Rigidity of entire self-shrinking solutions to K\"ahler-Ricci flow on complex plane

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by  Wenlong Wang

PROCEEDINGSOFTHE
AMERICANMATHEMATICALSOCIETY
Volume00,Number0,Pages000–000
S0002-9939(XX)0000-0
RIGIDITY OF ENTIRE SELF-SHRINKING SOLUTIONS TO
KA¨HLER-RICCI FLOW ON COMPLEX PLANE
6
1
0 WENLONGWANG
2
(Communicated byLeiNi)
t
c
O
Abstract. We show that every entire self-shrinking solution on C1 to the

4 K¨ahler-Ricciflowmustbegenerated fromaquadraticpotential.
2

]
G
1. Introduction
D
In this short note, we prove the following result.
.
h
t Theorem 1.1. Suppose that u(x) is an entire smooth subharmonic solution on Rn
a
m to the equation
1
[ (1.1) ln∆u= x Du u,

2 · −
2
then u is quadratic.
v
8
For n = 2, up to an additive constant, equation (1.1) is equivalent to the one-
6
dimensional case of the complex Monge-Amp`ere equation
3
0 1
0 (1.2) lndetui¯j = 2x·Du−u
.
2 on Cm. Any entire solution to (1.2) leads to an entire self-shrinking solution
0
6 x
v(x,t)= tu
1 − √ t
: (cid:18) − (cid:19)
v to a parabolic complex Monge-Amp`ere equation
i
X vt =lndet vi¯j
r
a onCm ( ,0), where zi =xi+√ 1xm+i(cid:0). N(cid:1)ote that aboveequation of v is the
× −∞ −
potentialequationoftheK¨ahler-Ricciflow∂tgi¯j =−Ri¯j. Infact,thecorresponding
metric ui¯j is a shrinking K¨aher-Ricci (non-gradient) soliton.
Assuming a certaindecay of ∆u–a specific completeness condition, Q.Ding and
(cid:0) (cid:1)
Y.L. Xin have proved Theorem (1.1) in [2]. Under the condition that the K¨ahler
metric ui¯j iscomplete,rigiditytheoremforequation(1.2)hasbeenobtainedbyG.
Drugan,P.LuandY.Yuanin[3]. Similarrigidityresultsforself-shrinkingsolutions
(cid:0) (cid:1)
toLagrangianmeancurvatureflowsinpseudo-Euclideanspacewereobtainedin[1],
[2], [4] and [5].
2010 Mathematics Subject Classification. Primary53C44,53C24.
Theauthor ispartiallysupportedbyCSC(ChinaScholarshipCouncil).
(cid:13)cXXXXAmericanMathematicalSociety
1
2 WENLONGWANG
Ourcontributionisremovingextraassumptionsfortherigidityofequation(1.1).
As in [3] and [2], the idea of our argument is still to prove the phase–ln∆u is
constant. Then the homogeneity of the self-similar term on the right-hand side
of equation (1.1) leads to the quadratic conclusion. However, it’s hard to con-
struct a barrier function as in [3] or to find a suitable integral factor as in [2]
withoutcompletenessassumption. Takingadvantageofthe conformalityofthelin-
earized equation (1.1), we establish a second order ordinary differential inequality
for M(r)=max ln∆u(x) in the sense of comparison function. Then we prove
|x|=r
that M(r) blows up in finite time by Osgood’s criterion unless ln∆u is constant.
2. proof
Proof. Define the phase by
1
φ(x)= x Du(x) u(x).
2 · −
Taking two derivatives and using (1.1), we have
eφ
(2.1) ∆φ= x Dφ.
2 ·
Define M(r):[0,+ ) R by
∞ →
M(r)= maxφ(x).
|x|=r
Assuming φ(x) is not a constant, we prove that M(r) blows up in finite time.
Since M is locally Lipschitz, it is differentiable a.e. in [0,+ ). For all r > 0,
there exists a corresponding angle θ Sn−1 satisfying ∞
r
∈
(2.2) M(r)=φ(r,θ ).
r
For r′ > 0 small enough, we have M(r+r′) φ(r +r′,θ ) and M(r r′)
r
≥ − ≥
φ(r r′,θ ). It follows that
r
−
M(r+r′) M(r) φ(r+r′,θ ) φ(r,θ )
r r
− −
r′ ≥ r′
and
M(r) M(r r′) φ(r,θ ) φ(r r′,θ )
− − r − − r .
r′ ≤ r′
Letting r′ 0 in above two equations, we have
→
∂φ
limM′ (r) (r,θ ) limM′ (r).
− ≤ ∂r r ≤ +
So if r >0 is a differential point of M, we have
∂φ
(2.3) M′(r)= (r,θ ).
r
∂r
Because of the maximality of φ(r,θr) among θ Sn−1, we have ∆Sn−1φ(r,θr) 0.
∈ ≤
Plugin this inequality into (2.1), we obtain
∂2φ n 1∂φ r ∂φ
(2.4) (r,θ )+ − (r,θ ) exp[φ(r,θ )] (r,θ ).
∂r2 r r ∂r r ≥ 2 r · ∂r r
Fixing a positive R0, for any r [0,R0], θ Sn−1, t [0,1], we have the
∈ ∈ ∈
following Taylor’s expansion
∂φ 1∂2φ
φ(r+t,θ) φ(r,θ) (r,θ ) t+ (r,θ ) t2 Ct3.
− ≥ ∂r r · 2∂r2 r · −
RIGIDITY OF ENTIRE SELF-SHRINKING SOLUTIONS TO KA¨HLER-RICCI FLOW ON COMPLEX PLANE3
HereCisaconstantdependsonlyonR0,infactwecanchooseC =max|x|≤R0+1|D3φ(x)|.
We evaluate above inequality at (r,θ ), where r is a differentiable point and θ
r r
is the corresponding critical angle. Using M(r+t) φ(r+t,θ ), (2.2), (2.3) and
r
≥
(2.4), we have the following inequality that only involves M, namely
1 2(n 1)
M(r+t) M(r) M′(r)t+ exp[M(r)]r − M′(r)t2 Ct3.
− ≥ 4 − r −
(cid:26) (cid:27)
Or equivalently,
(2.5)
1 1 2(n 1)
[M(r+t) M(r) M′(r)t] exp[M(r)]r − M′(r) Ct.
t2 − − ≥ 4 − r −
(cid:26) (cid:27)
Becauseinequality(2.5)holdsforr [0,R0]a.e.,wecanintegrateitwithrespect
∈
to r over any subinterval [a,b] [0,R0], and get the following inequality for every
⊂
t [0,1],
∈
1 b+t a+t
(2.6) M(r)dr M(r)dr [M(b) M(a)]t
t2 (Zb −Za − − )
1 b 2(n 1)
exp[M(r)]r − M′(r)dr C(b a)t.
≥4 − r − −
Za (cid:26) (cid:27)
Choosing differentiable points a and b and letting t 0 in (2.6), we have
→
1 b 2(n 1)
(2.7) M′(b) M′(a) exp[M(r)]r − M′(r)dr.
− ≥ 2 − r
Za (cid:26) (cid:27)
Since R0 can be arbitrarily large, in fact (2.7) holds for all differentiable points
a,b R+.
∈
We claim there exists l0 > 0 such that M′(r) > 0 at every differentiable point
in [l0,+ ). Otherwise, there exist an increasing sequence of differentiable points
rk R∞+, and a sequence of corresponding critical angles θk Sn−1 such that
{ }⊂ { }⊂
∂φ
M′(r )= (r ,θ ) 0, and lim r =+ .
k k k k
∂r ≤ k→∞ ∞
ThenaccordingtoHopf’slemma,weknowφ(x)isconstantinB (0). Sincer can
rk k
be arbitrarily large, φ(x) is in fact a constant on the whole Rn, which contradicts
our assumption.
Sothereexistsacertainl0 >0,suchthatM′(r)>0holdsa.e. in[l0,+ ). Then
∞
M(r)monotonicallyincreaseson[l0,+ ). Whena l1 ,l0+n+2exp[ M(l0)],
∞ ≥ −
we have
b 2(n 1) b
(2.8) exp[M(r)]r − M′(r)dr >2 exp[M(r)]M′(r)dr
− r
Za (cid:26) (cid:27) Za
=2 exp[M(b)] exp[M(a)] .
{ − }
Combining (2.7) and (2.8), we obtain
(2.9) M′(b) M′(a) exp[M(b)] exp[M(a)].
− ≥ −
Above inequality holds for all differentiable points a,b [l1,+ ). Choosing a
∈ ∞
differentiable point l2 l1, then M′(r) M′(l2)>0 holds a.e. in [l2,+ ). Thus
≥ ≥ ∞
M(r) + as r + .
→ ∞ → ∞
4 WENLONGWANG
Then according to Osgood’s criterion, M(r) blows up in finite time, which con-
tradicts the assumption that φ(x) is entire. Hence, we conclude φ(x) is constant.
Using φ(x)= 1x Du(x) u(x), we have
2 · −
1
x D[u(x)+φ(0)]=u(x)+φ(0).
2 ·
Finally, it follows from Euler’s homogeneous function theorem that smooth
u(x)+φ(0) is a homogeneous order 2 polynomial. (cid:3)
Remark 2.1. From the proof, it’s not hard to see that the theorem also holds for
∆u=f(x Du 2u)
· −
if f C1(R) is convex, monotone increasing, and f−1 L1([d,+ )) for a certain
d ∈R. Integrability condition for f−1 is necessary.∈Otherwis∞e, we have such
∈
counterexample: f(x) x and
≡
x1 1 s2 1 x2
u(x)= x21−1 0 s2(exp 2 −1)ds− x1(exp 21 −1)−x1.
Z
(cid:0) (cid:1)
Acknowledgement. I would like to sincerely thank Professor Yu Yuan for sug-
gesting this problem to me and for many heuristic discussions. The result was
obtained when I was visiting the University of Washington. I would also like to
thankCSC(ChinaScholarshipCouncil)foritssupportandtheUniversityofWash-
ington for its hospitality. I am grateful to the referees for useful suggestions.
References
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3. G.Drugan,P.LuandY.Yuan,Rigidityofcompleteentireself-shrinkingsolutionstoK¨ahler-
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School of Mathematical Sciences, Peking University, Science Building in Peking
University,No5. YiheyuanRoad, Beijing,P.R.China 100871
E-mail address: [email protected]