An Integral Formula for the Maslov-Index of a Symplectic Path

Stefan Bechtluft-Sachs

Department of Mathematics, NUI Maynooth, Co Kildare, Ireland e-mail: stefan@maths.nuim.ie

November 8, 2007

Abstract

The Maslov index of a not necessarily closed path M in the symplectic group Sp(2n) is expressed by an integral formula. We have an explicit formula for the integrand which is a rational 1-form on Sp(2n).

Keywords: Symplectic Group, Lagrangian Grassmannian, Maslov Index

1 Introduction

The homotopy class of a loop Λ(t) in the Lagrangian Grassmannian L(2n) in standard symplectic space ℝ^{2n} is detected by an integer, the Maslov index. Following Arnold, [1], this can be defined as an intersection number with a certain subset of L(2n), the Maslov cycle. In [6] this definition of the Maslov index was extended to not necessarily closed paths, and it is this index which enters in the Gutzwiller trace formula in semiclassical quantisation, [2], see also [4].

The group Sp(2n) of symplectic automorphisms of ℝ^{2n} acts transitively on the set of Lagrangians and thus L(2n) = Sp(2n)/ H(2n) is a homogeneous space. In applications the path Λ(t) is frequently given in the form M(t)Λ_{0} with a curve M(t) ∈ Sp(2n) of symplectic automorphisms and a fixed Lagrangian Λ_{0}. In view of calculations of the Maslov index as in [3] it seems desirable to compute the Maslov index μ(Λ) = μ(M) directly in terms of M.

In the present note we derive such a formula. We show that the Maslov index is the line integral over a certain differential form χ_{Sp(2n)} ∈ Ω^{1}(Sp(2n), ℝ) plus end point terms which are also given by an explicit function Φ on Sp(2n).

For unitary paths, i.e. paths M(t) ∈ U(n) ⊂ Sp(2n) in the unitary group the (complex) Trace Tr_{ℂ} on the Lie algebra 𝔲(n) is an invariant for the adjoint representation, and therefore provides a bi-invariant 1-form on U(n). On the symplectic group Sp(2n) however a nontrivial biinvariant 1-form does not exist. The 1-form χ_{Sp(2n)} and the function Φ we construct are invariant under the left action of Sp(2n) and under the right action of the stabilizer subgroup H(n) which contains O(n).

We formulate our main result in the next section 2. Then, in section 3, we compute the 1-form χ_{Sp(2n)} explicitely in terms of the entries of the symplectic matrix. Finally we prove our integral formula by verifying the axioms for the Maslov index in section 4.

2 A local formula for the Maslov index

We need to introduce some notation and recall some well known facts from symplectic geometry as may be found in [5], for instance. Let Λ_{0} be a Lagrangian in ℝ^{2n} and Λ_{0}^{⊥} the orthogonal complement with respect to the standard scalar product ⟨·| ·⟩. The symplectic structure ω on ℝ^{2n} can be written as

ω(x, y) = ⟨x| Jy⟩

where J is a complex structure, i.e. J ∈ O(2n) with J^{2} = −1_{2n} and we have JΛ_{0} = Λ_{0}^{⊥}.

Usually we choose coordinates such that Λ_{0} = ℝ^{n} × {0}, Λ_{0}^{⊥} = {0} × ℝ^{n} and so that the complex structure J becomes

J = ( \begin{matrix}0 & −1_{n} \\ 1_{n} & 0 \end{matrix} ) , 1 = ( \begin{matrix}1 & 0 & 0 \\ 0 & ... & 0 \\ 0 & 0 & 1 \end{matrix} ) ∈ Mat(n × n, ℝ) .  (1)

With respect to the orthogonal decomposition

ℝ^{2n} = Λ_{0} ⊕ Λ_{0}^{⊥}  (2)

we can write a (symplectic) matrix M as

M = ( \begin{matrix}a & b \\ c & d \end{matrix} )  (3)

where

a ∈ Hom(Λ_{0}, Λ_{0}), b ∈ Hom(Λ_{0}^{⊥}, Λ_{0}), c ∈ Hom(Λ_{0}, Λ_{0}^{⊥}), d ∈ Hom(Λ_{0}^{⊥}, Λ_{0}^{⊥}) .

All these will be identified with Mat(n × n; ℝ).

By means of the complex structure (1) we can read a complex (n×n)-matrix

Y as a real (2n × 2n)-matrix ( \begin{matrix}α & −β \\ β & α \end{matrix} ). The traces Tr_{ℂ} over ℂ respectively

Tr = Tr_{ℝ} over ℝ are related by

Tr_{ℂ}(Y) = \frac{1}{2} Tr(Y) − \frac{i}{2} Tr(JY) = Tr(α) + i Tr(β) .  (4)

The group Sp(2n) of symplectic automorphisms is

Sp(2n) = {M ∈ Gl(2n) | M^{t}JM = J}

and acts transitively on the Lagrangian Grassmannian L(2n). The stabilizer group, i.e. the isotropy group of this action at Λ_{0} = ℝ^{n} × {0} is

H(n) = \{( \begin{matrix}A & X \\ 0 & (A^{t})^{−1} \end{matrix} ) | A ∈ Gl(n) , AX ∈ Mat(n × n, ℝ) symmetric\}

We will also need the subgroup H^{+}(n) consisting of those matrices in H(n) with A upper triangular and positive. The unitary group U(n) and the orthogonal group O(n) are naturally identified as subgroups of Sp(2n),

U(n) = {M ∈ Sp(2n) | MJ = JM} = Sp(2n) ∩ O(2n)
O(n) = {M ∈ Sp(2n) | MJ = JM and MΛ_{0} = Λ_{0}}  (5)
= \{( \begin{matrix}T & 0 \\ 0 & T \end{matrix} ) | T ∈ O(n)\} = H(n) ∩ O(2n) ⊂ U(n) .

With these identifications we can represent the Lagrangian Grassmannian as homogeneous space,

L(2n) = Sp(2n)/ H(n) = U(n)/ O(n) .  (6)

The square of the (complex) determinant

L(2n) = U(n)/ O(n) → U(1), F ↦ det_{ℂ}(F)^{2}  (7)

induces an isomorphism of the fundamental groups

π_{1}(L(2n)) ≅ π_{1}(U(1)) = ℤ .

The Jordan-, or QR- decomposition, of the symplectic group is

Sp(2n) = U(n) · H^{+}(n)

and

H^{+}(n) = D^{+}(n) ⋉ Sym(n)

is the semidirect product of the group of positive upper triangular matrices with the vector group of symmetric matrices. In particular, we have a diffeomorphism and a homotopy equivalence, both equivariant under the right action of H^{+}(n),

F: Sp(2n) ≅ U(n) × H^{+}(n) ⟶^{≃} U(n) .  (8)

We will also need the map

H: Sp(2n) ≅ U(n) × H^{+}(n) ⟶^{≃} H^{+}(n) .  (9)

These are defined by first decomposing a given M ∈ Sp(2n), M = F H with F ∈ U(n), H ∈ H^{+}(n) uniquely determined by M. This can be done efficiently by any of the algorithms for the QR-decomposition. Then we set F(M) := F and H(M) := H

For the contribution of end points we need a function

Φ: Sp(2n) → ℝ .

which is defined as follows: Φ(M) is computed from the unitary part

F(M) = ( \begin{matrix}α & −β \\ β & α \end{matrix} )  (10)

alone. Since F^{t}F = 1_{2n} = F F^{t} we have

α^{t}α + β^{t}β = 1_{n}, α^{t}β = β^{t}α,
αα^{t} + ββ^{t} = 1_{n}, αβ^{t} = βα^{t}.

From these relations it follows that α^{t}α, β^{t}β and α^{t}β commute and can therefore be simultaneously diagonalized. Since Spec(β^{t}β) ⊂ [0, 1] we may set

X := arcsin ((β^{t}β)^{1/2}) ∈ Mat(n × n; ℝ) .  (11)

Note that X is nonegative symmetric and Spec X ⊂ [0, \frac{π}{2} ]. Now define T_{α}, T_{β} ∈ Mat(n × n; ℝ) as

T_{β} = β sin(X)^{−1} pr_{(ker X)^{⊥}} , T_{α} = α cos(X)^{−1} pr_{ker}_{(X− \frac{π}{2} 1_{n})}_{⊥}  (12)

where pr denotes the orthogonal projection on the subspace indicated. We then have

β = T_{β} sin(X), α = T_{α} cos(X), T_{β}^{t}T_{β} = pr_{(ker X)^{⊥}}, T_{α}^{t}T_{α} = pr_{ker}_{(X− \frac{π}{2} 1_{n})}_{⊥}

We define

Φ(M) := \frac{1}{π} Tr (T_{α}^{t}T_{β} (X − \frac{π}{2} 1_{n})) .  (13)

This is to arrange for d Φ = χ_{Sp(2n)} on the complement of the Maslov cycle (see Lemma 21 and (23) in the proof of that Lemma).

We can now write down an integral formula for the Maslov index:

Theorem Let M: [0, 1] → Sp(2n) be a differentiable path in the symplectic group of ℝ^{2n} and let Λ_{0} ⊂ ℝ^{2n} be a fixed Lagrangian subspace. Then the Maslov index μ(Λ) ∈ \frac{1}{2}ℤ of the path Λ: [0, 1] → L(2n), Λ(t) = M(t)Λ_{0}, in the Lagrangian-Grassmannian of ℝ^{2n} is given by

μ(M) = μ(Λ) = ∫_{M} χ_{Sp(2n)} − Φ(M(1)) + Φ(M(0))  (14)

where Φ is as defined in (13) and

χ_{Sp(2n)} = −\frac{1}{2π} Tr (JM^{t} d M) − Tr (J H(M)^{t} d(H(M)))

This form is explicitely computed in (17) in terms of the coefficients of M.

Recall that the stabilizer part H = H(M) ∈ H^{+}(n) ⊂ Mat(2n × 2n) is uniquely determined from M by the following conditions (see (9)):

1. M^{t}M = H^{t}H),

2. H = ( \begin{matrix}A & AX \\ 0 & (A^{t})^{−1} \end{matrix} ) w.r.t. the splitting ℝ^{2n} = Λ_{0} ⊕ Λ_{0}^{⊥},

3. A ∈ Mat(n × n, ℝ) is upper triangular with positive eigenvalues,

4. X ∈ Mat(n × n, ℝ) is symmetric.

3 The 1-form on the symplectic group

We will now derive explicit formulas for the integrand in (14). The form χ = \frac{1}{2πi}z^{−1} d z on U(1) = S^{1} ⊂ ℂ detects the homotopy type of a loop in U(1). To find the right 1-form for the Maslov index we first pull back χ to U(n) via the map det_{ℂ}^{2} in (7). Sine d(det_{ℂ} F) = det_{ℂ}(F) Tr_{ℂ}(F^{−1} d F) we get

χ_{U(n)} = (det_{ℂ}^{2})^{∗} χ = \frac{1}{πi} Tr_{ℂ} F^{−1} d F ∈ Ω^{1}(U(n)) .

This already gives a formula, similiar to that for the winding number, for the Maslov index of a unitary path F (t)Λ_{0}, F: [0, 1] → U(n),

μ(F) = \frac{1}{πi} ∫ Tr_{ℂ}(F^{−1} d F) − Φ(F (1)) + Φ(F (0))

The 1-form on Sp(2n) we are looking for is the pull back of χ_{U(n)} over the map F in (8). We compute this explicitely.

Lemma 15 Denote by a, b, c, d ∈ Mat(n×n, ℝ) the components of a symplectic matrix M, i.e the matrices such that

M = ( \begin{matrix}a & b \\ c & d \end{matrix} )  (16)

with respect to the decomposition (2). Then

χ_{Sp(2n)} = F^{∗}χ_{U(n)} = \frac{1}{πi} Tr_{ℂ} (F(M)^{−1} d F(M))

= \frac{−1}{2π} Tr (JM^{t} d M − JH^{t} d H)

= \frac{−1}{2π} Tr [−b^{t} d a − d^{t} d c + a^{t} d b + c^{t} d d

−(a^{t}a + c^{t}c) d [(a^{t}a + c^{t}c)^{−1}(a^{t}b + c^{t}d)]] (17)

Proof: Let M ∈ Sp(2n) and F = F(M) = MH^{−1} ∈ U(n) with H ∈ H^{+}(n). We get

F^{−1} d F = HM^{−1}(d M)H^{−1} + HM^{−1}M d(H^{−1}) ∈ Ω^{1}(Sp(2n), 𝔲(n)) .

Because of (4) and since H^{t}H = M^{t}M,

Tr_{ℂ} F^{−1} d F = \frac{1}{2i} Tr JF^{−1}dF = \frac{1}{2i} Tr (JHM^{−1}(d M)H^{−1} + JHM^{−1}M d (H^{−1}))

= \frac{1}{2i} Tr (H^{−1}JHM^{−1} d M − JHH^{−1}(d H)H^{−1})

= \frac{1}{2i} Tr (JH^{t}HM^{−1} d M − J(d H)H^{−1})

= \frac{1}{2i} Tr (JM^{t}MM^{−1} d M + J(d H)JH^{t}J)

= \frac{1}{2i} Tr (JM^{t} d M − JH^{t} d H)

(18)

In view of (16) we can simplify this further. First

Tr(JM^{t} d M) = Tr ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}a^{t} & c^{t} \\ b^{t} & d^{t} \end{matrix} ) ( \begin{matrix}d a & d b \\ d c & d d \end{matrix} )
= Tr(−b^{t} d a − d^{t} d c + a^{t} d b + c^{t} d d)

For the second summand in (18) let H = ( \begin{matrix}A & AX \\ 0 & (A^{t})^{−1} \end{matrix} ) ∈ H^{+}(n). We com-

pute

Tr (JH^{t} d H) = Tr ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}A^{t} & 0 \\ X^{t}A^{t} & A^{−1} \end{matrix} ) ( \begin{matrix}d A & d(AX) \\ 0 & d ((A^{t})^{−1}) \end{matrix} )
= Tr(−X^{t}A^{t} d A + A^{t} d(AX))
= Tr(A^{t}A d X)

since X is symmetric. As H^{t}H = M^{t}M we have A^{t}A = a^{t}a + c^{t}c and A^{t}AX = a^{t}b + c^{t}d which yields

Tr (JH^{t} d H) = Tr [(a^{t}a + c^{t}c) d [(a^{t}a + c^{t}c)^{−1}(a^{t}b + c^{t}d)]]

4 The axiomatic characterization of the Maslov index

For the proof of the Theorem, denote by μ̃(M) the right hand side of (14) for a path M(t) ∈ Sp(2n). In order to show that μ̃ coincides with the Maslov index, we will check that μ̃ satisfies the five properties which were shown to characterize the Maslov index by Theorem 4.1 in [6].

We need two Lemmas. The first will permit us to move the path by paths in O(n).

Lemma 19 The form χ_{Sp(2n)} ∈ Ω^{1}(Sp(2n)) from (17) and the function Φ of (13) are left-O(n) and right-H(n)-invariant. Thus if M(t) ∈ Sp(2n) and T (t) ∈ O(n), H(t) ∈ H(n) are smooth paths, then

μ̃(M) = μ̃(T MH) .

Proof: Let

M = F H^{+} and H = T_{1}H_{1}^{+} with F ∈ U(n), H^{+}, H_{1}^{+} ∈ H^{+}(n), T_{1} ∈ O(n) .

Then

F(T MH) = F(T F H^{+}T_{1}H_{1}^{+}) = F(T F T_{2}H_{2}^{+}) = T F T_{2}

with some T_{2} = F(H^{+}T_{1}H_{1}^{+}) ∈ O(n) and some H_{2}^{+} ∈ H^{+}(n), since

H(n) = O(n) · H^{+}(n) = H^{+}(n) · O(n).  (20)

Now

Tr_{ℂ} ((T F T_{2})^{−1} d(T F T_{2})) = Tr_{ℂ} ((T F T_{2})^{−1}T F (d T_{2}))
+ Tr_{ℂ} ((T F T_{2})^{−1}T (d F)T_{2})
+ Tr_{ℂ} ((T F T_{2})^{−1}(d T)F T_{2})
= Tr_{ℂ}(F^{−1} d F)

since T^{−1} d T lies in the Lie algebra of O(n) and therefore Tr_{ℂ}(T^{−1} d T) = 0.

For the invariance of the function Φ of (13) it suffices to prove left and right invariance under O(n), because of (20) as before. We also may assume M = F unitary since right-H^{+}(n)-invariance of Φ is clear from its definition. So let

F = ( \begin{matrix}α & −β \\ β & α \end{matrix} )

be unitary with real (n × n)-matrices α and β. We then have

T F = ( \begin{matrix}T & 0 \\ 0 & T \end{matrix} ) ( \begin{matrix}α & −β \\ β & α \end{matrix} ) = ( \begin{matrix}T α & −T β \\ T β & T α \end{matrix} )

From (12) and (11) we see that this replaces the matrices T_{α}, T_{β} used to define Φ by T T_{α}, T T_{β} while leaving X unchanged. By definition (13) Φ(T F) = Φ(F). Similiarly, multiplying T from the right, replaces α, β by αT, βT. Thus, by (12) and (11), T_{α}, T_{β}, X are replaced by T_{α}T, T_{β}T and T^{−1}XT, since sin(T^{−1}XT) = T^{−1} sin(X)T. By (13) and the conjugation invariance of the trace, Φ again is unchanged.

By the next Lemma μ̃ vanishes on a path M(t) such that M(t)Λ_{0} stays away from the Maslov cycle {Λ ∈ L(2n) | Λ ∩ Λ_{0} ≠ 0}. Thus, away from the Maslov cycle, the function Φ is an integral of χ_{Sp(2n)}.

Lemma 21 For M ∈ Sp(2n) let β(M), α(M) ∈ Mat(n × n; ℝ) be the matrices in the unitary part

F(M) = ( \begin{matrix}α(M) & −β(M) \\ β(M) & α(M) \end{matrix} )

of M as in (10). Then on the set Sp^{×}(2n) = {M | β(M) invertible } we have d Φ = χ_{Sp(2n)}.

In particular μ̃ vanishes on paths staying in Sp^{×}(2n).

The set Sp^{×}(2n) is the set of all M such that M(t)Λ_{0}∩Λ_{0} = 0. Thus Sp^{×}(2n) is the preimage of the complement of the Maslov cycle under the projection Sp(2n) → L(2n). Proof: It suffices to prove the Lemma on U^{×}(n) = U(n) ∩ Sp^{×}(2n) since both the form χ_{Sp(2n)} and Φ are right-H^{+}(n) invariant. We will now show that ∫_{F} χ_{^{(}2n)} = [Φ(F (t)]_{t=0}^{t=1} for arbitrary paths F: [0, 1] → U(n).

Let M = F ∈ U^{×}(n). Then the matrices X(F) and T_{β}(F) defined as in (11) and (12) are invertible and depend smoothly on F. Let

X_{s} := (1 − s)\frac{π}{2} 1_{n} + sX(F) for s ∈ [0, 1] and

α_{s} = T_{α(F)} cos(X_{s}) , β_{s} = T_{β(F)} sin(X_{s}) .

The path

F_{s} = ( \begin{matrix}α_{s} & −β_{s} \\ β_{s} & α_{s} \end{matrix} )  (22)

has F_{1} = F = M and

F_{0} = ( \begin{matrix}0 & −T_{β(M)} \\ T_{β(F)} & 0 \end{matrix} ) = J ( \begin{matrix}T_{β(F)} & 0 \\ 0 & T_{β(F)} \end{matrix} ) = JT_{β(F)} ∈ J O(n) .

Now by definition Φ vanishes on J O(n), so Φ(F_{0}) = 0. We integrate

∫_{F_{s}} χ_{Sp(2n)} = \frac{1}{π} ∫_{F_{s}} Tr (α_{s}^{t} d β_{s} − β_{s}^{t} d α_{s})

= \frac{1}{π} ∫_{F_{s}} Tr (cos(X_{s})T_{α(F)}^{t}T_{β(F)} cos(X_{s}) d X_{s} + sin(X_{s})T_{β(F)}^{t}T_{α(F)} sin(X_{s}) d X_{s})

= \frac{1}{π} ∫_{F_{s}} Tr (T_{α(F)}^{t}T_{β(F)} d X_{s})

= \frac{1}{π} Tr (T_{α(F)}^{t}T_{β(F)}(X_{1} − X_{0}))

= \frac{1}{π} Tr (T_{α(F)}^{t}T_{β(F)} (X(F) − \frac{π}{2} 1_{n}))

= Φ(F_{1}) = Φ(F)

(23)

by definition (13) of Φ. Here we have used that X_{s}, d X_{s}, T_{α(F)}^{t}T_{β(F)} commute, together with the identity cos^{2}(X_{s}) + sin^{2}(X_{s}) = 1_{n}.

Since F_{s} defined by (22) depends smoothly on F, we have a deformation retraction of U^{×}(n) (and Sp^{×}(2n)) on J O(n) ≅ O(n). Therefore any path F (t) ∈ U^{×}(n) is homotopic relative end points to the catenation a ∗ T ∗ b of two paths a, b of type (22) and a path T in J O(n). Since χ_{Sp(2n)} is the pull back of the closed form χ on U(1) it is closed as well and therefore line integrals over χ_{Sp(2n)} depend on the homotopy class relative end points only. For an arbitrary path F in U^{×}(n) we can therefore compute

∫_{F} χ_{Sp(2n)} = ∫_{a∗T ∗b} χ_{Sp(2n)} = ∫_{a} χ_{Sp(2n)} + ∫_{T} χ_{Sp(2n)} + ∫_{b} χ_{Sp(2n)} .

The integral over T vanishes since (JT)^{−1} d(JT) = T^{−1} d T has trace 0. The integrals over a and b have been evaluated in (23). Thus we get that

∫_{F} χ_{Sp(2n)} = Φ(F (1)) − Φ(F (0))

for any path F (t) in U^{×}(n) which proves the Lemma.

We now proceed to verify the axioms from [6] for the Maslov index for μ̃.

1. Homotopy: Two paths in Sp(2n) are homotopic relative end points if and only if the respective values of μ̃ coincide.

Since χ_{Sp(2n)} = ((det_{ℂ}^{2}) ∘ F )^{∗} χ is closed we have that μ̃ is homotopy invariant relative end points. The map (det_{ℂ}^{2}) ∘ F induces an injective map π_{1}(Sp(2n)) → π_{1}(U(1)) and the isomorphism π_{1}(U(1)) ≅ ℤ is given by integrating the form χ. For closed loops the end point terms (14) cancel and the claim follows.

2. Catenation: Let M_{i}(t), i = 0, 1 be paths in Sp(2n) with M_{0}(1) = M_{1}(0) and M = M_{0} ∗ M_{1} denote the path with M(t) = M_{0}(2t) for t ≤ 1/2 and M_{1}(2t − 1) for t ≥ 1/2. Then μ̃(M) = μ̃(M_{0}) + μ̃(M_{1}).

The contributions of the end points M_{0}(1) and M_{1}(0) cancel. The claim thus follows from the additivity of the integral.

3. Product: If M(t) = ( \begin{matrix}M_{1}(t) & 0 \\ 0 & M_{2}(t) \end{matrix} ) with M_{i}(t) ∈ Sp(2n_{i}), n_{1} +

n_{2} = n, then μ̃(M_{1}) + μ̃(M_{2}) = μ̃(M).

This follows from the analogous property of the trace.

4. Zero: Let M(t) be a path such that dim(M(t)Λ_{0} ∩ Λ_{0}) = k > 0 is constant for all t. Then μ̃(M) = 0.

By lemma 19 we may replace M(t) by M(t)H(t) with H(t) ∈ H(n) without changing μ̃. We can thus achieve that that

M(t) = F(M(t)) = ( \begin{matrix}α(t) & −β(t) \\ β(t) & α(t) \end{matrix} ) ∈ U(n)

is unitary. We also have ker β(t) = M(t)Λ_{0} ∩ Λ_{0} which is of constant dimension k by our assumption. Also α(t) maps ker β(t) isometrically into Λ_{0}. Therefore replacing M(t) by T_{1}(t)M(t)T_{2}(t) with suitable paths T_{1}(t), T_{2}(t) ∈ O(n), we may assume that V = ker β(t) is independent of t and that α(t) is the identity on V. Again by Lemma 19, we have μ̃(M) = μ̃(T_{1}MT_{2}).

Let W be the orthogonal complement of V in Λ_{0}. We then can split

ℝ^{2n} = (V ⊕ JV) ⊕ (W ⊕ JW)

as the orthogonal sum of two symplectic subspaces. With respect to this splitting we now have

M(t) = ( \begin{matrix}1_{V ⊕JV} & 0 \\ 0 & Q(t) \end{matrix} ) .

The path Q(t) ∈ Sp^{×}(W ⊕ JW) = Sp^{×}(2(n − k)) avoids the Maslov cycle. By Lemma 21, μ̃(M) = μ̃(1) + μ̃(Q) = 0.

5. Normalization: Let Y (t) ∈ Sym(n) be a path of symmetric matrices

and M(t) = ( \begin{matrix}1 & 0 \\ Y (t) & 1 \end{matrix} ) the corresponding path of symplectic shears. Then

μ̃(M) = [sign(Y (1)) − sign(Y (0))]/2 ,

where the signature sign(Y) is the number of positive eigenvalues of Y minus the number of negative eigenvalues.

To see this, let T (t) ∈ O(n) be such that

T (t)^{−1}Y (t)T (t) = ( \begin{matrix}λ_{1}(t) \\ & . & . & . \\ & & & & λ_{n}(t) \end{matrix} )

is diagonal. Then

T^{−1}(t)M(t)T (t) = ( \begin{matrix}T (t)^{−1} & 0 \\ 0 & T (t)^{−1} \end{matrix} ) ( \begin{matrix}1 & 0 \\ Y (t) & 1 \end{matrix} ) ( \begin{matrix}T (t) & 0 \\ 0 & T (t) \end{matrix} )

= ( \begin{matrix}1 & 0 \\ T (t)^{−1}Y (t)T (t) & 1 \end{matrix} ) .

By Lemma 19 and the product property,

μ̃(M) = μ̃(T^{−1}MT) = ∑_{k=0}^{n} μ̃ ( \begin{matrix}1 & 0 \\ λ_{k}(t) & 1 \end{matrix} ) .

It suffices therefore to verify the normalization property in the case n = 1,

Y (t) = λ(t) and M(t) = ( \begin{matrix}1 & 0 \\ λ(t) & 1 \end{matrix} ). The integral in (14), i.e. the integral

over the form (17), becomes

∫_{M} χ_{Sp(2)} = \frac{−1}{2π} ∫_{M} − d λ − (1 + λ^{2}) d ((1 + λ^{2})^{−1}λ) = \frac{[arctan(λ(t))]_{t=0}^{t=1}}{π}

For the end point term in (14) we compute

F = F ( \begin{matrix}1 & 0 \\ λ & 1 \end{matrix} ) = \frac{1}{\sqrt{1 + λ^{2}}} ( \begin{matrix}1 & −λ \\ λ & 1 \end{matrix} )

which gives

X = arcsin (\frac{|λ|}{\sqrt{1 + λ^{2}}} ) = arctan(|λ|) , T_{β} = sign(λ) , T_{α} = 1

in the notation of (11) and (12). From (13) we get

Φ(M) = \{ \begin{matrix}\frac{1}{π} arctan(λ) − \frac{1}{2} if & λ > 0 \\ 0 if & λ = 0 \\ \frac{1}{π} arctan(λ) + \frac{1}{2} if & λ < 0 \end{matrix} .

This yields μ̃(M) = [sign(λ(1)) − sign(λ(0))]/2 as required.

References

[1] V. I. Arnol’d On a characteristic class entering into conditions of quantization Functional Analysis 1 (1967), 1-8.

[2] M. C. Gutzwiller Chaos in classical and quantum mechanics Springer, New York, 1990

[3] M. Brack, M. Pletyukhov On the canonically invariant calculation of Maslov indices J. Phys. A 36 (2003), no. 36, 9449–9469.

[4] M. Brack, R. K. Bhaduri Semiclassical physics Frontiers in Physics, 96. Westview Press, Boulder, CO, 2003

[5] D. McDuff, D. Salamon Introduction to symplectic topology Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 1998

[6] J. Robbin, D. Salamon The Maslov Index for Paths Topology 32 no 4 (1993), 827-844