pairwise distinct great circles ofS2, with anglesαjk as requiredReasoning the same way, we find 4 LagrangiansL1, L2, L3 and L4 with prescribed

angles[ei2λjk , ei2µjk ], and since(l1, l2, l3) is positive we have:L4 = ϕ(L1) = ei δ2 . L1,

and therefore, asδ ≡ 0 (mod 2π), L4 = L1.The other cases are treated identically. �We now obtain the following classification theorem for triples of Lagrangian subspac

of C2.

Theorem 3.6 (Unitary classification of Lagrangian triples ofC2, second version).Giventwo triples(L1,L2,L3) and(L′

1,L′2,L

′3) of Lagrangian subspaces ofC2, a necessary an

sufficient condition for the existence of a unitary mapϕ ∈ U(2) such thatϕ(L1) = L′1,

ϕ(L2) = L′2 and ϕ(L3) = L′

3 is that meas(L1,L2) = meas(L′1,L

′2), meas(L2,L3) =

meas(L′2,L

′3), andmeas(L3,L1) = meas(L′

3,L′1).

Equivalently, the mapρ : (L(2) × L(2) × L(2))/U(2) → T2/S2 × T2/S2 × T2/S2is one-to-one and is therefore a homeomorphism from the orbit space(L(2) × L(2) ×L(2))/U(2) onto a closed subset of the measure spaceT2/S2 × T2/S2 × T2/S2.

Proof. It only remains to prove that the above conditions are sufficient. Let(L1,L2,L3)

and (L′1,L

′2,L

′3) be two Lagrangian triples such that meas(Lj ,Lk) = meas(L′

j ,L′k) for

all j , k. Then, the (generalized) triangles(b12, b23, b31) and(b′12, b

′23, b

′31) have the same

angles, so there exists a mapψ ∈ SU(2) such thatψ(bjk) = b′jk for all j , k. Since

moreoverδ = δ′, the triples(l1, l2, l3) and (l′1, l′2, l′3) have same sign and we therefoeven haveψ(lj ) = l′j for all j . Equivalently,p(ψ(Lj )) = ψ(p(Lj )) = ψ(lj ) = l′j =p(L′

j ). In particular, by Proposition 3.2, we haveL′1 = eiθ . ψ(L1) for someθ ∈ [0,π[.

Set ϕ = eiθ . ψ ∈ U(2). Then ϕ(L1) = L′1 and p(ϕ(L2)) = ϕ(p(L2)) = ψ(l2) = l′2

and meas(L′1, ϕ(L2)) = meas(ϕ(L1), ϕ(L2)) = meas(L1,L2) = meas(L′

1,L′2), hence, by

unicity in Proposition 3.4,ϕ(L2) = L′2. Likewise,p(ϕ(L3)) = l′3 and meas(L′

2, ϕ(L3)) =meas(L′

2,L′3), thereforeϕ(L3) = L′

3. �The above study suggests using trigonometry inCPn−1 to classify triples of Lagrangia

subspaces ofCn.

3.3. Equivalence of the two classification results

We now wish to explain why the two classification results that we have obta(Theorems 3.1 and 3.6) are indeed equivalent.

Let (L1,L2,L3) be a triple of Lagrangian subspaces ofC2. If one of the unitary map

ϕjk is of the formeiλId (for instance, ifL2 = eiλ . L1), and if (L′1,L

′2,L

′3) is a triple

of Lagrangian subspaces such that meas(L1,L2) = meas(L′1,L

′2) and meas(L1,L3) =

meas(L′1,L

′3) (or equivalently meas(L3,L1) = meas(L′

3,L′1)), we necessarily hav


′3), which proves that in this case the two classification res

are indeed the same.


re

Fig. 6. Relation between the two classification results.

If now each unitary mapϕjk has two distinct eigenvalueseiλjk and eiµjk , whereπ > λjk > µjk � 0, setd12 = Ru12 ⊂ L1 and d13 = Ru13 ⊂ L1 (whereu12 and u13are defined as earlier by means ofϕ12 and ϕ13), and let θ = meas(d12, d13) ∈ [0, π

2 ]be the measure of the non-oriented angle formed by the real linesd12 and d13 in theEuclidean spaceL1. Recall thatL1 = d12 ⊕ d⊥

12 = d13 ⊕ d⊥13, whered⊥

12 = Rv12 andd⊥

13 = Rv13, and observe thatθ is also the measure of the angle(d⊥12, d

⊥13). As earlier,

definebjk = [vjk] ∈ lj ∩ lk ⊂ CP1 S2.Observe now thatb31 ∈ l1 ∩ l3 is one of the two antipodal pointsa13 or b13 ∈ l1 ∩ l3.

One can then check the following remarks:

• The measure of the non-oriented angle formed by the two vectorsb12 and b13 ofS2 ⊂ R3 is 2θ ∈ [0,π] (in particular, two orthogonal vectors ofL1 project ontoantipodal points ofS2).

• If µ13 = 0 thenb31 = b13 and therefore meas(b12, b31) = 2θ . If µ13 �= 0 thenb31 = a13and therefore meas(b12, b31) = π − 2θ .

Let now(γ12, γ23, γ31) be the measures of the angles of the spherical triangle(b12, b23, b31)

(from the study of projective Lagrangians ofCP1, we know that either(γ12, γ23, γ31) =(α12, α23, α31) or (γ12, γ23, γ31) = (β12, β23, β31), whereαjk = λjk − µjk and βjk =π − αjk). Let η ∈ [0,π] be the measure of the non-oriented angle(b12, b31) (from thestudy above, we know that eitherη = 2θ or η = π − 2θ , see Fig. 6). Then we know fromspherical trigonometry that cosγ23 = sinγ12sinγ31cosη − cosγ12cosγ31.

The next proposition completes the explanation why our two classification results aindeed equivalent.

Proposition 3.7. Let (L1,L2,L3) be a triple of Lagrangian subspaces ofC2 suchthat ϕ12, ϕ23 and ϕ31 have distinct eigenvalues. Let(L′

1,L′2,L

′3) be a triple of

Lagrangian subspaces ofC2 such thatmeas(L1,L2) = meas(L′

1,L′2) andmeas(L1,L3) =

meas(L′1,L

′3) (this last condition being equivalent tomeas(L3,L1) = meas(L′

3,L′1)). Let

θ = meas(d12, d13) ∈ [0, π2 ] be the measure of the non-oriented angle(d12, d13) in L1 and

defineθ ′ = meas(d ′12, d

′13) ∈ [0, π

2 ] in L′1 similarly. Thenmeas(L2,L3) = meas(L′

2,L′3) if

and only ifθ = θ ′.


Proof. Assume first that meas(L2,L3) = meas(L′ ,L′ ). Since we also have meas(L1,L2)

rbitr

d

ne

s

e

dtic

e

2 3= meas(L′1,L

′2) and meas(L1,L3) = meas(L′

1,L′3), we getδ = δ′: the triples(l1, l2, l3)

and (l′1, l′2, l′3) have same sign. As a consequence, the spherical triangles(b12, b23, b31)

and(b′12, b

′23, b

′31) have the same angles:γjk = γ ′

jk ∈]0,π[ for all j , k. Since meas(L1,

L3) = meas(L′1,L

′3) we haveµ13 = µ′

13, therefore, by the above remarks, eitherb31 =b13 and b′

31 = b′13 (when µ13 and µ′

13 equal zero) orb31 = a13 and b′31 = a′

13 (whenµ13 = µ′

13 �= 0), so eitherη = 2θ andη′ = 2θ ′ or η = π − 2θ andη′ = π − 2θ ′. But thenfrom the relation from spherical trigonometry recalled above, since sinγjk �= 0 for all j, k,we have cosη = cosη′, and sinceη,η′ ∈ [0,π] we getη = η′, thereforeθ = θ ′.

Assume now thatθ = θ ′. Then, as in Proposition 3.1, there exists a unitaryψ ∈ U(2)

such thatψ(Lj ) = L′j for j = 1,2,3, so that meas(L′

2,L′3) = meas(L2,L3). �

4. Applications

4.1. Computation of the inertia index of a Lagrangian triple

4.1.1. Basic properties of the inertia indexIn contrast with the corresponding situation for pairs of Lagrangian subspaces, the o

of a triple (L1,L2,L3) of Lagrangian subspaces of a 2n-dimensional symplectic vectospace(V ,ω) under the diagonal action of the symplectic groupSp(V ) is not characterizeby the integersn12 = dim(L1 ∩ L2), n23 = dim(L2 ∩ L3), n31 = dim(L3 ∩ L1) andn0 = dim(L1 ∩ L2 ∩ L3), which are invariants of this action. To classify the orbits, ointroduces the notion ofinertia index(sometimes calledMaslov index, or simply index, orsignature) of a Lagrangian triple(L1,L2,L3). For the following definition and propertieof the inertia index, we refer to Kashiwara [9, p. 486 sqq].

Definition 4 (Inertia index). The inertia index of the Lagrangian triple(L1,L2,L3),denoted byτ (L1,L2,L3), is the signature of the quadratic formq defined on the 3n-dimensional vector spaceL1 ⊕ L2 ⊕ L3 by: q(x1, x2, x3) = ω(x1, x2) + ω(x2, x3) +ω(x3, x1).

In a suitable basis ofL1 ⊕ L2 ⊕ L3, one can representq by a diagonal matrix whosentries consist ofr terms+1, s terms−1 and 3n− r − s terms 0, the integersr ands beingindependent from the choice of the basis. What is calledsignatureof q here, and denotesgn(q) is the integer sgn(q) = r − s. From the definition, we see that for any symplecmapψ ∈ Sp(n), we haveτ (ψ(L1),ψ(L2),ψ(L3)) = τ (L1,L2,L3). We summarize hersome of the properties of the inertia index that we will need in the following.

Proposition 4.1. The inertia index has the following properties:

(i) τ (L1,L2,L3) ≡ n − (n12 + n23 + n31) mod 2Z;(ii) |τ (L1,L2,L3)| � n + 2n0 − (n12 + n23 + n31).


We may now state the theorem of symplecticclassification of triples of Lagrangian

l,

eryevery

ic

t

subspaces of(V ,ω), which is due to Kashiwara. Ford = (n0, n12, n23, n31, τ ) ∈ N4 × Z,we set:

Od = {(L1,L2,L3) ∈ L(V ) ×L(V ) ×L(V )

| dim(L1 ∩ L2 ∩ L3) = n0, dim(L1 ∩ L2) = n12, dim(L2 ∩ L3) = n23,

dim(L3 ∩ L1) = n31, τ (L1,L2,L3) = τ}.

Theorem 4.2 (Symplectic classification of Lagrangian triples, [9, p. 493]).Od is non-empty if and only ifd = (n0, n12, n23, n31, τ ) satisfies the conditions:

(i) 0 � n0 � n1, n2, n3 � n;(ii) n12 + n23 + n31 � n + 2n0;(iii) |τ | � n + 2n0 − (n12 + n23 + n31);(iv) τ ≡ n − (n12 + n23 + n31) mod 2Z.

If (L1,L2,L3) and(L′1,L

′2,L

′3) are two triples of Lagrangian subspaces ofV , there exists

a symplectic mapψ ∈ Sp(V ) such thatψ(L1) = L′1, ψ(L2) = L′

2 andψ(L3) = L′3 if and

only if n0 = n′0, n12 = n′

12, n23 = n′23, n31 = n′

31 andτ = τ ′.

Thus, the diagonal action ofSp(V ) on L(V ) × L(V ) × L(V ) has only finitely manyorbits and these orbits are theOd ’s, whered satisfies conditions (i) to (iv) above.

We now specialize to the case whereV = Cn, and make the following definition.

Definition 5. A triple (L1,L2,L3) of Lagrangian subspaces ofCn is said to be anexceptional tripleif the unitary mapsϕ12 andϕ13 have the same eigenspaces.

As can be seen from the casen = 2, a triple(L1,L2,L3) is generically not exceptionawhich justifies the terminology.

We now extract from Theorem 4.2 the following proposition, which will be vimportant to us. It shows the interest of the notion of exceptional Lagrangian triple:Lagrangian triple is symplectically equivalent to an exceptional triple.

Proposition 4.3. Let (L1,L2,L3) be a triple of Lagrangian subspaces ofCn. Then there

exists an exceptional triple(L′1,L

′2,L

′3) and a symplectic mapψ ∈ Sp(n) such that

L′j = ψ(Lj ) for j = 1,2,3; that is, each orbitOd of the diagonal action of the symplect

group Sp(n) onL(n) ×L(n) ×L(n) contains at least one exceptional triple.

4.1.2. From angles to inertia indexWe saw earlier (Proposition 3.3) that the quantityδ = (λ12 + µ12) + (λ23 + µ23) +

(λ31 + µ31) defined for Lagrangian subspaces ofC2, verifies δ ≡ 0 (mod π) andcontains information about the triple(L1,L2,L3). Namely, if δ ≡ 0 (mod 2π) the triple(L1,L2,L3) is positive (that is, settingϕ = ϕ31 ◦ ϕ23 ◦ ϕ12, we have detϕ = eiδ = 1 > 0),and if δ ≡ π (mod 2π) the triple is negative (that is detϕ = eiδ = −1 < 0). The interesof that notion was that the triple(l1, l2, l3) of projective Lagrangians ofCP1 had same


sign as(L1,L2,L3): if δ ≡ 0 (mod 2π) the transformationϕ = ϕ31 ◦ ϕ23 ◦ ϕ12 of CP1

if);, which

an

t

d

s

,,n

s

preserves the orientation onl1 (the triple (l1, l2, l3) is then said to be positive), andδ ≡ π (mod 2π) thenϕ reverses that orientation (the triple(l1, l2, l3) is said to be negativeand this enabled us to distinguish between positive and negative spherical triangleswas essential in order to determine the image of the mapρ : (L(2)×L(2)×L(2))/U(2) →T2/S2 × T2/S2 × T2/S2. But δ can actually be defined for a triple of Lagrangisubspaces ofCn for any integern. For such a triple(L1,L2,L3), sinceσ 2

Lj= Id, we have

the following relation:(σL1 ◦ σL3) ◦ (σL3 ◦ σL2) ◦ (σL2 ◦ σL1) = Id, and the determinanof this unitary map therefore is of the formei2δ with δ ≡ 0 (mod π). Whenn = 2, theeigenvalues of the unitary mapσLk ◦ σLj are ei2λjk and ei2µjk , so that we have indeeδ = (λ12 + µ12) + (λ23 + µ23) + (λ31 + µ31).

In the following, we shall consider a triple(L1,L2,L3) of Lagrangian subspaceof Cn, for arbitraryn. We shall denote the measures of the angles(L1,L2), (L2,L3)

and(L3,L1) by meas(L1,L2) = [ei2α1, . . . , ei2αn], meas(L2,L3) = [ei2β1, . . . , ei2βn] andmeas(L3,L1) = [ei2γ1, . . . , ei2γn], whereπ > α1 � · · · � αn � 0, π > β1 � · · · � βn � 0andπ > γ1 � · · · � γn � 0. We then haveδ = ∑n

j=1(αj + βj + γj ), whereei2δ = 1 isthe determinant of the unitary map(σL1 ◦ σL3) ◦ (σL3 ◦ σL2) ◦ (σL2 ◦ σL1) = Id, so thatδ ≡ 0 (mod π). Sinceδ, which we shall also denoteδ(L1,L2,L3) to avoid confusionis defined by means of the measures of the angles(Lj ,Lk) (that is, up to permutationthe eigenvalues of the unitary mapsσLk ◦ σLj ), δ is invariant under the diagonal actioof the unitary groupU(n) onL(n) ×L(n) ×L(n): if ϕ ∈ U(n), thenδ(ϕ(L1,L2,L3)) =δ(L1,L2,L3).

The next theorem is the main result of this subsection.

Theorem 4.4. Let (L1,L2,L3) be a triple of Lagrangian subspaces ofCn, and setnjk = dim(Lj ∩Lk), τ = τ (L1,L2,L3) andδ = δ(L1,L2,L3). Thenτ = 3n− 2δ

π−(n12+

n23 + n31).

Lemma 4.5. If c : t ∈ [0,1] �→ (L1(t),L2(t),L3(t)) ∈ L(n)×L(n)×L(n) is a continuousmap such that the dimensionsnjk(t) = dim(Lj (t) ∩ Lk(t)) of the pairwise intersectionare constant functions oft , then the mapδ : t �→ δ(L1(t),L2(t),L3(t)) is a constant map.

Observe that this result is also true for the inertia index (see [9, pp. 487–488]).

Proof. Since thenjk ’s remain constant along the deformation, the non-zeroαj (t), βj (t)

and γj (t) vary continuously. Therefore,δ(L1,L2,L3) varies continuously. Asδ(t) ≡0 (modπ), δ is a constant map.�Lemma 4.6. Let (L1,L2,L3) be a triple of Lagrangian subspaces ofCn and letψ ∈ Sp(n)

be a symplectic map. Thenδ(ψ(L1),ψ(L2),ψ(L3)) = δ(L1,L2,L3), that is: δ is asymplectic invariant.

Proof. Since the symplectic group is connected, there exists a continuous patht ∈ [0,1] �→ψt ∈ Sp(n) such thatψ0 = Id and ψ1 = ψ . For all t ∈ [0,1], set Lj (t) = ψt (Lj ) for


j = 1,2,3. As ψt is invertible,n12(t), n23(t) andn31(t) are constant, and by the above

d

x

ts a

e

n

lemma so isδ(t), so thatδ(ψ(L1),ψ(L2),ψ(L3)) = δ(1) = δ(0) = δ(L1,L2,L3). �Lemma 4.7. Let (L1,L2,L3) be an exceptional triple of Lagrangian subspaces ofCn

and let (u1, . . . , un) be an orthonormal basis forL1 formed of eigenvectors ofϕ12:ϕ12(uk) = eiαkuk , where[ei2α1, . . . , ei2αn] = meas(L1,L2). For all k, setdk

1 = L1 ∩ C(k),

dk2 = L2 ∩ C(k) anddk

3 = L3 ∩ C(k). Thendk1, dk

2 anddk3 are real lines ofC(k) and, if we

denote bymeas(dk1, dk

2), meas(d2k , d3

k ), meas(dk3, dk

1) ∈ [0,π[ the measures of the orienteangles(dk

1, dk2), (dk

2, dk3), (dk

3, dk1) in C

(k), then

δ(L1,L2,L3) =n∑

k=1

(meas

(dk

1, dk2

) + meas(d2k , d3

k

) + meas(dk

3, dk1

)).

Proof. Set meas(L1,L3) = [ei2ε1, . . . , ei2εn]. Observe first thatL1 intersects the compleline C(k) = Cuk becauseuk ∈ L1. Since (u1, . . . , un) is a basis ofL1 formed ofeigenvectors ofϕ12, and sinceϕ12 and ϕ13 have the same eigenspaces, there exispermutationg ∈ Sn such that, for allk ∈ {1, . . . , n}, ϕ13(uk) = eiεg(k)uk ∈ L3. Therefore,we haveeiαkuk ∈ L2 and eiεkuk ∈ L3, so thatC(k) also intersects bothL2 and L3.But if u ∈ Cn\{0} is contained in a Lagrangian subspaceL of Cn then L ∩ Cu = Ru.Indeed, if v ∈ L ∩ Cu then v = λu + µJu with λ,µ ∈ R, and sinceL is Lagrangianω(u, v) = 0. Butω(u, v) = λω(u,u) + µω(u,Ju) = µg(u,u) with g(u,u) �= 0, thereforev = λu ∈ Ru. Therefore, sinceϕ12(uk) = eiαkuk ∈ L2, we havedk

1 = L1 ∩ Cuk = Ruk

andd2k = L2 ∩ Cuk = R(eiαkuk) = eiαkdk

1, hence meas(dk1, dk

2) = αk ∈ [0,π[. Likewise,sinceeiεg(k)uk ∈ L3, we havedk

3 = eiεg(k)dk1, so that meas(dk

1, dk3) = εg(k), hence, setting

ξk = π − εg(k) mod π , meas(dk3, dk

1) = ξk ∈ [0,π[. Setting wk = eiεg(k)uk ∈ L3, wehave eiξkwk = ±ei(π−εg(k))wk = ±uk ∈ L1. The (eiξk ) therefore are the roots of thcharacteristic polynomialP(L3,L1) of the pair (L3,L1), hence [ei2ξ1, . . . , ei2ξn] =meas(L3,L1) = [ei2γ1, . . . , ei2γn], and sinceξk, γk ∈ [0,π[, there exists a permutatiog3 ∈ Sn such that, for allk, ξk = γg3(k). Similarly, setting vk = eiαkuk ∈ L2 andζk = (εg(k) − αk) mod π , we haveeiζk vk = ±eiεg(k)uk ∈ L1, hence[ei2ζ1, . . . , ei2ζn] =meas(L2,L3) = [ei2β1, . . . , ei2βn], and sinceζk, βk ∈ [0,π[, there existsg2 ∈ Sn suchthat, for allk, ζk = βg2(k). Furthermore, sincedk

2 = Rvk anddk3 = Reiεg(k)uk = Reiζk vk ,

we have meas(d2k , dk

3) = ζk. Hence

n∑k=1

(meas

(dk

1, dk2

) + meas(dk

2, dk3

) + meas(dk

3, dk1

))

=n∑

k=1

αk +n∑

k=1

ζk +n∑

k=1

ξk =n∑

k=1

αk +n∑

k=1

βg2(k) +n∑

k=1

γg3(k)

=n∑

k=1

(αk + βk + γk) = δ(L1,L2,L3). �


We now have all the material we need to relateδ to τ and show that the inertia index can

vest

t

rs

be computed from the measures of the Lagrangian angles(L1,L2), (L2,L3) and(L3,L1);that is, from the eigenvalues of the unitary mapsσLk ◦ σLj , whereσLj is the Lagrangianinvolution associated toLj .

Proof of Theorem 4.4. By Proposition 4.3, there exists a symplectic mapψ ∈ Sp(n) suchthat (ψ(L1), ψ(L2), ψ(L3)) is an exceptional triple. Since such a transformation leaτ , δ and thenjk ’s invariant, we may assume that(L1,L2,L3) is itself exceptional. Leus recall the notations meas(L1,L2) = [ei2α1, . . . , ei2αn] meas(L1,L3) = [ei2ε1, . . . , ei2εn]whereπ > α1 � · · · � αn � 0 andπ > ε1 � · · · � εn � 0. Then, since(L1,L2,L3) isexceptional, there exists an orthonormal basis(u1, . . . , un) for L1 and a permutationg ∈Sn such that(eiα1u1, . . . , e

iαnun) is an orthonormal basis forL2 and(eiε1u1, . . . , eiεnun) is

an orthonormal basis forL3. By abandoning the conditionπ > ε1 � · · · � εn � 0, we maysuppose thatg = Id. Setdk

1 = Ruk, dk2 = eiαkdk

1, dk3 = eiεkdk

1, andτk = τ (dk1, dk

2, dk3) in the

symplectic spaceCuk. Setδk = meas(dk1, dk

2) + meas(dk2, dk

3) + meas(dk3, dk

1) and set, asin Lemma 4.7,ζk = (εk − αk) modπ andξk = (π − εk) modπ , so thatδk = αk + ζk + ξk .Observe thatδk = δ(dk

1, dk2, dk

3) in the symplectic spaceCuk . In particular, this implies thaδk ≡ 0 modπ . If dk

1 = dk2 = dk

3, which happensn0 times, thenτk = 0 andδk = 0. If eitherdk

1 = dk2 �= dk

3 or dk2 = dk

3 �= dk1 or dk

3 = dk1 �= dk

2, which happens(n12− n0) + (n23− n0) +(n31− n0) times, thenτk = 0 and 0< δk = αk + ζk + ξk < 2π (since one of these numbeis 0 and since all of them are< π and two of them are non-zero), butδk ≡ 0 modπ soδk = π . If dk

1 �= dk2 �= dk

3 �= dk1, which happensn + 2n0 − (n12 + n23 + n31) times, then

eitherτk = 1 andδk = π or τk = −1 andδk = 2π , so thatτk = 3− 2δk

π(see Fig. 7).

Since(L1,L2,L3) is an exceptional triple, we have, by Proposition 4.7,δ = ∑nk=1 δk .

Likewise,τ = ∑nk=1 τk , so that we have:

τ =n+2n0−(n12+n23+n31)∑

k=1

(3− 2δk

π

)

= 3(n + 2n0 − (n12 + n23 + n31)

)− (2/π)

(δ − π

((n12 − n0) + (n23 − n0) + (n31 − n0)

))= 3n − 2δ/π − (n12 + n23 + n31). �

Fig. 7. Relation betweenδ andτ for exceptional triples of Lagrangians.


4.2. Two-dimensional unitary representations ofπ1(S2\{s1, s2, s3})

t

-e

g the

esult inn)as [4,ing

,

e

t

p

of

Let s1, . . . , sn be n distinct points of the Euclidean sphereS2. Then, for anyz ∈S2\{s1, . . . , sn}, the fundamental group of the sphere minus thesen points has finitepresentation

π1(S2\{s1, . . . , sn}, z

) = 〈g1, . . . , gn | gn · · ·g1 = 1〉where gk is the homotopy class of a loop aroundsk for k = 1, . . . , n. Giving atwo-dimensional unitary representationρ :π1(S

2\{s1, . . . , sn}) → U(2) of this group istherefore equivalent to givingn unitary matricesU1, . . . ,Un ∈ U(2) satisfyingUn · · ·U1 =Id and settingρ(gk) = Uk for all k. Two such representationsρ and ρ′ are equivalenif there exists a unitary transformationϕ ∈ U(2) of C2 such thatU ′

k ◦ ϕ = ϕ ◦ Uk ,or equivalentlyU ′

k = ϕ ◦ Uk ◦ ϕ−1 for k = 1, . . . , n. Determining and classifying twodimensional unitary representations ofπ1(S

2\{s1, . . . , sn}) up to equivalence thereforis equivalent to finding, given conjugacy classesC1, . . . ,Cn in the unitary groupU(2),necessary and sufficient conditions for the existence of unitary matricesUk ∈ Ck verifyingUn · · ·U1 = Id.

These conditions have been determined by Biswas in [4] after reformulatinquestion in terms of parabolic vector bundles of rank two over the projective lineCP1. Themethods presented in Section 3 above provide an elementary proof of the same rthe casen = 3. The general case (for arbitraryn and for representations in any dimensiois studied in [3,1,5] The relation between the result of existence obtained by Biswp. 524] and the result of existence of Theorem3.5 above is a consequence of the followproposition.

Proposition 4.8. Given three unitary matricesU12,U23,U31 ∈ U(2) verifyingU31U23U12 = Id, there exists a triple(L1,L2,L3) of Lagrangian subspaces ofC2 suchthat σLk ◦ σLj = Ujk for all j , k. If (L1,L2,L3) and (L′

1,L′2,L

′3) are two such triples

then there exists a unitary mapϕ ∈ U(2) such thatϕ(Lj ) = L′j for j = 1,2,3.

Lemma 4.9. LetA ∈ U(2) be a unitary matrix and let(ei2λ, ei2µ) be the eigenvalues ofA,whereπ > λ � µ � 0. Let v ∈ C2 be an eigenvector ofA with respect to the eigenvaluei2µ and such thath(v, v) = 1. Let l1 be a projective Lagrangian ofCP1 containing[v] ∈ CP1 S2, and let l2 be the projective Lagrangian image ofl1 by the rotation ofangle(λ − µ) ∈ [0,π[ around[v]. Then, given a LagrangianL1 ∈ p−1(l1), there exists aLagrangianL2 ∈ p−1(l2) such thatσL2 ◦ σL1 = A.

Proof. Let u be an eigenvector ofA with respect to the eigenvalueei2λ and such thah(u,u) = 1. SetL0 = Ru ⊕ Rv. Then(u, v) is a unitary basis ofC2 and thereforeL0 isLagrangian. Furthermore,[v] ∈ l0 ∩ l1, wherel0 = p(L0). Therefore,l1 is the image ofl0 by a rotationψ ∈ SO(3) around[v] ∈ S2, whereψ ∈ SU(2) is a special unitary mahavingu andv as eigenvectors (sinceψ is a rotation around[v] = −[u] ∈ R

3): ψ(u) = αu

andψ(v) = βv, whereα,β ∈ C. SetL = ψ(L0). ThenL is a Lagrangian subspaceC2 andp(L) = ψ(l0) = l1. Then, by Proposition 3.2,L1 = eiθL for someθ ∈ [0,π[. Setu12 = eiθψ(u) and v12 = eiθψ(v). Then (u12, v12) is an orthonormal basis forL1. Set


L2 = Reiλu12 ⊕ Reiµv12. ThenL2 is Lagrangian and, by Proposition 3.2,p(L2) is the

s

image ofp(L1) = l1 by the rotation of angle(λ − µ) around[v12] = [v], sop(L2) = l2.At last

σL2 ◦ σL1(u12) = σL2(u12) = σL2

(e−iλeiλu12

)= eiλσL2

(eiλu12

) = eiλeiλu12 = ei2λu12

so thatσL2 ◦σL1(eiθαu) = ei2λeiθαu hence, sinceσL2 ◦σL1 is holomorphic,σL2 ◦σL1(u) =

Au. For the same reasons,σL2 ◦ σL1(v) = Av, and thereforeσL2 ◦ σL1 = A. �Proof of Proposition 4.8. For fixedj , k, let (ujk, vjk) be a unitary basis ofC2 formedof eigenvectors ofUjk : we then haveUjkujk = ei2λjkujk andUjkvjk = ei2µjk vjk , whereπ > λjk � µjk � 0. There exists a great circlel2 of S2 containing both[v23] ∈ CP1 S2

and [v12] ∈ CP1 S2. Let l1 be the great circle ofS2 containing[v12] and such thatl2be the image ofl1 by the rotation of angle(λ12 − µ12) aroundv12. Fix a LagrangianL1 ∈ p−1(l1) arbitrarily. By the lemma above, there exists a LagrangianL2 ∈ p−1(l2)

such thatσL2 ◦ σL1 = U12. Now let l3 be the image ofl2 by the rotation of angle(λ23 − µ23) around[v23]. By the lemma above, there exists a LagrangianL3 ∈ p−1(l3)

such thatσL3 ◦ σL2 = U23. ThenσL1 ◦ σL3 = (σL3 ◦ σL1)−1 = (σL3 ◦ σL2 ◦ σL2 ◦ σL1)

−1 =(U23 ◦ U12)

−1 = U31. At last, if (L1,L2,L3) and(L′1,L

′2,L

′3) are two Lagrangian triple

verifying σLk ◦ σLj = Ujk = σL′k◦ σL′

jfor all j , k, then meas(Lj ,Lk) = meas(L′

j ,L′k) for

all j , k, hence, by Theorem 3.6, there exists a unitary mapϕ ∈ U(2) such thatϕ(Lj ) = L′j

for j = 1,2,3. �Corollary 4.10. Given three conjugacy classesC12,C23,C31 in the unitary groupU(2) (thatis, given eigenvalues[ei2λ12, ei2µ12], [ei2λ23, ei2µ23] and [ei2λ31, ei2µ31], whereπ > λjk �µjk � 0), there exists unitary matricesUjk ∈ Cjk verifyingU31U23U12 = Id if and only ifthere exists a triple(L1,L2,L3) of Lagrangian subspaces ofC2 such thatmeas(Lj ,Lk) =[ei2λjk , ei2µjk ] for all j , k, that is, settingδ = ∑

j,k(λjk + µjk), αjk = λjk − µjk andβjk = π − αjk , if and only if

δ ≡ π (mod 2π) and (α12, α23, α31) ∈ ∆,

or

δ ≡ 0 (mod 2π) and (β12, β23, β31) ∈ ∆.

Proof. Given such a triple(L1,L2,L3), the mapsσL2 ◦ σL1, σL3 ◦ σL2 andσL1 ◦ σL3 areunitary, have the prescribed eigenvalues, and verify(σL1 ◦σL3)◦ (σL3 ◦σL2)◦ (σL2 ◦σL1) =Id.

Conversely, given three unitary matricesU12, U23, U31 verifyingU31U23U12 = Id, thereexists, by the above proposition, a triple(L1,L2,L3) of Lagrangian subspaces ofC2 suchthatσLk ◦ σLj = Ujk for all j , k, so that meas(Lj ,Lk) = [ei2λjk , ei2µjk ].

The conditions onδ, theαjk ’s and theβjk ’s then follow from Theorem 3.5. �


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