Volumes of geodesic balls in Heisenberg group 

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理 學 博 士 學 位 論 文

2n+1차원 하이젠버그군에서 측지구의 부피

Volumes of geodesic balls in Heisenberg group 

^{  }

蔚山大學校 大學院 數學科

金 惠 娟

[UCI]I804:48009-200000286123 [UCI]I804:48009-200000286123

Volumes of geodesic balls in Heisenberg group  ^{  }

指導敎授 朴 健

이 論文을 理學博士學位 論文으로 提出함

2020年 2月

蔚山大學校 大學院 數學科

金 惠 娟

金惠娟의 理學博士學位 論文을 認准함

審査委員 강 태 호 (印) 審査委員 신 준 국 (印) 審査委員 추 상 목 (印) 審査委員 심 인 보 (印) 審査委員 박 건 (印)

蔚 山 大 學 校 大 學 院

2020年 2月

(ABSTRACT)

The geodesic and volume of a geodesic ball are well understood in simply connected abelian Lie groups having a left invariant metric, ^. A non-abelian Lie group which is most closely related to the abelian group ^ is known as a Heisenberg Group. Thus, understanding the geometric properties of a Heisenberg Group is essential in studies of differential geometry, and the calculation of the volume of a geodesic ball in a Heisenberg group with a left invariant metric has geometric significance.

Studies into calculating the volume of geodesic balls for 3D and 5D Heisenberg groups have been conducted, and have established the volume of a geodesic ball in the 2n + 1D Heisenberg group.

Chapter 2 provides an introduction to a 2-step nilpotent Lie group, which belongs to the Heisenberg group, as well as a general description of the Jacobi field necessary to calculate the volume of a geodesic ball.

Section 1 of Chapter 3 describes the results required to prove the main theorem of this study. Section 2 of Chapter 3 presents the proof of an equation for the volume of a geodesic ball in the 2n+1 D Heisenberg group. Section 3, as a special case of the equation for the volume of a geodesic ball in the 2n+1 D Heisenberg group where n =2, presents the equation of the volume of a geodesic ball of the 5D Heisenberg group, consistent with the results of previous studies.

< CONTENTS >

1. Introduction and main results ···1

2. Preliminaries ···3

3. Proof of Main Results ···7

3.1 Some Preliminary Lemmas ···7

3.2 Proof of Main Theorem ···13

3.3 Proof of Corollary ···44

References ···45

Chapter 1

Introduction and main results

LetN be a 2-step nilpotent Lie algebra with an inner product<, >and N be its unique simply connected 2-step nilpotent Lie group with the left invariant metric induced by<, >onN. LetZ be the center ofN. ThenN is represented by the direct sum ofZand its orthogonal complement Z^⊥.

For each Z ∈ Z, a skew symmetric linear transformation j(Z) : Z^⊥ → Z^⊥ is defined by j(Z)X= (adX)^∗Z forX ∈ Z^⊥. Or, equivalently,

< j(Z)X, Y >=<[X, Y], Z >

for all X, Y ∈ Z^⊥.

A 2-step nilpotent Lie algebra N is said to be an algebra of Heisenberg type if j(Z)²=−|Z|²id

for all Z ∈ Z. And a Lie group N is said to be a group of Heisenberg type if its Lie algebra N is an algebra of Heisenberg type. The Heisenberg groups are examples of Heisenberg type.

That is, letn≥1be any integer and{X₁,· · · , X_n, Y₁,· · · , Y_n}a basis of R²ⁿ=V.LetZ be an one dimensional vector space spanned by {Z}.Define

[Xi, Yi] =−[Yi, Xi] =Z

for any i= 1,2,· · · , n that all other brackets are zero. Give on N =V ⊕ Z the inner product such that the set of vectors {Xi, Yi, Z|i= 1,2,· · · , n} forms an orthonormal basis. Let N be the simply connected 2-step nilpotent group of Heisenberg type which is determined by N and equipped with a left-invariant metric induced by the inner product in N. The groupN is called the (2n+ 1)-dimensional Heisenberg group and denoted by H²ⁿ⁺¹. In this thesis, we calculate the volumes of the geodesic balls in the Heisenberg group H²ⁿ⁺¹:

Main Theorem. Let Be(R) be the geodesic ball with center e (the identity of H²ⁿ⁺¹) and radius R inH²ⁿ⁺¹.Then the following holds.

V ol(B_e(R))

= 2ⁿ⁺¹πA (2n+ 2)!n

(

u₀R+v_n,2n+1R²ⁿ⁺¹+

n−1

∑

k=0

(p_k(R) sin^2n−2k−1(R

2) cos(R 2) +qk(R) sin^2n−2k(R

2) +rk(R)

∫ ^R

2

0

sin^2n−2k−1tcost

t dt)

)

(1.0.1)

where A,u₀ andv_n,2n+1are constants,p_k(R)is a polynomial with2k+ 2≤deg(p_k(R)−p_k(0))≤ 2n, q_k(R) is a polynomial with 2k+ 1≤deg(q_k(R))≤ 2n+ 1 and r_k(R) is a polynomial with 2k+ 2≤deg(r_k(R))≤2n+ 2for0≤k≤n−1.

In special case, we obtained the volumes of geodesic balls in the Heisenberg group forn= 2.

Corollary ([?]). Let Be(R) be the geodesic ball with center e(the identity of H⁵) and radius R inH⁵.Then the following holds.

V ol(Be(R))

=4π² 6!

(

576R+ (−240−108R²−2R⁴) sinR+ (30 + 54R²+ 4R⁴) sin 2R + (−528R−116R³−2R⁵) cosR+ (132R+ 116R³+ 8R⁵) cos 2R + (−720R²−120R⁴−2R⁶)

∫ _R

0

sint

t dt+ (360R²+ 240R⁴+ 16R⁶)

∫ _2R

0

sint t dt

) .

Chapter 2

Preliminaries

LetN be a 2-step nilpotent Lie algebra with an inner product<, >and N be its unique simply connected 2-step nilpotent Lie group with the left invariant metric induced by <, >on N.The center ofN is denoted byZ.ThenN can be expressed as the direct sum ofZ and its orthogonal complement Z^⊥.

Recall that for Z ∈ Z, a skew symmetric linear transformation j(Z) :Z^⊥ → Z^⊥ is defined by j(Z)X= (adX)^∗Z forX ∈ Z^⊥. Or, equivalently,

⟨j(Z)X, Y⟩=⟨[X, Y], Z⟩

forX, Y ∈ Z^⊥. A 2-step nilpotent Lie group N is said to bea group of Heisenberg typeif j(Z)²=−|Z|²id

for all Z ∈ Z.

Let γ(t) be a curve in N such that γ(0) = e (identity element in N) and γ^′(0) = X₀ +Z₀ where X0 ∈ Z^⊥ and Z0 ∈ Z. Since exp :N → N is a diffeomorphism ([?]), the curve γ(t) can be expressed uniquely by γ(t) = exp [X(t) +Z(t)] with

X(t)∈ Z^⊥, X^′(0) =X₀, X(0) = 0 Z(t)∈ Z, Z^′(0) =Z0, Z(0) = 0.

A. Kaplan ([?],[?]) shows that the curve γ(t)is a geodesic in N if and only if X^′′(t) =j(Z₀)X^′(t),

Z^′(t) +¹₂[X^′(t), X(t)]≡Z0. The following lemma is useful in the later.

Lemma 2.0.1 ([?]). Let N be a simply connected 2-step nilpotent Lie group with a left invariant metric, and let γ(t) be a geodesic ofN with γ(0) =e and γ^′(0) =X0+Z0 where X0∈ Z^⊥ and Z₀ ∈ Z.Then one has

γ^′(t) =dl_γ(t)(X^′(t) +Z0), t∈R where X^′(t) =e^tj(Z0)X0 and l_γ(t) is the left translation by γ(t).

Throughout this thesis, different tangent spaces will be identified withN via a left transla- tion. So, in the above lemma, we can considerγ^′(t) as

γ^′(t) =X^′(t) +Z₀ =e^tj(Z0)X₀+Z₀.

Let H²ⁿ⁺¹ be the (2n+ 1)-dimensional Heisenberg group with a left invariant metric and H be its Lie algebra. Let γ(t) be a unit speed geodesic on H²ⁿ⁺¹ with γ(0) = e(the identity element of H²ⁿ⁺¹) and γ^′(0) =X0+Z0 whereX0 ∈ Z^⊥ and Z0 ∈ Z.Assume thatX0 ̸= 0and Z0 ̸= 0.Since

{X0+Z0,|Z0|

|X₀|X0−|X0|

|Z₀|Z0, 1

|Z₀||X₀|j(Z0)X0} is an orthonormal set in H,we can obtain an orthonormal basis B ={X₀+Z₀,_|^|_X^Z0|

0|X₀ −^|_|^X_Z0^0|_|Z₀,_|Z ¹

0||X0|j(Z₀)X₀, Y_k, _|Z¹

0|j(Z₀)Y_k|Y_k ∈ Z^⊥, k = 1,2,· · ·, n−1} by adding {Y_k,_|Z¹

0|j(Z0)Y_k|Y_k ∈ Z^⊥, k = 1,2,· · ·, n−1} to

{X0+Z0,|Z₀|

|X0|X0− |X₀|

|Z0|Z0, 1

|Z0||X0|j(Z0)X0}.

Let

e₁(t) = |Z₀|

|X0|X^′(t)−|X₀|

|Z0|Z₀, e₂(t) = 1

|Z0||X0|j(Z₀)X^′(t) and let

e_2k−1(t) =e^tj(Z0)Y_k, e_2k(t) = 1

|Z₀|e^tj(Z0)j(Z0)Y_k for each k= 2,3,· · · , n.

Then{γ^′(t), e_2k−1(t), e_2k(t)|k= 1,2,· · · , n}is an orthonormal frame alongγ(t)onH²ⁿ⁺¹(see[?]).

Proposition 2.0.2 ([?]). For each k = 1,2,· · ·, n, let J_2k−1(t) and J_2k(t) be the Jacobi fields with J2k−1(0) =J2k(0) = 0, J_2k^′ ₋₁(0) =e2k−1(0) and J_2k^′ (0) =e2k(0).Then we have

(1) for k= 1 [

J1(t) J₂(t) ]

=B₁(t) [

e1(t) e₂(t) ]

where

B1(t) = 1

|Z₀|³

[sin(|Z₀|t)−(1− |Z₀|²)|Z₀|t |Z₀|(cos(|Z₀|t)−1)

|Z0|(1−cos(|Z0|t)) |Z0|²sin(|Z0|t) ]

, (2) for k= 2,3,· · ·, n [

J_2k−1(t) J_2k(t)

]

=B_k(t) [

e_2k−1(t) e_2k(t)

]

where

B_k(t) =

[ ₁

|Z0|sin(|Z₀|t) |Z₀|(cos(|Z₀|t)−1)

|Z10|³(1−cos(|Z₀|t)) _|Z¹

0|sin(|Z₀|t) ]

.

Corollary 2.0.3 ([?],[?]). Let H²ⁿ⁺¹ be the (2n+ 1)-dimensional Heisenberg group and N be its Lie algebra. Let γ(t) be a unit speed geodesic onN with γ(0) =e(the identity element of N) and γ^′(0) =X₀+Z₀ where X₀ ∈ Z^⊥ andZ₀∈ Z.

(1) If Z0 ̸= 0, then all the conjugate points along γ are att∈ _|Z^2π_0|Z^∗∪A where

Z^∗={±1,±2, . . .}

and

A={t∈R− {0}|(1− |Z₀|²)|Z₀|t

2 = tan|Z₀|t 2 }. In particular, _|^2π_Z

0| is the first conjugate point of e alongγ.

(2) If Z₀ = 0, then there are no conjugate points along γ.

For the conjugate points of another type of Heisenberg groups, Quaternionic Heisenberg groups H⁴ⁿ⁺³, see [?].

G.Walschap([?]) showed that the first conjugate loci and the cut loci are equal in the case of the groups of Heisenberg type or the 2-step nilpotent groups with one-dimensional center. So, we consider the geodesic balls B_e(R) with the radius R ≤ 2π. In this thesis we calculate the volumes of geodesic balls inH²ⁿ⁺¹.

In ([?]), C.Jang, J. Park and K. Park obtained the formula of the volumes of geodesic balls in the Heisenberg groupH³ as the form of power series.

Theorem 2.0.4 ([?]). Let B_e(R) be the geodesic ball with center e and radius R in H³. Then the following holds.

V ol(Be(R)) = 4π (

R³ 3 + 2

∑∞ n=2

(−1)ⁿ R²ⁿ⁺¹

(2n+ 1)!(2n−1)(2n−3) )

.

In ([?]), S. Jeong and K. Park obtained the formula of the volumes of geodesic balls in the Heisenberg group H³.

Theorem 2.0.5([?]). Let0≤R≤2π andBe(R)be the geodesic ball with centere(the identity of H³) and radius R in H³. Then the following holds.

V ol(B_e(R))

=π 6

(

−16R+ (R²+ 6) sinR+ (R³+ 10R) cosR+ (R⁴+ 12R²)

∫ _R

0

sint t dt

) .

In ([?]), the author obtained the formula of the volumes of the geodesic balls in the Heisenberg group H⁵.

Theorem 2.0.6 ([?]). Let Be(R) be the geodesic ball with center e (the identity of H⁵) and radius R in H⁵.Then the following holds.

V ol(B_e(R))

=4π² 6!

(

576R+ (−240−108R²−2R⁴) sinR+ (30 + 54R²+ 4R⁴) sin 2R + (−528R−116R³−2R⁵) cosR+ (132R+ 116R³+ 8R⁵) cos 2R + (−720R²−120R⁴−2R⁶)

∫ _R

0

sint

t dt+ (360R²+ 240R⁴+ 16R⁶)

∫ _2R

0

sint t dt

) .

Chapter 3

Proof of Main Results

3.1 Some Preliminary Lemmas

Note that

det(B₁(t)) = 1

|Z₀|⁴{2(1−cos(|Z₀|t))−(1− |Z₀|²)|Z₀|tsin(|Z₀|t)} and

det(B_k(t)) = 2

|Z₀|²(1−cos(|Z0|t)) for each k= 2,3,· · ·, n.

Lemma 3.1.1 ([?]). For t >0,the following holds.

det (∏n

k=1

B_k(t) )

= 1

|Z0|⁴{2(1−cos(|Z0|t))−(1− |Z0|²)|Z0|tsin(|Z0|t)}{ 2

|Z0|²(1−cos(|Z0|t))}ⁿ⁻¹≥0.

Lemma 3.1.2 ([?]). Let n be a natural number and f : [0, x] → R has continuous n−th derivatives. Assume that the lim_t→0^{+ f}_t^(k)n−^(t)k exists fork= 0,1,· · ·, n−1 and ^f(n)_t^(t) is integrable on [0, x].Then _t^f(t)n+1 is integrable on [0, x]and the following holds.

∫ _x

0

f(t)

tⁿ⁺¹dt=−

n−1

∑

k=0

1

n(n−1)· · ·(n−k) [

f^(k)(t) t^n−k

]x⁻ 0⁺

+ 1 n!

∫ _x

0

f⁽ⁿ⁾(t) t dt

where [

f^(k)(t) t^n−k

]_x⁻

0⁺

= lim

t→x⁻

f^(k)(t)

t^n−k − lim

t→0⁺

f^(k)(t) t^n−k .

The following lemma is useful in the later.

Lemma 3.1.3 ([?]). Let n be a natural number and f : [0,1] → R has continuous n−th derivatives. (i) f^(k)(0) = 0 for k= 0,1,· · ·, n−1 and (ii) ^f(n)_t^(t) is integrable on [0,1]. Then

f(t)

tⁿ⁺¹ is integrable on[0,1] and

∫ ₁

0

f(t)

tⁿ⁺¹dt= 1 n!

(

−

n−1

∑

k=0

(n−k−1)!f^(k)(1) +f⁽ⁿ⁾(0)

∑n k=1

1 k+

∫ ₁

0

f⁽ⁿ⁾(t) t dt

) . Proof. By the L’Hopital’s theorem, we have

lim

t→0⁺

f(t) tⁿ = lim

t→0⁺

f⁽¹⁾(t) ntⁿ⁻¹

= lim

t→0⁺

f⁽²⁾(t) n(n−1)tⁿ⁻² ...

= lim

t→0⁺

f^(k)(t)

n(n−1)· · ·(n−k+ 1)t^n−k ...

=f⁽ⁿ⁾(0) n! . So, we get

tlim→0⁺

f^(k)(t)

t^n−k =n(n−1)· · ·(n−k+ 1)f⁽ⁿ⁾(0) n!

fork(0≤k≤n).By Lemma 3.1.2, we have

∫ ₁

0

f(t) tⁿ⁺¹dt

=−

n−1

∑

k=0

1

n(n−1)· · ·(n−k) (

lim

t→1⁻

f^(k)(t) t^n−k − lim

t→0⁺

f^(k)(t) t^n−k

) + 1

n!

∫ ₁

0

f⁽ⁿ⁾(t) t dt

=−

n−1

∑

k=0

1

n(n−1)· · ·(n−k) (

f^(k)(1)−n(n−1)· · ·(n−k+ 1)·f⁽ⁿ⁾(0) n!

) + 1

n!

∫ ₁

0

f⁽ⁿ⁾(t) t dt

=−

n−1

∑

k=0

f^(k)(1)

n(n−1)· · ·(n−k)+f⁽ⁿ⁾(0) n!

n−1

∑

k=0

1 n−k + 1

n!

∫ ₁

0

f⁽ⁿ⁾(t) t dt

=−

n−1

∑

k=0

f^(k)(1)

n(n−1)· · ·(n−k)+f⁽ⁿ⁾(0) n!

∑n k=1

1 k + 1

n!

∫ ₁

0

f⁽ⁿ⁾(t) t dt

=−

n−1

∑

k=0

(n−k−1)!f^(k)(1)

n(n−1)· · ·(n−k)(n−k−1)! +f⁽ⁿ⁾(0) n!

∑n k=1

1 k + 1

n!

∫ ₁

0

f⁽ⁿ⁾(t) t dt

=1 n!

(

−

n−1

∑

k=0

(n−k−1)!f^(k)(1) +f⁽ⁿ⁾(0)

∑n k=1

1 k+

∫ ₁

0

f⁽ⁿ⁾(t) t dt

) . This completes the proof.

Lemma 3.1.4 ([?]). Area element du on the unit sphere S²ⁿ is given by

du= 1

√1−(x²₁+x²₂+· · ·+x²_2n)dx1dx2· · ·dx2n

where dudenotes the canonical measure of S²ⁿ.

The n-dimension spherical coordinates are well known in Wikipedia the Free Encyclopedia.

Lemma 3.1.5 ([?]). (n-dimension spherical coordinates) Let









































x1 =rcosθ1

x2 =rsinθ1cosθ2

x₃ =rsinθ₁sinθ₂cosθ₃ x₄ =rsinθ₁sinθ₂sinθ₃cosθ₄

...

x_n−2=rsinθ₁sinθ₂sinθ₃· · ·sinθ_n−3cosθ_n−2

x_n−1=rsinθ₁sinθ₂sinθ₃· · ·sinθ_n−3sinθ_n−2cosθ_n−1 xn=rsinθ1sinθ2sinθ3· · ·sinθn−3sinθn−2sinθn−1

where r ≥0, 0≤θ_i≤π(i= 1,2,· · · , n−2), and 0≤θ_n−1 ≤2π. Then Jacobian determinant

∂(x1, x2,· · · , xn)

∂(r, θ₁, θ₂,· · · , θ_n−1) =rⁿ⁻¹sinⁿ⁻²θ₁sinⁿ⁻³θ₂· · ·sin²θ_n−3sinθ_n−2.

Lemma 3.1.6. For 0≤m≤n and a >0, we have d^2m

dx^2m(sin²ⁿ(ax)) =a^2m

∑m k=0

c_ka(k, m−k) sin^2n−2k(ax) where c_k= _(2n^(2n)!_−2k)!,

a(k, l) =







1 , l= 0

(−2²)^l∑

a1,a2,···,al∈{n,n−1,···,n−k} a1≥a2≥···≥al

(a1a2· · ·al)² , l≥1 (3.1.1) Proof. For the convenience of calculation, denotef_2n(x) = sin²ⁿ(ax) and we use mathematical induction for m. For m=0, we have

a⁰

∑0 k=0

c_ka(k,−k) sin^2n−2k(ax) = (2n)!

(2n−0)!a(0,0) sin²ⁿ(ax) = sin²ⁿ(ax).

The proposition is true for m=0. Next assume that for 0≤m < n, f_2n^(2m)(x) =a^2m

∑m k=0

c_ka(k, m−k) sin^2n−2k(ax).

Differentiating both sides the equation with respect tox, we get f_2n^(2m+1)(x) =a^2m+1

∑m k=0

(2n−2k)c_ka(k, m−k) sin^2n−2k−1(ax) cos(ax) and

f_2n^(2m+2)(x) =a^2m+2

∑m k=0

(2n−2k)(2n−2k−1)cka(k, m−k) sin^2n−2k−2(ax)

−a^2m+2

∑m k=0

(2n−2k)²cka(k, m−k) sin^2n−2k(ax).

(3.1.2)

Using

a^2m+2

∑m k=0

(2n−2k)(2n−2k−1)c_ka(k, m−k) sin^2n−2k−2(ax)

=a^2m+2

m+1∑

k=1

(2n−2k+ 2)(2n−2k+ 1)ck−1a(k−1, m−k+ 1)f2n−2k(x), we rewrite (3.1.2) as

f_2n^(2m+2)(x) =a^2m+2

m+1∑

k=1

(2n−2k+ 2)(2n−2k+ 1)c_k−1a(k−1, m−k+ 1)f_2n−2k(x)

−a^2m+2

∑m k=0

(2n−2k)²c_ka(k, m−k)f_2n−2k(x).

(3.1.3)

We consider k= 0,1≤k≤m and k=m+ 1in (3.1.3).

For k= 0,we get

−a^2m+2(2n−0)²c0a(0, m−0)f2n−0(x) =a^2(m+1)c0a(0, m+ 1) sin²ⁿ(ax). (3.1.4) In case1≤k≤m, we have

a^2m+2

∑m k=1

(2n−2k+ 2)(2n−2k+ 1)c_k−1a(k−1, m−k+ 1)f_2n−2k(x)

−a^2m+2

∑m k=1

(2n−2k)²c_ka(k, m−k)f_2n−2k(x)

=a^2m+2

∑m k=1

( (2n)!

(2n−2k)!a(k−1, m−k+ 1)−(2n−2k)² (2n)!

(2n−2k)!a(k, m−k) )

f_2n−2k(x)

=a^2(m+1)

∑m k=1

c_k (

a(k−1, m−k+ 1)−(2n−2k)²a(k, m−k) )

sin^2n−2k(ax).

(3.1.5) Using (3.1.1) in the last equation of (3.1.5), we have

a(k−1, m−k+ 1)

=(−2²)^m−k+1 ∑

a1,a2,···,am−k,am−k+1∈{n,n−1,···,n−k+1} a1≥a2≥···≥am−k≥am−k+1

(a₁a₂· · ·a_m−ka_m−k+1)² (3.1.6)

and

−(2n−2k)²a(k, m−k)

=(−2²)^m−k+1 ∑

a1,a2,···,am−k∈{n,n−1,···,n−k+1,n−k} a1≥a2≥···≥am−k

(n−k)²(a₁a₂· · ·a_m−k)². (3.1.7)

LetSbe event of (3.1.6). EventSrepresents choice of(m−k+ 1)elementsa₁, a₂,· · · , a_m−kand am−k+1 witha1≥a2 ≥ · · · ≥am−k≥am−k+1 exceptn−kfrom a collection of (k+ 1)elements n, n−1,· · · , n−k+ 1,andn−k.LetDbe event of (3.1.7). Puttinga_m−k+1=n−k,eventD represents choice of (m−k) elements a₁, a₂,· · ·, a_m−k witha₁ ≥a₂ ≥ · · · ≥ a_m−k including at leastam−k+1=n−kfrom a collection of(k+ 1)elementsn, n−1,· · · , n−k+ 1,and n−k.We see that the complement of event S is event D. The outcomes of an event and the outcomes of the complement make up the entire sample space. Adding (3.1.6) and (3.1.7) represents choice of (m−k+ 1) elements a1, a2,· · ·, a_m−k, a_m−k+1 witha1 ≥a2 ≥ · · · ≥a_m−k ≥a_m−k+1 from a collection of (k+ 1)elements n, n−1,· · · , n−k+ 1,and n−k.

That is, we have

a(k−1, m−k+ 1)−(2n−2k)²a(k, m−k)

=(−2²)^m−k+1 ∑

a1,a2,···,am−k,am−k+1∈{n,n−1,···,n−k+1,n−k} a1≥a2≥···≥am−k+1

(a₁a₂· · ·a_m−ka_m−k+1)²

=a(k, m−k+ 1).

Then we obtain a^2m+2

∑m k=1

c_k (

a(k−1, m−k+ 1)−(2n−2k)²a(k, m−k) )

sin^2n−2k(ax)

=a^2(m+1)

∑m k=1

c_ka(k, m−k+ 1) sin^2n−2k(ax).

For k=m+ 1,we get

a^2m+2(2n−2(m+ 1) + 2)(2n−2(m+ 1) + 1)c_ma(m,0)f_2n−2(m+1)(x)

=a^2(m+1)c_m+1a(m+ 1,0) sin^2n−2(m+1)(ax).

(3.1.8) From (3.1.4), (3.1.5) and (3.1.8), we consider (3.1.3) as

f_2n^(2m+2)(x) =a^2(m+1)

m+1∑

k=0

c_ka(k, m−k+ 1) sin^2n−2k(ax).

Therefore, it is true that for 0≤m+ 1< n, d^2(m+1)

dx^2(m+1)(sin²ⁿ(ax)) =a^2(m+1)

m+1∑

k=0

c_ka(k, m+ 1−k) sin^2n−2k(ax).

This completes the proof.

Lemma 3.1.7. Let nbe a natural number. For R>0 the following holds.

∫ _Rx

0

(1−cost)ⁿdt= 2ⁿ⁺¹

( (2n)!

(n!2ⁿ)² ·Rx

2 − 1

2n+ 1

n−1

∑

k=0

δ_ksin^2n−2k−1Rx

2 cosRx 2

)

where

δ_k = 2n+ 1

2n ·2n−1

2n−2· · ·2n+ 1−2k 2n−2k =

∏k l=0

2n+ 1−2l 2n−2l . Proof. We make the substitutionRx=a, so that

∫ _a

0

(1−cost)ⁿdt=

∫ _a

0

2ⁿsin²ⁿ t 2dt

=2ⁿ⁺¹

∫ ^a

2

0

sin²ⁿθdθ.

We calculate∫^a₂

0 sin²ⁿθdθ as follows. Let∫^a₂

0 sin²ⁿθdθ=I2n,thenI0 = ^a₂ and I_2n=− 1

2nsin²ⁿ⁻¹ a 2cosa

2+ 2n−1 2n I_2n−2

=− 1

2nsin²ⁿ⁻¹ a 2cosa

2+ 2n−1 2n

(

− 1

2n−2sin²ⁿ⁻³a 2cosa

2 +2n−3 2n−2I_2n−4

)

= ...

=− 1

2nsin²ⁿ⁻¹ a 2cosa

2+ (

− 2n−1

2n(2n−2)sin²ⁿ⁻³a 2cosa

2 )

+ (

− (2n−1)(2n−3)

2n(2n−2)(2n−4)sin²ⁿ⁻⁵a 2cosa

2 )

+· · · +

(

− (2n−1)(2n−3)· · ·3

2n(2n−2)(2n−4)· · ·2sin¹ a 2cosa

2 )

+ (2n−1)(2n−3)· · ·3 2n(2n−2)(2n−4)· · ·2 ·I₀

=− 1

2nsin²ⁿ⁻¹ a 2cosa

2− 2n−1

2n(2n−2)sin²ⁿ⁻³ a 2cosa

2 − (2n−1)(2n−3)

2n(2n−2)(2n−4)sin²ⁿ⁻⁵a 2cosa

2

− · · · − (2n−1)(2n−3)· · ·3

2n(2n−2)(2n−4)· · ·2sin¹a 2cosa

2 + (2n)!

(n!2ⁿ)² ·a 2. Therefore,

I_2n= (2n)!

(n!2ⁿ)² ·a

2− 1

2n+ 1

n−1

∑

k=0

δ_ksin^2n−2k−1a 2cosa

2 where

δ_k = 2n+ 1

2n ·2n−1

2n−2· · ·2n+ 1−2k 2n−2k =

∏k l=0

2n+ 1−2l 2n−2l .

Since we invent the new variablea, we need to convert back to the original variableRx. We get

∫ _Rx

0

(1−cost)ⁿdt= 2ⁿ⁺¹

( (2n)!

(n!2ⁿ)² ·Rx

2 − 1

2n+ 1

n−1

∑

k=0

δ_ksin^2n−2k−1 Rx

2 cosRx 2

) . This completes the proof.

3.2 Proof of Main Theorem

We introduce the volume formula of geodesic balls in Riemannian manifolds, which is well- known. For example, see [?]. Let M be a Riemannian manifold with a metric g and p ∈ M.

Take an orthonormal basis {u1, u2,· · ·, un} of TpM and let (x1, x2,· · · , xn) be the coordinates determined by {u₁, u₂,· · · , u_n}. This local coordinate system is called the normal coordinate system at p. It is easy to show that _∂x^∂

im = (dexp_p)^∑n

i=1xiui(ui) where m = exp_p(∑_n

i=1xiui).

Then the volume form v_g on U_p is given by v_g =

√ det

( g( ∂

∂xi

, ∂

∂xj

) )

dx₁∧dx₂∧ · · · ∧dx_n

where gij is the metric coefficients of g inUp. Therefore, the volume of the geodesic ballBp(r) is given by

V ol(B_p(r)) =

∫

exp⁻p¹(Bp(r))

exp^∗_pv_g.

Let γ(t) be the unit speed geodesic in M with γ(0) = p, γ^′(0) = u1 and let Ji(t) be the Jacobi field with Ji(0) = 0 and J_i^′(0) =ui for each i= 2,3,· · · , n.Then we get

(dexp_p)tu1u1 =γ^′(t) and

(dexp_p)_tu1u_i = 1 tJ_i(t) for each i= 2,3,· · ·, n.So, we see that

√ det

( g( ∂

∂x_i, ∂

∂x_j) )

=t⁻⁽ⁿ⁻¹⁾

√

det(g(Ji(t), Jj(t))).

Hence, we have

exp^∗_pv_g =t⁻⁽ⁿ⁻¹⁾

√

det(g(J_i(t), J_j(t)))dx₁dx₂· · ·dx_n=

√

det(g(J_i(t), J_j(t)))dtdu

wheredudenote the canonical measure of the unit sphereSⁿ⁻¹.Therefore, by Fubini’s Theorem, we get

V ol(Bp(r)) =

∫

Sⁿ⁻¹

∫ _r

0

√

det(g(Ji(t), Jj(t)))dtdu.