Leray J. Lagrangian Analysis and Quantum Mechanics: A Mathematical Structure Related to Asymptotic Expansions and the Maslov Index

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I,§2,3

Since 1r1 [S U(1)] and no [S U(1)] are trivial in the exact sequence (3.3), p is

an isomorphism. Now

9G1[S1]3FH-

f d-

is an isomorphism.

The composition of p, which is induced by (3.2), with this isomorphism

is an isomorphism n1 [U(1)] -+ Z, which clearly is defined by (3.1).

LEMMA 3.2.

1 °) The composition of the natural isomorphism n

[A(1)]

n1 [W(1)] [cf. (2.3)] and the morphism

nl[j'1'(l)]-3

ddetw)EZ

(3.5)

7ri

is a natural isomorphism (Arnold [1]):

[A(1)] ^' Z.

2°) The fibration Sp(l)/St(l) = A(1) defines a monomorphism

p : Z '=' it1 [Sp(1)] - nl (A(1)] = Z,

(3.6)

which is multiplication by 2 on Z.

Proof of 1 °).

The homeomorphism (2.3) allows us to define

det ). = det w e S 1; (3.7)

the mapping

A(l)a2"detA c- S' (3.8)

is clearly an epimorphism. By (2.2) we have

det A = det

u, if i = uX *.

Hence for all u e U(1)

det(uA) = det 2 u det A.

'(3.9)

The mapping (3.8) thus defines a fibration on which U(1) permutes the

fibers. The fibration is

A(1)/SA(l) = S 1,

I§23 35

where SA(1) denotes the variety in A(l) defined by the equation

SA(l) : det w = 1.

The exactness of the homotopy sequence of this fibration proves the

exactness of the sequence

71,

[SA(l)] -4 n, [A(l)] A n, [S1].

(3.10)

Since

p: n, [A(l)] - n,

[S1]

is induced by

det:W(l) - Si,

p is an epimorphism. Let us compute n, [SA(l)].

SU(1) acts transitively on SA(1) and X * E SA(l). The stabilizer of X * in

SU(1) is SO(1), the connected component of the identity element of O(l),

and so we have the fibration

SU(1)/SO(l) = SA(l).

The exactness of its homotopy sequence implies the exactness of

n, [SU(l)] 4 n, [SA(1)] 4 no[SO(l)],

where n, [SU(1)] and no [SO(1)] are trivial, by (3.4) and the fact that SO(1)

is connected. Thus n, [SA(1)] is trivial, and so pin (3.10) is an isomorphism

Now p is induced by the mapping (3.8). Taking its composition with the

isomorphism

7t, [S'] 3Fr-,

dzEZ

2ni z

we obtain an isomorphism 7r, [A(1)] - Z, defined by (3.5).

Proof of 2°). In (3.6) the isomorphism Z ^- n, [Sp(l)] is the composition

of the isomorphisms defined by parts 2 and 3 of lemma 3.1. The isomor-

phism n, [A(l)] Z is the composition of the isomorphism (3.5) and the

one that induces the homeomorphism of A(l) and W(1). Proving part 2 of

lemma 3.2 is thus the same as proving the following:

The fibration U(1)/0(1) = W(l) induces a morphism

1,§2,3

P: Z

i i[U(1)] - n [W(l)] "' Z (3.11)

that is multiplication by 2 on Z.

This fibration is defined by the mapping

U(1) 9 u "_w = u `u, which satisfies det w = (det u)2.

Since the isomorphisms entering into (3.11) are defined by (3.1) and (3.5),

we have

Yd(det u)1--.

2ni det u 27ci

fd(detu)'cZ.

(det u)

This morphism p: Z - Z is evidently multiplication by 2.

The two preceding lemmas have the following theorem as an immediate

consequence. It is clearly independent of the hermitian structure on Z(1)

used above.

Definition 3.

a and Q are the generators of n,[Sp(l)] and n, [A(l)] whose

natural images in Z are 1.

THEOREM 3.

1°) Sp(l) has a unique covering group, Spq(l), of order q

(q = 1, 2, ... , .x) [namely: the number of points having the same projection

onto Sp(l) is q]; a acts on Spq(I); a' does not act as the identity on Spq(1)

unless r = 0 mod q.

2°) A(l) has a unique covering space, Aq(l), of order q; fi acts on A,(1);

(lr does not act

as the identity on Aq(I) unless r = 0 mod q.

3°) Spq(l) acts transitively on A2,(1):

(aSq)) .2q = Sq(Q2A2q) = f32(SgA2q), where A2q E A2q(I),

Sq E Spq(1). (3.12)

Example 3.1.

A2(0 is the set of oriented (in the euclidean sense) lagran-

gian subspaces of Z(1); Sp(l) acts on A2(1).

Example 3.2.

Sp2(1) acts on A4(l). This result is essential for the theory of

asymptotic expansions (chapter II).

Notation 3.

Denote by s the projection of Sc Spq(l) onto Sp(l); by A the

projection of Aq c- NO onto A(l); by A2 the projection of A2q E A2,(1)

onto A2(1); by e the identity element of Sp(l); and by E the identity element

of Spq(1).

I,§2,3-I,§2,4

Let us choose an element X of A,,(1) projecting onto X * in A(l); XQ

denotes its projection onto A9(1).

4. Indices of Inertia

Definition of the index of inertia Inert(,, A', ).") of a triple (A, A', A") of

elements of A(1) pairwise transverse. We have

A0+A'=A'

A"=A" $+A=Z(1).

The conditions

Z E A,

Z' E A',

Z" E

Z + Z' + Z" = 0

therefore define three isomorphisms

1 %

A" 3 Z" a--I

Z' E A'

whose product is the identity. By (4.1)

[z, z'] = [z', z"] = [z", Z];

(4.3)

this number is the value of a quadratic form at z e A (at z' E A' or at z" e A').

The isomorphisms (4.2) transform each one of these forms into the others.

Thus they all have the same index of inertia, which will be denoted

Inert(A, A',1").

LEMMA. The quadratic form

A3z -+ [z, z'] ER

is nondegenerate (that is, has no zero eigenvalues).

Proof.

Take a second triple

C E A,

C' E A',

C" e A" such that C + C' + C"

= 0.

Since

A', and A" are lagrangian, the bilinear form

(Z, r) i- [Z, 4'] = [yr', Z"] = [Z", 4]

= [C, Z'] = [Z ,

y r"]

= [S", Z]

is symmetric and, hence, is the polar form of the quadratic form

Z H [Z, Z'].

1,§2,4

If this were degenerate, then there would exist z -A 0 such that

zeA, [z, C] = 0

VC'eA,

that is,

zeA n A'.

contrary to hypothesis.

Thus Inert(-, , ) has the following properties:

Inert(., A") = Inert(A', A", 1) = l - Inert(2,).", A'). (4.4)

Inert( , , ), which is defined when its arguments are pairwise transverse,

is locally constant on its domain of definition.

Inert(sA, sA', s2") = Inert(A, 2', A")

Vs E Sp(l).

(4.5)

Formulas (2.11) and (2.14) of 1,§1 defined Inert(s, s', s") for

s, s', s" E Sp(l)\Esp, s s' s" = e.

(4.6)

The following relation exists between these two indices of inertia:

THEOREM 4.

Under assumption (4.6),

Inert(s, s', s") = Inert(sX *, X *, s"-1 X *).

(4.7)

Remark 4.1.

The condition s 0 Esp is equivalent to the following: sX*

and X* are transverse.

Proof. We return to the notation of I,§1,2 (specifically that of lemma

2.3), setting

s = SA,

s = SA ,

Sri = SA,. ;

sA : (x', p') h-. (x, p) means p = Px - `Lx',

p' = Lx - Qx';

sA., : (x, p) i - (x", p") means p" = P"x" - `L"x,

p = L"x" - Q"x.

The equations of the subspaces

A=sX*, J' = X*, s"-1X*

are then

1: p = Px,

2':x, = 0, p

_Qx

I,§2,4

Condition (4.1),

Z = (x, p) E A,

Z" _ (x", p") E A",

Z + Z" E A',

may be written

z = (x, Px),

z" = (-X, Q"x),

from which follows

[z", z] = <(P + Q")x, x>.

Hence, by definition (2.14) in §1,

Inert(., Z, A") = Inert(P + Q") = Inert,,(A + A' + A")

= Inert(s, s', s").

The two lemmas below express the index of inertia Inert in terms of the

Kronecker index KI. After we have defined the Maslov index m in terms

of the Kronecker index in section 5, lemma 6.1 will deduce the relation

between the indices of inertia and the Maslov index from these lemmas.

LEMMA 4.1.

Sp(l) acts transitively on the set of pairs of transverse elements

of A(l).

Remark 4.2.

Therefore every triple of pairwise transverse elements of

A(l) has the form

(sl, sX, sX *).

We recall that

Inert(sA, sX, sX*) = Inert(A, X, X*).

Proof.

Since Sp(l) acts transitively on A(l), it suffices to show that the

stabilizer St(l) of X* acts transitively on the set of elements of A(1) trans-

verse to X*. The element s of St(l) defined by (2.6) transforms

X:p' = 0into sX:p = `si ls2si lx,

where s2 : X

X* is symmetric. Now the condition that the equation

p = s3x define an element of A(1) is clearly that s3 : X - X* be symmetric.

Notation 4.

Recall that the spectrum of u e U(1), sp(u), is a 0-chain in

S1. Let

I,§2,4

SP(1) = sp(w),

(4.8)

where w is the image of) under the natural homeomorphism (2.3).

Denote by (exp iB) the point in S1 with the coordinate exp iO (0 E R)

and by l(exp iO) the 0-chain consisting of this point with multiplicity

1(1 e Z). We have, for example,

sp(X*) = sp(e),

sp(X) = sp(-e),

(4.9)

sp(e) = 1(1),

sp(-e) = l(-1).

Let a denote a 1-chain in S1, Jul its support, a6 its boundary, and KI

the Kronecker index (see Lefschetz [9] or any theory of chain intersection).

Recall that the integer KI[a1, 60]

is defined when a, and a0 are a 1-chain and a 0-chain in S1 such that

I6oInIaa,I='0'

is zero if I ao I n I6, I = 0,

is linear in 60 and a,,

is equal to 1 if 6, is a positively oriented arc and 6o is an interior point

of this arc,

satisfies KI[a1, aft] _ -KI[61, aa1].

LEMMA 4.2.

Let a and a* be two 1-chains in S1 such that

a6 = sp(ti) - sp(X*), au* = sp(ti) - sp(X), (4.10)

6 - 6* belongs to the half-circle in S1, where Im(z) > 0. (4.11)

Then

Inert(;., X, X*) = KI[a,(-1)] - KI[a*, (1)].

(4.12)

Remark 4.3. Lemma 5.1 will use this decomposition of Inert.

Proof.

Let

Z = x + iy e 2,

Z' E X,

Z" E X*

be such that

z+z'+z"=0, where x, y e X.

Clearly

I,§2,4

i = - x,

z" = - iy,

[z,, z"] = Im(z' I z") = - (x

Let w be the natural image of A under (2.3). By (2.5)

y= i

e+ w

(e is the identity).

e - w

Since w is unitary and symmetric, i(e + w)/(e - w) is real and symmetric.

Hence Inert(1, X, X*) is the index of inertia of the quadratic form

X=3xi-' -I

ie + w

x ER,

e - w

that is, the number of positive eigenvalues of the real symmetric matrix

i(e + w)/(e - w).

Now the positive real axis is the image of the half-circle in S', where

Im z < 0, under the homographic transformation

S 3 (exp i0)

1 + exp iO

i--*

- E R. (4.13)

1 - exp i0

We orient it positively: it forms a chain x in S'. Since (4.13) transforms the

eigenvalues of w into those of i(e + w)/(e - w),

Inert(1, X, X*) = KI[y, sp(w)].

Let r be a chain in S' with boundary

ar = sp(w)

- sp(ie).

Then

Inert(;., X, X*) = KI[x, ar] _ -KI[r, OZ]

= KI[r, (-1)] - KI[r, (1)].

Let a and a* be 1-chains in S' such that a

- r and a* - r are defined as

follows:

a(a - r) = sp(ie)

- sp(e),

a - r belongs to the arc 0 E [0, n/2] in S'

(exp i0) t;

a(Q* - r) = sp(ie)

- sp(-e),

a* - r belongs to the arc 0 E [n/2, n] in S'.

I,§2,4-I,§2,5

Since

KI[ r - z, (- 1)] = KI[Q* - T, (1)] = 0,

the preceding expression for Inert(A, X, X*) is equivalent to (4.12). Now

or and rr* satisfy conditions (4.10) and (4.11) and are determined by them

up to the addition of 1-cycles y and y* in S' such that y - y* belongs to

the half-circle in S', where Im z > 0. Since y and y* are then homologous,

KI[y, (-1)] = KI[;,*, (01

Thus (4.12) holds for any pair (a, a*) satisfying (4.10) and (4.11).

5. The Maslov Index m on A(1)

We are going to supplement Arnold's definition [1], but first we shall use

the following preliminary definition.

Definition 5.1 of the index M on A4(1). Let (v,, v') be an element of

A,,(l) x A,,(1) = A',(1), that is, a pair of elements of A.,(1). Let v and

w(v) be the natural images of v., in A(1) and W(l) [cf. (2.3)], and let v'

and w' = w(v') be the images of v.. By lemma 2.1, 3°), the condition that

v and v' be transverse can be expressed as

(1) 0 sp(ww'-');

sp(ww'-1), which we denote sp(v, v'), is a 0-chain in S' (see Lefschetz [9]).

The mapping

F n (vim, vim)

- sp(v,

maps the arc r onto a 1-chain in S1 denoted sp(r). If

ar = 0, tip)

- (µ., Nx),

then

asp(r) = sp(, ti') - sp(µ, µ').

Suppose

,Z and 2' are transverse, u and µ' are transverse;

then by lemma 2.1, 3°),

(1) Ojasp(r)I.

1,§2,5

Thus KI[sp(r), (1)] is defined. It is an integer depending only on the

homotopy class of r, that is, on OF. We denote it

M[A.

p ., p.] = KI[sp(r), (1)] e Z.

(5.2)

Remark.

By Remark 1, M is independent of the choice of the hermitian

structure on Z(l) used in its definition.

Properties of M. M is defined under hypothesis (5.1). M is locally con-

stant on its domain of definition. M has the obvious additive property

+ M[ux,

v ,

ti's] =

vti> v ],

(5.3)

which implies

Ar] = 0.

M has the invariance property

M[S%c,SA':'; i

,p' ] =

dS E

(5.5)

If /3 is the generator of n, [A (1)] (cf. definition 3), then

fl`Ate; µx, µx] = M[A,;x;µ.,µ'] + r - r'

dr,r'EZ.

(5.6)

Proof of (5.5).

Assuming hypothesis (5.1), s1. and sip' are transverse, and

the mapping

S r-4 M[S)., S;",; per,, ux] e Z

is defined and locally constant. Therefore it is constant since Sp,c (1) is

connected.

Proof of (5.6).

Choose the arc r to be differentiable and such that

Clearly one can define 1 continuous functions

O:F=(v,x,vj- U;(vz,,v.)ER,

j = 1, 1,

such that

sp(v, v') (expi9,); O (/1'A, /x'A ) = O A) mod 27r.