7 Spectrum of the Laplacian on Quaternionic
Kähler Manifolds
Shengli Kong, Peter Li∗and Detang Zhou†
Abstract
Let M4n be a complete quaternionic Kähler manifold with scalar
curvature bounded below by −16n(n+2). We get a sharp estimate for
the first eigenvalue λ1(M) of the Laplacian which is λ1(M) ≤ (2n+1)
If the equality holds, then either M has only one end, or M is diffeo-
morphic to R × N with N given by a compact manifold. Moreover,
if M is of bounded curvature, M is covered by the quaterionic hy-
perbolic space QHn and N is a compact quotient of the generalized
Heisenberg group. When λ1(M) ≥
8(n+2)
, we also prove that M must
have only one end with infinite volume.
0 Introduction
Let Mn be a complete n-dimensional Riemannian manifold whose Ricci cur-
vature bounded below by −(n − 1). It is well known from Cheng [Ch] that
the first eigenvalue λ1(M) satisfies
λ1(M) ≤
(n− 1)2
In [LW3], Li and Wang proved an analogous theorem for complete Käh-
ler manifolds. They showed that if M2n is a complete Kähler manifold of
complex dimension n with holomorphic bisectional curvature BKM bounded
below by −1, then the first eigenvalue λ1(M) satisfies
λ1(M) ≤ n
∗Research partially supported by NSF grant DMS-0503735
†Research partially supported by CAPES and CNPq of Brazil.
http://arxiv.org/abs/0704.1851v1
Here BKM ≥ −1 means that
Rīijj̄ ≥ −(1 + δij)
for any unitary frame e1, . . . , en.
In this paper, we prove the corresponding Laplacian comparison theorem
for a quaterionic Kähler manifold M4n. As an application we get the sharp
estimate λ1(M) for a complete quaterionic Kähler manifold M
4n with scalar
curvature bounded below by −16n(n+ 2) as
λ1(M) ≤ (2n + 1)
It is an interesting question to ask what one can say about those manifolds
when the above inequalities are realized as equalities. In works of Li and
Wang [LW1] and [LW2], the authors obtained the following theorems. The
first was a generalization of the theory of Witten-Yau [WY], Cai-Galloway
[CG], and Wang [W] for conformally compact manifolds. The second was to
answer the aforementioned question.
Theorem 0.1. Let Mn be a complete Riemannian manifold of dimension
n ≥ 3 with Ricci curvature bounded below by −(n − 1). If λ1(M) ≥ n − 2,
then either
(1) M has only one infinite volume end; or
(2) M = R×N with warped product metric of the form
ds2M = dt
2 + cosh2 t ds2N ,
where N is an (n − 1)-dimensional compact manifold of Ricci curvature
bounded below by λ1(M).
Theorem 0.2. Let Mn be a complete Riemannian manifold of dimension
n ≥ 2 with Ricci curvature bounded below by −(n − 1). If λ1(M) ≥
(n−1)2
then either
(1) M has no finite volume end; or
(2) M = R×N with warped product metric of the form
ds2M = dt
2 + e2t ds2N ,
where N is an (n − 1)-dimensional compact manifold of nonnegative Ricci
curvature.
In [LW3] and [LW5], Li and Wang also consider the Kähler case. They
proved the following theorems.
Theorem 0.3. Let Mn be a complete Kähler manifold of complex dimension
n ≥ 1 with Ricci curvature bounded below by
RicM ≥ −2(n + 1).
If λ1(M) >
, then M must have only one infinite volume end.
Theorem 0.4. Let Mn be a complete Kähler manifold of complex dimension
n ≥ 2 with holomorphic bisectional curvature bounded by
BKM ≥ −1.
If λ1(M) ≥ n
2, then either
(1) M has only one end; or
(2) M = R × N with N being a compact manifold. Moreover the metric on
M is of the form
ds2M = dt
2 + e4t ω22 + e
ω2i ,
where {ω2, ω3, . . . , ω2n} are orthonormal coframe of N with Jdt = ω2. If M
has bounded curvature, then we further conclude that M is covered by CHn
and N is a compact quotient of the Heisenberg group.
In [LW5], the authors pointed out that the assumption on the lower bound
of λ1(M) in Theorem 0.3 is sharp, since one can construct M of the form
M = Σ×N satisfying
RicM ≥ −2(n + 1) (0.1)
λ1(M) =
n + 1
(0.2)
with N being a compact Kähler manifold and Σ being a complete surface
with at least two infinite volume ends. However, it is still an open question to
characterized all those complete Kähler manifolds satisfying conditions ( 0.1)
and ( 0.2).
In sections 4 and 5, we will prove the following quaternionic Kähler ver-
sions of the above theorems.
Theorem 0.5. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature satisfying
SM ≥ −16n(n + 2).
If λ1(M) ≥
8(n+2)
, then M must have only one infinite volume end.
Theorem 0.6. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature satisfying
SM ≥ −16n(n + 2).
If λ1(M) ≥ (2n+ 1)
2, then either
(1) M has only one end, or
(2) M is diffeomorphic to R×N where N is a compact manifold. Moreover,
the metric is given by the form
ds2M = dt
2 + e4t
ω2p + e
where {ω2, . . . , ω4n} are orthonormal coframes for N. If M is of bounded
curvature then we further conclude that M is covered by the quaterionic hy-
perbolic space QHn and N is a compact quotient of the generalized Heisenberg
group.
Remark 0.1. It is known that a horosphere in QHn is isometric to a certain
generalized Heisenberg group with three-dimensional center and left-invariant
Riemannian metric. Such generalized Heisenberg groups have compact quo-
tients. For an explicit construction see for instance Example 2.6 in [G]. We
don’t have an example to show that the bounded curvature condition in The-
orem 0.6 is necessary. If such an example exists, its curvature should decay
at exponentially in some directions.
Perhaps it is interesting to restrict our attention to the special case when
M4n = QHn/Γ is given by the quotient of the quaternionic hyperbolic space
n with a discrete group of isometies Γ. In particular, it is instructional to
compare with previous results by Corlette [C2] and Corlette-Iozzi [CI] where
Lie group theoretic approach was used in understanding these manifolds. For
example, in [CI], the authors proved a Patterson-Sullivan type formula for
λ1(M) in terms of the Hausdorff dimension δ(Γ) of the limit set of Γ. More
specifically, they proved that if Γ is geometrically finite, then for δ(Γ) ≥ 2n+1
one has
λ1(M) = δ(Γ)((4n+ 2)− δ(Γ)).
Hence in this case, the condition in Theorem 0.6 on λ1(M) = (2n + 1)
equivalent to the condition δ(Γ) = 2n+ 1.
In [C2] (Theorem 4.4), Corlette also pointed out that by a result of
Kostant λ1(M) = 0 or λ1(M) ≥ 8n. On the other hand, it was also shown
in [CI] that if Γ is geometrically finite and torsion free, then M = QHn/Γ
must have at most one end with infinite volume. These two statements give
an interesting comparison to Theorem 0.5 stated above.
We would also like to point out to the interested readers that in [LW4] and
[LW5] Li and Wang considered a more general class of manifolds satisfying a
weighted Poincaré inequality. However, since quaternionic Kähler manifolds
are automatically Einstein, the same type of questions are not interesting for
this class of manifolds.
Acknowledgement. This work was done when the third author was
visiting the University of California, Irvine. He wishes to thank the institu-
tion for its hospitality. He also would like to thank Professor J. Berndt for
pointing out the paper of [G] to him.
1 Preliminaries on quaternionic Kähler man-
ifolds
In this section, we will recall basic properties of quaternionic Kähler manifolds
that will be needed in the sequel. These properties were proved by Berger
[B] and Ishihara [I] (also see [Be]).
Let (Mn, g) be a Riemannian manifold, TM the tangent space of M and
∇ the Levi-Civita connection. The Riemannian curvature R : TM ⊗ TM ⊗
TM −→ TM is defined by
R(X, Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z
If {e1, · · · , en} is an orthonormal basis of TM , the components of curvature
tensor is defined by
Rijkl = 〈R(ei, ej)el, ek〉,
the Ricci curvature is defined by
RicM(X, Y ) =
〈R(X, ei)ei, Y 〉,
and the scalar curvature is defined by
i,j=1
〈R(ei, ej)ej , ei〉.
Definition 1.1. A quaternionic Kähler manifold (M, g) is a Riemannian
manifold with a rank 3 vector bundle V ⊂ End(TM) satisfying
(a) In any coordinate neighborhood U ofM , there exists a local basis {I, J,K}
of V such that
I2 = J2 = K2 = −1
IJ = −JI = K
JK = −KJ = I
KI = −IK = J
〈IX, IY 〉 = 〈JX, JY 〉 = 〈KX,KY 〉 = 〈X, Y 〉
for all X, Y ∈ TM .
(b) If φ ∈ Γ(V ), then ∇Xφ ∈ Γ(V ) for all X ∈ TM .
Remark 1.1. It follows from (a) that dimM = 4n. A well known fact about
4n-dimensional Riemannian manifold is that it is quaternionic Kähler if and
only if its restricted holonomy group is contained in Sp(n)Sp(1).
The 4-dimensional Riemannian manifolds with holonomy Sp(1)Sp(1) are
simply the oriented Riemanian manifolds, naturally we only consider those
when n ≥ 2.
Notice that in general I, J,K are not defined everywhere on M . For
example, the canonical quaternionic projective space QP n admits no almost
complex structure.
On the other hand, the vector space generated by I, J,K is well defined
at each point of M and this 3-dimensional subbundle V of End(TM) is in
fact “globally parallel” under the Levi-Civita connection ∇ of g. A basic fact
about the connection is the following lemma.
Lemma 1.1. The condition (b) is equivalent to the following condition:
∇XI = c(X)J − b(X)K,
∇XJ = −c(X)I + a(X)K,
∇XK = b(X)I − a(X)J,
where a, b, c are local 1-forms.
Definition 1.2. Let (M, g) be a quaternionic Kähler manifold. We can
define a 4-form by
Ω = ω1 ∧ ω1 + ω2 ∧ ω2 + ω3 ∧ ω3,
where
ω1 = 〈·, I·〉,
ω2 = 〈·, J ·〉,
ω3 = 〈·, K·〉.
Let {e1, Ie1, Je1, Ke1, · · · , en, Ien, Jen, Ken} be an orthonormal basis of
TM and {θ1, Iθ1, Jθ1, Kθ1, · · · , θn, Iθn, Jθn, Kθn, } the dual basis. It follows
θi ∧ Iθi + Jθi ∧Kθi
θi ∧ Jθi +Kθi ∧ Iθi
θi ∧Kθi + Iθi ∧ Jθi
θi ∧ Iθi ∧ θj ∧ Iθj + θi ∧ Jθi ∧ θj ∧ Jθj + θi ∧Kθi ∧ θj ∧Kθj
Jθi ∧Kθi ∧ Jθj ∧Kθj +Kθi ∧ Iθi ∧Kθj ∧ Iθj
+ Iθi ∧ Jθi ∧ Iθj ∧ Jθj
θi ∧ Iθi ∧ Jθj ∧Kθj + θi ∧ Jθi ∧Kθj ∧ Iθj
+ θi ∧Kθi ∧ Iθj ∧ Jθj
Lemma 1.2. The condition (b) is equivalent to the following condition:
∇Xω1 = c(X)ω2 − b(X)ω3,
∇Xω2 = −c(X)ω1 + a(X)ω3,
∇Xω3 = b(X)ω1 − a(X)ω2.
where a, b, c are local 1-forms.
Proof: It follows from the identities
(∇Xω1)(Y, Z) = 〈Y, (∇XI)Z〉,
(∇Xω2)(Y, Z) = 〈Y, (∇XJ)Z〉,
(∇Xω3)(Y, Z) = 〈Y, (∇XK)Z〉.
Using this lemma, we have that
Theorem 1.1. The condition (b) is equivalent to that Ω is parallel, that is
∇XΩ = 0
for any X ∈ TM .
In the following, we shall study the curvature of quaternionic Kähler
manifold. First we have the following lemma.
Lemma 1.3. If (M4n, g) is a quaternionic Kähler manifold, then
[R(X, Y ), I] = γ(X, Y )J − β(X, Y )K,
[R(X, Y ), J ] = −γ(X, Y )I + α(X, Y )K,
[R(X, Y ), K] = β(X, Y )I − α(X, Y )J,
where α, β and γ are local 2-forms given by
α = da+ b ∧ c,
β = db+ c ∧ a,
γ = dc+ a ∧ b.
Corollary 1.1. If (M4n, g) is a quarternionic Kähler manifold, then
〈R(X, Y )Z, IZ〉+ 〈R(X, Y )JZ,KZ〉 = α(X, Y ) |Z|2,
〈R(X, Y )Z, JZ〉+ 〈R(X, Y )KZ, IZ〉 = β(X, Y ) |Z|2,
〈R(X, Y )Z,KZ〉+ 〈R(X, Y )IZ, JZ〉 = γ(X, Y ) |Z|2.
The following lemma is the key for quaternionic Kähler manifolds.
Lemma 1.4. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2, then
α(X, IY ) = β(X, JY ) = γ(X,KY ) = −
n + 2
RicM(X, Y ). (1.1)
As applications of the above lemma, one can show the following two main
theorems on curvature of quaternionic Kähler manifolds.
Theorem 1.2. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2,
then (M4n, g) is Einstein, that is, there is a constant δ such that
RicM(g) = 4(n+ 2)δg.
Theorem 1.3. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2,
(1) For any tangent vector X, the sectional curvature satisfies
〈R(X, IX)IX,X〉+ 〈R(X, JX)JX,X〉
+〈R(X,KX)KX,X〉 = 12δ |X|4.
(2) For any tangent vector Y satisfying
〈Y,X〉 = 〈Y, IX〉 = 〈Y, JX〉 = 〈Y,KX〉 = 0,
the sectional curvature satisfies
〈R(X, Y )Y,X〉+ 〈R(X, IY )IY,X〉+
〈R(X, JY )JY,X〉+ 〈R(X,KY )KY,X〉 = 4δ |X|2 |Y |2,
where 4(n+ 2)δ is the Einstein constant.
Finally, we end this section with the following lemma.
Lemma 1.5. Let γ : [a, b] → M be a geodesic with unit speed. If S =
16n(n + 2)δ, and XI(t), XJ(t), XK(t) are parallel vector fields along γ such
that XI(a) = Iγ
′(a), XJ(a) = Jγ
′(a), XK(a) = Kγ
′(a), then
K(γ′(t), XI(t)) +K(γ
′(t), XJ(t)) +K(γ
′(t), XK(t)) = 12δ,
for all t and γ.
Let Y be a tangent vector at γ(a) satisfying 〈γ′(a), Y 〉 = 0, 〈Iγ′(a), Y 〉 =
0, 〈Jγ′(a), Y 〉 = 0, and 〈Kγ′(a), Y 〉 = 0. If we denote the parallel vector fields
Y (t), YI(t), YJ(t), and YK(t) along γ with initial data Y (a) = Y , YI(a) = IY,
YJ(a) = JY , and YK(a) = KY , respectively, then
K(γ′(t), Y (t)) +K(γ′(t), YI(t)) +K(γ
′(t), YJ(t)) +K(γ
′(t), YK(t)) = 4δ,
for all t and γ.
Proof. By the discussion above, we know the 3-dimensional vector
space E(t) spanned by X(t), Y (t), Z(t) does not depend on the choice of
I, J,K. Hence it is parallel under the Levi-Civita connection. We consider
〈R(·, γ′(t))γ′(t), ·〉 as a symmetric bilinear form on E(t). ThenK(γ′(t), X(t))+
K(γ′(t), Y (t)) + K(γ′(t), Z(t)) is its trace on E(t) which independent of the
choice of orthonormal basis. By the computation above it is equal to 12δ.
The same argument also applies to the second part of the lemma. �
2 Laplacian comparison theorem
For a complete Riemannian manifold M and p ∈ M , let us denote the cut
locus with respect to p by Cut(p).
Theorem 2.1. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature SM ≥ 16n(n+2)δ and let r(x) be the distance function to a
fixed point p ∈M . Then, for x /∈ Cut(p),
∆r(x) ≤
6 coth 2r(x) + 4(n− 1) coth r(x) when δ = −1
(4n− 3)r−1(x) when δ = 0
6 cot 2r(x) + 4(n− 1) cot r(x) when δ = 1.
(2.1)
Proof. Let γ be the minimizing geodesic joining p to x. At x, we choose
{e1, e2, · · · , en}, and two local almost complex structures I, J and K = IJ
such that e1 = ∇r and
{e1, Ie1, Je1, Ke1, e2, Ie2, Je2, Ke2, · · · , en, Ien, Jen, Ken}
is an orthonormal frame. By parallel translating along γ we obtain an or-
thonormal frame with e1 = ∇r. For convenience sake, we denote this frame
by {ε1, ε2, · · · , ε4n}. Since |∇r|
2 = 1 on M\Cut(p), by taking covariant
derivative of this equation, we have
0 = |∇r|2kl
rikril + 2
ririkl, (2.2)
for each k, l = 2, · · · , 4n. Since
rikl = rkli +
Rjkilrj,
with Rijkl = 〈R(εi, εj)εl, εk〉, and r1 = 1, rj = 0, j = 2, · · · , 4n, we have
rikril + rkl1 +R1k1l = 0. (2.3)
In particular, if k = l, we have
r2ik + rkk1 +K(ε1, εk) = 0, (2.4)
where K(ε1, εk) = R1k1k is the sectional curvature of the 2-plane section
spanned by ε1, εk. Using the inequality
r2ik ≥
and setting f(t) =
k=2 rkk, ( 2.4) implies that
f ′(t) +
f 2(t) +
K(ε1, εk) ≤ 0. (2.5)
By Lemma 1.5, we have
f ′(t) +
f 2(t) + 12δ ≤ 0. (2.6)
Since a smooth Riemannian metric is locally Euclidean, then limt→0 tf(t) = 3.
By a standard comparison argument for ordinary differential equations, we
conclude that
f(t) ≤
6 cot 2t when δ = 1
3t−1 when δ = 0
6 coth 2t when δ = −1.
(2.7)
Similarly, using the inequality
k=4i+1
r2ik ≥
k=4i+1
for 1 ≤ i ≤ n− 1, and setting hi(t) =
∑4i+4
k=4i+1 rkk, ( 2.4) implies that
h′i(t) +
h2i (t) +
k=4i+1
K(ε1, εk) ≤ 0. (2.8)
Together with Lemma 1.5 asserting that
k=4i+1
K(ε1, εk) = 4δ,
we have
h′i(t) +
h2i (t) + 4δ ≤ 0. (2.9)
Hence, as before, we conclude that
hi(t) ≤
4 cot t when δ = 1
4t−1 when δ = 0
4 coth t when δ = −1.
(2.10)
The result follows from the equation ∆r(x) = f(r(x)) +
i=1 hi(r(x)). �
Remark 2.1. The estimate in Theorem 2.1 is sharp since the right hand
sides are exactly the Laplacian of the distance functions of quaternionic hy-
perbolic space QHn, quaternionic Euclidean space Qn and quaternionic pro-
jective space QPn respectively.
Remark 2.2. We actually proved the estimate for Hessian of the distance
function. In particular,
rkk ≤
6 cot 2t when δ = 1
3t−1 when δ = 0
6 coth 2t when δ = −1.
(2.11)
Also for 1 ≤ i ≤ n− 1, we have
k=4i+1
rkk ≤
4 cot 2t when δ = 1
4t−1 when δ = 0
4 coth 2t when δ = −1.
(2.12)
Corollary 2.1. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ −16n(n + 2). Then for any point x ∈ M and
r > 0, the area A(r) of the geodesic spheres centered at x satisfies
A′(r)
≤ 6 coth 2r + 4(n− 1) coth r. (2.13)
In particular, A(r) ≤ C(sinh 2r)3(sinh r)4(n−1) ≤ Ce(4n+2)r.
Corollary 2.2. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ −16n(n + 2). Then for any point x ∈ M and
0 < r1 ≤ r2, the volume of the geodesic balls centered at x satisfies
Vx(r2)
Vx(r1)
VQHn(r2)
VQHn(r1)
, (2.14)
where VQHn(r) denotes the volume of the geodesic ball of radius r in QH
n. In
particular, λ1(M) ≤ (2n+ 1)
Corollary 2.3. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ 16n(n+2) . Then it is compact, and the diameter
d(M) ≤ π
, which is the diameter of the model space QPn. Moreover, the
volume of M is bounded by
V (M) ≤ V (QPn), (2.15)
where VQPn is the volume of QP
3 Quaternionic harmonicity
In this section we will derive an over-determined system of harmonic func-
tions with finite Dirichlet integral on a manifold with a parallel form. This
result was first proved by Siu [S] for harmonic maps in his proof of the rigidity
theorem for Kähler manifolds. Corlette [C1] gave a more systematic approach
for harmonic map with finite energy from a finite-volume quaternionic hy-
perbolic space or Cayley hyperbolic plane to a manifold with nonpositive
curvature. In [L], the second author generalized Siu’s argument to harmonic
functions with finite Dirichlet integral on a Kähler manifold. We will pro-
vide an argument that generalizes Corlette’s argument to harmonic functions
with finite Dirichlet integral on a complete manifold with a parallel form. We
believe that it should be of independent interest.
Theorem 3.1. Let M be a complete Riemannian manifold with a parallel
p-form Ω. Assume that f is a harmonic function with its Dirichlet integral
over geodesic balls centered at o of radius R satisfying the growth condition
Bo(R)
|∇f |2dv = o(R2)
as R → ∞, then f satisfies
d ∗ (df ∧ Ω) = 0. (3.1)
Before we prove the theorem, let us first recall the following operators
and some of the basic properties. For an oriented real vector space V with
an inner product, we have the Hodge star operator
∗ : ∧pV → ∧n−pV.
For any θ ∈ ∧1V and v ∈ V , we also have exterior multiplication and interior
product operators
ε(θ) : ∧pV → ∧p+1V,
ℓ(v) : ∧pV → ∧p−1V.
For θ ∈ ∧1V and v ∈ V is the dual of θ by the inner product, if ξ ∈ ∧pV we
list the following identities among the operators:
1. ∗ ∗ ξ = (−1)p(n−p)ξ,
2. ∗ε(θ)ξ = (−1)pℓ(v) ∗ ξ,
3. ε(θ) ∗ ξ = (−1)p−1 ∗ ℓ(v)ξ,
4. ∗ε(θ) ∗ ξ = (−1)(p−1)(n−p)ℓ(v)ξ,
5. ℓ(v)ε(θ′)ξ + ε(θ)ℓ(v′)ξ = 0, where v⊥v′,
6. ℓ(v)ε(θ)ξ + ε(θ)ℓ(v)ξ = ξ.
We are now ready to prove Theorem 3.1.
Proof of Theorem 3.1. Let η : [0,+∞) → R be a smooth function satis-
fying η′(t) ≤ 0, and
η(t) =
1 when t ∈ [0, 1]
0 when t ∈ [2,+∞].
For R ≥ 1, we define the cut-off function φR(x) = η(r(x)/R), where r(x)
is the distance function from a fixed point o ∈ M , then there is a positive
constant C1 depending on η and C such that
|∇φR(x)| ≤ C1R
Since d2 = 0, then
φ2R ∗ (df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
d(φ2R) ∧ ∗(df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
φ2R d ∗ (df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω). (3.2)
We claim that
∗ d ∗ (df ∧ Ω) = (−1)n−1d ∗ (df ∧ ∗Ω). (3.3)
In fact, for any point x ∈ M , we can choose an orthonormal tangent basis
i=1 in a neighborhood of x such that ∇eiej(x) = 0. Denote by {θ
i}mi=1
the dual basis of {ei}
i=1. Then for ω ∈ ∧
p(T ∗M) we have
dω = ε(θi)∇eiω.
Hence
d ∗ (df ∧ ∗Ω) = d ∗ ε(df) ∗ Ω
= (−1)(p−1)(m−p)d[ℓ(∇f)Ω]
= (−1)(p−1)(m−p)
ε(θi)∇ei(ℓ(∇f)Ω)
= (−1)(p−1)(m−p)
i,j=1
ε(θi)(∇ei∇ejf)(ℓ(ej)Ω)
= (−1)(p−1)(m−p)
i,j=1
fijε(θi)(ℓ(ej)Ω),
where fij = ∇ei∇ejf and the facts Ω is parallel and ∇eiej(x) = 0 have been
used. On the other hand,
∗d ∗ (df ∧ Ω) = ∗d ∗ ε(df)Ω
ε(θi)∇ei(∗ε(
fjθj)d[ℓ(∇f)Ω]
i,j=1
fijε(θi) ∗ ε(θj)(Ω)
= (−1)p(m−p−1)
i,j=1
fijℓ(ei)ε(θj)Ω
= (−1)p(m−p−1)
fiiℓ(ei)ε(θi)Ω +
i 6=j
fijℓ(ei)ε(θj)Ω
= (−1)p(m−p−1)
fii[Ω− ε(θi)ℓ(ei)Ω]−
i 6=j
fijε(θj)ℓ(ei)Ω
= (−1)p(m−p−1)
i,j=1
fijε(θi)(ℓ(ej)Ω), (3.4)
where we used fij = fji and
i=1 fii = 0. So the claim is proved. By ( 3.2),
we have
φ2R |d ∗ (df ∧ Ω)|
= (−1)m
d(φ2R) ∧ ∗(df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
φ2R |d ∗ (df ∧ Ω)|
|dφR|
2| ∗ (df ∧ Ω)|2dv
(3.5)
On the other hand, ( 3.3) and the fact that ω is bounded imply that there
exists a constant C2 > 0, such that
| ∗ (df ∧ Ω)| ≤ C2 |df |
|d ∗ (df ∧ ∗Ω)| = |d ∗ (df ∧ Ω)|.
Hence combining with ( 3.5) and using the definition of φR we conclude that
Bo(R)
|d ∗ (df ∧ Ω)|2dv ≤ C1R
Bo(2R)
|df |2dv.
The assumption on the growth of the Dirichlet integral of f implies that the
right hand side tends to zero as R → ∞. Therefore d ∗ (df ∧Ω) = 0, and the
proof is complete. �
Lemma 3.1. Let (M4n, g) be a quarternionic Kähler manifold and n ≥ 2. If
f is a function on M satisfying
d ∗ (df ∧ Ω) = 0 (3.6)
for the 4-form Ω determined by the quaternionic Kähler structure, then f is
quaternionic harmonic, namely, for any nonzero tangent vector X,
fX,X + fIX,IX + fJX,JX + fKX,KX = 0
where fX,X = ∇df(X,X).
Proof: Let
A=1 = {e1, e2, · · · , en, Ie1, Ie2, · · · , Ien,
Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
be an orthonormal basis of TM and {ωA} the dual basis with e1 =
. Since
Ω is parallel, by ( 3.4) and ( 3.6), we have
(∇eAdf) ∧ ℓ(eA)Ω
A,B=1
feA,eB ωB ∧ ℓ(eA)Ω
where we have used the fact that f is a harmonic function. Hence equation
( 3.6) implies
A,B=1
feA,eB ωB ∧ ℓ(eA)Ω = 0
Comparing the coefficient of ωi∧Iωi∧Jωi∧Kωi on both sides by the explicit
formula for Ω given before, we obtain that
6 (fei,ei + fIei,Iei + fJei,Jei + fKei,Kei) = 0
for all ei, (1 ≤ i ≤ n). So the proof is complete. �
The following corollary is an immediate consequence of the lemma.
Corollary 3.1. Let M4n be a complete quaternionic Kähler manifold. As-
sume that f is a harmonic function with its Dirichlet integral satisfying the
growth condition
Bo(R)
|∇f |2dv = o(R2)
as R → ∞, then f must satisfy
d ∗ (df ∧ Ω) = 0, (3.7)
where Ω is the parallel 4-form determined by the quaternionic Kähler struc-
ture. Moreover, f is quaternionic harmonic.
4 Uniqueness of infinite volume end
Recall that for any complete manifold if λ1(M) > 0 then M must be non-
parabolic. In particular,M must have at least one nonparabolic ends. It was
also proved in [LW1] that under the assumption that λ1(M) > 0, an end is
nonparabolic if and only if it has infinite volume.
Let us assume that M has at least two nonparabolic ends, E1 and E2.
A construction of Li-Tam [LT] asserts that one can construct a nonconstant
bounded harmonic function with finite Dirichlet integral. The harmonic func-
tion f can be obtained by taking a convergent subsequence of the harmonic
functions fR, as R→ +∞, satisfying
∆fR = 0 on B(R),
with boundary conditions
fR = 1 on ∂B(R) ∩ E1
fR = 0 on ∂B(R) \ E1.
It follows from the maximum principle that 0 ≤ fR ≤ 1, hence 0 ≤ f ≤ 1.
We need the following estimates from [LW1](Lemma 1.1 and 1.2 in [LW1]),
and [LW3](Lemma 5.1 in [LW3]).
Lemma 4.1. Let M be a complete Riemannian manifold with λ1(M) > 0.
Suppose M has at least two nonparabolic ends and E be an end of M . Then
for the harmonic function f constructed above, it must satisfy the following
growth conditions:
1. There exists a constant a such that f − a ∈ L2(E). Moreover, the
function f − a must satisfy the decay estimate
E(R+1)\E(R)
(f − a)2 ≤ C exp(−2
λ1(E)R)
for some constant C > 0 depending on f , λ1(E) and the dimension of
2. The Dirichlet integral of the function f must satisfy the decay estimate
E(R+1)\E(R)
|∇f |2 ≤ C exp(−2
λ1(E)R),
exp(−2
λ1(E)r(x))|∇f |
2 ≤ CR
for R sufficiently large.
Lemma 4.2. Let M be a complete Riemannian manifold with at least two
nonparabolic ends and λ1(M) > 0. Then for the harmonic function f con-
structed above, for any t ∈ (inf f, sup f) and (a, b) ⊂ (inf f, sup f),
L(a,b)
|∇f |2 = (b− a)
|∇f |,
where
l(t) = {x ∈M |f(x) = t},
L(a, b) = {x ∈M |a < f(x) < b}.
Moreover,
|∇f | =
|∇f |.
We are now ready to prove Theorem 0.5.
Proof of Theorem 0.5: Suppose to the contrary that there exist two ends
E1 and E2 with infinite volume. The assumption that λ1(M) > 0 implies that
they are nonparabolic. By the construction above, there exists a harmonic
function f with finite energy such that
lim inf
x→∞, x∈E1
f(x) = 1
lim inf
x→∞, x∈E2
f(x) = 0.
The Bochner formula implies that
∆|∇f |2 = RicM(∇f,∇f) + |∇
2f |2. (4.1)
We now choose an orthonormal basis {eA}
A=1 satisfying
{e1, e2, · · · , en, Ie1, Ie2, · · · , Ien, Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
with e1 =
|∇f |
. Corollary 3.1 implies that
f(in+1)(in+1) = 0.
Therefore, applying the arithmetic-geometric means, we have
|∇2f |2 =
A,B=1
f 2AB
≥ f 211 +
f 2(in+1)(in+1) + 2
f 21A
≥ f 211 +
f(in+1)(in+1))
2 + 2
f 21A
|∇|∇f ||2, (4.2)
hence combining with ( 4.1) we obtain
∆|∇f |2 ≥ −4(n + 2)|∇f |2 +
|∇|∇f ||2. (4.3)
If we write u = |∇f |
3 , then
∆u ≥ −
8(n+ 2)
u. (4.4)
We want to prove that the above inequality is actually an equality. The
argument follows from that in [LW4] after making suitable modification to
fit our situation. For any compactly supported smooth function φ on M , we
8(n+ 2)
φu〈∇u,∇φ〉 −
φ2|∇u|2 + λ1(M)
(φu)2
φu〈∇u,∇φ〉 −
φ2|∇u|2 +
|∇(φu)|2
|∇φ|2u2. (4.5)
Let us choose φ = ψχ to be the product of two compactly supported
functions. For any ε ∈ (0, 1
), we define
χ(x) =
0 on L(0, σε) ∪ L(1− ε
(log 2)−1(log f − log( ε
)) on L( ε
, ε) ∩ (M \E1)
(log 2)−1(log(1− f)− log( ε
)) on L(1− ε, 1− ε
) ∩ E1
1 otherwise.
For R > 1 we define
1 on B(R− 1)
R− r on B(R) \B(R− 1)
0 on M \B(R).
Applying to the right hand side of ( 4.5), we obtain
|∇φ|2u2 ≤ 2
|∇ψ|2χ2|∇f |
3 + 2
|∇χ|2ψ2|∇f |
3 . (4.6)
Since RicM ≥ −4(n+2), then the local estimate of Cheng-Yau [CY] (see also
[LW2]) implies that there exists a constant depending on n such that
|∇f |(x) ≤ C|1− f(x)|.
On E1, the first term of ( 4.6) satisfies
|∇ψ|2χ2|∇f |
|∇f |2
, (4.7)
where Ω = E1 ∩ (B(R) \B(R − 1)) ∩ (L(1− ε, 1−
) ∪ L( ε
, ε). Since
1 ≤ 4
(1− f)2
(1− f)2
≤ 4Cε−2 exp(−2
λ1R),
where in the last inequality we have used Lemma 4.1. Again by Lemma 4.1,
from ( 4.7) we have
|∇ψ|2χ2|∇f |
3 ≤ Cε−
3 exp(−2
λ1R). (4.8)
For the second term of ( 4.6) we have
|∇χ|2ψ2|∇f |
≤ (log 2)−2
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |
+2(1− f)−2
≤ C(log 2)−2
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |2(1− f)−
Using the co-area formula and Lemma 4.2 we have
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |2(1− f)−
∫ 1− ε
(1− t)−
l(t)∩E1∩B(R)
|∇f |dAdt
|∇f |dA
∫ 1− ε
(1− t)−
= −3C[(1− t)
|∇f |dA
= 3Cε
|∇f |dA.
Combining the above inequality with ( 4.8) we have
|∇φ|2u2 ≤ C(ε
3 exp(−2
λ1R) + ε
3 ). (4.9)
A similar argument using f instead of 1 − f on the other end yields the
estimate
|∇φ|2u2 ≤ C(ε
3 exp(−2
λ1R) + ε
Letting R → ∞ and ε→ 0, we have
∆u = −
8(n + 2)
u (4.10)
with λ1(M) =
8(n+2)
, since f is nonconstant and u cannot be identically zero.
Therefore all the inequalities used to prove ( 4.4) are equalities. Thus there
exists a function µ, such that,
(fAB) =
, (4.11)
where D1 and D2 are n× n matrices defined by
· · ·
· · ·
Since f1α = 0 for α 6= 1 implies that |∇f | is constant along the level set of
f . Moreover, regularity of the equation ( 4.10) implies that |∇f | can never
be zero. Hence M must be diffeomorphic to R×N , where N is given by the
level set of f . Also N must be compact since we assume that M has at least
2 ends.
Fix a level set N0 of f , consider (−ε, ε)×N0 ⊂M . Note that {eA} is an
orthonormal basis of TM such that e1 is the normal vector to N0 and {eα}
are the tangent vectors of N0. We shall compute the sectional curvature
K(e1, eα) = 〈R(e1, eα)eα, e1〉.
We claim that
∇e1e1 = 0.
Indeed it suffices to prove all integral curves η(t) of the vector field e1 =
|∇f |
emanating from N0 are geodesics. For any point η(t0), let γ be the geodesic
realizing the distance between η(t0) and N0. Then γ is perpendicular to every
level set Nt. So γ
′ is parallel to e1 along γ. This implies γ coincides with the
integral curve of e1.
Let (hαβ) with 2 ≤ α, β ≤ 4n be the second fundamental form of the level
set of f . Then
hαβ f1 = −fαβ , (4.12)
∇eαe1 = −
hαβeβ.
By the definition of curvature tensor, we have
〈R(e1, eα)e1, eα〉 = 〈∇e1∇eαe1 −∇eα∇e1e1 −∇[e1,eα]e1, eα〉
= 〈∇e1∇eαe1, eα〉 − 〈∇[e1,eα]e1, eα〉
= 〈∇e1∇eαe1, eα〉 − 〈∇∇e1eα−∇eαe1e1, eα〉
= 〈∇e1∇eαe1, eα〉 −
〈∇e1eα, eβ〉〈∇eβe1, eα〉
〈∇eαe1, eβ〉〈∇eβe1, eα〉
〈∇e1(hαβeβ), eα〉+
hαβ〈∇e1eα, eβ〉+
〈(e1hαβ)eβ , eα〉 −
hαβ〈∇e1eβ, eα〉
hαβ〈∇e1eα, eβ〉+ h
= −e1hαα + 2
hαβ〈∇e1eα, eβ〉+ h
Therefore
K(e1, eα) = e1hαα − 2
hαβ〈∇e1eα, eβ〉 −
h2αβ . (4.13)
Since hαβ is diagonal, this implies that
K(e1, eα) = e1hαα − h
Combining with ( 4.11) and ( 4.12), we conclude that
K(e1, e2) = K(e1, Ie2) = K(e1, Je2) = K(e1, Ke2) = 0
which implies M is Ricci flat by Theorem 1.3. This contradicts to the as-
sumption that λ1 >
8(n+2)
> 0. Therefore M must have only one end with
infinite volume. �
5 Maximal first eigenvalue
In this section, we will consider the case when λ1(M) is of maximal value.
Proof of Theorem 0.6: According to Theorem 0.5, we know that M has
exactly one nonparabolic end. Suppose thatM has more than one end. Then
there must exist at least an end with finite volume. We divide the rest of the
proof into several parts. The first part follows exactly as that in the proof of
the corresponding theorem in the Kähler case (Theorem 3.1) in [LW5]. For
completeness sake, we will give a quick outline of it.
Part 1. Assume that E1 is such an end with finite volume given by
M \ Bp(1). Then we can choose a ray η : [0,+∞) such that η(0) = p and
η[1,+∞) ⊂ E1. The Busemann function corresponding to γ is defined by
β(x) = lim
[t− d(x, η(t))].
The Laplacian comparison theorem, Theorem 2.1, asserts that
∆β ≥ −2(2n+ 1)
in the sense of distribution. We define the function f = exp((2n+ 1)β), and
using the fact that |∇β| = 1 almost everywhere, we have
∆f = (2n+ 1) exp((2n+ 1)β)∆β + (2n+ 1)2
≥ −(2n + 1)2f.
Similar to the proof of above theorem, we conclude that for any compactly
supported function φ,
(∆f + (2n+ 1)2f)fφ2
f 2|∇φ|2.
By choosing the function φ to be
1, on Bp(R);
2R−r(x)
, on Bp(2R) \Bp(R);
0, on M \Bp(2R);
we obtain
f 2|∇φ|2
(Bp(2R)\Bp(R))∩E1
(Bp(R+i)\Bp(R+i−1))∩E1
(VE1(R + i)− VE1(R + i− 1)) exp(2(2n+ 1)(R+ i))
where VE1(R+ i) denotes the volume of the set E1∩Bp(R+ i). On the other
hand, the volume estimate in Theorem 1.4 of [LW1] implies that
VE1(∞)− VE1(R) ≤ C exp(−2(2n+ 1)R),
hence
VE1(R + i)− VE1(R + i− 1)
= VE1(∞)− VE1(R + i− 1)− (VE1(∞)− VE1(R + i))
≤ C exp(−2(2n+ 1)(R + i)).
Therefore, we conclude that
f 2|∇φ|2 ≤
Let us now denote E2 = M \ (Bp(1) ∪ E1) to be the other end of M .
When x ∈ E2, following the argument in Theorem 3.1 of [LW4], we have
β(x) ≤ −d(p, x) + 2.
Therefore
f 2|∇φ|2 ≤
(Bp(2R)\Bp(R))∩E2
(Bp(2R)\Bp(R))∩E2
exp(−2(2n+ 1)(r − 2))
Letting R → +∞, we conclude that
∆f + (2n+ 1)2f = 0, (5.1)
and all inequalities used are indeed equalities and f is smooth by regularity
of the equation ( 5.1). Moreover, |∇β| = 1, and
∆β = −2(2n+ 1).
This implies that M must be diffeomorphic to R × N , where N is given by
the level set of β. We choose an orthonormal basis {ei}
i=1 as follows
{e1, e2, · · · , en, Ie1, Ie2, · · · , Ien, Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
with e1 = ∇β. Applying the Bochner formula to β, we get
∆|∇β|2
i,j=1
β2ij + RicM(∇β,∇β) +
βi(∆β)i
i,j=1
β2ij − 4(n+ 2).
By the comparison theorem, we have,
β(in+1)(in+1) = −6.
Hence
(βαβ) =
where D1 and D2 are n× n matrices defined by
· · ·
· · ·
Part 2. For a fix level set N0 of β, we consider (−ε, ε)×N0 ⊂M . Note
that {ei} is an orthonormal basis of TM such that e1 is the normal vector
to N0 and {eα}, for 2 ≤ α ≤ 4n, are the tangent vectors of N0. We shall
compute the sectional curvature
K(e1, eα) = 〈R(e1, eα)eα, e1〉.
Since ∇e1e1 = 0 implies that the integral curves of e1 are geodesics. Let (hαγ)
be the second fundamental form of the level set of ∇β. Then
hαγ = 〈∇eαeγ, e1〉
= 〈∇eαeγ,∇β〉
= −βαγ
∇eαe1 = −
hαγeγ. (5.2)
By ( 4.13) in the proof of Theorem 0.5 we have
〈R(e1, eα)e1, eα〉 = −e1hαα + 2
hαγ〈∇e1eγ, eβ〉+
h2αγ .
Since (hαγ) are constant and diagonal, then
K(e1, eα) = −h
In particular, we have
K(e1, eα) =
−4 when α = in+ 1, i = 1, 2, 3
−1 otherwise.
On the other hand, we also have
K(en+1, e2n+1) +K(en+1, e3n+1) = −12−K(e1, en+1) = −8
K(en+1, e2n+1) +K(e3n+1, e2n+1) = −8
K(e3n+1, e2n+1) +K(en+1, e3n+1) = −8,
hence
K(en+1, e2n+1) = K(en+1, e3n+1) = K(e2n+1, e3n+1) = −4.
Since for α = 2, 3, · · · , n,
K(Ie1, eα) = −〈R(Ie1, eα)Ie1, eα〉
= −〈IR(Ie1, eα)Ie1, Ieα〉
= 〈R(Ie1, eα)e1, Ieα〉
= 〈R(e1, Ieα)Ie1, eα〉
= K(e1, Ieα)
= −1,
and K(Je1, eα) = K(Ke1, eα) = −1, we have
K(ein+1, eα) = −1,
for all i = 0, 1, 2, 3 and α 6= 1, n+ 1, 2n+ 1, 3n+ 1.
Let KN(eα, eγ) denote the sectional curvature of the level set with induced
metric. By Gaussian equation,
KN (eα, eγ)−K(eα, eγ) = hααhγγ,
it is straightforward to obtain
KN (en+1, e2n+1) = K
N (en+1, e3n+1) = K
N(e2n+1, e3n+1) = 0,
KN(ein+1, eα) = 1, (5.3)
for all i = 1, 2, 3 and α 6= 1, n+ 1, 2n+ 1, 3n+ 1.
Part 3. There is a natural map ϕt between the level sets N0 and Nt given
by the gradient flow of β. Since the integral curves are geodesics, dϕt(X) are
Jacobi fields along corresponding curves. Let (N, g0) = N0 with the induced
metric. We can consider ϕ as a flow on N . We claim that
dϕt|V1 = e
2t id
dϕt|V2 = e
t id,
where TN = V1 ⊕ V2, V1 = span{Ie1, Je1, Ke1} and V2 = V
1 . Indeed for
any point q ∈ N0, denote e1(t) = ∇β(ϕ(t)) and {εα(t)}
α=2 to be the parallel
transport of the orthonormal base {eα}
α=2 of N0 at q along ϕt(q). Since both
V1 and V2 are ϕ-invariant, we have, in particular,
〈∇e1(t)εα, εγ〉 = 0, (5.4)
when α ∈ {n+ 1, 2n+ 1, 3n+ 1}, and γ /∈ {n + 1, 2n+ 1, 3n+ 1}.
Now we can compute R1α1γ . Then
〈R(e1, εα)e1, εγ〉
= 〈∇e1∇εαe1 −∇εα∇e1e1 −∇[e1,εα]e1, εγ〉
= 〈∇e1∇εγe1, εα〉 − 〈∇[e1,εα]e1, εγ〉
= 〈∇e1∇εαe1, εγ〉 − 〈∇∇e1εα−∇εαe1e1, εγ〉
= 〈∇e1∇εαe1, εγ〉 −
〈∇e1εα, ετ〉〈∇ετ e1, εγ〉
〈∇εαe1, ετ 〉〈∇ετe1, εγ〉
〈∇e1(hατετ ), εγ〉+
hγτ 〈∇e1εα, ετ 〉+
hατhτγ
= −e1hαγ −
hατ 〈∇e1ετ , εγ〉
hγτ 〈∇e1εα, ετ 〉+
hατhτγ. (5.5)
We see that (hαγ) is diagonal and
hαα =
2, when α = n + 1, 2n+ 1, 3n+ 1;
1, otherwise.
Therefore, when α 6= γ,
R1α1γ = −hαα〈∇e1εα, εγ〉+ hγγ〈∇e1εα, εγ〉
= (hγγ − hαα)〈∇e1εα, εγ〉.
Since hαα = hγγ when α, γ ∈ {n+ 1, 2n+ 1, 3n+ 1} and α, γ /∈ {n+ 1, 2n+
1, 3n+ 1}, using ( 5.4), we have
R1α1γ = 0, for all α 6= γ.
Define
Jα(t) =
e−2tεα, when α ∈ {n+ 1, 2n+ 1, 3n+ 1};
e−tεα, when α /∈ {n+ 1, 2n+ 1, 3n+ 1}.
Since
dϕt(eα)|t=0 = [e1, eα] = −∇eαe1,
then we see that Jα satisfies the Jacobi equation and initial conditions Jα(0) =
eα and J
α(0) = eα = ∇ ∂
dϕt(eα)|t=0. By the uniqueness theorem for the
Jacobi equations, we have dϕt(eα) = Jα. The claim is proved.
Part 4. We have now a family of metrics on N written as
ds2t = e
ω2in+1 + e
ω2in+α,
and the metric of M can rewritten as
ds2 = dt2 + e4t
ω2p + e
ω2α (5.6)
where {ω2, ω3, ω4, . . . , ω4n} is the dual coframe to {e2, e3, e4, . . . , e4n} at N0.
We also choose that Ie4s−3 = e4s−2, Je4s−3 = e4s−1, and Ke4s−3 = e4s for
s = 1, . . . , n, with e1 =
. In particular, the second fundamental form on Nt
must be a diagonal matrix when written in terms of the basis {ei}
i=2 with
eigenvalues given by
(〈∇eiej, e1〉) =
2I3 0
0 I4(n−1)
, (5.7)
where Ik denotes the k× k identity matrix. Also, the sectional curvatures of
the sections containing e1 are given by
K(e1, ep) = −4 for 2 ≤ p ≤ 4
K(e1, eα) = −1 for 5 ≤ α ≤ 4n.
The Guass curvature equation also asserts that
Rijkl = R̄ijkl + hlihkj − hkihlj,
where R̄ijkl is the curvature tensor on Nt. In particular,
Rijkl =
R̄ijkl + δliδkj − δkiδlj if 5 ≤ i, j, k, l ≤ 4n
R̄ijkl + 4δliδkj − 4δkiδlj if 2 ≤ i, j, k, l ≤ 4
R̄ijkl + 2 if 2 ≤ i = l ≤ 4 and 5 ≤ k = j ≤ 4n
R̄ijkl + 2 if 2 ≤ k = j ≤ 4 and 5 ≤ i = l ≤ 4n
R̄ijkl − 2 if 2 ≤ i = k ≤ 4 and 5 ≤ j = l ≤ 4n
R̄ijkl − 2 if 2 ≤ j = l ≤ 4 and 5 ≤ i = k ≤ 4n
R̄ijkl − 2 if 2 ≤ k = i ≤ 4 and 5 ≤ j = l ≤ 4n
R̄ijkl otherwise.
(5.8)
We will now use ( 5.6) to compute the curvature tensor of M and hence
N0. Using the orthonormal coframe
η1 = ω1 = dt,
ηp = e
ηα = e
for 2 ≤ p ≤ 4 and 5 ≤ α ≤ 4n, we obtain the first structural equations
dη1 = 0, (5.9)
dηp = 2e
2t ω1 ∧ ωp + e
ωpq ∧ ωq + e
ωpα ∧ ωα
= −2ηp ∧ η1 +
ωpq ∧ ηq + e
ωpα ∧ ηα, (5.10)
dηα = e
t ω1 ∧ ωα + e
ωαp ∧ ωp + e
ωαβ ∧ ωβ
= −ηα ∧ η1 + e
ωαp ∧ ηp +
ωαβ ∧ ηβ, (5.11)
where ωij are the connection forms of N0. In the above and all subsequent
computations, we will adopt the convention that 5 ≤ α, β ≤ 4n, 2 ≤ i, j ≤
4n, 2 ≤ o, p, q, r ≤ 4, 2 ≤ s, t ≤ n, and 1 ≤ A,B ≤ 4n.
Note that using the endomorphism I and the fact that ∇I = cJ − bK,
we have
ωij(X) = 〈∇̄Xej, ei〉
= 〈I∇Xej , Iei〉
= 〈∇XIej, Iei〉+ c(X) 〈Jej, Iei〉 − b(X) 〈kej , Iei〉
= 〈∇XIej, Iej〉+ c(X) 〈ej, Kei〉+ b(X)〈ej , Jei〉
for any tangent vector X to N0, where ∇̄ denotes the connection on N0.
Hence we conclude that
ωij = ωIiIj + c 〈ej, Kei〉+ b 〈ej, Jei〉, (5.12)
where Ii denotes the index corresponding to Iei = eIi. Similarly, we have
ωij = ωJiJj + c 〈ej, Kei〉+ a 〈ej, Iei〉,
ωij = ωKiKj + b 〈ej, Jei〉+ a 〈ej , Iei〉.
Together with ( 5.7), we conclude that
ω2(4s−1)(e4s) = −1 = −ω2(4s)(e4s−1),
ω2(4s−3)(e4s−2) = −1 = −ω2(4s−2)(e4s−3),
for all 2 ≤ s ≤ n, and
ω2α(eβ) = 0 otherwise.
Similarly,
ω2α(ep) = 〈∇epeα, e2〉
= −〈∇epIeα, e1〉
These identities imply that
ω2(4s−3) = −ω4s−2,
ω2(4s−2) = ω4s−3,
ω2(4s−1) = −ω4s,
ω2(4s) = ω4s−1. (5.13)
A similar calculation using the endomorphisms J and K yield
ω3(4s−3) = −ω4s−1,
ω3(4s−2) = ω4s,
ω3(4s−1) = ω4s−3,
ω3(4s) = −ω4s−2, (5.14)
ω4(4s−3) = −ω4s,
ω4(4s−2) = −ω4s−1,
ω4(4s−1) = ω4s−2,
ω4(4s) = ω4s−3. (5.15)
We claim that the connection forms are given by
η1p = −ηp1
= 2ηp for 2 ≤ p ≤ 4, (5.16)
η1α = −ηα1
= ηα for 5 ≤ α ≤ 4n, (5.17)
ηpq = −ηqp = ωpq, (5.18)
ηpα = −ηαp = e
t ωpα, (5.19)
η(4s)β = −ηβ(4s)
ω(4s)β − (1−e
−2t) η2 if β = 4s− 1
ω(4s)β + (1−e
−2t) η3 if β = 4s− 2
ω(4s)β − (1−e
−2t) η4 if β = 4s− 3
ω(4s)β if β 6= 4s− 1, 4s− 2, or 4s− 3,
(5.20)
η(4s−1)β = −ηβ(4s−1)
ω(4s−1)β + (1−e
−2t) η2 if β = 4s
ω(4s−1)β − (1−e
−2t) η4 if β = 4s− 2
ω(4s−1)β − (1−e
−2t) η3 if β = 4s− 3
ω(4s−1)β if β 6= 4s, 4s− 2, or 4s− 3,
(5.21)
η(4s−2)β = −ηβ(4s−2)
ω(4s−2)β − (1−e
−2t) η3 if β = 4s
ω(4s−2)β + (1−e
−2t) η4 if β = 4s− 1
ω(4s−2)β − (1−e
−2t) η2 if β = 4s− 3
ω(4s−2)β if β 6= 4s, 4s− 1, or 4s− 3,
(5.22)
η(4s−3)β = −ηβ(4s−3)
ω(4s−3)β + (1−e
−2t) η4 if β = 4s
ω(4s−3)β + (1−e
−2t) η3 if β = 4s− 1
ω(4s−3)β + (1−e
−2t) η2 if β = 4s− 2
ω(4s−3)β if β 6= 4s, 4s− 1, or 4s− 2.
(5.23)
Indeed, if we substitute ( 5.16− 5.23) into the first structural equations
dηA = ηA1 ∧ η1 +
ηAq ∧ ηq +
ηAβ ∧ ηβ
we obtain ( 5.9), ( 5.10), and ( 5.11).
To compute the curvature, we consider the second structural equations.
In particular,
dη1p − η1q ∧ ηqp − η1α ∧ ηαp
= 2dηp − 2ηq ∧ ηqp − ηα ∧ ηαp
= −4ηp ∧ η1 + ηα ∧ ηαp
= −4ηp ∧ η1 + e
t ωpα ∧ ηα.
Hence using ( 5.13− 5.15), we have
R1p1p = −4,
R12(4s−1)(4s) =− 2 = −R12(4s)(4s−1),
R12(4s−3)(4s−2) =− 2 = −R12(4s−2)(4s−3),
R13(4s)(4s−2) =− 2 = −R13(4s−2)(4s),
R13(4s−1)(4s−3) =2 = −R13(4s−3)(4s−1),
R14(4s)(4s−3) =2 = −R14(4s−3)(4s),
R14(4s−1)(4s−2) =2 = −R14(4s−2)(4s−1),
R1pAB = 0, otherwise.
Also,
dη1α − η1q ∧ ηqα − η1β ∧ ηβα
= dηα − 2ηq ∧ ηqα − ηβ ∧ ηβα
= −ηα ∧ η1 + e
t ωqα ∧ ηq,
hence
R1α1α = −1,
R1(4s)(4s−1)2 =− 1 = −R1(4s−1)(4s)2,
R1(4s)(4s−2)3 =1 = −R1(4s−2)(4s)3,
R1(4s)(4s−3)4 =− 1 = −R1(4s−3)(4s)4,
R1(4s−1)(4s−3)3 =− 1 = −R1(4s−3)(4s−1)3,
R1(4s−1)(4s−2)4 =− 1 = −R1(4s−2)(4s−1)4,
R1(4s−2)(4s−3)2 =− 1 = −R1(4s−3)(4s−2)2,
R1αAB = 0 otherwise.
Similarly,
dηpq − ηp1 ∧ η1q − ηpr ∧ ηrq − ηpβ ∧ ηβq
= dωpq + 4ηp ∧ ηq − ωpr ∧ ωrq − e
2tωpβ ∧ ωβq
= Ω̄pq + (1− e
2t)ωpβ ∧ ωβq + 4ηp ∧ ηq,
where
Ω̄pq =
R̄pqijωj ∧ ωi
is the curvature form of N0. In particular, this implies that
Rpqro =
−4+e−4t R̄pqpq if r = p and o = q
4+e−4t R̄pqqp if r = q and o = p
e−4t R̄pqro otherwise,
(5.24)
R23(4s)(4s−3) = e
−2t R̄23(4s)(4s−3) − 2(e
−2t − 1),
R23(4s−1)(4s−2) = e
−2t R̄23(4s−1)(4s−2) − 2(e
−2t − 1),
R24(4s)(4s−2) = e
−2t R̄24(4s)(4s−2) − 2(e
−2t − 1),
R24(4s−1)(4s−3) = e
−2t R̄24(4s−1)(4s−3) + 2(e
−2t − 1),
R34(4s)(4s−1) = e
−2t R̄34(4s)(4s−1) − 2(e
−2t − 1),
R34(4s−2)(4s−3) = e
−2t R̄34(4s−2)(4s−3) − 2(e
−2t − 1),
Rpqαβ = e
−2t R̄pqαβ, otherwise.
We now continue with our curvature computation and consider
dηpα − ηp1 ∧ η1α − ηpq ∧ ηqα − ηpβ ∧ ηβα
= d(et ωpα) + 2ηp ∧ ηα − ωpq ∧ e
t ωqα − e
t ωpβ ∧ ηβα
= et η1 ∧ ωpα +
etR̄pαijωj ∧ ωi + 2ηp ∧ ηα + e
t ωpβ ∧ (ωβα − ηβα),
where R̄pαij is the curvature tensor of N0. Using ( 5.13− 5.15) and ( 5.20−
5.23), we have
R2(4s)ABηB ∧ ηA
= η1 ∧ η(4s−1) +
etR̄2(4s)ij ωj ∧ ωi − 2η(4s) ∧ η2 + (1− e
−2t) η(4s) ∧ η2
+ (1− e−2t) η(4s−3) ∧ η3 + (1− e
−2t) η(4s−2) ∧ η4
= η1 ∧ η(4s−1) +
etR̄2(4s)ij ωj ∧ ωi − (1 + e
−2t) η(4s) ∧ η2
+ (1− e−2t) η(4s−3) ∧ η3 + (1− e
−2t) η(4s−2) ∧ η4.
R2(4s−1)ABηB ∧ ηA
= −η1 ∧ η(4s) +
etR̄2(4s−1)ij ωj ∧ ωi − (1 + e
−2t) η(4s−1) ∧ η2
+ (1− e−2t)η(4s−2) ∧ η3 − (1− e
−2t) η(4s−3) ∧ η4.
R2(4s−2)ABηB ∧ ηA
= η1 ∧ η4s−3 +
etR̄2(4s−2)ij ωj ∧ ωi − (1 + e
−2t) η(4s−2) ∧ η2
+ (1− e−2t) η(4s−1) ∧ η3 − (1− e
−2t) η(4s) ∧ η4.
R2(4s−3)ABηB ∧ ηA
= −η1 ∧ η4s−2 +
etR̄2(4s−3)ij ωj ∧ ωi − (1 + e
−2t) η(4s−3) ∧ η2
− (1− e−2t) η(4s) ∧ η3 + (1− e
−2t) η(4s−1) ∧ η4.
Similar formulas for the curvature tensors of the form R3αAB and R4αAB .
Continuing with our computation of the second structural equations using
( 5.13− 5.15), we have
dη(4s−1)(4s) − η(4s−1)1 ∧ η1(4s) − η(4s−1)q ∧ ηq(4s) − η(4s−1)β ∧ ηβ(4s)
= dω(4s−1)(4s) + 2e
−2t η1 ∧ η2 + (1− e
−2t) dη2
+ η(4s−1) ∧ η(4s) − e
2t ω(4s−1)q ∧ ωq(4s)
− (ω(4s−1)(4s−2) − (1− e
−2t) η4) ∧ (ω(4s−2)(4s) − (1− e
−2t) η3)
− (ω(4s−1)(4s−3) − (1− e
−2t) η3) ∧ (ω(4s−3)(4s) + (1− e
−2t) η4)
R̄(4s−1)(4s)ij ωj ∧ ωi + (1− e
2t)ω(4s−1)q ∧ ωq(4s) + 2η1 ∧ η2
+ (1− e−2t)ω2q ∧ ηq + e
t(1− e−2t)ω2β ∧ ηβ − η(4s) ∧ η(4s−1)
+ (1− e−2t)ω(4s−1)(4s−2) ∧ η3 + (1− e
−2t) η4 ∧ ω(4s−2)(4s)
+ (1− e−2t) η3 ∧ ω(4s−3)(4s) − (1− e
−2t)ω(4s−1)(4s−3) ∧ η4
+ 2(1− e−2t)2 η3 ∧ η4
R̄(4s−1)(4s)ij ωj ∧ ωi + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2)
+ 2η1 ∧ η2 + (1− e
−2t)ω2q ∧ ηq + 2(1− e
−2t) η(4r−3) ∧ η(4r−2)
+ 2(1− e−2t) η(4r−1) ∧ η(4r) + (1− e
−2t) (ω(4s−1)(4s−2) − ω(4s−3)(4s)) ∧ η3
− (1− e−2t) (ω(4s−2)(4s) + ω(4s−1)(4s−3)) ∧ η4 + 2(1− e
−2t)2 η3 ∧ η4.
(5.25)
Note that ( 5.22) asserts that
(1− e−2t)ω2q ∧ ηq = (1− e
−2t) (−ω14 ∧ η3 + c ∧ η3 + ω13 ∧ η4 − b ∧ η4)
= (1− e−2t) (4e−2t η3 ∧ η4 + c ∧ η3 − bη4),
(1− e−2t) (ω(4s−1)(4s−2) − ω(4s−3)(4s)) ∧ η3
= −(1− e−2t) c ∧ η3,
−(1− e−2t) (ω(4s−2)(4s) + ω(4s−1)(4s−3)) ∧ η4
= (1− e−2t) b ∧ η4.
Hence substituting into ( 5.25), we obtain
R(4s−1)(4s)AB ηB ∧ ηA
R̄(4s−1)(4s)ij ωj ∧ ωi + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2) + 2η1 ∧ η2
+ 2(1− e−2t) η(4r−3) ∧ η(4r−2) + 2(1− e
−2t) η(4r−1) ∧ η(4r) + 2(1− e
−4t) η3 ∧ η4
= 2η1 ∧ η2 +
R̄(4s−1)(4s)pq e
−4t ηq ∧ ηp + 2(1− e
−4t) η3 ∧ η4 + R̄(4s−1)(4s)pα e
−3t ηα ∧ ηp
R̄(4s−1)(4s)αβ e
−2t ηβ ∧ ηα + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2)
+ 2(1− e−2t) η(4r−3) ∧ η(4r−2) + 2(1− e
−2t) η(4r−1) ∧ η(4r).
A similar computation yields the curvature tensor of the formR(4s−1)(4s−2)AB ,
R(4s−1)(4s−3)AB , R(4s−2)(4s−3)AB , R(4s−2)(4s)AB , and R(4s−3)(4s)AB . It remains to
compute
R(4s−3)(4r)AB ηB ∧ ηA
= dη(4s−3)(4r) − η(4s−3)1 ∧ η1(4r) − η(4s−3)q ∧ ηq(4r) − η(4s−3)β ∧ ηβ(4r)
= dω(4s−3)(4r) + η(4s−3) ∧ η(4r) − e
2t ω(4s−3)q ∧ ωq(4r) − η(4s−3)β ∧ ηβ(4r)
R̄(4s−3)(4r)ij ωj ∧ ωi + (1− e
2t)ω(4s−3)q ∧ ωq(4r) − (1− e
−2t) (η4 ∧ ω(4s)(4r)
+ η3 ∧ ω(4s−1)(4r))− (1− e
−2t) (η2 ∧ ω4s−2)(4r) + ω(4s−3)(4r−1) ∧ η2
− ω(4s−3)(4r−2) ∧ η3 + ω(4s−3)(4r−3) ∧ η4) + η(4s−3) ∧ η(4r). (5.26)
Using ( 5.12− 5.14), we can write
ω(4s−3)q ∧ ωq(4r) = −η(4s− 2) ∧ η(4r−1) + η(4s−1) ∧ η(4r−2) − η(4s) ∧ η(4r−3)).
Also using ( 5.12) asserts that
ω(4s−3)(4r−1) = ω(4s−2)(4r),
ω(4s−3)(4r−2) = −ω((4s−1)(4r)
ω((4s−3)(4r−3) = ω((4s)(4r).
Hence ( 5.26) becomes
R(4s−3)(4r)AB ηB ∧ ηA
R̄(4s−3)(4r)pq e
−4t ηq ∧ ηp + R̄(4s−3)(4r)pβ e
−3t ηβ ∧ ηp +
R̄(4s−3)(4r)αβ e
−2t ηβ ∧ ηα
− (1− e2t) η(4s−2) ∧ η(4r−1) + (1− e
−2t) η(4s−1) ∧ η(4r−2)
− (1− e−2t) η(4s) ∧ η(4r−3) + η(4s−3) ∧ η4r.
So we have determined all curvature tensors of M . Note that the quater-
nionic curvatures satisfy
K(e1, e2) +K(e1, e3) +K(e1, e4) = −12
K(e2, e1) +K(e2, e3) +K(e2, e4) = −12 + e
−2t (KN(e2, e3) +K
N (e2, e4))
K(e3, e1) +K(e3, e2) +K(e3, e4) = −12 + e
−2t (KN(e3, e2) +K
N (e3, e4))
K(e4, e1) +K(e4, e2) +K(e4, e3) = −12 + e
−2t (K̄(e4, e2) +K
N (e4, e3)).
In particular, this implies that
KN(e2, e3) = K
N (e2, e4) = K
N (e3, e4) = 0.
Also, for 2 ≤ p ≤ 4, we have
K(e1, e(4s−i)) = −4
K(ep, e(4s−i) = −4 + e
−2t (
K̄(ep, e(4s−i))− 4)
implying
KN (ep, e(4s−i)) = 4.
We also have
K(e(4s), e(4s−1)) = −12 + e
−2t (
KN(e(4s), e(4s−i)) + 9)
implying
KN (e(4s), e(4s−i)) = −9.
Lastly,
K(e(4s), e(4r−i)) = −4 + e
KN(e(4s), e(4r−i))
implying
KN (e(4s), e(4r−i)) = 0.
The above computation determined the whole curvature tensor for M
and N0. In particular, if M has bounded curvature, then from the formulas
about the components of curvature tensors of M , all curvature components
are determined as those of QHn. So it must be covered by QHn. �
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Shengli Kong
Department of Mathematics
http://arxiv.org/abs/math/0701865
University of California, Irvine
Irvine, CA92697-3875, USA
email: skong@math.uci.edu
Peter Li
Department of Mathematics
University of California, Irvine
Irvine, CA92697-3875, USA
email:pli@math.uci.edu
Detang Zhou
Departamento de Geometria
Insitituto de Matematica
Universidade Federal Fluminense- UFF
Centro, Niterói, RJ 24020-140, Brazil
email: zhou@impa.br
	Preliminaries on quaternionic Kähler manifolds
	Laplacian comparison theorem
	Quaternionic harmonicity
	Uniqueness of infinite volume end
	Maximal first eigenvalue

  Let $M^{4n}$ be a complete quaternionic K\"ahler manifold with scalar
curvature bounded below by $-16n(n+2)$. We get a sharp estimate for the first
eigenvalue $\lambda_1(M)$ of the Laplacian which is $\lambda_1(M)\le (2n+1)^2$.
If the equality holds, then either $M$ has only one end, or $M$ is
diffeomorphic to $\mathbb{R}\times N$ with N given by a compact manifold.
Moreover, if $M$ is of bounded curvature, $M$ is covered by the quaterionic
hyperbolic space $\mathbb{QH}^n$ and $N$ is a compact quotient of the
generalized Heisenberg group. When $\lambda_1(M)\ge \frac{8(n+2)}3$, we also
prove that $M$ must have only one end with infinite volume.

Introduction
Let Mn be a complete n-dimensional Riemannian manifold whose Ricci cur-
vature bounded below by −(n − 1). It is well known from Cheng [Ch] that
the first eigenvalue λ1(M) satisfies
λ1(M) ≤
(n− 1)2
In [LW3], Li and Wang proved an analogous theorem for complete Käh-
ler manifolds. They showed that if M2n is a complete Kähler manifold of
complex dimension n with holomorphic bisectional curvature BKM bounded
below by −1, then the first eigenvalue λ1(M) satisfies
λ1(M) ≤ n
∗Research partially supported by NSF grant DMS-0503735
†Research partially supported by CAPES and CNPq of Brazil.
http://arxiv.org/abs/0704.1851v1
Here BKM ≥ −1 means that
Rīijj̄ ≥ −(1 + δij)
for any unitary frame e1, . . . , en.
In this paper, we prove the corresponding Laplacian comparison theorem
for a quaterionic Kähler manifold M4n. As an application we get the sharp
estimate λ1(M) for a complete quaterionic Kähler manifold M
4n with scalar
curvature bounded below by −16n(n+ 2) as
λ1(M) ≤ (2n + 1)
It is an interesting question to ask what one can say about those manifolds
when the above inequalities are realized as equalities. In works of Li and
Wang [LW1] and [LW2], the authors obtained the following theorems. The
first was a generalization of the theory of Witten-Yau [WY], Cai-Galloway
[CG], and Wang [W] for conformally compact manifolds. The second was to
answer the aforementioned question.
Theorem 0.1. Let Mn be a complete Riemannian manifold of dimension
n ≥ 3 with Ricci curvature bounded below by −(n − 1). If λ1(M) ≥ n − 2,
then either
(1) M has only one infinite volume end; or
(2) M = R×N with warped product metric of the form
ds2M = dt
2 + cosh2 t ds2N ,
where N is an (n − 1)-dimensional compact manifold of Ricci curvature
bounded below by λ1(M).
Theorem 0.2. Let Mn be a complete Riemannian manifold of dimension
n ≥ 2 with Ricci curvature bounded below by −(n − 1). If λ1(M) ≥
(n−1)2
then either
(1) M has no finite volume end; or
(2) M = R×N with warped product metric of the form
ds2M = dt
2 + e2t ds2N ,
where N is an (n − 1)-dimensional compact manifold of nonnegative Ricci
curvature.
In [LW3] and [LW5], Li and Wang also consider the Kähler case. They
proved the following theorems.
Theorem 0.3. Let Mn be a complete Kähler manifold of complex dimension
n ≥ 1 with Ricci curvature bounded below by
RicM ≥ −2(n + 1).
If λ1(M) >
, then M must have only one infinite volume end.
Theorem 0.4. Let Mn be a complete Kähler manifold of complex dimension
n ≥ 2 with holomorphic bisectional curvature bounded by
BKM ≥ −1.
If λ1(M) ≥ n
2, then either
(1) M has only one end; or
(2) M = R × N with N being a compact manifold. Moreover the metric on
M is of the form
ds2M = dt
2 + e4t ω22 + e
ω2i ,
where {ω2, ω3, . . . , ω2n} are orthonormal coframe of N with Jdt = ω2. If M
has bounded curvature, then we further conclude that M is covered by CHn
and N is a compact quotient of the Heisenberg group.
In [LW5], the authors pointed out that the assumption on the lower bound
of λ1(M) in Theorem 0.3 is sharp, since one can construct M of the form
M = Σ×N satisfying
RicM ≥ −2(n + 1) (0.1)
λ1(M) =
n + 1
(0.2)
with N being a compact Kähler manifold and Σ being a complete surface
with at least two infinite volume ends. However, it is still an open question to
characterized all those complete Kähler manifolds satisfying conditions ( 0.1)
and ( 0.2).
In sections 4 and 5, we will prove the following quaternionic Kähler ver-
sions of the above theorems.
Theorem 0.5. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature satisfying
SM ≥ −16n(n + 2).
If λ1(M) ≥
8(n+2)
, then M must have only one infinite volume end.
Theorem 0.6. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature satisfying
SM ≥ −16n(n + 2).
If λ1(M) ≥ (2n+ 1)
2, then either
(1) M has only one end, or
(2) M is diffeomorphic to R×N where N is a compact manifold. Moreover,
the metric is given by the form
ds2M = dt
2 + e4t
ω2p + e
where {ω2, . . . , ω4n} are orthonormal coframes for N. If M is of bounded
curvature then we further conclude that M is covered by the quaterionic hy-
perbolic space QHn and N is a compact quotient of the generalized Heisenberg
group.
Remark 0.1. It is known that a horosphere in QHn is isometric to a certain
generalized Heisenberg group with three-dimensional center and left-invariant
Riemannian metric. Such generalized Heisenberg groups have compact quo-
tients. For an explicit construction see for instance Example 2.6 in [G]. We
don’t have an example to show that the bounded curvature condition in The-
orem 0.6 is necessary. If such an example exists, its curvature should decay
at exponentially in some directions.
Perhaps it is interesting to restrict our attention to the special case when
M4n = QHn/Γ is given by the quotient of the quaternionic hyperbolic space
n with a discrete group of isometies Γ. In particular, it is instructional to
compare with previous results by Corlette [C2] and Corlette-Iozzi [CI] where
Lie group theoretic approach was used in understanding these manifolds. For
example, in [CI], the authors proved a Patterson-Sullivan type formula for
λ1(M) in terms of the Hausdorff dimension δ(Γ) of the limit set of Γ. More
specifically, they proved that if Γ is geometrically finite, then for δ(Γ) ≥ 2n+1
one has
λ1(M) = δ(Γ)((4n+ 2)− δ(Γ)).
Hence in this case, the condition in Theorem 0.6 on λ1(M) = (2n + 1)
equivalent to the condition δ(Γ) = 2n+ 1.
In [C2] (Theorem 4.4), Corlette also pointed out that by a result of
Kostant λ1(M) = 0 or λ1(M) ≥ 8n. On the other hand, it was also shown
in [CI] that if Γ is geometrically finite and torsion free, then M = QHn/Γ
must have at most one end with infinite volume. These two statements give
an interesting comparison to Theorem 0.5 stated above.
We would also like to point out to the interested readers that in [LW4] and
[LW5] Li and Wang considered a more general class of manifolds satisfying a
weighted Poincaré inequality. However, since quaternionic Kähler manifolds
are automatically Einstein, the same type of questions are not interesting for
this class of manifolds.
Acknowledgement. This work was done when the third author was
visiting the University of California, Irvine. He wishes to thank the institu-
tion for its hospitality. He also would like to thank Professor J. Berndt for
pointing out the paper of [G] to him.
1 Preliminaries on quaternionic Kähler man-
ifolds
In this section, we will recall basic properties of quaternionic Kähler manifolds
that will be needed in the sequel. These properties were proved by Berger
[B] and Ishihara [I] (also see [Be]).
Let (Mn, g) be a Riemannian manifold, TM the tangent space of M and
∇ the Levi-Civita connection. The Riemannian curvature R : TM ⊗ TM ⊗
TM −→ TM is defined by
R(X, Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z
If {e1, · · · , en} is an orthonormal basis of TM , the components of curvature
tensor is defined by
Rijkl = 〈R(ei, ej)el, ek〉,
the Ricci curvature is defined by
RicM(X, Y ) =
〈R(X, ei)ei, Y 〉,
and the scalar curvature is defined by
i,j=1
〈R(ei, ej)ej , ei〉.
Definition 1.1. A quaternionic Kähler manifold (M, g) is a Riemannian
manifold with a rank 3 vector bundle V ⊂ End(TM) satisfying
(a) In any coordinate neighborhood U ofM , there exists a local basis {I, J,K}
of V such that
I2 = J2 = K2 = −1
IJ = −JI = K
JK = −KJ = I
KI = −IK = J
〈IX, IY 〉 = 〈JX, JY 〉 = 〈KX,KY 〉 = 〈X, Y 〉
for all X, Y ∈ TM .
(b) If φ ∈ Γ(V ), then ∇Xφ ∈ Γ(V ) for all X ∈ TM .
Remark 1.1. It follows from (a) that dimM = 4n. A well known fact about
4n-dimensional Riemannian manifold is that it is quaternionic Kähler if and
only if its restricted holonomy group is contained in Sp(n)Sp(1).
The 4-dimensional Riemannian manifolds with holonomy Sp(1)Sp(1) are
simply the oriented Riemanian manifolds, naturally we only consider those
when n ≥ 2.
Notice that in general I, J,K are not defined everywhere on M . For
example, the canonical quaternionic projective space QP n admits no almost
complex structure.
On the other hand, the vector space generated by I, J,K is well defined
at each point of M and this 3-dimensional subbundle V of End(TM) is in
fact “globally parallel” under the Levi-Civita connection ∇ of g. A basic fact
about the connection is the following lemma.
Lemma 1.1. The condition (b) is equivalent to the following condition:
∇XI = c(X)J − b(X)K,
∇XJ = −c(X)I + a(X)K,
∇XK = b(X)I − a(X)J,
where a, b, c are local 1-forms.
Definition 1.2. Let (M, g) be a quaternionic Kähler manifold. We can
define a 4-form by
Ω = ω1 ∧ ω1 + ω2 ∧ ω2 + ω3 ∧ ω3,
where
ω1 = 〈·, I·〉,
ω2 = 〈·, J ·〉,
ω3 = 〈·, K·〉.
Let {e1, Ie1, Je1, Ke1, · · · , en, Ien, Jen, Ken} be an orthonormal basis of
TM and {θ1, Iθ1, Jθ1, Kθ1, · · · , θn, Iθn, Jθn, Kθn, } the dual basis. It follows
θi ∧ Iθi + Jθi ∧Kθi
θi ∧ Jθi +Kθi ∧ Iθi
θi ∧Kθi + Iθi ∧ Jθi
θi ∧ Iθi ∧ θj ∧ Iθj + θi ∧ Jθi ∧ θj ∧ Jθj + θi ∧Kθi ∧ θj ∧Kθj
Jθi ∧Kθi ∧ Jθj ∧Kθj +Kθi ∧ Iθi ∧Kθj ∧ Iθj
+ Iθi ∧ Jθi ∧ Iθj ∧ Jθj
θi ∧ Iθi ∧ Jθj ∧Kθj + θi ∧ Jθi ∧Kθj ∧ Iθj
+ θi ∧Kθi ∧ Iθj ∧ Jθj
Lemma 1.2. The condition (b) is equivalent to the following condition:
∇Xω1 = c(X)ω2 − b(X)ω3,
∇Xω2 = −c(X)ω1 + a(X)ω3,
∇Xω3 = b(X)ω1 − a(X)ω2.
where a, b, c are local 1-forms.
Proof: It follows from the identities
(∇Xω1)(Y, Z) = 〈Y, (∇XI)Z〉,
(∇Xω2)(Y, Z) = 〈Y, (∇XJ)Z〉,
(∇Xω3)(Y, Z) = 〈Y, (∇XK)Z〉.
Using this lemma, we have that
Theorem 1.1. The condition (b) is equivalent to that Ω is parallel, that is
∇XΩ = 0
for any X ∈ TM .
In the following, we shall study the curvature of quaternionic Kähler
manifold. First we have the following lemma.
Lemma 1.3. If (M4n, g) is a quaternionic Kähler manifold, then
[R(X, Y ), I] = γ(X, Y )J − β(X, Y )K,
[R(X, Y ), J ] = −γ(X, Y )I + α(X, Y )K,
[R(X, Y ), K] = β(X, Y )I − α(X, Y )J,
where α, β and γ are local 2-forms given by
α = da+ b ∧ c,
β = db+ c ∧ a,
γ = dc+ a ∧ b.
Corollary 1.1. If (M4n, g) is a quarternionic Kähler manifold, then
〈R(X, Y )Z, IZ〉+ 〈R(X, Y )JZ,KZ〉 = α(X, Y ) |Z|2,
〈R(X, Y )Z, JZ〉+ 〈R(X, Y )KZ, IZ〉 = β(X, Y ) |Z|2,
〈R(X, Y )Z,KZ〉+ 〈R(X, Y )IZ, JZ〉 = γ(X, Y ) |Z|2.
The following lemma is the key for quaternionic Kähler manifolds.
Lemma 1.4. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2, then
α(X, IY ) = β(X, JY ) = γ(X,KY ) = −
n + 2
RicM(X, Y ). (1.1)
As applications of the above lemma, one can show the following two main
theorems on curvature of quaternionic Kähler manifolds.
Theorem 1.2. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2,
then (M4n, g) is Einstein, that is, there is a constant δ such that
RicM(g) = 4(n+ 2)δg.
Theorem 1.3. If (M4n, g) is a quaternionic Kähler manifold and n ≥ 2,
(1) For any tangent vector X, the sectional curvature satisfies
〈R(X, IX)IX,X〉+ 〈R(X, JX)JX,X〉
+〈R(X,KX)KX,X〉 = 12δ |X|4.
(2) For any tangent vector Y satisfying
〈Y,X〉 = 〈Y, IX〉 = 〈Y, JX〉 = 〈Y,KX〉 = 0,
the sectional curvature satisfies
〈R(X, Y )Y,X〉+ 〈R(X, IY )IY,X〉+
〈R(X, JY )JY,X〉+ 〈R(X,KY )KY,X〉 = 4δ |X|2 |Y |2,
where 4(n+ 2)δ is the Einstein constant.
Finally, we end this section with the following lemma.
Lemma 1.5. Let γ : [a, b] → M be a geodesic with unit speed. If S =
16n(n + 2)δ, and XI(t), XJ(t), XK(t) are parallel vector fields along γ such
that XI(a) = Iγ
′(a), XJ(a) = Jγ
′(a), XK(a) = Kγ
′(a), then
K(γ′(t), XI(t)) +K(γ
′(t), XJ(t)) +K(γ
′(t), XK(t)) = 12δ,
for all t and γ.
Let Y be a tangent vector at γ(a) satisfying 〈γ′(a), Y 〉 = 0, 〈Iγ′(a), Y 〉 =
0, 〈Jγ′(a), Y 〉 = 0, and 〈Kγ′(a), Y 〉 = 0. If we denote the parallel vector fields
Y (t), YI(t), YJ(t), and YK(t) along γ with initial data Y (a) = Y , YI(a) = IY,
YJ(a) = JY , and YK(a) = KY , respectively, then
K(γ′(t), Y (t)) +K(γ′(t), YI(t)) +K(γ
′(t), YJ(t)) +K(γ
′(t), YK(t)) = 4δ,
for all t and γ.
Proof. By the discussion above, we know the 3-dimensional vector
space E(t) spanned by X(t), Y (t), Z(t) does not depend on the choice of
I, J,K. Hence it is parallel under the Levi-Civita connection. We consider
〈R(·, γ′(t))γ′(t), ·〉 as a symmetric bilinear form on E(t). ThenK(γ′(t), X(t))+
K(γ′(t), Y (t)) + K(γ′(t), Z(t)) is its trace on E(t) which independent of the
choice of orthonormal basis. By the computation above it is equal to 12δ.
The same argument also applies to the second part of the lemma. �
2 Laplacian comparison theorem
For a complete Riemannian manifold M and p ∈ M , let us denote the cut
locus with respect to p by Cut(p).
Theorem 2.1. Let (M4n, g) be a complete quaternionic Kähler manifold with
scalar curvature SM ≥ 16n(n+2)δ and let r(x) be the distance function to a
fixed point p ∈M . Then, for x /∈ Cut(p),
∆r(x) ≤
6 coth 2r(x) + 4(n− 1) coth r(x) when δ = −1
(4n− 3)r−1(x) when δ = 0
6 cot 2r(x) + 4(n− 1) cot r(x) when δ = 1.
(2.1)
Proof. Let γ be the minimizing geodesic joining p to x. At x, we choose
{e1, e2, · · · , en}, and two local almost complex structures I, J and K = IJ
such that e1 = ∇r and
{e1, Ie1, Je1, Ke1, e2, Ie2, Je2, Ke2, · · · , en, Ien, Jen, Ken}
is an orthonormal frame. By parallel translating along γ we obtain an or-
thonormal frame with e1 = ∇r. For convenience sake, we denote this frame
by {ε1, ε2, · · · , ε4n}. Since |∇r|
2 = 1 on M\Cut(p), by taking covariant
derivative of this equation, we have
0 = |∇r|2kl
rikril + 2
ririkl, (2.2)
for each k, l = 2, · · · , 4n. Since
rikl = rkli +
Rjkilrj,
with Rijkl = 〈R(εi, εj)εl, εk〉, and r1 = 1, rj = 0, j = 2, · · · , 4n, we have
rikril + rkl1 +R1k1l = 0. (2.3)
In particular, if k = l, we have
r2ik + rkk1 +K(ε1, εk) = 0, (2.4)
where K(ε1, εk) = R1k1k is the sectional curvature of the 2-plane section
spanned by ε1, εk. Using the inequality
r2ik ≥
and setting f(t) =
k=2 rkk, ( 2.4) implies that
f ′(t) +
f 2(t) +
K(ε1, εk) ≤ 0. (2.5)
By Lemma 1.5, we have
f ′(t) +
f 2(t) + 12δ ≤ 0. (2.6)
Since a smooth Riemannian metric is locally Euclidean, then limt→0 tf(t) = 3.
By a standard comparison argument for ordinary differential equations, we
conclude that
f(t) ≤
6 cot 2t when δ = 1
3t−1 when δ = 0
6 coth 2t when δ = −1.
(2.7)
Similarly, using the inequality
k=4i+1
r2ik ≥
k=4i+1
for 1 ≤ i ≤ n− 1, and setting hi(t) =
∑4i+4
k=4i+1 rkk, ( 2.4) implies that
h′i(t) +
h2i (t) +
k=4i+1
K(ε1, εk) ≤ 0. (2.8)
Together with Lemma 1.5 asserting that
k=4i+1
K(ε1, εk) = 4δ,
we have
h′i(t) +
h2i (t) + 4δ ≤ 0. (2.9)
Hence, as before, we conclude that
hi(t) ≤
4 cot t when δ = 1
4t−1 when δ = 0
4 coth t when δ = −1.
(2.10)
The result follows from the equation ∆r(x) = f(r(x)) +
i=1 hi(r(x)). �
Remark 2.1. The estimate in Theorem 2.1 is sharp since the right hand
sides are exactly the Laplacian of the distance functions of quaternionic hy-
perbolic space QHn, quaternionic Euclidean space Qn and quaternionic pro-
jective space QPn respectively.
Remark 2.2. We actually proved the estimate for Hessian of the distance
function. In particular,
rkk ≤
6 cot 2t when δ = 1
3t−1 when δ = 0
6 coth 2t when δ = −1.
(2.11)
Also for 1 ≤ i ≤ n− 1, we have
k=4i+1
rkk ≤
4 cot 2t when δ = 1
4t−1 when δ = 0
4 coth 2t when δ = −1.
(2.12)
Corollary 2.1. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ −16n(n + 2). Then for any point x ∈ M and
r > 0, the area A(r) of the geodesic spheres centered at x satisfies
A′(r)
≤ 6 coth 2r + 4(n− 1) coth r. (2.13)
In particular, A(r) ≤ C(sinh 2r)3(sinh r)4(n−1) ≤ Ce(4n+2)r.
Corollary 2.2. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ −16n(n + 2). Then for any point x ∈ M and
0 < r1 ≤ r2, the volume of the geodesic balls centered at x satisfies
Vx(r2)
Vx(r1)
VQHn(r2)
VQHn(r1)
, (2.14)
where VQHn(r) denotes the volume of the geodesic ball of radius r in QH
n. In
particular, λ1(M) ≤ (2n+ 1)
Corollary 2.3. Let (M4n, g) be a complete quaternionic Kähler manifold
with scalar curvature SM ≥ 16n(n+2) . Then it is compact, and the diameter
d(M) ≤ π
, which is the diameter of the model space QPn. Moreover, the
volume of M is bounded by
V (M) ≤ V (QPn), (2.15)
where VQPn is the volume of QP
3 Quaternionic harmonicity
In this section we will derive an over-determined system of harmonic func-
tions with finite Dirichlet integral on a manifold with a parallel form. This
result was first proved by Siu [S] for harmonic maps in his proof of the rigidity
theorem for Kähler manifolds. Corlette [C1] gave a more systematic approach
for harmonic map with finite energy from a finite-volume quaternionic hy-
perbolic space or Cayley hyperbolic plane to a manifold with nonpositive
curvature. In [L], the second author generalized Siu’s argument to harmonic
functions with finite Dirichlet integral on a Kähler manifold. We will pro-
vide an argument that generalizes Corlette’s argument to harmonic functions
with finite Dirichlet integral on a complete manifold with a parallel form. We
believe that it should be of independent interest.
Theorem 3.1. Let M be a complete Riemannian manifold with a parallel
p-form Ω. Assume that f is a harmonic function with its Dirichlet integral
over geodesic balls centered at o of radius R satisfying the growth condition
Bo(R)
|∇f |2dv = o(R2)
as R → ∞, then f satisfies
d ∗ (df ∧ Ω) = 0. (3.1)
Before we prove the theorem, let us first recall the following operators
and some of the basic properties. For an oriented real vector space V with
an inner product, we have the Hodge star operator
∗ : ∧pV → ∧n−pV.
For any θ ∈ ∧1V and v ∈ V , we also have exterior multiplication and interior
product operators
ε(θ) : ∧pV → ∧p+1V,
ℓ(v) : ∧pV → ∧p−1V.
For θ ∈ ∧1V and v ∈ V is the dual of θ by the inner product, if ξ ∈ ∧pV we
list the following identities among the operators:
1. ∗ ∗ ξ = (−1)p(n−p)ξ,
2. ∗ε(θ)ξ = (−1)pℓ(v) ∗ ξ,
3. ε(θ) ∗ ξ = (−1)p−1 ∗ ℓ(v)ξ,
4. ∗ε(θ) ∗ ξ = (−1)(p−1)(n−p)ℓ(v)ξ,
5. ℓ(v)ε(θ′)ξ + ε(θ)ℓ(v′)ξ = 0, where v⊥v′,
6. ℓ(v)ε(θ)ξ + ε(θ)ℓ(v)ξ = ξ.
We are now ready to prove Theorem 3.1.
Proof of Theorem 3.1. Let η : [0,+∞) → R be a smooth function satis-
fying η′(t) ≤ 0, and
η(t) =
1 when t ∈ [0, 1]
0 when t ∈ [2,+∞].
For R ≥ 1, we define the cut-off function φR(x) = η(r(x)/R), where r(x)
is the distance function from a fixed point o ∈ M , then there is a positive
constant C1 depending on η and C such that
|∇φR(x)| ≤ C1R
Since d2 = 0, then
φ2R ∗ (df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
d(φ2R) ∧ ∗(df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
φ2R d ∗ (df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω). (3.2)
We claim that
∗ d ∗ (df ∧ Ω) = (−1)n−1d ∗ (df ∧ ∗Ω). (3.3)
In fact, for any point x ∈ M , we can choose an orthonormal tangent basis
i=1 in a neighborhood of x such that ∇eiej(x) = 0. Denote by {θ
i}mi=1
the dual basis of {ei}
i=1. Then for ω ∈ ∧
p(T ∗M) we have
dω = ε(θi)∇eiω.
Hence
d ∗ (df ∧ ∗Ω) = d ∗ ε(df) ∗ Ω
= (−1)(p−1)(m−p)d[ℓ(∇f)Ω]
= (−1)(p−1)(m−p)
ε(θi)∇ei(ℓ(∇f)Ω)
= (−1)(p−1)(m−p)
i,j=1
ε(θi)(∇ei∇ejf)(ℓ(ej)Ω)
= (−1)(p−1)(m−p)
i,j=1
fijε(θi)(ℓ(ej)Ω),
where fij = ∇ei∇ejf and the facts Ω is parallel and ∇eiej(x) = 0 have been
used. On the other hand,
∗d ∗ (df ∧ Ω) = ∗d ∗ ε(df)Ω
ε(θi)∇ei(∗ε(
fjθj)d[ℓ(∇f)Ω]
i,j=1
fijε(θi) ∗ ε(θj)(Ω)
= (−1)p(m−p−1)
i,j=1
fijℓ(ei)ε(θj)Ω
= (−1)p(m−p−1)
fiiℓ(ei)ε(θi)Ω +
i 6=j
fijℓ(ei)ε(θj)Ω
= (−1)p(m−p−1)
fii[Ω− ε(θi)ℓ(ei)Ω]−
i 6=j
fijε(θj)ℓ(ei)Ω
= (−1)p(m−p−1)
i,j=1
fijε(θi)(ℓ(ej)Ω), (3.4)
where we used fij = fji and
i=1 fii = 0. So the claim is proved. By ( 3.2),
we have
φ2R |d ∗ (df ∧ Ω)|
= (−1)m
d(φ2R) ∧ ∗(df ∧ Ω) ∧ d ∗ (df ∧ ∗Ω)
φ2R |d ∗ (df ∧ Ω)|
|dφR|
2| ∗ (df ∧ Ω)|2dv
(3.5)
On the other hand, ( 3.3) and the fact that ω is bounded imply that there
exists a constant C2 > 0, such that
| ∗ (df ∧ Ω)| ≤ C2 |df |
|d ∗ (df ∧ ∗Ω)| = |d ∗ (df ∧ Ω)|.
Hence combining with ( 3.5) and using the definition of φR we conclude that
Bo(R)
|d ∗ (df ∧ Ω)|2dv ≤ C1R
Bo(2R)
|df |2dv.
The assumption on the growth of the Dirichlet integral of f implies that the
right hand side tends to zero as R → ∞. Therefore d ∗ (df ∧Ω) = 0, and the
proof is complete. �
Lemma 3.1. Let (M4n, g) be a quarternionic Kähler manifold and n ≥ 2. If
f is a function on M satisfying
d ∗ (df ∧ Ω) = 0 (3.6)
for the 4-form Ω determined by the quaternionic Kähler structure, then f is
quaternionic harmonic, namely, for any nonzero tangent vector X,
fX,X + fIX,IX + fJX,JX + fKX,KX = 0
where fX,X = ∇df(X,X).
Proof: Let
A=1 = {e1, e2, · · · , en, Ie1, Ie2, · · · , Ien,
Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
be an orthonormal basis of TM and {ωA} the dual basis with e1 =
. Since
Ω is parallel, by ( 3.4) and ( 3.6), we have
(∇eAdf) ∧ ℓ(eA)Ω
A,B=1
feA,eB ωB ∧ ℓ(eA)Ω
where we have used the fact that f is a harmonic function. Hence equation
( 3.6) implies
A,B=1
feA,eB ωB ∧ ℓ(eA)Ω = 0
Comparing the coefficient of ωi∧Iωi∧Jωi∧Kωi on both sides by the explicit
formula for Ω given before, we obtain that
6 (fei,ei + fIei,Iei + fJei,Jei + fKei,Kei) = 0
for all ei, (1 ≤ i ≤ n). So the proof is complete. �
The following corollary is an immediate consequence of the lemma.
Corollary 3.1. Let M4n be a complete quaternionic Kähler manifold. As-
sume that f is a harmonic function with its Dirichlet integral satisfying the
growth condition
Bo(R)
|∇f |2dv = o(R2)
as R → ∞, then f must satisfy
d ∗ (df ∧ Ω) = 0, (3.7)
where Ω is the parallel 4-form determined by the quaternionic Kähler struc-
ture. Moreover, f is quaternionic harmonic.
4 Uniqueness of infinite volume end
Recall that for any complete manifold if λ1(M) > 0 then M must be non-
parabolic. In particular,M must have at least one nonparabolic ends. It was
also proved in [LW1] that under the assumption that λ1(M) > 0, an end is
nonparabolic if and only if it has infinite volume.
Let us assume that M has at least two nonparabolic ends, E1 and E2.
A construction of Li-Tam [LT] asserts that one can construct a nonconstant
bounded harmonic function with finite Dirichlet integral. The harmonic func-
tion f can be obtained by taking a convergent subsequence of the harmonic
functions fR, as R→ +∞, satisfying
∆fR = 0 on B(R),
with boundary conditions
fR = 1 on ∂B(R) ∩ E1
fR = 0 on ∂B(R) \ E1.
It follows from the maximum principle that 0 ≤ fR ≤ 1, hence 0 ≤ f ≤ 1.
We need the following estimates from [LW1](Lemma 1.1 and 1.2 in [LW1]),
and [LW3](Lemma 5.1 in [LW3]).
Lemma 4.1. Let M be a complete Riemannian manifold with λ1(M) > 0.
Suppose M has at least two nonparabolic ends and E be an end of M . Then
for the harmonic function f constructed above, it must satisfy the following
growth conditions:
1. There exists a constant a such that f − a ∈ L2(E). Moreover, the
function f − a must satisfy the decay estimate
E(R+1)\E(R)
(f − a)2 ≤ C exp(−2
λ1(E)R)
for some constant C > 0 depending on f , λ1(E) and the dimension of
2. The Dirichlet integral of the function f must satisfy the decay estimate
E(R+1)\E(R)
|∇f |2 ≤ C exp(−2
λ1(E)R),
exp(−2
λ1(E)r(x))|∇f |
2 ≤ CR
for R sufficiently large.
Lemma 4.2. Let M be a complete Riemannian manifold with at least two
nonparabolic ends and λ1(M) > 0. Then for the harmonic function f con-
structed above, for any t ∈ (inf f, sup f) and (a, b) ⊂ (inf f, sup f),
L(a,b)
|∇f |2 = (b− a)
|∇f |,
where
l(t) = {x ∈M |f(x) = t},
L(a, b) = {x ∈M |a < f(x) < b}.
Moreover,
|∇f | =
|∇f |.
We are now ready to prove Theorem 0.5.
Proof of Theorem 0.5: Suppose to the contrary that there exist two ends
E1 and E2 with infinite volume. The assumption that λ1(M) > 0 implies that
they are nonparabolic. By the construction above, there exists a harmonic
function f with finite energy such that
lim inf
x→∞, x∈E1
f(x) = 1
lim inf
x→∞, x∈E2
f(x) = 0.
The Bochner formula implies that
∆|∇f |2 = RicM(∇f,∇f) + |∇
2f |2. (4.1)
We now choose an orthonormal basis {eA}
A=1 satisfying
{e1, e2, · · · , en, Ie1, Ie2, · · · , Ien, Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
with e1 =
|∇f |
. Corollary 3.1 implies that
f(in+1)(in+1) = 0.
Therefore, applying the arithmetic-geometric means, we have
|∇2f |2 =
A,B=1
f 2AB
≥ f 211 +
f 2(in+1)(in+1) + 2
f 21A
≥ f 211 +
f(in+1)(in+1))
2 + 2
f 21A
|∇|∇f ||2, (4.2)
hence combining with ( 4.1) we obtain
∆|∇f |2 ≥ −4(n + 2)|∇f |2 +
|∇|∇f ||2. (4.3)
If we write u = |∇f |
3 , then
∆u ≥ −
8(n+ 2)
u. (4.4)
We want to prove that the above inequality is actually an equality. The
argument follows from that in [LW4] after making suitable modification to
fit our situation. For any compactly supported smooth function φ on M , we
8(n+ 2)
φu〈∇u,∇φ〉 −
φ2|∇u|2 + λ1(M)
(φu)2
φu〈∇u,∇φ〉 −
φ2|∇u|2 +
|∇(φu)|2
|∇φ|2u2. (4.5)
Let us choose φ = ψχ to be the product of two compactly supported
functions. For any ε ∈ (0, 1
), we define
χ(x) =
0 on L(0, σε) ∪ L(1− ε
(log 2)−1(log f − log( ε
)) on L( ε
, ε) ∩ (M \E1)
(log 2)−1(log(1− f)− log( ε
)) on L(1− ε, 1− ε
) ∩ E1
1 otherwise.
For R > 1 we define
1 on B(R− 1)
R− r on B(R) \B(R− 1)
0 on M \B(R).
Applying to the right hand side of ( 4.5), we obtain
|∇φ|2u2 ≤ 2
|∇ψ|2χ2|∇f |
3 + 2
|∇χ|2ψ2|∇f |
3 . (4.6)
Since RicM ≥ −4(n+2), then the local estimate of Cheng-Yau [CY] (see also
[LW2]) implies that there exists a constant depending on n such that
|∇f |(x) ≤ C|1− f(x)|.
On E1, the first term of ( 4.6) satisfies
|∇ψ|2χ2|∇f |
|∇f |2
, (4.7)
where Ω = E1 ∩ (B(R) \B(R − 1)) ∩ (L(1− ε, 1−
) ∪ L( ε
, ε). Since
1 ≤ 4
(1− f)2
(1− f)2
≤ 4Cε−2 exp(−2
λ1R),
where in the last inequality we have used Lemma 4.1. Again by Lemma 4.1,
from ( 4.7) we have
|∇ψ|2χ2|∇f |
3 ≤ Cε−
3 exp(−2
λ1R). (4.8)
For the second term of ( 4.6) we have
|∇χ|2ψ2|∇f |
≤ (log 2)−2
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |
+2(1− f)−2
≤ C(log 2)−2
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |2(1− f)−
Using the co-area formula and Lemma 4.2 we have
L(1−ε,1− ε
)∩E1∩B(R)
|∇f |2(1− f)−
∫ 1− ε
(1− t)−
l(t)∩E1∩B(R)
|∇f |dAdt
|∇f |dA
∫ 1− ε
(1− t)−
= −3C[(1− t)
|∇f |dA
= 3Cε
|∇f |dA.
Combining the above inequality with ( 4.8) we have
|∇φ|2u2 ≤ C(ε
3 exp(−2
λ1R) + ε
3 ). (4.9)
A similar argument using f instead of 1 − f on the other end yields the
estimate
|∇φ|2u2 ≤ C(ε
3 exp(−2
λ1R) + ε
Letting R → ∞ and ε→ 0, we have
∆u = −
8(n + 2)
u (4.10)
with λ1(M) =
8(n+2)
, since f is nonconstant and u cannot be identically zero.
Therefore all the inequalities used to prove ( 4.4) are equalities. Thus there
exists a function µ, such that,
(fAB) =
, (4.11)
where D1 and D2 are n× n matrices defined by
· · ·
· · ·
Since f1α = 0 for α 6= 1 implies that |∇f | is constant along the level set of
f . Moreover, regularity of the equation ( 4.10) implies that |∇f | can never
be zero. Hence M must be diffeomorphic to R×N , where N is given by the
level set of f . Also N must be compact since we assume that M has at least
2 ends.
Fix a level set N0 of f , consider (−ε, ε)×N0 ⊂M . Note that {eA} is an
orthonormal basis of TM such that e1 is the normal vector to N0 and {eα}
are the tangent vectors of N0. We shall compute the sectional curvature
K(e1, eα) = 〈R(e1, eα)eα, e1〉.
We claim that
∇e1e1 = 0.
Indeed it suffices to prove all integral curves η(t) of the vector field e1 =
|∇f |
emanating from N0 are geodesics. For any point η(t0), let γ be the geodesic
realizing the distance between η(t0) and N0. Then γ is perpendicular to every
level set Nt. So γ
′ is parallel to e1 along γ. This implies γ coincides with the
integral curve of e1.
Let (hαβ) with 2 ≤ α, β ≤ 4n be the second fundamental form of the level
set of f . Then
hαβ f1 = −fαβ , (4.12)
∇eαe1 = −
hαβeβ.
By the definition of curvature tensor, we have
〈R(e1, eα)e1, eα〉 = 〈∇e1∇eαe1 −∇eα∇e1e1 −∇[e1,eα]e1, eα〉
= 〈∇e1∇eαe1, eα〉 − 〈∇[e1,eα]e1, eα〉
= 〈∇e1∇eαe1, eα〉 − 〈∇∇e1eα−∇eαe1e1, eα〉
= 〈∇e1∇eαe1, eα〉 −
〈∇e1eα, eβ〉〈∇eβe1, eα〉
〈∇eαe1, eβ〉〈∇eβe1, eα〉
〈∇e1(hαβeβ), eα〉+
hαβ〈∇e1eα, eβ〉+
〈(e1hαβ)eβ , eα〉 −
hαβ〈∇e1eβ, eα〉
hαβ〈∇e1eα, eβ〉+ h
= −e1hαα + 2
hαβ〈∇e1eα, eβ〉+ h
Therefore
K(e1, eα) = e1hαα − 2
hαβ〈∇e1eα, eβ〉 −
h2αβ . (4.13)
Since hαβ is diagonal, this implies that
K(e1, eα) = e1hαα − h
Combining with ( 4.11) and ( 4.12), we conclude that
K(e1, e2) = K(e1, Ie2) = K(e1, Je2) = K(e1, Ke2) = 0
which implies M is Ricci flat by Theorem 1.3. This contradicts to the as-
sumption that λ1 >
8(n+2)
> 0. Therefore M must have only one end with
infinite volume. �
5 Maximal first eigenvalue
In this section, we will consider the case when λ1(M) is of maximal value.
Proof of Theorem 0.6: According to Theorem 0.5, we know that M has
exactly one nonparabolic end. Suppose thatM has more than one end. Then
there must exist at least an end with finite volume. We divide the rest of the
proof into several parts. The first part follows exactly as that in the proof of
the corresponding theorem in the Kähler case (Theorem 3.1) in [LW5]. For
completeness sake, we will give a quick outline of it.
Part 1. Assume that E1 is such an end with finite volume given by
M \ Bp(1). Then we can choose a ray η : [0,+∞) such that η(0) = p and
η[1,+∞) ⊂ E1. The Busemann function corresponding to γ is defined by
β(x) = lim
[t− d(x, η(t))].
The Laplacian comparison theorem, Theorem 2.1, asserts that
∆β ≥ −2(2n+ 1)
in the sense of distribution. We define the function f = exp((2n+ 1)β), and
using the fact that |∇β| = 1 almost everywhere, we have
∆f = (2n+ 1) exp((2n+ 1)β)∆β + (2n+ 1)2
≥ −(2n + 1)2f.
Similar to the proof of above theorem, we conclude that for any compactly
supported function φ,
(∆f + (2n+ 1)2f)fφ2
f 2|∇φ|2.
By choosing the function φ to be
1, on Bp(R);
2R−r(x)
, on Bp(2R) \Bp(R);
0, on M \Bp(2R);
we obtain
f 2|∇φ|2
(Bp(2R)\Bp(R))∩E1
(Bp(R+i)\Bp(R+i−1))∩E1
(VE1(R + i)− VE1(R + i− 1)) exp(2(2n+ 1)(R+ i))
where VE1(R+ i) denotes the volume of the set E1∩Bp(R+ i). On the other
hand, the volume estimate in Theorem 1.4 of [LW1] implies that
VE1(∞)− VE1(R) ≤ C exp(−2(2n+ 1)R),
hence
VE1(R + i)− VE1(R + i− 1)
= VE1(∞)− VE1(R + i− 1)− (VE1(∞)− VE1(R + i))
≤ C exp(−2(2n+ 1)(R + i)).
Therefore, we conclude that
f 2|∇φ|2 ≤
Let us now denote E2 = M \ (Bp(1) ∪ E1) to be the other end of M .
When x ∈ E2, following the argument in Theorem 3.1 of [LW4], we have
β(x) ≤ −d(p, x) + 2.
Therefore
f 2|∇φ|2 ≤
(Bp(2R)\Bp(R))∩E2
(Bp(2R)\Bp(R))∩E2
exp(−2(2n+ 1)(r − 2))
Letting R → +∞, we conclude that
∆f + (2n+ 1)2f = 0, (5.1)
and all inequalities used are indeed equalities and f is smooth by regularity
of the equation ( 5.1). Moreover, |∇β| = 1, and
∆β = −2(2n+ 1).
This implies that M must be diffeomorphic to R × N , where N is given by
the level set of β. We choose an orthonormal basis {ei}
i=1 as follows
{e1, e2, · · · , en, Ie1, Ie2, · · · , Ien, Je1, Je2, · · · , Jen, Ke1, Ke2, · · · , Ken}
with e1 = ∇β. Applying the Bochner formula to β, we get
∆|∇β|2
i,j=1
β2ij + RicM(∇β,∇β) +
βi(∆β)i
i,j=1
β2ij − 4(n+ 2).
By the comparison theorem, we have,
β(in+1)(in+1) = −6.
Hence
(βαβ) =
where D1 and D2 are n× n matrices defined by
· · ·
· · ·
Part 2. For a fix level set N0 of β, we consider (−ε, ε)×N0 ⊂M . Note
that {ei} is an orthonormal basis of TM such that e1 is the normal vector
to N0 and {eα}, for 2 ≤ α ≤ 4n, are the tangent vectors of N0. We shall
compute the sectional curvature
K(e1, eα) = 〈R(e1, eα)eα, e1〉.
Since ∇e1e1 = 0 implies that the integral curves of e1 are geodesics. Let (hαγ)
be the second fundamental form of the level set of ∇β. Then
hαγ = 〈∇eαeγ, e1〉
= 〈∇eαeγ,∇β〉
= −βαγ
∇eαe1 = −
hαγeγ. (5.2)
By ( 4.13) in the proof of Theorem 0.5 we have
〈R(e1, eα)e1, eα〉 = −e1hαα + 2
hαγ〈∇e1eγ, eβ〉+
h2αγ .
Since (hαγ) are constant and diagonal, then
K(e1, eα) = −h
In particular, we have
K(e1, eα) =
−4 when α = in+ 1, i = 1, 2, 3
−1 otherwise.
On the other hand, we also have
K(en+1, e2n+1) +K(en+1, e3n+1) = −12−K(e1, en+1) = −8
K(en+1, e2n+1) +K(e3n+1, e2n+1) = −8
K(e3n+1, e2n+1) +K(en+1, e3n+1) = −8,
hence
K(en+1, e2n+1) = K(en+1, e3n+1) = K(e2n+1, e3n+1) = −4.
Since for α = 2, 3, · · · , n,
K(Ie1, eα) = −〈R(Ie1, eα)Ie1, eα〉
= −〈IR(Ie1, eα)Ie1, Ieα〉
= 〈R(Ie1, eα)e1, Ieα〉
= 〈R(e1, Ieα)Ie1, eα〉
= K(e1, Ieα)
= −1,
and K(Je1, eα) = K(Ke1, eα) = −1, we have
K(ein+1, eα) = −1,
for all i = 0, 1, 2, 3 and α 6= 1, n+ 1, 2n+ 1, 3n+ 1.
Let KN(eα, eγ) denote the sectional curvature of the level set with induced
metric. By Gaussian equation,
KN (eα, eγ)−K(eα, eγ) = hααhγγ,
it is straightforward to obtain
KN (en+1, e2n+1) = K
N (en+1, e3n+1) = K
N(e2n+1, e3n+1) = 0,
KN(ein+1, eα) = 1, (5.3)
for all i = 1, 2, 3 and α 6= 1, n+ 1, 2n+ 1, 3n+ 1.
Part 3. There is a natural map ϕt between the level sets N0 and Nt given
by the gradient flow of β. Since the integral curves are geodesics, dϕt(X) are
Jacobi fields along corresponding curves. Let (N, g0) = N0 with the induced
metric. We can consider ϕ as a flow on N . We claim that
dϕt|V1 = e
2t id
dϕt|V2 = e
t id,
where TN = V1 ⊕ V2, V1 = span{Ie1, Je1, Ke1} and V2 = V
1 . Indeed for
any point q ∈ N0, denote e1(t) = ∇β(ϕ(t)) and {εα(t)}
α=2 to be the parallel
transport of the orthonormal base {eα}
α=2 of N0 at q along ϕt(q). Since both
V1 and V2 are ϕ-invariant, we have, in particular,
〈∇e1(t)εα, εγ〉 = 0, (5.4)
when α ∈ {n+ 1, 2n+ 1, 3n+ 1}, and γ /∈ {n + 1, 2n+ 1, 3n+ 1}.
Now we can compute R1α1γ . Then
〈R(e1, εα)e1, εγ〉
= 〈∇e1∇εαe1 −∇εα∇e1e1 −∇[e1,εα]e1, εγ〉
= 〈∇e1∇εγe1, εα〉 − 〈∇[e1,εα]e1, εγ〉
= 〈∇e1∇εαe1, εγ〉 − 〈∇∇e1εα−∇εαe1e1, εγ〉
= 〈∇e1∇εαe1, εγ〉 −
〈∇e1εα, ετ〉〈∇ετ e1, εγ〉
〈∇εαe1, ετ 〉〈∇ετe1, εγ〉
〈∇e1(hατετ ), εγ〉+
hγτ 〈∇e1εα, ετ 〉+
hατhτγ
= −e1hαγ −
hατ 〈∇e1ετ , εγ〉
hγτ 〈∇e1εα, ετ 〉+
hατhτγ. (5.5)
We see that (hαγ) is diagonal and
hαα =
2, when α = n + 1, 2n+ 1, 3n+ 1;
1, otherwise.
Therefore, when α 6= γ,
R1α1γ = −hαα〈∇e1εα, εγ〉+ hγγ〈∇e1εα, εγ〉
= (hγγ − hαα)〈∇e1εα, εγ〉.
Since hαα = hγγ when α, γ ∈ {n+ 1, 2n+ 1, 3n+ 1} and α, γ /∈ {n+ 1, 2n+
1, 3n+ 1}, using ( 5.4), we have
R1α1γ = 0, for all α 6= γ.
Define
Jα(t) =
e−2tεα, when α ∈ {n+ 1, 2n+ 1, 3n+ 1};
e−tεα, when α /∈ {n+ 1, 2n+ 1, 3n+ 1}.
Since
dϕt(eα)|t=0 = [e1, eα] = −∇eαe1,
then we see that Jα satisfies the Jacobi equation and initial conditions Jα(0) =
eα and J
α(0) = eα = ∇ ∂
dϕt(eα)|t=0. By the uniqueness theorem for the
Jacobi equations, we have dϕt(eα) = Jα. The claim is proved.
Part 4. We have now a family of metrics on N written as
ds2t = e
ω2in+1 + e
ω2in+α,
and the metric of M can rewritten as
ds2 = dt2 + e4t
ω2p + e
ω2α (5.6)
where {ω2, ω3, ω4, . . . , ω4n} is the dual coframe to {e2, e3, e4, . . . , e4n} at N0.
We also choose that Ie4s−3 = e4s−2, Je4s−3 = e4s−1, and Ke4s−3 = e4s for
s = 1, . . . , n, with e1 =
. In particular, the second fundamental form on Nt
must be a diagonal matrix when written in terms of the basis {ei}
i=2 with
eigenvalues given by
(〈∇eiej, e1〉) =
2I3 0
0 I4(n−1)
, (5.7)
where Ik denotes the k× k identity matrix. Also, the sectional curvatures of
the sections containing e1 are given by
K(e1, ep) = −4 for 2 ≤ p ≤ 4
K(e1, eα) = −1 for 5 ≤ α ≤ 4n.
The Guass curvature equation also asserts that
Rijkl = R̄ijkl + hlihkj − hkihlj,
where R̄ijkl is the curvature tensor on Nt. In particular,
Rijkl =
R̄ijkl + δliδkj − δkiδlj if 5 ≤ i, j, k, l ≤ 4n
R̄ijkl + 4δliδkj − 4δkiδlj if 2 ≤ i, j, k, l ≤ 4
R̄ijkl + 2 if 2 ≤ i = l ≤ 4 and 5 ≤ k = j ≤ 4n
R̄ijkl + 2 if 2 ≤ k = j ≤ 4 and 5 ≤ i = l ≤ 4n
R̄ijkl − 2 if 2 ≤ i = k ≤ 4 and 5 ≤ j = l ≤ 4n
R̄ijkl − 2 if 2 ≤ j = l ≤ 4 and 5 ≤ i = k ≤ 4n
R̄ijkl − 2 if 2 ≤ k = i ≤ 4 and 5 ≤ j = l ≤ 4n
R̄ijkl otherwise.
(5.8)
We will now use ( 5.6) to compute the curvature tensor of M and hence
N0. Using the orthonormal coframe
η1 = ω1 = dt,
ηp = e
ηα = e
for 2 ≤ p ≤ 4 and 5 ≤ α ≤ 4n, we obtain the first structural equations
dη1 = 0, (5.9)
dηp = 2e
2t ω1 ∧ ωp + e
ωpq ∧ ωq + e
ωpα ∧ ωα
= −2ηp ∧ η1 +
ωpq ∧ ηq + e
ωpα ∧ ηα, (5.10)
dηα = e
t ω1 ∧ ωα + e
ωαp ∧ ωp + e
ωαβ ∧ ωβ
= −ηα ∧ η1 + e
ωαp ∧ ηp +
ωαβ ∧ ηβ, (5.11)
where ωij are the connection forms of N0. In the above and all subsequent
computations, we will adopt the convention that 5 ≤ α, β ≤ 4n, 2 ≤ i, j ≤
4n, 2 ≤ o, p, q, r ≤ 4, 2 ≤ s, t ≤ n, and 1 ≤ A,B ≤ 4n.
Note that using the endomorphism I and the fact that ∇I = cJ − bK,
we have
ωij(X) = 〈∇̄Xej, ei〉
= 〈I∇Xej , Iei〉
= 〈∇XIej, Iei〉+ c(X) 〈Jej, Iei〉 − b(X) 〈kej , Iei〉
= 〈∇XIej, Iej〉+ c(X) 〈ej, Kei〉+ b(X)〈ej , Jei〉
for any tangent vector X to N0, where ∇̄ denotes the connection on N0.
Hence we conclude that
ωij = ωIiIj + c 〈ej, Kei〉+ b 〈ej, Jei〉, (5.12)
where Ii denotes the index corresponding to Iei = eIi. Similarly, we have
ωij = ωJiJj + c 〈ej, Kei〉+ a 〈ej, Iei〉,
ωij = ωKiKj + b 〈ej, Jei〉+ a 〈ej , Iei〉.
Together with ( 5.7), we conclude that
ω2(4s−1)(e4s) = −1 = −ω2(4s)(e4s−1),
ω2(4s−3)(e4s−2) = −1 = −ω2(4s−2)(e4s−3),
for all 2 ≤ s ≤ n, and
ω2α(eβ) = 0 otherwise.
Similarly,
ω2α(ep) = 〈∇epeα, e2〉
= −〈∇epIeα, e1〉
These identities imply that
ω2(4s−3) = −ω4s−2,
ω2(4s−2) = ω4s−3,
ω2(4s−1) = −ω4s,
ω2(4s) = ω4s−1. (5.13)
A similar calculation using the endomorphisms J and K yield
ω3(4s−3) = −ω4s−1,
ω3(4s−2) = ω4s,
ω3(4s−1) = ω4s−3,
ω3(4s) = −ω4s−2, (5.14)
ω4(4s−3) = −ω4s,
ω4(4s−2) = −ω4s−1,
ω4(4s−1) = ω4s−2,
ω4(4s) = ω4s−3. (5.15)
We claim that the connection forms are given by
η1p = −ηp1
= 2ηp for 2 ≤ p ≤ 4, (5.16)
η1α = −ηα1
= ηα for 5 ≤ α ≤ 4n, (5.17)
ηpq = −ηqp = ωpq, (5.18)
ηpα = −ηαp = e
t ωpα, (5.19)
η(4s)β = −ηβ(4s)
ω(4s)β − (1−e
−2t) η2 if β = 4s− 1
ω(4s)β + (1−e
−2t) η3 if β = 4s− 2
ω(4s)β − (1−e
−2t) η4 if β = 4s− 3
ω(4s)β if β 6= 4s− 1, 4s− 2, or 4s− 3,
(5.20)
η(4s−1)β = −ηβ(4s−1)
ω(4s−1)β + (1−e
−2t) η2 if β = 4s
ω(4s−1)β − (1−e
−2t) η4 if β = 4s− 2
ω(4s−1)β − (1−e
−2t) η3 if β = 4s− 3
ω(4s−1)β if β 6= 4s, 4s− 2, or 4s− 3,
(5.21)
η(4s−2)β = −ηβ(4s−2)
ω(4s−2)β − (1−e
−2t) η3 if β = 4s
ω(4s−2)β + (1−e
−2t) η4 if β = 4s− 1
ω(4s−2)β − (1−e
−2t) η2 if β = 4s− 3
ω(4s−2)β if β 6= 4s, 4s− 1, or 4s− 3,
(5.22)
η(4s−3)β = −ηβ(4s−3)
ω(4s−3)β + (1−e
−2t) η4 if β = 4s
ω(4s−3)β + (1−e
−2t) η3 if β = 4s− 1
ω(4s−3)β + (1−e
−2t) η2 if β = 4s− 2
ω(4s−3)β if β 6= 4s, 4s− 1, or 4s− 2.
(5.23)
Indeed, if we substitute ( 5.16− 5.23) into the first structural equations
dηA = ηA1 ∧ η1 +
ηAq ∧ ηq +
ηAβ ∧ ηβ
we obtain ( 5.9), ( 5.10), and ( 5.11).
To compute the curvature, we consider the second structural equations.
In particular,
dη1p − η1q ∧ ηqp − η1α ∧ ηαp
= 2dηp − 2ηq ∧ ηqp − ηα ∧ ηαp
= −4ηp ∧ η1 + ηα ∧ ηαp
= −4ηp ∧ η1 + e
t ωpα ∧ ηα.
Hence using ( 5.13− 5.15), we have
R1p1p = −4,
R12(4s−1)(4s) =− 2 = −R12(4s)(4s−1),
R12(4s−3)(4s−2) =− 2 = −R12(4s−2)(4s−3),
R13(4s)(4s−2) =− 2 = −R13(4s−2)(4s),
R13(4s−1)(4s−3) =2 = −R13(4s−3)(4s−1),
R14(4s)(4s−3) =2 = −R14(4s−3)(4s),
R14(4s−1)(4s−2) =2 = −R14(4s−2)(4s−1),
R1pAB = 0, otherwise.
Also,
dη1α − η1q ∧ ηqα − η1β ∧ ηβα
= dηα − 2ηq ∧ ηqα − ηβ ∧ ηβα
= −ηα ∧ η1 + e
t ωqα ∧ ηq,
hence
R1α1α = −1,
R1(4s)(4s−1)2 =− 1 = −R1(4s−1)(4s)2,
R1(4s)(4s−2)3 =1 = −R1(4s−2)(4s)3,
R1(4s)(4s−3)4 =− 1 = −R1(4s−3)(4s)4,
R1(4s−1)(4s−3)3 =− 1 = −R1(4s−3)(4s−1)3,
R1(4s−1)(4s−2)4 =− 1 = −R1(4s−2)(4s−1)4,
R1(4s−2)(4s−3)2 =− 1 = −R1(4s−3)(4s−2)2,
R1αAB = 0 otherwise.
Similarly,
dηpq − ηp1 ∧ η1q − ηpr ∧ ηrq − ηpβ ∧ ηβq
= dωpq + 4ηp ∧ ηq − ωpr ∧ ωrq − e
2tωpβ ∧ ωβq
= Ω̄pq + (1− e
2t)ωpβ ∧ ωβq + 4ηp ∧ ηq,
where
Ω̄pq =
R̄pqijωj ∧ ωi
is the curvature form of N0. In particular, this implies that
Rpqro =
−4+e−4t R̄pqpq if r = p and o = q
4+e−4t R̄pqqp if r = q and o = p
e−4t R̄pqro otherwise,
(5.24)
R23(4s)(4s−3) = e
−2t R̄23(4s)(4s−3) − 2(e
−2t − 1),
R23(4s−1)(4s−2) = e
−2t R̄23(4s−1)(4s−2) − 2(e
−2t − 1),
R24(4s)(4s−2) = e
−2t R̄24(4s)(4s−2) − 2(e
−2t − 1),
R24(4s−1)(4s−3) = e
−2t R̄24(4s−1)(4s−3) + 2(e
−2t − 1),
R34(4s)(4s−1) = e
−2t R̄34(4s)(4s−1) − 2(e
−2t − 1),
R34(4s−2)(4s−3) = e
−2t R̄34(4s−2)(4s−3) − 2(e
−2t − 1),
Rpqαβ = e
−2t R̄pqαβ, otherwise.
We now continue with our curvature computation and consider
dηpα − ηp1 ∧ η1α − ηpq ∧ ηqα − ηpβ ∧ ηβα
= d(et ωpα) + 2ηp ∧ ηα − ωpq ∧ e
t ωqα − e
t ωpβ ∧ ηβα
= et η1 ∧ ωpα +
etR̄pαijωj ∧ ωi + 2ηp ∧ ηα + e
t ωpβ ∧ (ωβα − ηβα),
where R̄pαij is the curvature tensor of N0. Using ( 5.13− 5.15) and ( 5.20−
5.23), we have
R2(4s)ABηB ∧ ηA
= η1 ∧ η(4s−1) +
etR̄2(4s)ij ωj ∧ ωi − 2η(4s) ∧ η2 + (1− e
−2t) η(4s) ∧ η2
+ (1− e−2t) η(4s−3) ∧ η3 + (1− e
−2t) η(4s−2) ∧ η4
= η1 ∧ η(4s−1) +
etR̄2(4s)ij ωj ∧ ωi − (1 + e
−2t) η(4s) ∧ η2
+ (1− e−2t) η(4s−3) ∧ η3 + (1− e
−2t) η(4s−2) ∧ η4.
R2(4s−1)ABηB ∧ ηA
= −η1 ∧ η(4s) +
etR̄2(4s−1)ij ωj ∧ ωi − (1 + e
−2t) η(4s−1) ∧ η2
+ (1− e−2t)η(4s−2) ∧ η3 − (1− e
−2t) η(4s−3) ∧ η4.
R2(4s−2)ABηB ∧ ηA
= η1 ∧ η4s−3 +
etR̄2(4s−2)ij ωj ∧ ωi − (1 + e
−2t) η(4s−2) ∧ η2
+ (1− e−2t) η(4s−1) ∧ η3 − (1− e
−2t) η(4s) ∧ η4.
R2(4s−3)ABηB ∧ ηA
= −η1 ∧ η4s−2 +
etR̄2(4s−3)ij ωj ∧ ωi − (1 + e
−2t) η(4s−3) ∧ η2
− (1− e−2t) η(4s) ∧ η3 + (1− e
−2t) η(4s−1) ∧ η4.
Similar formulas for the curvature tensors of the form R3αAB and R4αAB .
Continuing with our computation of the second structural equations using
( 5.13− 5.15), we have
dη(4s−1)(4s) − η(4s−1)1 ∧ η1(4s) − η(4s−1)q ∧ ηq(4s) − η(4s−1)β ∧ ηβ(4s)
= dω(4s−1)(4s) + 2e
−2t η1 ∧ η2 + (1− e
−2t) dη2
+ η(4s−1) ∧ η(4s) − e
2t ω(4s−1)q ∧ ωq(4s)
− (ω(4s−1)(4s−2) − (1− e
−2t) η4) ∧ (ω(4s−2)(4s) − (1− e
−2t) η3)
− (ω(4s−1)(4s−3) − (1− e
−2t) η3) ∧ (ω(4s−3)(4s) + (1− e
−2t) η4)
R̄(4s−1)(4s)ij ωj ∧ ωi + (1− e
2t)ω(4s−1)q ∧ ωq(4s) + 2η1 ∧ η2
+ (1− e−2t)ω2q ∧ ηq + e
t(1− e−2t)ω2β ∧ ηβ − η(4s) ∧ η(4s−1)
+ (1− e−2t)ω(4s−1)(4s−2) ∧ η3 + (1− e
−2t) η4 ∧ ω(4s−2)(4s)
+ (1− e−2t) η3 ∧ ω(4s−3)(4s) − (1− e
−2t)ω(4s−1)(4s−3) ∧ η4
+ 2(1− e−2t)2 η3 ∧ η4
R̄(4s−1)(4s)ij ωj ∧ ωi + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2)
+ 2η1 ∧ η2 + (1− e
−2t)ω2q ∧ ηq + 2(1− e
−2t) η(4r−3) ∧ η(4r−2)
+ 2(1− e−2t) η(4r−1) ∧ η(4r) + (1− e
−2t) (ω(4s−1)(4s−2) − ω(4s−3)(4s)) ∧ η3
− (1− e−2t) (ω(4s−2)(4s) + ω(4s−1)(4s−3)) ∧ η4 + 2(1− e
−2t)2 η3 ∧ η4.
(5.25)
Note that ( 5.22) asserts that
(1− e−2t)ω2q ∧ ηq = (1− e
−2t) (−ω14 ∧ η3 + c ∧ η3 + ω13 ∧ η4 − b ∧ η4)
= (1− e−2t) (4e−2t η3 ∧ η4 + c ∧ η3 − bη4),
(1− e−2t) (ω(4s−1)(4s−2) − ω(4s−3)(4s)) ∧ η3
= −(1− e−2t) c ∧ η3,
−(1− e−2t) (ω(4s−2)(4s) + ω(4s−1)(4s−3)) ∧ η4
= (1− e−2t) b ∧ η4.
Hence substituting into ( 5.25), we obtain
R(4s−1)(4s)AB ηB ∧ ηA
R̄(4s−1)(4s)ij ωj ∧ ωi + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2) + 2η1 ∧ η2
+ 2(1− e−2t) η(4r−3) ∧ η(4r−2) + 2(1− e
−2t) η(4r−1) ∧ η(4r) + 2(1− e
−4t) η3 ∧ η4
= 2η1 ∧ η2 +
R̄(4s−1)(4s)pq e
−4t ηq ∧ ηp + 2(1− e
−4t) η3 ∧ η4 + R̄(4s−1)(4s)pα e
−3t ηα ∧ ηp
R̄(4s−1)(4s)αβ e
−2t ηβ ∧ ηα + (2− e
−2t) η(4s−1) ∧ η(4s) − 2(1− e
−2t) η(4s−3) ∧ η(4s−2)
+ 2(1− e−2t) η(4r−3) ∧ η(4r−2) + 2(1− e
−2t) η(4r−1) ∧ η(4r).
A similar computation yields the curvature tensor of the formR(4s−1)(4s−2)AB ,
R(4s−1)(4s−3)AB , R(4s−2)(4s−3)AB , R(4s−2)(4s)AB , and R(4s−3)(4s)AB . It remains to
compute
R(4s−3)(4r)AB ηB ∧ ηA
= dη(4s−3)(4r) − η(4s−3)1 ∧ η1(4r) − η(4s−3)q ∧ ηq(4r) − η(4s−3)β ∧ ηβ(4r)
= dω(4s−3)(4r) + η(4s−3) ∧ η(4r) − e
2t ω(4s−3)q ∧ ωq(4r) − η(4s−3)β ∧ ηβ(4r)
R̄(4s−3)(4r)ij ωj ∧ ωi + (1− e
2t)ω(4s−3)q ∧ ωq(4r) − (1− e
−2t) (η4 ∧ ω(4s)(4r)
+ η3 ∧ ω(4s−1)(4r))− (1− e
−2t) (η2 ∧ ω4s−2)(4r) + ω(4s−3)(4r−1) ∧ η2
− ω(4s−3)(4r−2) ∧ η3 + ω(4s−3)(4r−3) ∧ η4) + η(4s−3) ∧ η(4r). (5.26)
Using ( 5.12− 5.14), we can write
ω(4s−3)q ∧ ωq(4r) = −η(4s− 2) ∧ η(4r−1) + η(4s−1) ∧ η(4r−2) − η(4s) ∧ η(4r−3)).
Also using ( 5.12) asserts that
ω(4s−3)(4r−1) = ω(4s−2)(4r),
ω(4s−3)(4r−2) = −ω((4s−1)(4r)
ω((4s−3)(4r−3) = ω((4s)(4r).
Hence ( 5.26) becomes
R(4s−3)(4r)AB ηB ∧ ηA
R̄(4s−3)(4r)pq e
−4t ηq ∧ ηp + R̄(4s−3)(4r)pβ e
−3t ηβ ∧ ηp +
R̄(4s−3)(4r)αβ e
−2t ηβ ∧ ηα
− (1− e2t) η(4s−2) ∧ η(4r−1) + (1− e
−2t) η(4s−1) ∧ η(4r−2)
− (1− e−2t) η(4s) ∧ η(4r−3) + η(4s−3) ∧ η4r.
So we have determined all curvature tensors of M . Note that the quater-
nionic curvatures satisfy
K(e1, e2) +K(e1, e3) +K(e1, e4) = −12
K(e2, e1) +K(e2, e3) +K(e2, e4) = −12 + e
−2t (KN(e2, e3) +K
N (e2, e4))
K(e3, e1) +K(e3, e2) +K(e3, e4) = −12 + e
−2t (KN(e3, e2) +K
N (e3, e4))
K(e4, e1) +K(e4, e2) +K(e4, e3) = −12 + e
−2t (K̄(e4, e2) +K
N (e4, e3)).
In particular, this implies that
KN(e2, e3) = K
N (e2, e4) = K
N (e3, e4) = 0.
Also, for 2 ≤ p ≤ 4, we have
K(e1, e(4s−i)) = −4
K(ep, e(4s−i) = −4 + e
−2t (
K̄(ep, e(4s−i))− 4)
implying
KN (ep, e(4s−i)) = 4.
We also have
K(e(4s), e(4s−1)) = −12 + e
−2t (
KN(e(4s), e(4s−i)) + 9)
implying
KN (e(4s), e(4s−i)) = −9.
Lastly,
K(e(4s), e(4r−i)) = −4 + e
KN(e(4s), e(4r−i))
implying
KN (e(4s), e(4r−i)) = 0.
The above computation determined the whole curvature tensor for M
and N0. In particular, if M has bounded curvature, then from the formulas
about the components of curvature tensors of M , all curvature components
are determined as those of QHn. So it must be covered by QHn. �
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Shengli Kong
Department of Mathematics
http://arxiv.org/abs/math/0701865
University of California, Irvine
Irvine, CA92697-3875, USA
email: skong@math.uci.edu
Peter Li
Department of Mathematics
University of California, Irvine
Irvine, CA92697-3875, USA
email:pli@math.uci.edu
Detang Zhou
Departamento de Geometria
Insitituto de Matematica
Universidade Federal Fluminense- UFF
Centro, Niterói, RJ 24020-140, Brazil
email: zhou@impa.br
	Preliminaries on quaternionic Kähler manifolds
	Laplacian comparison theorem
	Quaternionic harmonicity
	Uniqueness of infinite volume end
	Maximal first eigenvalue
