Normalized Ricci flow on nonparabolic surfaces
NORMALIZED RICCI FLOW ON NONPARABOLIC SURFACES
HAO YIN
Abstract. This paper studies normalized Ricci flow on a nonparabolic sur-
face, whose scalar curvature is asymptotically −1 in an integral sense. By a
method initiated by R. Hamilton, the flow is shown to converge to a metric of
constant scalar curvature −1. A relative estimate of Green’s function is proved
as a tool.
1. Introduction
Let (M, g) be a Riemannian manifold of dimension 2. The normalized ricci flow
= (r −R)gij ,
where R is the scalar curvature and r is some constant. For compact surface, r is
the average of scalar curvature. In this case, Hamilton [4] and Chow [2] proved the
normalized Ricci flow from any initial metric will exist for all time and converge
to a metric of constant curvature. It’s therefore nature to ask if such result holds
for non-compact surfaces. Recently, a preprint of Ji and Sesum [14] generalized
the above result to complete surfaces with logarithmic ends. Such surfaces have
infinities like hyperbolic cusps. In particular, they have finite volume, therefore are
parabolic, in the sense that there exists no positive Green’s function. One of their
result shows that the normalized Ricci flow from such a metric will exist for all time
and converge to hyperbolic metric. In this paper, we study nonparabolic complete
surfaces, i.e. surfaces admitting positive Green’s function. In contrast to [14],
such surfaces have at least one nonparabolic end and have infinite volume. For a
discussion of parabolic and nonparabolic ends and their geometric characterization,
see Li’s survey paper [6].
Here we choose r = −1 because if the flow converges, the limit metric will be
of constant curvature r. Since we are considering noncompact surfaces, r can’t
be positive. If r = 0, the limit will be flat R2 or its quotient. However, it’s
well known that these flat surfaces are parabolic. On the other hand, whether a
surface is parabolic or nonparabolic is invariant under quasi-isometries. Since if
the normalized Ricci flow converges, then the limit metric will be quasi-isometric
to the initial one, we know r can’t be zero.(For the definition of quasi-isometry, see
also [6].) If r < 0, we can always assume r = −1 by a scaling.
The main result of this paper is
Theorem 1.1. Let (M, g) be a nonparabolic surface with bounded curvature. If the
infinity is close to a hyperbolic metric in the sense that
|R+ 1| dV < +∞.
http://arxiv.org/abs/0704.0853v2
2 HAO YIN
Then, the normalized Ricci flow will converge to a metric of constant scalar curva-
ture −1.
As in [14], we try to apply the above result to prove results along the line
of Uniformization theorem. That amounts to prove the existence of a complete
hyperbolic metric within a given conformal class of a noncompact surface. In [14],
the authors proved that there is a uniformization theorem for Riemann surfaces
obtained from compact Riemann surface by removing finitely many points and
remarked that similar result should be true for Riemann surfaces obtained from
compact ones by removing finitely many disjoint disks and points. Our theorem
can be used to prove the same result in the case there is at least one disk removed.
In fact, we will give a unified proof, which includes and simplifies the proof of [14].
Precisely, we will show
Corollary 1.2. Let M be a Riemann surface obtained from compact Riemann sur-
face by removing finitely many disjoint disks and/or points. If no disk is removed,
then we further assume that the Euler number of M is less than zero. Then there
exists on M a complete hyperbolic metric compatible with the conformal structure.
The proof of Theorem 1.1 is along the same line as [14]. The method was initiated
by Hamilton in [4]. There, Hamilton considered only compact case. for the purpose
of generalizing this method to complete case, we need to overcome some analytic
difficulties. Precisely, one need to solve Poisson equations and obtain estimates for
the solutions, for all t. Those growth estimates for the solution are needed to apply
the maximum principle. As for the maximum principle, there are many versions of
maximum principle on complete manifolds. Since we will be working on complete
manifold with a changing metric, the closest version for our need is in [1]. We still
need a little modification.
Theorem 1.3. Suppose g(t) is a smooth family of complete metrics defined on M ,
0 ≤ t ≤ T with Ricci curvature bounded from below and
∣ ≤ C on M × [0, T ].
Suppose f(x, t) is a smooth function defined on M × [0, T ] such that
△tf −
whenever f(x, t) > 0 and
exp(−ar2t (o, x))f2+(x, t)dVt < ∞
for some a > 0. If f(x, 0) ≤ 0 for all x ∈ M , then f ≤ 0 on M × [0, T ].
Although there is no detail in [1], one can prove it using the method of Ecker
and Huisken in [3] and Ni and Tam in [12].
To solve the Poisson equation△u = R+1 for t = 0. We use a result of Ni[10], See
Theorem 3.1. That’s the reason why we assume
|R+ 1| dV < +∞. Moreover,
we prove a growth estimate of the solution under the further assumption that Ricci
curvature bounded from blow. This result is true for all dimensions. For the growth
estimate, an estimate of Green’s function is proved under the assumption that Ricci
curvature bounded from below. This estimate may be of independent interests, see
the discussion in Section 2.
Instead of solving △tu(x, t) = R(x, t) + 1 for later t. We solve an evolution
equation for u. Thanks to the recent preprint of Chau, Tam and Yu [1], we can
RICCI FLOW ON SURFACES 3
solve this evolution equation with a changing metric. Following a method in [11],
we show that u, |∇u| and △u satisfy the growth estimate like in equation (1). With
these preperation, we proceed to show that u(x, t) is indeed the potential functions
we need. Now the Theorem 1.1 follows from the approach of Hamilton and repeated
use of Theorem 1.3.
The paper is organized as follows: In Section 2, we prove the crucial estimate of
Green’s function needed for the growth estimate. In Section 3, we solve the Poisson
equation and prove the relevant growth estimates. In the last section, we prove
Theorem 1.1 and discuss results related to Uniformization theorem.
2. An estimate of Green’s function
In this section we prove that
Theorem 2.1. Let (M, g) be a complete noncompact manifold with Ricci curvature
bounded from below by −K. Assume that M admits a positive Green’s function
G(x, y). Let x0 be a fixed point in M . Then there exists constant A > 0 and B > 0,
which may depend on M and x0, so that
{G(x,y)>eAr(y,x0)}
G(x, y)dx ≤ BeAr(y,x0),
where r(y, x0) is the distance from y to x0.
Remark 2.2. It’s impossible to get an estimate of this kind with constant depending
only on K. Considering a family of nonparabolic manifolds Mi, which are becoming
less and less ’nonparabolic’, i.e. their infinities are closing up. For any A,B > 0,
there exists Mi and some xi ∈ Mi such that
{Gi(x,xi)>A}
Gi(x, xi)dx > B.
See [8].
Remark 2.3. To the best of the author’s knowledge, known estimates on Green’s
function in terms of volume of balls require Ricci curvature to be non-negative,
See [9]. There could be one estimate of such type for Ricci curvature bounded from
below, in light of [1]. If so, our relative estimate should be a corollary. The following
proof is a direct one.
We begin with a lemma,
Lemma 2.4. There is a constant C depending only on K and the dimension, such
that if Ricci curvature on B(x, 1) is bounded from below by −K and G(x, y) is the
Dirichlet Green’s function on B(x, 1), then
B(x,1)
G(x, y)dy < C.
Proof. Let H(x, y, t) be the Dirichlet heat kernel of B(x, 1). It’s easy to see
B(x,1)
H(x, y, t)dy ≤ 1,
for all t > 0.
4 HAO YIN
Now we prove that H(x, y, 2) is bounded from above. The proof is Moser itera-
tion, which has appeared several times. Here we follow computations in [17]. Since
we have Dirichlet boundary condition, we don’t need cut off function of space.
Let 0 < τ < 2 and 0 < δ ≤ 1/2 be some positive constants, σk = (1− (1/2)kδ)τ
and ηi be smooth function on [0,∞) such that 1) ηi = 0 on [0, σi], 2) ηi = 1 on
[σi+1,∞) and 3) η′i ≤ 2i+3(δτ)−1. Let pi = (1 + 2n )
i. Since H is a solution to the
heat equation, it’s easy to know Hp is a subsolution to the heat equation for p > 1.
−△y)Hp(x, y, t) ≤ 0.
Multiply by η2iH
pi and integrate
B(x,1)
−△y)Hpidydt ≤ 0.
Routine computation gives
B(x,1)
|∇yHpi |2 dydt+
B(x,1)
H2pi(x, y, T )dy ≤ 2i+3(τδ)−1
B(x,1)
H2pidydt.
The sobolev inequality in [13] implies
B(x,1)
(Hpi)
n−2 dy
≤ CV −2/n
B(x,1)
|∇yHpi |2 +H2pidy,
where V is the volume of B(x, 1). By Hölder inequality,
B(x,1)
H2pi+1dy ≤
B(x,1)
(Hpi)
n−2 dy
B(x,1)
H2pidy)2/n
≤ (CV −2/n
B(x,1)
|∇yHpi |2 +H2pidy)(
B(x,1)
H2pidy)2/n.
By (2), integrate over time
B(x,1)
H2pi+1dydt ≤ CV −2/nci+30 (στ)−(1+2/n)(
B(x,1)
H2pidydt)1+
where c0 = 2
1+2/n. A standard Moser iteration gives
t∈[τ,2]
y∈B(x,1)
H2(x, y, t) ≤ CV −1(στ)−
(1−δ)τ
B(x,1)
H2(x, y, t)dydt.
An iteration process as given in [7] implies the L1 mean value inequality. In par-
ticular,
y∈B(x,1)
H(x, y, 2) ≤ CV −1
B(x,1)
H(x, y, t)dydt ≤ CV −1.
Hence,
B(x,1)
H2(x, y, 2)dy ≤ CV −1.
RICCI FLOW ON SURFACES 5
Due to a Poincaré inequality in [7],
B(x,1)
H2(x, y, t)dy =
B(x,1)
2H△yHdy
B(x,1)
|∇yH |2 dy
B(x,1)
H2(x, y, t)dy.
This differential inequality implies
B(x,1)
H2(x, y, t)dy ≤
B(x,1)
H2(x, y, 2)dy × e−C(t−2)
≤ CV −1e−C(t−2).
Hölder inequality shows
B(x,1)
H(x, y, t) ≤ V (B(x, 1))
B(x,1)
H2(x, y, t)dy ≤ Ce−C(t−2),
for t ≥ 2. The lemma follows from
B(x,1)
G(x, y)dy =
B(x,1)
H(x, y, t)dydt.
Now let’s turn to the proof of Theorem 2.1.
Proof. The key tool in the proof is Gradient estimate for harmonic function. Recall
that if u is a positive harmonic function on B(x, 2R), then
B(x,R)
|∇ log u(x)|2 ≤ C1K + C2R−2
This is to say outside B(x, 0.1), the Green function as a function of y decays or
increases at most exponentially with a factor
C1K + 100C2.
(1) Consider G(x0, y), Set
p = max
y∈∂B(x0,1)
G(x0, y).
As pointed out in Li and Tam, in the paper constructing Green function, G(x0, y) ≤
p for y /∈ B(x0, 1). Since the Green function is symmetric, for any point y far out
in the infinity, G(y, x0) ≤ p.
(2) If the theorem is not true, then for any big A and B, there is a point y (far
away) so that
{G(x,y)>eAr(y,x0)}
G(x, y) > BeAr(y,x0).
We will derive a contradiction with (1).
Claim: {x|G(x, y) > eAr} ⊂ B(y, 1) is not true.
If true, then consider the Dirichlet Green function G1(z, y) on B(y, 1). It’s well
known that G(z, y) − G1(z, y) is a harmonic function. Notice that this harmonic
function has boundary value less than eAr. Therefore, its integration on B(y, 1) is
less than eAr × V ol(B(y, 1)). Since we assume Ricci lower bound, V ol(B(y, 1)) is
less than a universal constant depending on K.
6 HAO YIN
Therefore,
{G(x,y)>eAr}
G(x, y)dx ≤
B(y,1)
G(x, y)dx
≤ V ol(B(y, 1))× eAr +
B(y,1)
G1(x, y)dx
≤ C(K,n)× eAr,
where we used Lemma 2.4 for the last inequality. If we choose B to be any number
larger than C(K,n) in the above equation, then the choice of y gives an contradic-
tion and implies that the claim is true.
(3) There is a z ∈ {x|G(x, y) > eAr} so that d(z, y) = 1 because the set
{G(x, y) > eAr} is connected. This follows from the maximum principle and the
construction of Green’s function.
(3.1) If |d(y, x0)− d(z, x0)| < 0.3, then
Let σ be the minimal geodesic connecting z and x0.
Claim: the nearest distance from y to σ is no less than 0.1.
If not, let w be the point in σ such that d(y, w) < 0.1. Since d(y, z) > 1, we
d(w, z) > 0.9
Now, w is on the minimal geodesic from z to x0, so
d(w, x0) ≤ d(z, x0)− 0.9
d(y, x0) < d(w, x0) + d(y, w) < d(z, x0)− 0.8
This is a contradiction , so the claim is true.
We can use the gradient estimate along the segment σ. (Notice that d(z, x0) <
r(x, x0) + 1)
G(y, x0) >
G(y, z)
C1K + 100C2(r + 1))
This is a contradiction if we choose A >>
C1K + 100C2.
(3.2) If d(z, x0) ≤ d(y, x0)− 0.3, then
The distance from y to the minimal geodesic connecting z and x0 will be larger
than 0.1. The above argument gives a contradiction.
(3.3) If d(z, x0) ≥ d(y, x0) + 0.3, then
Since G(z, y) > eAr, we move the center to z, by symmetry of Green function.
G(y, z) > eAr(y,x0)) > eA
′r(z,x0). This is case (3.2). We get a contradiction at z.
This finishes the proof of estimate of Green function. �
3. Poisson equations △u = R+ 1
This section is divided into two parts. The first part solves the Poisson equation
for t = 0. The second part solves for t > 0 before the maximum time using an
indirect way.
First, we use Theorem 2.1 to obtain an growth estimate of the solution of the
Poisson equation △u = R + 1 for t = 0. The existence part without curvature
restriction and boundedness of f of the following theorem is due to Lei Ni in [10].
RICCI FLOW ON SURFACES 7
Theorem 3.1. Let M be a complete nonparabolic manifold with Ricci curvature
bounded from below by −K. For non-negative bounded continuous function f the
Poisson equation
△u = −f
has a non-negative solution u ∈ W 2,nloc (M) ∩ C
loc (M)(0 < α < 1) if f ∈ L1(M).
Moreover, for any fixed x0 ∈ M , there exists A > 0 and C > 0 such that
u(x) ≤ CeAr(x,x0).
Proof. Let G(x, y) be the positive Green’s function.
G(x, y)f(y)dy =
{G(x,y)≤eAr(x,x0)}
G(x, y)f(y)dy
{G(x,y)>eAr(x,x0)}
G(x, y)f(y)dy
≤ CeAr(x,x0).
For the first term, we use the assumption that f is integrable, for the second term,
we use the boundedness of f and the Theorem 2.1. The estimate above shows the
Poisson equation is solvable with the required estimate. �
Corollary 3.2. Let M be a surface satisfying the assumptions in Theorem 1.1.
There exists a solution u0 to the equation △u0 = R(x) + 1 satisfying
exp(−ar2(x, x0))u20(x)dV < ∞
exp(−br2(x, x0)) |∇u0|2 (x)dV < ∞
where a and b are some positive constants.
Proof. Solve the Poisson equation for the positive part and the negative part of
R+ 1 respectively. Then subtract the solutions. The first integral estimate follows
from the pointwise growth estimate and volume comparison.
Let R > 1. Choose a cut-off function ϕ such that
ϕ(x) =
1 x ∈ B(x0, R)
0 x /∈ B(x0, 2R)
|∇ϕ|2 ≤ C1ϕ.
Multiply the equation by ϕu0 and integrate over M ,
ϕu0△u0dV =
(R + 1)ϕu0dV,
which implies
ϕ |∇u0|2 dV +
u0∇ϕ · ∇u0dV = −
(R+ 1)ϕu0dV.
8 HAO YIN
Hence
(ϕ− |∇ϕ|
) |∇u0|2 dV ≤ C
B(x0,2R)
u20dV + C
B(x0,2R)
|u0| dV.
B(x0,2R)
u20dV + CV ol(B(x0, 2R)).
From the integration estimate of u0,
B(x0,2R)
u20dV ≤ Ce4aR
By choice of ϕ,
B(x0,R)
|∇u0|2 dV ≤ CeãR
From here, it’s not difficult to see the estimate we need. �
Now let’s look at the case of t > 0. In fact, it’s not difficult to show the above
method can be used for t > 0. This amounts to show that M is still nonparabolic
for t > 0 and
|R+ 1| dV is still finite. The first claim is trivial and the second
follows from the evolution equation and maximum principle. Assume the solutions
are u(t). We have trouble in deriving the evolution equation for u(t), due to the
possible existence of nontrivial harmonic functions. This explains why we use the
following indirect way.
Lemma 3.3. Assume the normalized Ricci flow exists for t ∈ [0, Tmax). The
following equation has a solution u(x, t) (0 ≤ t < Tmax) with initial value u0,
= △u− u,
where △ is the Laplace operator of metric g(t). Moreover, there exists a > 0
depending on T such that for any T < Tmax
exp(−ar2(x, x0))u2(x, t)dVt < ∞.
Similar estimates hold for |∇u| and △u with different constants.
Remark 3.4. Since g(0) and g(t) are equivalent up to a constant depending on T ,
it doesn’t matter whether we estimate ∇u or ∇tu and whether we use r to stand
for distance at g(0) or g(t) if t ∈ [0, T ].
Proof. In [1], the authors considered a class of evolution equation with changing
metric. ∂u
= △u− u with the underling metric evolving by normalized Ricci flow
is in this class. They proved, among other things, that the fundamental solution
Z(x, t; y, s) has a Gaussian upper bound, i.e
Z(x, t; y, x) ≤ C
t− s)
r2(x,y)
D(t−s) .
These constants depends on the solution of normalized Ricci flow and T . See
Corollary 5.2 in [1]. For simplicity, denote Z(x, t; y, 0) by H(x, y, t), then to solve
the equation, it suffices to show the following integral converges,
u(x, t) =
H(x, y, t)u0(y)dy.
RICCI FLOW ON SURFACES 9
Bt(x,1)
H(x, y, t)u0(y)dy
≤ CeAr(x,x0),
because the integral of H on Bt(x, 1) is less than 1 and u0 grows at most exponen-
tially by Theorem 2.1.
M\Bt(x,1)
H(x, y, t)u0(y)dy
M\Bt(x,1)
r2(x,y)
Dt |u0(y)| dy.
By volume comparison,
Vx(1) ≥ C1e−A1r(x,x0)Vx0(1)
t) ≥ C2e−A1r(x,x0) min(1, tn/2).
Therefore
M\Bt(x,1)
H(x, y, t)u0(y)dy
M\Bt(x,1)
CeA1r(x,x0)e−
r2(x,y)
2DT |u0(y)| dy
M\Bt(x,1)
CeA2r(x,x0)eAr(x,y)e−
r2(x,y)
2DT dy
≤ CeA2r(x,x0).
In summary,
|u(x, t)| ≤ CeAr(x,x0),
where A means a different constant. Volume comparison then implies
exp(−ar2(x, x0))u2(x, t)dVt < ∞.
For estimates on derivatives, note first that etu(x, t) is a solution of heat equation
(with evolving metric)
with initial value u0. Since we allow constants depend on T , it’s equivalent to prove
estimates for etu(x, t). Therefore, from now on, to the end of this proof, we assume
u(x, t) is a solution of heat equation. Then
(2) (△− ∂
)u2 = 2 |∇u| .
Assume that ϕ : R+ → R+ satisfies
1) ϕ(x) = 1 for x ≤ 1;
2) ϕ(x) = 0 for x ≥ 2.
Choose the cut-off function ϕ(
r(x,x0)
)(R > 1). Multiplying this to the equation
(2) and integrate,
r(x, x0)
) |∇u|2 dVtdt ≤
r(x, x0)
)u2dVtdt.
△ϕ( r
) = div(ϕ′(
= ϕ′′(
|∇r|2
+ ϕ′(
10 HAO YIN
By definition of ϕ, we know ϕ( r
) vanishes unless R ≤ r(x, x0) ≤ 2R. Laplacian
comparison implies (curvature is bounded from below −k)
△r ≤ (n− 1)
kcoth(
kr) ≤ C.
Therefore,
ϕ△u2dVt ≤ C
B(x0,2R)
u2dVt.
Let dVt = e
FdV0,
(ϕu2)eF dV0dt ≥
(ϕu2eF )dtdV0 −
Cϕu2dVtdt
ϕu2(x, T )dVT −
ϕu2(x, 0)dV0 − C
ϕu2dVtdt
ϕu20(x)dV0 − C
ϕu2dVtdt.
Here we have used the fact that ∂e
is bounded. Combined with equation (3) and
B(x0,R)
|∇u|2 dVtdt ≤ C
B(x0,2R)
u2dVtdt+
B(x0,2R)
u20(x)dV0.
From here it’s easy to see the type of estimate in Theorem 1.3. For △u, it suffices
to consider
∣. The Bochner formula in this case is (remember we have assumed
that u is a solution of the heat equation),
(△− ∂
) |∇u|2 = 2
2 − |∇u|2 .
The same argument as before works for
Lemma 3.5. For t ∈ [0, Tmax),
△tu(x, t) = R(x, t) + 1.
Proof. We know for t = 0 it’s true. Calculation shows
(△tu−R(t)− 1) = (R+ 1)△tu+△t(△tu− u)−△tR−R(R+ 1)
= △t(△tu−R(t)− 1) +R(△tu−R− 1)
By previous lemma, we have growth estimate for △tu−R(t)−1. If △tu−R−1 ≥ 0,
−△t)(△tu−R(t)− 1) ≤ C(△tu−R− 1).
If △tu−R− 1 ≤ 0, then
−△t)(△tu−R(t)− 1) ≥ C(△tu−R− 1).
Apply maximum principle for △tu − R − 1, which is zero at t = 0. We know it’s
zero forever. �
RICCI FLOW ON SURFACES 11
4. Proof of the main theorem and the corollary
Assume we have a surface satisfying the assumptions of Theorem 1.1. Short
time existence is known, see [15]. The long time existence and convergence follows
exactly by an argument of Hamilton in [4]. For completeness, we outline the steps.
Solve Poisson equations△tu(x, t) = R(x, t)+1 as we did. Consider the evolution
equation for H = R+ 1 + |∇u|2,
H = △H − 2 |M |2 −H,
where M = ∇∇u − 1
△f · g. Since we have growth estimate for H , maximum
principle says
R+ 1 ≤ H ≤ Ce−t.
Therefore, after some time R will be negative everywhere. Applying maximum
principle again to the evolution equation of scalar curvature
R = △R+ R(R+ 1)
will prove Theorem 1.1.
Next, we discuss the application of the above theorem to Uniformization theorem.
Let S be a compact Riemann surface. Let p1, · · · , pk be k different points in S
and D1, · · · , Dl be l domains on S such that all of them are disjoint and Di is
diffeomorphic to disk. Denote S \ ∪iDi \ {p1, · · · , pk} by M . The aim is to show
there exists a complete hyperbolic metric on M compatible with the conformal
structure.
The approach is to construct an initial metric g0 on M compatible with the
conformal structure so that the normalized Ricci flow starting from g0 will converge
to a hyperbolic metric. Assume there is metric h in the given conformal class of S.
Note that h is incomplete as a metric on M .
For pi, there is an isothermal coordinate (x, y) around pi. By a conformal change
of h, one can ask g0 to be
(x2 + y2) log2(x2 + y2)
(dx2 + dy2)
in a small neighborhood Ui of pi.
Remark 4.1. This is called hyperbolic cusp metric in [14] and it has scalar curva-
ture −1.
For Dj, let r be the distence to ∂Dj on M with respect to h. Let Vj be a
neighborhood of ∂Dj in M . Let (r, θ) be the Fermi coordinates for ∂Dj so that
h0 = dr
2 +A(r, θ)dθ2.
We will find ρ = ρ(r, θ) such that
1) ρ = 0 on ∂Dj;
2) dρ 6= 0 on ∂Dj;
12 HAO YIN
is asymptoticly hyperbolic in high order. Let K and K0 be the Gaussian curvature
of h and g0 respectively. We have the formula,
K0 = ρ
2(△h log ρ+K).
In order that K0 = −1,
1− |∇ρ|2 + ρ△ρ+ ρ2K = 0.
In terms of r and θ,
|∇ρ|2 = (∂ρ
)2 +A−1(r, θ)(
△ρ = ∂
Here A, B, C and D are smooth functions of r and θ. The equation now becomes
(5) ρ
+ 1− (
)2 −A−1(
)2 + ρ2K = 0.
If equation (5) is true at r = 0, then
(r, θ) = 1.
Here we used that fact that ρ > 0.
Set η(r, θ) = ρ
. Equation (6) implies η(0, θ) = 1. Equation (5) becomes
+Brη +Br2 ∂η
+ Cr2 ∂η
+Dr2 ∂
+ 1−η
)2 −A−1 r
)2 + ηr2K = 0
For the convinience of formal calculation, this equation is rewritten as
(7) (D2 −D − 2)η + F [r, η] = 0,
where D = r ∂
F [r, η] = Brη +Br2
+ Cr2
(1− η)2
)2 −A−1 r
)2 + ηr2K.
Equation (7) is a very typical form of Fuchsian type PDE. Formal solutions of
this kind of equation has been discussed many times. For example, Kichenassamy
[5] and Yin [16]. We will only outline the main steps here, for details see [5] and
[16].
Consider formal solution with the following expansion,
(8) η(r, θ) = 1 +
aij(θ)r
i(log r)j .
We will call the sum
j=0 aijr
i(log r)j the i-level of the expansion. Note that D
maps i-level to i-level. Details on formal calculation could be find in [5] and [16].
A common feature of all terms in F [r, η], which is crutial in obtaining a formal
solution, is that the k-level of F [r, η] could be calculated with knowledge of only
l-level of η with l < k. For example, consider (1 − η)2/η. It’s the multiplication
of three formal series, two 1 − η and 1/η. In order the k-level of η appears in the
RICCI FLOW ON SURFACES 13
k-level of (1 − η)2/η, the only possibility is that two of the three series contribute
zero level and one k-level. However, the zero level of 1− η vanishes.
The only thing we need is that there exists a formal solution and furthermore
due to Borel’s Lemma as in [16], there is an approximate solution so that
(D2 −D − 2)η + F [r, η] = o(rk)
for any k. In terms of ρ,
(9) K0 + 1 = 1 + ρ
2(△h log ρ+K) = o(ρk)
for any k. This metric g0 near ∂Dj has Gaussian curvature -1 asymptotically. By
a scaling, we assume it has scalar curvature -1 asymptotically.
We construct g0 by doing the above to every point Pi and disk Dj. If there is
at least one disk removed, we know M is nonparabolic.
|R+ 1| dV
is finite because of equation (9). Therefore, Theorem 1.1 proves the Uniformization
in this case.
If there is no disk removed, i.e. M = S \ {p1, · · · , pk} and M has negative
Euler number, then it’s proved in [14] that there exists a hyperbolic metric in the
conformal class. A large part of [14] is devoted to solve
(10) △g0u = Rg0 + 1
with |∇u| < ∞.
Observe that the above equation is equivalent to
(11) △hu =
(Rg0 + 1).
Since every end of (M, g0) is a hyperbolic cusp, Gauss-Bonnet theorem says
Rg0dV0 = 2πχ(M) < 0.
There exists a function f of compact support on M such that the volume of
(M, efg0) is −2πχ(M), because (M, g0) has finite volume. Denote efg0 by g0,
since the infinity is not changed, equation (12) is still true. Now, the volume of
(M, g0) is −2πχ(M). This implies
(Rg0 + 1)dV0 = 0.
Therefore
(Rg0 + 1)dVh = 0.
By construction of g0, we know
(Rg0+1) is zero near Pi. So
(Rg0+1) is a smooth
function on S. Therefore, equation (11) is solvable. Since u is a smooth function
on compact surface S, u has bounded gradient with respect to h. The relation of
h and g0 near Pi is explicit. It’s straight forward to check u has bounded gradient
as a function of (M, g0). This simplifies the proof in [14].
14 HAO YIN
Remark 4.2. In the case that there is at least one disk removed, by construction
of g0, Rg0 +1 vanishes at high order near ∂Dj. Then one can extend the definition
(Rg0 + 1) to S so that
(Rg0 + 1)dVh = 0.
The rest is the same as in the previous case.
This method of solving Poisson equation depends on the conformal structure of
M , therefore Theorem 3.1 and Theorem 1.1 are not coverd by the above discussion.
References
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	1. Introduction
	2. An estimate of Green's function
	3. Poisson equations u=R+1
	4. Proof of the main theorem and the corollary
	References
