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# Trieste Summer School june

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Niveau: Supérieur, Doctorat, Bac+8
Trieste Summer School, june 2005 Ricci flow with surgery Laurent Bessieres and Gerard Besson Institut Fourier, 100, rue des maths, 38402 Saint Martin d'Heres,France. , These notes provides some details on the lectures 2,3,4 on the Ricci flow with surgery. They are not complete and probably contains some inaccuracies. In- terested readers can find most exhaustives explanations on the Perelman's pa- pers in [KL]. 1 Lecture 2: classification of ?-solutions The aim of these lecture is to give the classification and the description of 3- dimensional ?-solutions. Let ? > 0 and (Mn, g(t)) a solution of the Ricci flow. Mn is supposed oriented. definition 1.1. (M, g(t)) is a ?-solution if • g(t) is an ancient solution of the Ricci flow ∂ ∂tg(t) = ?2Ricg(t), ?∞ < t ≤ 0. • for each t, g(t) is a complete, non flat metric of bounded curvature and non negative curvature operator. • for each t, g(t) is ?-noncollapsed on all scales, i.e. if |Rm(g(t))| ≤ 1r2 on B = B(p, t, r), then volg(t)(B) rn ≥ ? 1

• positive curvature

• asymptotic soliton

• dimensional ?-solution

• round flow

• local splitting

• now consider

• s2 ?r

Sujets

##### Constant curvature

Informations

 Publié par Nombre de lectures 9 Langue English

Exrait

Trieste Summer School, june 2005
Ricci ﬂow with surgery
Laurent Bessi`eres and G´erard Besson
Institut Fourier, 100, rue des maths, 38402 Saint Martin
d’H`eres,France.
Laurent.Bessieres@ujf-grenoble.fr, G.Besson@ujf-grenoble.fr
These notes provides some details on the lectures 2,3,4 on the Ricci ﬂow with
surgery. They are not complete and probably contains some inaccuracies. In-
terested readers can ﬁnd most exhaustives explanations on the Perelman’s pa-
pers in [KL].
1 Lecture 2: classiﬁcation of κ-solutions
The aim of these lecture is to give the classiﬁcation and the description of 3-
ndimensional κ-solutions. Letκ>0 and (M ,g(t)) a solution of the Ricci ﬂow.
nM is supposed oriented.
deﬁnition 1.1. (M,g(t)) is a κ-solution if
• g(t) is an ancient solution of the Ricci ﬂow

g(t)=−2Ric , −∞<t≤ 0.g(t)
∂t
• for each t, g(t) is a complete, non ﬂat metric of bounded curvature and
non negative curvature operator.
1• for eacht, g(t) is κ-noncollapsed on all scales, i.e. if|Rm(g(t))|≤ on2r
B =B(p,t,r), then
vol (B)g(t)
≥κ
nr
13 2Exemples: S and S ×R with their standard ﬂow are κ-solutions for some
2 1κ> 0. ButS ×S with the standard ﬂow is not aκ-solution. It isκ-collapsed
at very negative times.
Some properties of κ-solutions:
• All curvatures ofg(t) at x are controlled by the scalar curvatureR(x,t).
• For each point x in M, R(x,t) is nondecreasing.
It’s a consequence of the trace Harnack inequality [H93] (compare with
Carlo Sinestrari notes [S05] (6.6)
∂R
+2<X,∇R> +2Ric(X,X)≥ 0,
∂t
where X is an arbitrary vector ﬁeld. Thus
sup R(.,.) =supR(.,0)<∞
MM×]∞,0]
and all curvatures are uniformly bounded onM×]−∞,0].
• R(x,t)>0 for any (x,t).
It follows from the integrated version of the Harnack Inequality,

2d (x ,x1 2t1R(x ,t )≥exp − R(x ,t ),2 2 1 1
2(t −t2 1
for any t < t . Indeed, if R(x ,t ) = 0 for some point (x ,t ), then1 2 2 2 2 2
R(x ,t ) = 0 for any point (x ,t ) with t <t . Thus g(t) would be ﬂat1 1 1 1 1 2
for any t.
Tools: compactness theorem, asymptotic solitons, split-
ting
3compactness theoremGiven anyκ-solution(M ,g(t))and(x ,t )∈M×]−0 0
∞,0], one deﬁnes the normalized κ-solution at (x ,t ) by0 0
t
g (t) =R(x ,t )g(t + ).0 0 0 0
R(x ,t )0 0
WehavedoneashiftintimeandaparabolicrescalingsuchthatR (x ,0)= 1.g 00
The motivation is :
2theorem 1.2 ([P03]I.11.7, [KL]40). For any κ> 0, the set of pointed nor-
malized κ-solutions
{(M,g(.),x),R(x,0)=1}
is compact.
The same result holds with the normalization R(x,0)∈ [c ,c ], 0<c ≤c <1 2 1 2
∞.
AsymptoticsolitonsPerelmandeﬁnesanasymptoticsoliton(M ,g ,x )−∞ −∞ −∞
of an n-dimensional κ-solution (M,g(t) as follows. Pick a sequence t →−∞.k
1theorem 1.3 ([P03]I.11.2). there exists x ∈ M such that (M, g(t −k k−tk
t t),x )(sub)convergetoanonﬂatgradientshrinkingsoliton(M ,g ,x ),k k −∞ −∞ −∞
called an asymptotic soliton of the κ-solution.
Recall that a Ricci ﬂow (M,g(t)) on (a,b), a < 0< b, is a gradient shrinking
soliton if there exists a decreasing function α(t), diﬀeomorphisms of M ψt
generated by∇ f such thatg(t) t
∗g(t)=α(t)ψ g(0), ∀t∈ (a,b).t
The proof strongly uses the reduced length and reduced volume introduced in
[P03]ch.7.
corollary 1.4 (of the compactness theorem). Any 3-dimensional asymp-
totic soliton is a κ-solution.
Proof: The sequence τ R(x ,t ) has a limit R(x ,0)∈ (0,+∞). Thus thek k k −∞
asymptotic soliton is a parabolic rescaling of the limit of (M,R(x ,t )g(t +k k k
t ),x ), a κ-solution. Thus a 3-asymptotic solitons are particular κ-kR(x ,t )k k
solutions. Due to their self-similarity, they are much easier to classify.
Strong maximum principle the following will give splitting arguments
3theorem1.5 ([H86]). Let (M ,g(t)) a Ricci ﬂow on [0,T) such that sectional
curvatures of g(a) are≥ 0.Then precisely one of the following holds
a) For every t∈ (0,T), g(t) is ﬂat.
2 2b) For everyt∈ (0,T), g(t) has a local isometric splittingR×N , where N
is a surface with positive curvature.
3c) For every t∈]a,b[, g(t) has > 0 curvature.
2In case b), the universal covering is isometric R×N .
classiﬁcation of 3-asymptotic solitons
2 2proposition 1.6. The only asymptotic solitons areS ×R,S × R where theZ2
3Z -action is given by the relation (x,s)∼ (−x,−s), and ﬁnite quotients of S ,2
with their standard ﬂows.
Proof: Consider an asymptotic soliton (M ,g ,x ) = (M,g(t),x) of a−∞ −∞ −∞
κ-solution. By the strong maximum principle 1.5 and the non ﬂatness, either
g(t) has strictly positive curvature either it splits locally.
Consider the non compact case. The strictly positive curvature is ruled out by
theorem1.7([P03]II.1.2). There is no complete oriented 3-dimensional non
compact κ-noncollapsed gradient shrinking soliton with bounded (strictly) pos-
itive curvature.
2 2˜Thus (M,g(t)) has a local splitting and (M,g˜(t)) = (N ×R,h(t)+dx ). As
2the splitting is preserved by the ﬂow (N ,h(t)) is a Ricci ﬂow with strictly
positive curvature. It is an exercice to check that it is a κ-solution.
Now there is
theorem1.8([P02]I.11.2). thereisonlyoneoriented2-dimensionalκ-solution
- the round sphere.
2proof: (heuristic). Suppose that N is compact. It can be shown that the
2asymptotic soliton N is also compact (same arguments as in [CK04], prop−∞
29.23), thus diﬀeomorphic to S . By [H88], a metric with positive curvature
2on S gets more rounder under the Ricci ﬂow. More precisely, the curva-
tures pinching - the ratio of the minimum scalar curvature and the maximum-
2improves, i.e. converge to 1. On the other hand (N ,h (t)) evolves by dif-−∞∞
feomorphims and dilations hence the curvatures pinching is constant. Thus
for any t≤ 0, h (t) has constant curvature. Now the curvatures pinching of−∞
2(N ,h(t)) improves under the ﬂow as t→ 0 and is arbitrary close to 1 when
2t → −∞, as the asymptotic “initial condition” (N ,h ) is the round−∞(0)−∞
4sphere. The non compact case is ruled out by [KL][.37]. In fact, they give a
proof of 1.8 without solitons. 2.
2˜Thus (M,g˜(t)) =S ×R with a round cylindrical ﬂow. The only non compact
32 3oriented quotient isS × R=RP −B .Z2
Now consider the compact case. If (M,g(t)) has strictly positive curvature, by
3[H82]M isdiﬀeomorphictoaroundS /Γandg(t)getsmorerounderunderthe
ﬂow. By self-similarity of the metric, it is the round one, as above. We cannot
have a local splitting because the only oriented isometric compact quotients of
3 32 2 1 2 1S ×R, S ×S andS × S =RP #RP , are not κ-solutions. 2Z
classiﬁcation of κ-solutions We have the following
theorem 1.9. Any κ-solution (M,g(t)) is diﬀeomorphic to one of the follow-
ing.
2 2 3 3a S ×R or S × R=RP −B , and g(t) is the round cylindrical ﬂow.Z2
3b R and g(t) has strictly positive curvature.
3c A ﬁnite isometric quotient of the round S and g(t) has positive curvature.
Moreover, g(t) is round if and only if the asymptotic soliton is compact.
3If the asymptotic soliton is non compact, M is diﬀeomorphic to S or
3
RP .
Proof of theorem 1.9: Apply again the strong maximum principle to
the κ-solution (M,g(t)). If g(t) locally splits, we have the same classiﬁcation
as for asymptotic soliton. Suppose g(t) has strictly positive curvature. If it
3is compact, M is diﬀeomorphic to a ﬁnite quotient of the round S . If its
asymptotic solitonM is compact, it is the round ﬂow on a ﬁnite quotient of−∞
3S by the above classiﬁcation. Thus the asymptotic initial condition is round
and(M,g(t)isitself around ﬂow. In anoncompact caseM isdiﬀeomorphic to
3R by a theorem of Gromoll and Meyer [GM89]. The cases of strictly positive
curvature needs more geometrical control. The proof will be ﬁnished below.
More on κ-solutions
We describe the geometry of κ-solutions, which is useful for non round ﬂows.
We ’ll see that large parts of these κ-solution looks like round cylinders.
5deﬁnition 1.10. Let B(x,t,r) denotes the open metric ball of radius r, with
respect to g(t).
r 1Fix some ε > 0. A ball B(x,t, ) is an ε-neck, if after rescaling by , it is2ε r
−1[ε ]ε-close in the C topology to the corresponding subset of the standard neck
2 1 1 2S ×(− , ), where S has constant scalar curvature one. One says that x is
ε ε
the center of the ε-neck.
2 2For example, any point in S ×R is center of an ε-neck but (x,0)∈S × RZ2
is not center of an ε-neck.
deﬁnition 1.11. Let (M,g(.)) be aκ-solution. For everyε>0 and timet, let
M (t) (= M ) be the set of points which are not center of an ε-neck at timeε ε
t.
The geometry of the κ-solutions is described by the
proposition 1.12 ([KL]42.1, strong version of [P02]I.11.8). For all κ>
0, for 0 < ε < ε , there exists α = α(ε,κ) with the property that for any0
κ-solution(M,g(.)), and any time t precisely one of the following holds,
2A. M =∅ and (M,g(.))=S ×R is the round cylindrical ﬂow. So every pointε
at every time is center of an ε-neck for all ε>0.
2B. M =∅,M is noncompactandforallx,y∈M , wehaveR(x)d (x,y)<α.ε ε
C. M = ∅, M is compact and there is a pair of points x,y ∈ M such thatε ε
2R(x)d (x,y)>α,
−1/2 −1/2M ⊂B(x,αR(x) )∪B(y,αR(y) ),ε
2and every z∈M\M satisﬁes R(z)d (z,xy)<α.ε
2D. M =∅,M is compactand there is a pointx∈M such thatR(x)d (x,z)<ε ε
α for any z∈M.
Preliminary lemmas
A useful fact is
6
666lemma 1.13. Let (M,g(.)) be a κ-solution which contains a line for some t.
2Then M =S ×R and g(t)) is the round cylindrical ﬂow.
proof: Apply the Toponogov splitting theorem ([BBI]10.5.1). If there is a line
2at some time t, there is a splitting (M,g(t)) = (N (t)×R) and the result
follows from the classiﬁcation of 3-dim. κ-solutions.2
Wegive some consequences ofthe compactness theorem 1.2. Roughlythe ratio
of the scalar curvature at two pointsx,y of anyκ-solution is controlled by the
2normalized distanceR(x,t)d (x,y). Note that this expression is invariant by
g(t)
space dilation.
lemma 1.14. There exists α : [0,+∞[→ [1,+∞[ depending only on κ such
that for any κ-solution (M,g(.)), for each x,y in M,
R(y,t)−1 2 2α R(x,t)d (x,y) ≤ ≤α R(x,t)d (x,y)g(t) g(t)R(x,t)
Proof: One can deﬁne α on each [n,n+1[, n∈ N. Suppose that’s not true
for some integern. There is a sequence (M ,g (.)) ofκ-solutions, timest andk k k
R(y ,t )2 k kpointsx ,y inM such thatn≤R(x ,t )d (x,y))<n+1 and → 0k k k k k g(t ) R(x ,t )k k k
R(y ,t )k k tor ( → +∞). Normalize g (.) in g˜(t) =R(x ,t )g (t + ). Onek k k k k kR(x ,t ) R(x ,t )k k k k
˜obtains asequence ofpointedκ-solutions (M ,g˜(.),x ) such thatR(x ,0)= 1k k k k
2˜andd (x ,y )<n+1. Bythecompacitytheorem,onecanextractaconvergentk k
subsequence to aκ-solution (M ,g (.),x ). Lety ∈M be the limit ofy .∞ ∞ ∞ ∞ ∞ k
˜Then R (y ,0)=limR(y ,0)∈{0,∞} and we have a contradiction. 2∞ ∞ k
One can give another formulation (see [KL]36.1.5)
lemma 1.15. There exists β : [0,+∞[→ [0,+∞[, continuous, depending only
on κ such that lim β(s) = +∞, and for every κ-solution (M,g(.)) ands→+∞
2 2x,y∈M, we have R(y)d (x,y)≥β(R(x)d (x,y)).
Proof: exercice.
remark 1.16. in [P02] and [KL], these results are established before the com-
pactness theorem. Here we use the compactness theorem as a black box. We
have not the time for a proof.
The pattern to use the compactness theorem is the following. You want to
show that some points inκ-solutions have a nice geometry. Suppose they have
7not. Consider a sequence of bad points. Take a limit. Show that the limit
contains a line. Thus the limit is the round cylindrical ﬂow and the geometry
is controlled. So it is just before the limit. Let ε be a ﬁx small constant, say0
1ε = .0 10000
lemma 1.17 ([KL]42.2). For all κ > 0, for 0 < ε < ε , there exists0
α = α(ε,κ) with the followings property. Suppose (M,g(.)) is any κ-solution,
2x,y,z ∈ M and at time t we have x,y ∈ M and R(x)d (x,y)≥ α. Then atε
2 2 2time t either R(x)d (x,z)<α or R(y)d (y,z)<α or (R(z)d (z,xy)<α and
z∈/ M ).ε
Proof: Suppose not for some κ, ε. There exists a sequence of κ-solutions
(M ,g (.)),t ∈]−∞,0],x ,y ,z ∈M,x ,y ∈M suchthat,withquantitiesk k k k k k k k ε
2computed at time t , R(x )d (x ,y )→+∞ andk k k k
2 2 2R(x )d (x ,z )→+∞,R(y )d (y ,z )→ +∞and(R(z )d (z ,x y )→+∞orz ∈M ).k k k k k k k k k k k ε
2Consider ﬁrst the case where R(z )d (z ,x y ) → ∞ (up to a subsequence).k k k k
′Wedeﬁnez ∈x y asapointclosestfromz . Wewanttoprovethatx y con-k k k k kk
′verge to a line in the limit space of the (renormalized) sequence (M ,g (.),z ).k k k
′ 2 ′Claim: R(z )d (z ,x )→+∞.kk k
′ 2 ′Ifnot,supposethatR(z )d (x ,z )≤cforasubsequence. Normalize(i.e. shiftkk k
′time + parabolic rescaling)g (.) such thatR(z ,0)=1. Here we use the samek k
2 ′notation for the normalized metric. Thus we haved (x ,z )≤c. On the otherk k
R(x )k 2 2 ′hand, as the ratio is controlled, d (x ,y ) → +∞, d (z ,y ) → +∞′ k k kkR(z )
k
2 ′ ′and d (z ,z ) → +∞. Extract a subsequence such that (M ,g (.),z ) con-k k kk k
′ ′verge to aκ-solution (M ,g (.),z ). Thus the segments x y and z ,z con-∞ ∞ k k k∞ k
′′verge to rays x ξ and z η, where z ∈ x ξ. Note that angle ′ (ξ,η) =∞ ∞ z∞ ∞ ∞
′ π ′limangle (y ,z )≥ where angle is the comparison angle.′ k kz 2k
′ ′Nowwesaythatthereexistsr ≥ 0suchthateveryu∈z ξ withd(z ,u)≥r0 0∞ ∞
′is the center of an ε-neck. If not, consider a sequence u ∈ z ξ such thatk ∞
′ ′ 2 ′ 2 ′d(z ,u )→∞. ThusR(z )d (z ,u )→∞. Bylemma1.15,R(u )d (z ,u )→k k k k∞ ∞ ∞ ∞
+∞ also. Consider a sequence of normalizedκ-solution (M ,g (.),u ) such∞ ∞,k k
′that R(u ,0) = 1. Thus there is a convergent subsequence and the ray z ξk ∞
converge to a line in the limit. Thus the limit is the round cylindrical ﬂow by
1.13 andu is the center of an ε-neck for largek.k
′Letu ∈z ξ such thatevery pointuinu ξ is thecenter ofanε-neck. Onecan0 0∞
8′takeu far enough such thatz is not in theε-neck centered atu . Indeed, as0 0∞
ε is small, the length of this ε-neck is approximatively0
′2 2 d(z ,u )0′ ∞p = p d(z ,u )≤0∞
′ 10ε R(u ) ε R(u )d(z ,u )0 0 0∞
p
′ ′if R(u )d(z ,u ) is suﬃciently large. Clearly, z is in none of the ε-neck0 0∞ ∞
centered onu ξ. The point u is included in an embedded 2-sphere S , image0 0 0
2of a sphere S ×{∗} by the ε-approximation with the standard neck. Now
′every curve from u∈u ξ to z must exit from all ε-neck centered on u ξ on0 0∞
′the left side - the side of u which is closer to z - and thus must intersect S .0 0∞
′That means that S separates M . Moreover M has at least two ends z ξ0 ∞ ∞ ∞
′and z η. Thus (M ,g (0)) has a line. Indeed, one can consider a sequence∞ ∞∞
of geodesic segments with extremities in each end and extract a convergent
′subsequence with the help ofthe intersection withS . Thus (M ,g (.),z ) is0 ∞ ∞ ∞
the round cylindrical ﬂow. Thus x is the center of an ε-neck for largek, con-k
tradicting the hypothesis. That proves the claim. The same argument shows
′ 2 ′that R(z )d (z ,y )→+∞.kk k
′ ′The normalized sequence (M ,g (.),z ) converges to(M ,g (.),z )andx yk k ∞ ∞ k kk ∞
converge to a line in (M ,g (0)). Thus the limit is the round cylindrical ﬂow.∞ ∞
′The segmentz z is orthogonaltox y . Thus its limit is othogonalto the line,k k kk
2 ′′hence is a segment z z of bounded length. Thus R(z )d (z ,z ) remains∞ k k∞ k
bounded, proving the ﬁrst case.
Now it is clear thatthere isα such thatz ∈/ M . If not, the same constructionk ε
as above produces a limit z is in a round cylindrical ﬂow, thus z ∈/ M for∞ k ε
largek.
Proof of proposition 1.12
Let κ,ε>0 and (M,g(.)) a κ-solution.
Case 1 M =∅, i.e. every point is center of an ε-neck. Fix some x ∈M, anε 0
2 −1 1 2√ √ε-neck U ∼S ×[ , ] and let S be the image of S ×{0} . One0
R(x )ε R(x )ε0 0
shows that ifS separatesM, (M,g(.)) is the round cylindrical ﬂow. The other
Suppose that S separates. Choose point x in the left side of ∂U . There is1 0
an ε-neck U centered at x , thus one can choose x in the left side of ∂U1 1 2 1
(the side not in U ). Repeating the argument, one deﬁne a sequence (x ,U )0 k k
on the left of U and a sequence (y ,V ) on the right. Every segment x y0 k k k k
9cross all U ,...,U ,V ...V . Now, the length of each neck U is roughly0 k−1 1 k−1 l
2 k√ √. Either R(x) ≤ c for all integer l and then d(x ,x ) ≥ → ∞.l k 0 2 cεR(x )εl
2Either R(x) → ∞ for a subsequence and then R(x)d (x,x ) → ∞ thusl l l 0
2R(x )d (x ,x) → ∞. Using the same argument on the right, one conclude0 0 l
thatℓ(x y )→∞. Asallsegmentsx y intersectsU ,thereexistsaconvergentk k k k 0
subsequence and the limit is the line. Thus (M,g(.)) is the round cylindrical
ﬂow as in A.
˜ ˜Suppose that S does not separate. Let M the universal cover of M and S a
˜lift of S. We claim that S separates. Using a segment between sides of ∂U ,0
one can take a loop γ in M intersecting S transversally in one point. Thus γ
˜is homotopically non trivial. There is a lift γ˜ of γ intersecting S transversally
˜in one point, with extremities x = x . If S does not separate, there is a1 2
˜ ˜curve disjoint from S between x ,x . Thus there is a loop in M intersecting1 2
˜ ˜S transversally in one point. This is not possible, thus M separates. By the
˜˜previous argument, (M,g(.)) with the universal ﬂow (which is a κ-solution) is
2the round cylindrical ﬂow. Thus (M,g(t)) is a quotient ofS ×R by a group of
isometries, which contains translations, as S must separate. Thus (M,g(.)) is
2 1covered byS ×S with the round cylindrical ﬂow, but this is not aκ-solution.
2Case 2 M = ∅ and there exists x,y ∈ M sucht that R(x)d (x,y) ≥ α.ε ε
−1/2By the previous lemma, for z ∈ M either we have z ∈ B(x,αR(x) )∪
−1/2 2B(y,αR(y) ) either R(z)d (z,xy)<α and z∈/ M . Thus we have C.ε
2Case 3M =∅ and for anyx,y∈M , we have R(x)d (x,y)<α. If M is nonε ε
compact, we have B.
Ifwe suppose thatM compact, we want tohave D.We argueby contradiction.
2Fix a point x∈ M . Let z the point of M such that R(z)d (x,z) is maximalε
2and suppose that R(z)d (z,x) ≥ α. Thus z ∈/ M , that is z is center of anε
ε-neck. Consider the middle sphere S of the neck. Either S separates, either
S does not.
IfS separates,M is on one side. Indeed, if there were points on each side, anyε
geodesic joigning opposite points should intersect S, thus would have length
2α√≥ , which is not possible in our case. Now if M is on one side, z is notε
R(z)
maximal.
˜ ˜If S does not separate, then all lift S ⊂M separates, as in case 1. There is a
non trivial loopγ∈M, which intersectsS transversally in one point. As a lift
˜γ˜ hits an inﬁnite number of S on both sides, it is easy to construct a line in
˜(M,g˜(t)). Thus (M,g(.)) is covered by the round cylindrical ﬂow and we get
a contradiction as in case 1. 2
2 3endoftheproofof1.9Notethatκ-solutionsonS × RandB aredescribedZ2
by case B of proposition 1.12. Recall that we suppose the κ-solution compact
10
666

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