Table Of ContentCURVATURE TENSOR UNDER THE RICCI-HARMONIC FLOW
ANQIANGZHU,LIANGCHENG1
1
1
0 Abstract. Weprovethat ifthe Riccicurvatureisuniformlybounded under
2 theRicci-Harmonicflowforalltimest∈[0,T),thenthecurvaturetensorhas
tobeuniformlyboundedaswell.
n
a
J
6
1. introduction
]
G The Ricci flow has been a useful tool in the study of geometry problem. It
D played an important part in the proof of the Poincare Conjecture and Geometric
Conjecture. Since then, geometric flows attract more attention. In [5], B.List
.
h studied an extended Ricci flow
t
a ∂ g =−2R +αφ φ , t∈[0,T);
m (cid:26) ∂∂tφij=∆φ, ij i j t∈[0,T).
∂t
[
In his dissertation,B.List said this flow has some applications in generalrelativity.
1 The nonlocal collapse and monotonicity of some entropy have been proved. Some
v gradient estimate and local existence of the flow are obtained. As long as the
4
Riemannian curvature tensor is bounded, the flow will exists and the gradient and
4
Hessionof φ are also bounded, [5]. R.Muller also introduceda new geometric flow,
1
1 called Ricci-Harmonic flow,
.
1 ∂ g =−2R +αφ φ , t∈[0,T);
∂t ij ij i j
0 (cid:26) ∂ φ=τ φ, t∈[0,T).
1 ∂t g(t)
1 where φ : (M,g(t)) → (N,h) is a map between two Riemannian manifold, τ φ =
g
: trace∇dφ. The extended Ricci flow of B.List is a special case of Ricci-Harmonic
v
i flow. Similar to the extended Ricci flow, this flow will also exist if the Riemannian
X
curvature tensor is bounded.
r On the other hand, there are many results about controlling the Riemannian
a
curvature tensorunder Ricci flow. In [9], by a blow up argument,N.Susem showed
that Ricci curvature uniformly bounded on M ×[0,T), where T < ∞ is enough
to control the norm of Riemannian curvature tensor on closed manifold. Then
L.MaandL.Cheng improveedthe resultto the Ricciflowoncomplete noncompact
manifold. R.Ye[12]andB.Wang[13],bydifferentarguments,provedthatthenorm
ofthe Riemanniancurvaturetensorcanbe controlledby the bound ofkRmk =
n+2
2
( T |Rm|n+22dµdt)n+22. In this paper, we will prove that if the Ricci curvature
0 M
isRunRiformly bounded, the flow can continue.
The following are the main results of this paper.
1991 Mathematics Subject Classification. 35K15,35K55,53A04.
Key words and phrases. extended Ricciflow,Harnackinequality,conjugateheatequation.
1=correspondingauthor;TheresearchispartiallysupportedbytheNationalNaturalScience
FoundationofChina10631020 andSRFDP20090002110019.
1
2 ANQIANGZHU,LIANGCHENG1
Theorem 1. Let (M,g(t),φ(t)),t ∈ [0,T) be a solution to the Ricci Harmonic
flow on a Riemannian manifold. Suppose T <∞ and Ricci curvature is uniformly
bounded under the flow. Then the Riemannian curvature tensor |Rm| stays uno-
formly bounded under the flow.
In the following, we will prove 1, Since the Extended Ricci flow is a special case
of the Ricci Harmonic flow. Note that the proof doesn’t depend on the compact of
the manifold. We just take a closed manifold for simple.
If the manifold is compact, we also have the following result.
Theorem 2. Let (M,g(t),φ(t)),t ∈ [0,T), where T < ∞, be a solution to the
Ricci-Harmonic flow on a closed manifold satisfying
T
n+2
kRkn+2 =( |R| 2 dµdt)<∞,
2 Z0 ZM
and
T
n+2
kWkn+2 =( |W| 2 dµdt)<∞,
2 Z0 ZM
Then sup kRmk<∞.
M×[0,T)
In the following, we denote S =R −αφ φ , S =R−α|∇φ|2.
ij ij i j
2. Preliminary
Lemma 3. Under the same condition of the theorem 1, we have
|∇φ|(t)<C,t∈[0,T).
where C is independent of t.
Proof. Since Ric curvature is uniformly bounded along the flow at t ∈ [0,T), the
Scale curvature is also uniformly bounded. From [4], we have
(∂ −∆)S =2|S |2+2α|τ φ(t)|2.
t ij g
So the minimum of S(t) is nondecreasing along the flow.
S(t)=R−α|∇φ|2 ≥C.
Since R is uniformly bounded, |∇φ(t)|2 ≤C, where C is independent of t.
(cid:3)
Since Ric curvature and |∇φ(t)|2 is uniformly bouhnded, we will control the
variation of the distance along the Extended Ricci flow.
Lemma 4. [5],[14] Under the same condition of theorem 1. For all δ > 0, there
exists an η >0, such that if |t−t |<η,t,t ∈[0,T), then
0 0
′ ′ ′
|d (q,q )−d (q,q )|≤δd (q,q )
g(t) g(t0) g(t0)
′
for all q,q ∈M.
CURVATURE TENSOR UNDER THE RICCI-HARMONIC FLOW 3
Proof.
t ∂
|logg(t)(V,V)−logg(t )(V,V)| = | logg(s)(V,V)ds|
0 Z ∂t
t0
t (−2Ric(V,V)+2α<∇φ,V >2)
= | ds|
Z |V|2
t0 g(s)
t 2C|V|2 +2α|∇φ|2 |V|2
≤ gs g(s) g(s)ds
Z |V|2
t0 gs
≤ C(t−t )
0
So we have
e−C|t−t0||V|2 ≤|V|2 ≤eC|t−t0||V|2
g(t0) g(t) g(t0)
′
Suppose γ is a minimal geodesic from p to p with respect to metric g(t ).
1 0
d (p,p′)≤ |γ˙ | (s)ds≤ eC|t−t0|/2|γ˙ | (s)ds=eC|t−t0|/2d (p,p′).
g(t) Z 1 g(t) Z 1 g(t0) g(t0)
′
Similarly, Let γ be a minimal geodesic from p to p with respect to metric g(t).
2
Then
d (p,p′)= |γ˙ | (s)ds≥ e−C|t−t0|/2|γ˙ | (s)ds≥e−C|t−t0|/2d (p,p′).
g(t) Z 2 g(t) Z 2 g(t0) g(t0)
|d (p,p′)−d (p,p′)|≤(eC|t−t0|/2−1)d (p,p′).
g(t) g(t0) gt0
The lemma will be satisfied if we take η = 2ln(δ+1). (cid:3)
C
Corollary 5. Under the condition of theorem 1. For all ρ>0,
B (0,r(t)ρ)⊂B (0,ρ),
g(t) g(0)
B (0,r(t)ρ)⊂B (0,ρ),
g(0) g(t)
where r(t)=e−Ct/2.
Proof. Supposex∈B (o,r(t)ρ). Thenr(t)ρ≥d (o,x)≥e−Ct/2d (o,x). So
g(t) g(t) g(0)
d (o,x)≤ρ. Similarly for B (0,r(t)ρ)⊂B (0,ρ). (cid:3)
g(0) g(0) g(t)
Lemma 6. Under the same condition of theorem 1. For all ǫ>0, there exists an
δ(ǫ,C)>0, such that if t∈[t ,t +δ] then
0 0
r
Volg(t)Bg(t)(x,r)≥(1−ǫ)n2Volg(t0)Bg(t0)(x,1+ǫ).
Proof. By the corollary 5, for all ǫ > 0, there exists δ > 0, such that if t ∈
1
[t ,t +δ ], we have
0 0 1
r
B (x, )⊂B (x,r).
g(t0) 1+ǫ g(t)
Since the Ricci curvature is uniformly bounded in [0,T), by lemma 3, we have
|S|<C is uniformly bounded.
d
dv = −Sdv
dtZBg(t0)(x,1+rǫ) ZBg(t0)(x,1+rǫ)
i.e.
d r r
| Vol B (x, )|<CVol B (x, ).
dt g(t) g(t0) 1+ǫ g(t) g(t0) 1+ǫ
(cid:3)
4 ANQIANGZHU,LIANGCHENG1
We have
r r
Vol B (x, )≥e−C(t−t0)Vol B (x, )
g(t) g(t0) 1+ǫ g(t0) g(t0) 1+ǫ
There exists a δ2 >0, such that e−C(t−t0) ≥(1−ǫ)n2, if t∈[t0,t0+δ2].
Take δ =min{δ ,δ }, for t∈[t ,t +δ]
1 2 0 0
r
Volg(t)Bg(t)(x,r)≥(1−ǫ)n2Volg(t0)Bg(t0)(x,1+ǫ).
Lemma 7. Let {(Mn,g (t),φ (t),x )}∞ ,t ∈ [0,T] be a sequence of the Ricci-
i i i i i=1
Harmonic flow on complete manifolds such that sup |Rm(g (t))| ≤ 1,
Mi×[0,T] i gi(t)
sup |φ (x,0)| < C where C is independent of i. Let ψ = exp be the
x∈M i i xi,gi(0)
exponential map with respect to metric g (0) and B(o ,π) ⊂ T M equipped with
. i i 2 xi i
metric g˜(t) = ψ∗g (t),φ˜ = φ ◦ψ . Then (B(o ,π),g˜(t),φ˜(t),o ) subconverges
i i i i i i i 2 i i i
to a Ricci-Harmonic flow (B(o, π),g˜(t),φ˜,o) in C∞ sense, where B(o,π) ⊂ Rn
2 2
equipped with metric g˜(t).
.
Proof. Since g˜(t) = ψ∗g (t), we have sup |Rm(g˜(t))| ≤ 1. On
i i i B(oi,π2)×[0,T] i g˜i(t)
B(o ,π), we have inj(o ,g˜(0)) ≥ π [7]. Then the lemma follows from the com-
i 2 i i 2
pactness theorem in [14],[5]. (cid:3)
3. Blow up analysis
Proof. By the Maximum principle, we have
inf φ(x,0)≤φ(x,t)≤ sup φ(x,0)
x∈M x∈M
fort∈[0,T). Inthefollowing,weconsiderthepullbackmetricsontangentspaces.
Suppose the Ricci-Harmonic flow blows up at a finite time T. That means there
exist sequences t →T and p ∈M such that
i i
Q =|Rm|(p ,t )≥C−1 max |Rm|(x,t).
i i i
M×[0,ti]
whereC >1andQ →∞asi→∞. Bylemma7,thescaledflow(B(o , π ),g (t),φ (t))
i i 2C i i
willconvergetoacompleteancientsolution(B(o, π ),g(t),φ)oftheRicci-Harmonic
2C
flow, where g (t)=Q g(t + t ),φ (t)=φ(t + t ). Since S(t)=R(t)−α|∇φ(t)|2
i i i Qi i i Qi
is uniformly bounded, S (t)= 1 (R(g(t + t ))−α|∇φ|2(g(t + t ))). As i→∞,
i Qi i Qi i Qi
S (t) → 0. Our ancient solution (B(o, π ),g(t),φ) has S = 0. The evolution of S
i 2C
is
(∂ −∆)S =2|S |2+2α|∆φ|2.
t ij
This imply that |S |=0,|∆φ|=0. i.e. Ric=α∇φ ∇φ≥0.
ij
In the following, we will analyze the volume of thNe limit (B(o, π ),g(t),φ).
2C
Take r >0, since the convergence is C∞, the volume is convergence,
VolB(p,r) Vol B (p ,r)
i i i
= lim
rn i→∞ rn
wherethevolumeandtheballB(p,r)ontheLHSareconsideredinmetricg,while
the RHS are considered in metric g (0)=Q g(t ). So
i i i
VolB(p,r) = lim Volg(ti)Bg(ti)(pi,rQi−1/2)
rn i→∞ (Q−1/2r)n
i
CURVATURE TENSOR UNDER THE RICCI-HARMONIC FLOW 5
For ∀ǫ>0, there exists N >0, such that if i>N, we have
VolB(p,r) ≥ Volg(ti)Bg(ti)(pi,rQi−1/2) − ǫ.
rn (Q−1/2r)n 2
i
For such ǫ > 0, there exists a δ > 0 from lemma 6. We take t > T −δ. For i
0
sufficiently large, 0<t −t <δ,
i 0
VolB(p,r) ≥(1−ǫ)n/2Volg(t0)Bg(t0)(pi,1+1ǫrQi−1/2) − ǫ.
rn (Q−1/2r)n 2
i
Note that volume has the expansion within the injective radius of p(see [2] the-
orem 3.98),
R(p)
VolB(p,r)=ω rn(1− r2+o(r2)).
n 6(n+2)
Let i→∞, we have
VolB(p,r) (1−ǫ)n/2 ǫ
≥ω − .
rn n (1+ǫ)n 2
Let ǫ→0, we have
VolB(p,r)
≥ω .
rn n
Since Ric≥0, by the Bishop-Gromov volume comparison, we have (N,g) is a flat
manifold, which is contradict to the |Rm|(p,0)=1.
(cid:3)
Proof. From Theorem 8.6 in [5], since M is compact, the limit solution is in fact a
ancient solution to Ricci flow. The proof is the similar to the proof in [8]. (cid:3)
Remark 8. TheblowupcanalsobeappliedtothecompletenoncompactRiemann-
ian manifold as in [8].
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6 ANQIANGZHU,LIANGCHENG1
Anqiang Zhu, Department of Mathematical Sciences, WuHan University, WuHan
430072,P. R.China
E-mail address: [email protected]
Liang Cheng, Department of mathematics, Huazhong Normal University, WuHan
430079,P. R.China
E-mail address: [email protected]
