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

I,§2,5

Definition (5.2) of M gives

AlA.,2,]

2nfd6'

jrd(0j)

1 ddet(ww'-1)

2ni J

det(ww - 1)

d(det w) 1 r

d(det w')f

2iri

det w

2ni J

- r

-d-et w'

by (3.5) and definition 3. Hence (5.6) follows by the additive property

(5.3) of M.

In order to recover assumption (4.11) of lemma 4.2, we need the following

definitions.

Definition 5.2

X., is the point of A,,,(1) that projects onto X in A(1) and

that may be joined to X* by an arc y of Ax (l) whose spectrum sp(y)

belongs to the half-circle in S', where lm(z) > 0.

Definition 5.3.

The Maslov index is the function m given by

m(;,, 4) = M(2,, 4; X *, X.).

This allows us to formulate lemma 4.2 as follows.

LEMMA 5.1.

For any triple ,) , A of elements of A,,,(I),

Inert(2,

2', 2")

= m(A.,

m(2., 1

) + m(2,, 2

(5.7)

Proof.

By (5.6), the right-hand side of (5.7) depends only on (2, A', 2"). By

Remark 4.2 and the invariance property (5.5) it suffices to establish the

following special case of (5.7):

Inert(2, X, X*) = m(;.,, X.) - m(1.,, X*) + m(X,,, X*). (5.8)

Definition 5.3 of m and the additive property (5.3) of M give

m(1.,,, X.) = X*, X.]

m(2,

, X*) - m(X., X*) = M(1.z, X*; X*].

Definition 5.1 of these two values of M makes use of two arcs r and

r* of A'(1) such that

iar = (2,w, Xx) - (X*, XJ,

OF* = (A,, X*) - (X', X*).

1,§2,5

We choose these arcs as cartesian products,

f=y x x,,,,,

r*=y* x x*,,,

where y and y* are then arcs of AW(1) such that

ny=gym

oy* = L -X".

We have

a(y-y*)=XX

Definition 5.2 of X,,allows us to choose y and y* such that sp(y - y*)

belongs to the half-circle in S', where Im(z) ? 0.

Denote

a = sp(y),

a* = sP(y*)

Since the homeomorphism defined by (2.3) has the values

w(X) _ -e, w(X*) = e,

definition 5.1 of M gives

M[A ,

X*, Xj = KI[a, (-1)],

M[A, X**; X., X*] = KI[a*, (1)]

The formula (5.8) to be proved is thus identical to formula (4.12), which

holds because or - a* = sp(y - y*) satisfies condition (4.11).

LEMMA 5.2. We have

m(L, X.) + m(7.'', n.") = 1. (5.9)

Proof.

Substitute the expression (5.7) for Inert into (4.4).

The following lemma supplements lemma 5.1.

LEMMA 5.3. Every function

n:Oq, A) i-+ n(2q, Aq)

defined for Aq, Aq e Aq(1),

and A' transverse,

with values in an abelian group,

locally constant on its domain of definition,

such that, for pairwise transverse

., A', A",

1,§2,5

).q) + n()

, )) = 0,

(5.10)

is identically zero.

Proof.

By (5.10), n is constant in a neighborhood of any pair (:tq, 2q),

transverse or not. Therefore n is constant on Aq (l), which is connected.

By (5.10), its value is 0.

We have proved the following theorem.

THEOREM 5.

1°) The Maslov index (definitions 5.1-5.3) is the only function

m:(A ),2x)F-'m().

M ( ; .,

)EZ

defined on pairs of transverse elements of A. (1) that is locally constant on its

domain and makes the following decomposition of the index of inertia

possible:

Inert(A, )',.2.") = m(A,,, 2x) - m(2,°, Ate) + m(A', A

(5.7)

2°) Taking into account definition 5.2, this function has the following

properties:

m(da A') + m(2', A,°) = 1;

(5.9)

m(s), SA') = m(Am, Ax)

VS E Sp. (1); (5.11)

m(l'2,,, /I''A') = m(J.., %;°) + r - r' Vr, r' E Z; (5.12)

m(X*,, Xj = 0, Xm) = 1. (5.13)

Proof of 1 °).

Lemmas 5.1 and 5.3.

Proof of (5.9).

Lemma 5.2.

Proof of (5.11) and (5.12).

Definition 5.3 of m, (5.5), and (5.6).

Proof of (5.13). Definition 5.3 of m and (5.4); (5.9).

Remark 5.1.

Formula (5.7) clearly implies the following: If A, A', 2"'

are four pairwise transverse elements of A(l), then

Inert(A, 2', 2") - Inert(., A', 2) + Inert(., %", ti") - Inert(,', A", A"') = 0.

Remark 5.2.

We call any transverse pair (p., µ,) E AI(1) such that

M[2x.,

; p, , µz] = m(1.

, A')

VA, A'

E Ax(1)

I,§2,5-1,§2,6

a basis. For example,

(X *, X_) is a basis. By the additive property (5.3) of

M, the condition that (pm, i') be a basis can be expressed as

m(pm, t4) = 0.

6. The Jump of the Maslov Index m(),,,, , d' ) at a Point (2, A'), Where

dim 2n2'= 1

Maslov [11] defined his index, up to an additive constant, by the expression

of its jump across the hypersurface YA2 in A'(1), which is the set of pairs

(L ,

).) of nontransverse elements of A.(1). Theorem 6 will make the

expression for this jump explicit.

First of all let us establish some properties of U(1). By y we denote a

differentiable arc of U(1):

7:1- 1, 119 0 F- U (O) E U (I),

ue = du 00.

LEMMA 6.1.

Let exp(i/(0)) be a simple eigenvalue of u(O) and let z(O) be a

corresponding eigenvector. Then

IIZ(e)I

% = (u_1(o)uZ(o)Iz(o)).

Remark 6.1.

Recall that if U is a Lie group with generic element u,

infinitesimal transformations Xk, and Maurer-Cartan forms CO, then

u-1 du (hence, in particular, u-1ue) may be written

u-1 du = Iwk(u, du)Xk;

see E. Cartan [4].

Remark 6.2.

Since U(1) is the unitary group, (1/i)u-1(O)u9 is an arbitrary

I x l self-adjoint matrix characterizing the vector ug tangent to U(1) at

u(0).

Proof.

Since exp(i>!i(O)) is a simple eigenvalue, k(0) is a differentiable

function of 0, and the vector z(0), which is defined up to a scalar factor, may

be chosen to be a differentiable function of 6. Then differentiating the

relation

u(0)z(0) = z(0)exp(ii(0))

I,§2,6

and multiplying on the left by (1/i)u-'(0), we obtain

luez

zo = -u- '

zo exp(io) + z(6)Oe

t i

Taking the scalar product with z eliminates ze and gives (6.1) because

(u-ize exp(ii)

I z) = (Z e I exp(- i/i) uz) _ (z e l z)

since u is unitary.

Denote by Eu the set of u e U(1) such that (1) e sp(u); recall that (1)

denotes the point of S' c C with coordinate 1.

The following lemma has the sole purpose of interpreting the assumption

of lemma 6.3 geometrically.

LEMMA 6.2. If u(o) E Eu, (1) is a simple value of sp u(o), and z(o) is a

corresponding eigenvector, then the condition that uo define a direction

tangent to Eu at u(o) takes the form

= 0.

(6.2)

(u-'ouozozo))

Proof.

u(o) is a regular point of E,,,. By lemma 6.1, every direction

tangent to Eu at u(o) satisfies (6.2). The hyperplane of directions satisfying

(6.2) thus contains the tangent plane to Eu at u(o); but Eu is a hypersurface

in U(l).

LEMMA 6.3.

If the arc y in U(1) intersects E. only at the point u(o), if the

eigenvalue I of u(o) is simple, and if the corresponding eigenvector z(o)

satisfies

(U'O)UZOZO))

(i.e., if y is not tangent to Eu), then

KI[sp y, (1)] = sign I

( U-'(o)uoz(o) I

z(o)

. (6.3)

Proof.

Let exp(iir(O)) be the eigenvalue of u(O) near 1. Evidently

KI[spy,(1)] = sign/e(o)

if Le(o) j4 0,

so (6.3) follows from (6.1).

I,§2,6

Notation.

EA2 denotes the set of nontransverse (1t, X):

EA3

AZ(1).

F denotes a differentiable arc in A2(1):

[-1, 1[3 0 i-- (,(6), A'(B)) e A2(1).

w(9) and w'(B) denote the natural images of ,(O) and )'(O) E A(1) in W(1):

see (2.3).

LEMMA 6.4.

If the arc F in A2(1) intersects EA2 only at the point (L(o),

A'(o)) and if

dim ),(o) n ).'(o) = 1,

z(o) E ).(o) n ti (o),

(w'-'(o)z(o)lz(o))

9 I

nw''-' (o)z(o) z(o)

then

KI[sp T', (1)] = sign

WOW-1(o)z(o) I z(o))

l f

_ (ww'-'o)zo)Izo))}.

(6.4)

Proof.

The image of the arc r E A2(1) in U(1) is the arc

y : [-1, 11 -3 B

u(O) = w(O)w' 1(B).

By definition,

KI[spF, (1)] = KI[spy, (1)].

By lemma 2.1,3°),

intersects Ev only at the point A = o, and 1 is a

simple value of sp u(o). By lemma 2.1,2°),

z(o) + w(o)z(o) = o,

z(o) + w'(o)z(o) = o,

z(o) - u(o)z(o).

Now

u-1ue

u-1wew-lu

- wew'-1;

thus, because u is unitary,

(u-'(o)uez(o)I z(o)) = (wew-'(o)z(o)j z(o)) - (wBw'-'(o)z(o)Iz(o))

Hence, by lemma 6.3,

I,§2,6

KI[spy,(1)] = sign{(WW

'(o)z(o) I

z(o)) -

(ww'1(o)z(o)

Iz(o)l,

and (6.4) follows.

LEMMA 6.5.

Let

0 Hz(0)

be differentiable mappings of [-1, 11 into A(l) and Z(l) such that

Z(O) E 2(B).

Then

(

Iwew-1zIz)

= [ze,z(0)]

Proof.

By lemma 2.1,2°),

z+wa=o,

zo + wee + wee = o.

Hence

(ze I z) -

(wow-1 z I z)

+ (wee I z) = 0,

where

(wze I z) = (Ze I

w-1z)

_ -(291 Z) _ -(Ze II Z).

From (1.1), (6.5) follows.

The left-hand side of (6.4) can be expressed in terms of the Maslov index

by (5.2) and definition 5.3; the right-hand side of (6.4) can be expressed

in terms of the symplectic form [ , ] by lemma 6.5. Hence the following

theorem, which will allow us to establish theorem 3.2 of §3.

I,§2,6-I,§2,7

THEOREM 6.

Let

[-1, 1]30(Z,z')eZ2(l)

be two differentiable mappings such that

zEA,

z'E)

for each 0;

z = z' : 0 for 0 = 0;

:i and 2' are transverse for 0 : 0;

the function

[-1, 1] -3 0H[z,z']ER

vanishes only at 0 = 0 and vanishes only to first order. Then there exists a

constant c e Z such that

for [z, z'] < 0

1 +cfor[z,z']>0.

7. The Maslov Index on Spm(1); the Mixed Inertia

Let

S, S', S" E Sp.(l)\Esp" be such that

SS'S" = E (the identity element);

let s, s', s" be their projections onto Sp(l). Formula (4.7) of theorem 4.1,

which relates the definitions of the inertia on A(l) and Sp(l), and formula

(5.7) of theorem 5, which relates the inertia on A(l) to the Maslov index,

yield, by the invariance property of the Maslov index (5.11),

Inert(s, s', s") = m(SX*, X*) - m(S'-1X*, X*) + m(S"X*, X*).

We rewrite this formula as (7.3), by virtue of the following definition:

Definition 7.1 of the Maslov index on Sp,(l). If S e SpJ)\Esp. we define

m(S) = m(SX*, X*).

(7.2)

This definition makes sense by Remark 4.1: S

Esp. is equivalent to

the condition that SX* and X* be transverse.

Let us supplement this formula with a lemma analogous to lemma 5.3:

1,§2,7

LEMMA 7.

Every function

n:Si-+n(S)

defined for S e Spq(l)\Espg,

with values ip an abelian group,

locally constant on its domain of definition,

such that

n(S) - n(S'-1) + n(S") = 0 when SS'S" = E,

is identically zero.

This lemma, combined with theorem 5 and (3.12) gives the following

theorem.

THEOREM 7.1.

1°) The Maslov index defined by (7.2) is the only function

that is locally constant on its domain and that makes the following decom-

position of the index of inertia possible: under hypothesis (7.1),

Inert(s, s', s") = m(S) - m(S'-1) + m(S").

(7.3)

2°) This function has the following properties:

M(S) + m(S- `) = 1, (7.4)

m(a'S) = m(S) + 2r,

(7.5)

where a is the generator of 7r, [Sp(l)] (definition 3).

Definition 7.2 of the mixed inertia. Let

s e Sp(l)\Esp; A, 2' a A(l), transverse to X* and such that 2 = s2'.

(7.6)

The mixed inertia is defined by the formula

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

(7.7)

the last two terms being equal because of the invariance (4.5) of the inertia.

The following theorem is evident.

THEOREM 7.2.

Under assumption (7.6),

I,§2,7-I,§2,8

Inert(s-', A', A) = I - Inert(s, A, i.'),

(7.8)

) + m(A', X*o)

(7.9)

Inert(s, A, A') = m(S) - m(A.z X*

if A,, = SA',, and if s is the projection of S E Sp,,(1).

Now we express the mixed inertia in terms of the inertia of a quadratic

form, as we did for Inert(s, s', s") (§l,definition 2.4) and Inert(., A', A")

(§2,4). This result will be used in section 10.

THEOREM 7.3.

Under assumption (7.6), we have s = sA(§1,1) and, by (2.5),

the equation of A' is p' = cp;,,(x'), where cp' is a quadratic form on X. Then

Inert(s, A, A') = Inert(A(o, ) + (p'( )).

(7.10)

Proof.

By (7.7) and the definition of Inert(., A', A") (section 4),

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

is the inertia of the quadratic form

x' i--

[z, z']

for z = (x, p), i = (x', p'),

(X, P) E X*,

S(X', P) E X*, (X + X', p + p')

Relations (7.11) may be rewritten

x=o,

p' =-A(o,x'),

p+p'=cp.,,(x+x'),

whence

x = o,

p = Ax.(o, x') +

This implies

[z, z'] = <p, x'> = 2A(o, x') + 2N'(x'),

and therefore (7.10).

8. Maslov Indices on A,(/) and SpJ1)

Let Aq, A' E Aq(1) and S E Spq(1) be the projections of A,, A, E A,,(1) and

S., E Sp,,(1). By (5.12) and (7.5) (where a and /i are the generators of

n' [Sp(l)] and n, [A(1)] ), the relations

m(Aq, Xq) = m(tir,

mod q,

m(S) = m(S,,,) mod2q,

(8.1)

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

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