Chapter 4. INVARIANT DEGENERATE DIFFUSION
ON COTANGENT BUNDLES
1. Curvilinear Ornstein-Uhlenbeck process and stochastic geodesic flow
In this chapter we apply the theory developed in the previous chapter to
the investigation of invariant degenerate diffusions on manifolds. We confine
ourselves to the case of a regular degenerate diffusion of rank one. Since in the
conditions of the regularity of a Hamiltonian the linearity of some coefficient
in the second variable y is included, one has to suppose when constructing an
invariant object that this second variable lives in a linear space. Therefore, an
invariant operator ought to be defined on a vector bundle over some manifold:
coordinates y in fibres and coordinate x on a base. We reduce ourselves to the
most commonly used vector bundle, namely to the cotangent bundle T ? M of a
compact n-dimensional manifold M . In local coordinates, a regular Hamiltonian
H of a degenerate diffusion of rank one has form (2.3.4), where the matrix g is
positive definite and α is non-degenerate. The corresponding diffusion equation
(3.1.2) has the form
µ
¶
∂u
∂
∂
h2
∂2u
h
= Lu = H x, y, −h , −h
=
gij
∂t
∂x
∂y
2
∂yi ∂yj
+h(ai (x)+αij (x)yj )
∂u
1
∂u
+h(bi (x)+βij (x)yj + γijl (x)yj yl )
−V (x, y)u. (1.1)
∂xi
2
∂yi
In this section, we give the complete description of the invariant operators of
that kind on T ? M . Let us recall that a tensor γ of type (q, p) on a manifold M
i ...i
is by definition a set of np+q smooth functions γj11 ...jpq (x) on x that under the
change of coordinates x 7→ x̃ changes by the law
i ...i
k ...k
γ̃j11 ...jpq (x̃) = γl11...lq p (x)
∂ x̃i1 ∂ x̃ip ∂xl1 ∂xlq
...
...
.
∂xk1 ∂xkp ∂ x̃j1 ∂ x̃jq
To each tensor of the type (0, p) corresponds the polylinear function on the
cotangent bundle T ? M defined by the formula γ(x, y) = γ i1 ...ip (x)yi1 ...yip .
Theorem 1.1 Suppose the following objects are given on M :
(i) Riemanian metric, which in local coordinates x on M is given by a
positive definite matrix g(x), x ∈ M ;
(ii) non-degenerate tensor α = {αij (x)} of the type (0, 2) (non-degeneracy
means that the matrix α is non-degenerate everywhere) and a tensor a = {ai (x)}
of the type (0, 1) (i.e. a vector field); these tensors obviously define a quadratic
function f (x, y) = αij (x)yi yj + ai (x)yi on T ? M ;
(iii) tensors b, β, γ of the types (1, 0), (1, 1), (1, 2) respectively;
(iv) the sum V of tensors of the types (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), which
defines a bounded from below function V (x, y) on T ? M .
137
Then the second order differential operator
L=
1
∂2
∂f
∂
∂f
∂
gij (x)
+
(x, y)
−
(x, y)
2
∂yi ∂yj
∂yi
∂xi
∂xi
∂yi
µ
+ bi (x) +
βij (x)yj
1
+ γikl (x)yk yl
2
¶
∂
− V (x, y)
∂yi
(1.2)
is an invariant operator on T ? M , which is a regular diffusion of the rank one.
Conversely, each such operator has this form.
Proof. Under the change of the variables x 7→ x̃(x) the moments change by
the rule ỹ = y ∂x
∂ x̃ . Therefore,
∂u
∂u ∂ ỹj
=
,
∂yi
∂ ỹj ∂yi
∂2u
∂ 2 u ∂ ỹm ∂ ỹk
=
,
∂yi ∂yj
∂ ỹk ∂ ỹm ∂yj ∂yi
∂u
∂u ∂ x̃j
∂u ∂ ỹj
=
+
,
i
∂x
∂ x̃j ∂xi
∂ ỹj ∂xi
∂xj
∂ ỹk
=
.
∂yj
∂ x̃k
(1.3)
(1.4)
It follows, in particular, that under the change (x, y) 7→ (x̃, ỹ), the second order
part of (1.1), the first order part of (1.1), and the zero order part of (1.1) transforms to second order, first order, and zero order operators respectively, and
consequently, if the operator (1.1) is invariant, then its second order part, its
first order part, and its zero order part must be invariant. In order that the zero
order term V (x, y)u was invariant it is necessary and sufficient that V (x, y) is
invariant and therefore V (x, y) is a function. From the invariance of the second
order part one has
gij (x)
= gij (x)
∂2u
∂ 2 u ∂ ỹm ∂ ỹk
= gij (x)
∂yi ∂yj
∂ ỹk ∂ ỹm ∂yj ∂yi
∂ 2 u ∂xj ∂xi
∂2u
=
g̃
(x̃)
,
km
∂ ỹk ∂ ỹm ∂ x̃m ∂ x̃k
∂ ỹk ∂ ỹm
and consequently, the invariance of the second order part is equivalent to the
requirement that g is a tensor, and therefore defines a riemannian metric. Let
us write now the condition of the invariance of the first order part of operator
(1.1). Changing the variable (x, y) 7→ (x̃, ỹ) in the first order part of (1,1) one
has
µ
¶
¡ i
¢ ∂u
1 jl
∂u
j
a (x) + αij (x)yj
+
b
(x)
+
β
(x)y
+
γ
(x)y
y
i
j
j l
i
∂xi
2 i
∂yi
µ
¶µ
¶
∂ x̃m
∂u ∂ x̃l
∂u ∂ ỹl
i
ij
= a (x) + α (x)ỹm j
+
∂x
∂ x̃l ∂xi
∂ ỹl ∂xi
µ
¶
∂ x̃m
1 jl
∂ x̃m ∂ x̃p ∂u ∂ ỹq
j
+ bi (x) + βi (x)ỹm j + γi (x)ỹm ỹp j
.
∂x
2
∂x ∂xl ∂ ỹq ∂yi
138
Therefore, the invariance of this first order part is equivalent to the following
two equations:
¶
µ
∂ x̃m ∂ x̃i
l
lj
,
ã (x̃) + α̃ (x̃)ỹj = a (x) + α (x)ỹm j
∂x
∂xl
i
ij
(1.5)
and
µ
¶ µ
¶
1 jl
∂ x̃m ∂ ỹi
j
l
lj
b̃i (x̃) + β̃i (x̃)ỹj + γ̃i (x̃)ỹj ỹl = a (x) + α (x)ỹm j
2
∂x
∂xl
µ
¶
∂ x̃m
∂ x̃m ∂ x̃p ∂ ỹi
1
+ bq (x) + βqj (x)ỹm j + γqjl (x)ỹm ỹp j
.
∂x
2
∂x ∂xl ∂yq
(1.6)
From (1.5) one obtains that a and α are tensors, as is required. Next,
∂ ỹi
∂ x̃p ∂xm
=
−ỹ
,
p
∂xl
∂xl ∂xm ∂ x̃i
Therefore, equating in (1.6) the terms which do not depend on ỹ, the terms
depending on ỹ linearly, and the terms depending on ỹ quadratically, one gets
that b is a tensor of the type (1, 0), and that the law of the transformation of β
and γ has the form
γ̃imp (x̃) = γqjl (x)
∂ x̃m ∂ x̃p ∂xq
∂ x̃m ∂ 2 x̃p ∂xq
− 2αlj (x) j
,
j
l
i
∂x ∂x ∂ x̃
∂x ∂xl ∂xq ∂ x̃i
β̃ip (x̃) = βqj (x)
Since
∂ 2 x̃p ∂xm
∂ x̃p ∂xq
l
−
a
(x)
.
∂xj ∂ x̃i
∂xl ∂xm ∂ x̃i
∂ãp
∂aj ∂xq ∂ x̃p
∂ 2 x̃p ∂xm
(x̃) =
+ al (x) l m
,
i
q
i
j
∂ x̃
∂x ∂ x̃ ∂x
∂x ∂x ∂ x̃i
∂αlj ∂xq ∂ x̃m ∂ x̃p
∂ 2 x̃p ∂xq ∂ x̃m
∂ α̃mp
(x̃) =
+ 2αlj (x) q l
,
i
q
i
l
j
∂ x̃
∂x ∂ x̃ ∂x ∂x
∂x ∂x ∂ x̃i ∂xj
mp
p
p
∂a
it follows that {γimp + ∂α
∂xi } and {βi + ∂xi } are tensors of the types (1, 2) and
(1, 1) respectively. Denoting these tensors again by γ and β respectively, yields
representation (1.2). The proof is complete.
Let us write the stochastic differential equation for the diffusion process
corresponding to the operator (1.2) with vanishing V . Let r : M 7→ RN be
an embedding of the Riemanian manifold M in the Euclidean space (as is well
known, such embedding always exists). The operator (1.2) stands for the diffusion on T ? M defined by the stochastic system
(
dx =
∂f
∂y
dyi =
∂f
− ∂x
i
dt
dt + (bi (x) + βij (x)yj + 12 γikl (x)yk yl ) dt +
∂r j
∂xi
dwj ,
(1.7)
139
where w is the standard N -dimensional Wiener process. This statement follows
from the well known formula for the Riemanian metric
N
X
∂rk ∂rk
gij (x) =
∂xi ∂xj
k=1
and the Ito formula. It is interesting to note that though system (1.7) depends
explicitly on the embedding r, the corresponding operator L defining the transition probabilities for diffusion process (1.7) depends only on the Riemanian
structure.
One sees that system (1.7) describes a curvilinear version of the classical
Ornstein-Uhlenbeck process (see e.g. [Joe] for an invarian definition) defined
originally (see, e.g.[Nel1]) by the system (x, y ∈ Rn )
½
ẋ = y
(1.8)
dy = − ∂V
∂x dt − βy dt + dw(t)
as a model of Brownian motion, where β ≥ 0 is some constant and V (x) is
some (usually bounded from below) function (potential). System (1.8) defines
a Newton particle (Hamiltonian system with the Hamiltonian V (x) + y 2 /2) disturbed by the friction force βy and by the white noise random force dw. System
(1.7) describes a Hamiltonian system (defined by the Hamiltonian function f
which is quadratic in momentum but with varying coefficients) with additional
deterministic force (defined by the 1-form b), the friction βij (x)yj + 21 γikl (x)yk yl
(which can depend on the first and second degree of the velocity) and the white
noise force depending on the position of the particle.
In the case of vanishing b, β, γ system (1.7) is a stochastic Hamiltonian
system with non-homogeneous singular random Hamiltonian f (x, y) + r(x)ẇ,
which describes the deterministic Hamiltonian flow disturbed by the white noise
force:
(
dx = ∂f
∂y dt
(1.9)
∂
dy = − ∂f
∂x dt + ∂x (r, dw).
The ”plane” stochastic Hamiltonian systems, i.e. (1.9) for M = Rn , were investigated recently in connection with their application to the theory of stochastic
partial differential equation, see [K1], [TZ1], [TZ2].
The mostly used example of the Hamiltonian system on the cotangent bundle T ? M of a Riemanian manifold is of course the geodesic flow, which stands
for the Hamiltonian function f = (G(x)y, y)/2, where G(x) = g −1 (x). For this
f , system (1.9) takes the form
½
ẋ = G(x)y
(1.10)
∂
∂
dy = − 21 ∂x
(G(x)y, y) dt + ∂x
(r, dw),
This system was called in [K1] the stochastic geodesic flow. The investigation of
its small time asymptotics was begun in [AHK2]. Corresponding Hamiltonian
(2.3.4) of the stochastic geodesic flow is
µ
¶
1
1 ∂
H = (g(x)q, q) − (G(x)y, p) +
(G(x)y, y), q
(1.11)
2
2 ∂x
140
and the invariant diffusion equation is
µ
¶ µ
¶
µ
¶
∂u
h
∂2u
∂u
1 ∂
∂u
= Lu = tr g(x)
+ G(x)y,
−
(G(x)y, y),
.
∂t
2
∂yi ∂yj
∂x
2 ∂x
∂y
(1.12)
It depends only on the Riemanian structure and therefore its property should
reflect the geometry of M , which explain more explicitly in the next sections.
2. Small time asymptotics for stochastic geodesic flow
The stochastic geodesic flow is a good example for performing the general
results of the previous chapter. Using these results we present now the calculation of the main terms of the small time asymptotics for the Green function of
equation (1.12), i. e. its solution with the initial data
uG (0, x, y; x0 , y 0 ) = δ(x − x0 )δ(y − y 0 )
(2.1)
in a neighbourhood of the point (x0 , y 0 ) ∈ T ? M .
All calculations will be carried out in normal coordinates around x0 (see,
e.g. [CFKS]), in which x0 = 0,
1 kl k l
gij (x) = δij + gij
x x + O(|x|3 ),
2
(2.2)
and det g(x) = 1 identically. These conditions imply that
n
X
kl
gii
=0
∀k, l
(2.3)
i=1
and that the Gaussian (or scalar) curvature in x0 is equal to
X
ik
.
R=
gik
(2.4)
i,k
Remark. Some authors do not include the requirement det g = 1 in the
definition of normal coordinates. Notice however that if a system of coordinates
x on a n-dimensional riemanian manifold M satisfies all other conditions of
normality but for the condition det g = 1, then the coordinates x̃ defined by the
formula
Z x1
√
1
x̃ =
g(s, x2 , ..., xn ) ds, x̃i = xi , i ≥ 2,
0
satisfies all the conditions of normality given above, as one checks easily (see
[CFKS]).
Moreover, from (2.2) one gets obviously the expansions
1 kl k l
x x + O(|x|3 ),
Gij = δij − gij
2
(2.5)
141
for the inverse matrix G(x) = g −1 (x), and also
∂Gij
kl
(x) = −gij
+ O(|x|2 ).
∂xk
(2.6)
To find the asymptotics of the two-point function one should solve the main
equation (3.2.12), which for the case of Hamiltonian (1.11) takes the form
µ
¶
∂Σ
∂Σ ξ + G(tξ + x̃)(y + ỹ) − G(x̃)ỹ ∂Σ
−
− g(x̃)q̃,
∂t
t
∂ξ
∂y
+
1
2
·µ
¶ µ
¶¸
∂G
∂G
∂Σ
(tξ + x̃)(y + ỹ), y + ỹ −
(x̃)ỹ, ỹ
∂x
∂x
∂y
µ
¶
∂Σ ∂Σ
1
g(tξ + x̃)
,
= 0.
+
2
∂y ∂y
(2.7)
Using (2.2), (2.3) one concludes that
x̃ = x0 − ty 0 + O(t3 ),
ỹ = y 0 + O(t2 ),
q̃ = O(t2 ),
(2.8)
and then one rewrites (2.7) in the coordinate form (using now low indices for
both ξ and y:
∂Σ (ξ + y)i −
−
∂t
·
−
t2 kl
2 gij [(ξk
− yk0 )(ξl − yl0 )(yj + yj0 ) − yk0 yl0 yj0 ] + O(t3 ) ∂Σ
t
∂ξi
¸
t kl
∂Σ
gij [(ξl − yl0 )(yi + yi0 )(yj + yj0 ) + yi0 yj0 yl0 ] + O(t2 )
2
∂yk
1
t2 kl
∂Σ ∂Σ
+ (1 + gij
(ξk − yk0 )(ξl − yl0 ) + O(t3 ))
= 0.
2
2
∂yi ∂yj
(2.9)
Following the arguments of Sect.2 of the previous chapter one looks for the
solution of this equation in form (3.2.6), where Σ−1 is a positive quadratic form
and Σ0 (0, 0) = 0. For Σ−1 one gets equation (3.2.10) with α0 and g0 being unit
matrices. Its solution is given by (3.2.16). For Σ0 one finds then the equation
−(y + ξ)
∂Σ0
∂Σ0
+ (6ξ + 4y)
= 0,
∂ξ
∂y
whose solution vanishes, due to Proposition 3.7.1. Furthermore, for Σ1 one
obtains the equation
Σ1 − (y + ξ)i
∂Σ1
∂Σ1
+ (6ξ + 4y)i
∂ξi
∂yi
kl
+gij
[(ξk − yk0 )(ξl − yl0 )(yj + yj0 ) − yk0 yl0 yj0 ](6ξ + 3y)i
142
kl
−gij
[(ξl − yl0 )(yi + yi0 )(yj + yj0 ) + yi0 yj0 yl0 ](3ξ + 2y)k
kl
+gij
(ξk − yk0 )(ξl − yl0 )(3ξ + 2y)i (3ξ + 2y)j = 0.
Opening the brackets one presents this equation in the form
Σ1 − (y + ξ)i
∂Σ1
∂Σ1
+ (6ξ + 4y)i
= F (ξ, y),
∂ξi
∂yi
(2.10)
where F is the sum F2 + F3 + F4 of the homogeneous polynomials of degree 2,3,4
given by the formulas
kl
[(12ξi ξk −4yi yk )yj0 yl0 −(18ξi yj +7yi yj +9ξi ξj )yk0 yl0 +(3ξk ξl +2ξk yl )yi0 yj0 ],
F2 = gij
(2.11)
kl
F3 = gij
[(−6ξi ξk ξl + 3ξk ξl yi + 4ξk yi yl )yj0
+(36ξi ξk yj + 11ξk yi yj − 2yi yj yk + 18ξi ξj ξk )yl0 ],
(2.12)
kl
F4 = gij
[2ξk yi yj yl − 4ξk ξl yi yj − 18ξi ξk ξl yj − 9ξi ξj ξk ξl ].
(2.13)
The solution of this equation is the sum of the solutions Σ21 , Σ31 , Σ41 corresponding
to F2 , F3 , and F4 in the r.h.s. of (2.10). These solutions can be calculated by
formula (3.7.10). For instance, Σ21 is given by (3.7.15) with F2 being equal to
(2.11). These calculations are rather long, but the form of the solution is clear:
kl
Σ1 = gij
Rijkl (ξ, y, y 0 ),
(2.14)
where Rijkl are homogeneous polynomials of degree 4 in the variables ξ, y, y 0 .
Similarly one sees that the other terms Σj are homogeneous polynomials in
ξ, y, y 0 of degree j + 3, which is important to know when making the estimates
uniform in y 0 .
Let us find now the first nontrivial term of the asymptotic solution of the
transport equation. In the case of Hamiltonian (1.11), the general equation
(3.3.3) takes the form
∂Ψ α
ξ + G(ξt + x̃)(y + ỹ) − G(x̃)ỹ ∂Ψ
− Ψ−
∂t
t
t
∂ξ
·µ
¶ µ
¶¸
1
∂G
∂G
∂Ψ
+
(tξ + x̃)(y + ỹ), y + ỹ −
(x̃)ỹ, ỹ
2
∂x
∂x
∂y
µ
¶ µ
¶
µ
¶
2
∂Ψ
∂Σ ∂Ψ
1
∂ Σ
− g(x̃)q̃,
+ g(tξ + x̃)
,
+ Ψ tr g(tξ + x̃) 2 = 0, (2.15)
∂y
∂y ∂y
2
∂y
where
Ψ(t, ξ, y) = tα φ(t, tξ + x̃, y + ỹ; 0, y 0 ).
(2.16)
From (3.3.5) one finds α = 2n. Looking for the solution of (2.15) in the form
Ψ = 1 + tΨ1 + t2 Ψ2 + ...
143
one gets comparing the terms at t0 the following equation (since Σ0 = 0):
µ
∂Ψ1
Ψ1 − ξ + y,
∂ξ
¶
µ
¶
∂Ψ1
+ 6ξ + 4y,
= 0.
∂y
Due to Proposition 3.7.1, Ψ1 vanishes. Comparing the coefficients at t yields
µ
¶ µ
¶
∂Ψ2
∂Ψ2
Ψ2 − ξ + y,
+ 6ξ + 4y,
∂ξ
∂y
µ
+tr
1 ∂ 2 Σ1
+ g kl (ξk − yk0 )(ξl − yl0 )
2 ∂y 2
¶
= 0.
(2.17)
It is again the equation of type (2.10) with the polynomials of degree 0,1,2 in
the r.h.s. The solution of this equation is therefore given by Proposition 3.7.3.
Again the calculations are rather long but the form of the solution is clear:
Ψ2 =
X
ii
Pkl + gilik Qkl + Gkl
gkl
ii Rkl ,
(2.18)
i
where Pkl , Qkl , Rkl are some homogeneous polynomials in ξ, y, y 0 of degree 2.
3. The trace of the Green function and geometric invariants
It turns out that similarly to the case of non-degenerate diffusion on a
compact manifold (see, e.g. [Gr],[Roe]), the resolving operator for the Cauchy
problem for equation (1.12) belongs to the trace class, i.e. the trace
Z
tre−tL =
T ?M
uG (t, x, y; x, y) dxdy
(3.1)
exists. Moreover, this integral can be developed in asymptotic power series in t
with coefficients being the invariants of the Riemanian manifold. For brevity, let
us put h = 1. The following result was announced in [AHK2] and its complete
proof will be published elsewhere. We shall sketch here only the main line of
necessary calculations using the technique developed in Section 3.7.
Theorem 3.1. Integral (3.1) exists and has the asymptotical expansion for
small time in the form
(2πt3 )−n/2 (V ol M + a3 t3 + a4 t4 + ...),
the first nontrivial
coefficient a3 beingR proportional to the Gaussian curvature
R
G(M ) = M R dx of M and V ol M = M dx being the Riemanian volume.
Sketch of the Proof. The existence of the expansion follows from the asymptotic formula for the Green function obtained above. Let us show how to prove
144
the last statement, indicating as well the main steps of the exact calculation of
a3 . From (3.2.2),(3.2.4) it follows that
S(t, x0 , y 0 ; x0 , y 0 ) = Σ(t,
x0 − x̃ 0
, y − ỹ).
t
Therefore in normal coordinate around the point x0 = 0 one has
S=
1
t
µ 2
¶
µ
¶
x̃
x̃
x̃
6 2 − 6 (y 0 − ỹ) + 2(y 0 − ỹ)2 + tΣ1 − , y 0 − ỹ + O(t2 ).
t
t
t
Using (2.3.5), (2.3.14) and expansion (2.2), (2.5),(2.6) let us make formulas (2.8)
more precise:
(
ij 0 0 0
kl
− 12 gkl
)yj yk yl + O(t4 )
x̃i = −tyi0 + 61 t3 (gij
(3.3)
1 2 ij 0 0 0
0
ỹi = yi + 4 t gkl yj yk yl + O(t3 ).
Therefore
S=
6 X 0 1 2 kl 1 ij 0 0 0 2 3t 0 ij 0 0 0
(y − t (gij − gkl )yj yk yl ) − yi gkl yk yl yj + tΣ1 (y 0 , 0) + O(t2 ).
t i i
6
2
2
Consequently,
S=
6 0 0
5 kl 0 0 0 0
yi yj yk yl + tΣ1 (y 0 , 0) + O(t2 ).
(y , y ) − tgij
t
2
(3.4)
Therefore, to get the first nontrivial term of the expansion of S one needs the
solution of (2.10) at y = 0, ξ = y 0 .
Similarly, we have
x̃
φ(t, 0, y 0 ; 0, y 0 ) = t−2n Ψ(t, − , y 0 − ỹ)
t
x̃
= t−2n (1 + t2 Ψ2 (− , y 0 − ỹ) + O(t3 ) = t−2n (1 + t2 Ψ2 (y 0 , 0) + O(t3 )),
t
(3.5)
and therefore we need the solution of (2.17) also only at y = 0, ξ = y 0 . From
(2.14) and (2.18) it follows that
kl 0 0 0 0
Σ1 (y 0 , 0) = σgij
yi yj yk yl ,
Ψ2 (y 0 , 0) =
X
ij
ik 0 0
kk
(βgkk
+ γgij
+ δgjk
)yi yj
k
with some constants σ, β, γ, δ.
The key point in the proof of the theorem is the following fact.
Lemma 3.1. In formula (3.6), one has σ = 52 .
(3.6)
(3.7)
145
Proof. To simplify calculations let us first note that formula (7.31) will not
kl
change if we take instead of the tensor gij
its symmetrisation, and therefore,
0
when calculating Σ1 (y , 0) from equation (7.18) we can consider the coefficients
kl
gij
in the expression for F to be completely symmetric (with respect to any
change of the order of its indices i, j, k, l. In particular, it means that instead of
F2 and F3 from (2.11), (2.12) we can take
kl 0 0
F̃2 = (6ξi ξj − 16ξi yj − 11yi yj )gij
yk yl ,
F̃3 = (12ξi ξj ξk + 39ξi ξk yj + 15ξi yj yk −
kl 0
2yi yj yk )gij
yl .
(3.8)
(3.9)
Next, clearly
1 ∂ 2 Σ21 i j
1 ∂ 3 Σ31
1
∂ 4 Σ41
y0 y0 +
y0i y0j y0k +
yi yj yk yl ,
2 ∂ξi ∂ξj
3! ∂ξi ∂ξj ∂ξk
4! ∂ξi ∂ξj ∂ξk ∂ξl 0 0 0 0
(3.10)
where Σp1 , p = 2, 3, 4, denote the corresponding homogeneous part of Σ1 . Now
kl
taking into consideration the assumed symmetricity of the coefficients of gij
one
gets from (3.7.15) and (3.8) that
µ
¶
1 ∂ 2 Σ21 0
6
9
4
kl 0 0
kl 0 0
(y , 0) = − × 11 +
× 16 + × 6 gij
yk yl , = 6gij
yk yl ,
2 ∂ξi ∂ξj
5
10
5
Σ1 (y 0 , 0) =
from (3.7.16) and (3.9) that
µ
¶
9 kl 0
54
39
61
113
1 ∂ 3 Σ31
kl 0
yl = − gij
yl ,
=
×2+
× 15 −
× 39 +
× 12 gij
3! ∂ξi ∂ξj ∂ξk
35
35
70
140
2
and from (3.7.17) and (2.13) that
µ
¶
∂ 4 Σ41
263
51
92
38
1
kl
kl
.
= gij
= −
×9−
×2+
× 18 −
× 4 gij
4! ∂ξi ∂ξj ∂ξk ∂ξl
315
35
105
35
Subsituting these formulas to (3.10) yields
Σ1 (y 0 , 0) = (6 −
9
5 kl 0 0 0 0
kl 0 0 0 0
+ 1)gij
yi yj yk yl = gij
yi yj yk yl ,
2
2
and the Lemma is proved.
End of the proof of the Theorem. Due to the Lemma, the sum of the second
and third terms in the expression (3.4) for S vanishes. Therefore, due to (3.4),
(3.6), (3.7), and to the fact that the oddR degrees of y 0 do not contribute to the
integral, one concludes that the integral u(t, 0, y; 0, y) dy is equal to
à √ !n Z
X
3
ij
−6y 2 /t
2
kk
ik
(βgkk
+ γgij
+ δgjk
)yi yj + O(t3 |y|6 ) + O(t4 |y|4 )] dy.
e
[1
+
t
πt2
k
Due to (2.3), (2.4), this is equal to
à √ !n µ ¶
n/2
tπ
1
3
[1 + t3 δR + O(t4 )].
=
2
πt
6
12
Integrating this expression over M obviously gives (3.2) with a3 = δG(M )/12.