Course of Undergraduate Young Talent Project: 4-manifolds and Ricci FlowNotes, Part I. The proof of Hamilton’s 3-manifolds with positive Ricci curvature theorem

For lectures: Monday, March 26 - Wednesday, April 18, 2018

The basic Riemannian geometry section has been moved to the Appendix (last section).Hamilton’s papers are collected in the book [1].Hamilton’s papers will be cited as [3man] (1982), [4man] (1986).Perelman’s papers will be cited as [GP1], [GP2], and [GP3].

1 Variation formulas

Reference: §7 of [3man].

Lemma 1 If g (s) is a 1-parameter family of metrics with ∂∂sgij = vij (v is a symmetric covariant

2-tensor), then∂

∂sΓkij =

1

2gkℓ (∇ivjℓ +∇jviℓ −∇ℓvij) . (1)

Proof. The derivation of this formula illustrates a nice trick in computing evolutions of variousgeometric quantities such as the connection and the curvatures. We compute at an arbitrarily chosenpoint p ∈ M in normal coordinates {xi} centered at p so that Γk

ij (p) = 0. Note that ∂∂xi gjk (p) = 0.

In such coordinates, ∇kaj1···jqi1···ir (p) =

∂∂xka

j1···jqi1···ir (p) for any (r, q)-tensor a. Thus, at p we have

∂

∂sΓkij =

1

2gkℓ(

∂

∂xi

∂

∂sgjℓ +

∂

∂xj

∂

∂sgiℓ −

∂

∂xℓ

∂

∂sgij

)and hence (1) follows since

∇ivjℓ (p) =∂

∂xivjℓ (p)− Γm

ij (p) vmℓ − Γmiℓ (p) vjm =

∂

∂xivjℓ (p) .

Finally we note that since both sides of (1) are the components of tensors, equation (1) in fact holdsas a tensor equation, that is, it is true for any coordinate system, not just normal coordinates. Hereused the hint that the difference of two connections is a tensor and hence the variation of a family ofconnections is a tensor.

In the following one can skip the ∂∂s

Rm calculations if one just wants to prove Hamilton’s 3-manifold theorem because we will compute the evolution of the Einstein 2-tensor, which gives thesectional curvatures in dimension 3.

Lemma 2 If ∂∂sgij = vij, then

∂

∂sRℓ

ijk =1

2gℓp

{∇i∇jvkp +∇i∇kvjp −∇i∇pvjk

−∇j∇ivkp −∇j∇kvip +∇j∇pvik

}(2)

=1

2gℓp (∇i∇kvjp −∇i∇pvjk −∇j∇kvip +∇j∇pvik) (3)

− 1

2gℓp (Rijkqvqp +Rijpqvkq) .

1

Proof. FromRℓ

ijk = ∂iΓℓjk − ∂jΓ

ℓik + Γp

jkΓℓip − Γp

ikΓℓjp

we compute

∂

∂sRℓ

ijk = ∇i

(∂

∂sΓℓjk

)−∇j

(∂

∂sΓℓik

)=

1

2∇i

(∇jv

ℓk +∇kv

ℓj −∇ℓvjk

)− 1

2∇j

(∇iv

ℓk +∇kv

ℓi −∇ℓvik

),

which is the first formula. The second formula follows from this and the commutator formula

∇i∇jvℓk −∇j∇iv

ℓk = −Rq

ijkvℓq +Rℓ

ijqvqk.

Lemma 3 Suppose ∂∂sgij = vij. Then:

1.∂

∂sRij = −1

2

(∆Lvij +∇i∇jV −∇i (div v)j −∇j (div v)i

),

where the Lichnerowicz Laplacian is defined by

∆Lvij + ∆vij + 2Rkijℓvkℓ −Rikvjk −Rjkvik. (4)

2.∂

∂sR = ∇ℓ∇iviℓ −∆V − vij ·Rij.

Proof. (1) We have∂

∂sRij = ∇p

(∂

∂sΓpij

)−∇i

(∂

∂sΓppj

).

This follows from computing at the center in normal coordinates. Substituting the formula for ∂∂sΓkij

into this, we obtain∂

∂sRij =

1

2∇ℓ (∇ivjℓ +∇jviℓ −∇ℓvij)−

1

2∇i∇jV. (5)

Finally, commute derivatives.(2) Taking the trace, we obtain

∂

∂sR = gij

(∂

∂sRij

)− ∂

∂sgij ·Rij (6)

= ∇ℓ∇iviℓ −∆V − vij ·Rij.

We say that a 1-parameter family of metrics g(t) is a solution to the Ricci flow if ∂∂tgij = −2Rij.

Lemma 4 Let g(t) be a solution to the Ricci flow. Then:

2

1.∂R

∂t= ∆R + 2 |Rc|2 .

2.∂

∂tRij = ∆LRij = ∆Rij + 2RkijℓRkℓ − 2RikRjk.

3.

∂

∂tRijkℓ = ∆Rijkℓ + 2 (Bijkℓ −Bijℓk +Bikjℓ −Biℓjk) (7)

− (RipRpjkℓ +RjpRipkℓ +RkpRijpℓ +RℓpRijkp) ,

whereBijkℓ + −gprgqsRipjqRkrℓs = −RpijqRqℓkp. (8)

Proof. (1) and (2) follow immediately from the previous exercise.(3) We shall show the equivalent formula that under the Ricci flow, the (3, 1)-Riemann curvature

tensor evolves by

∂

∂tRℓ

ijk = ∆Rℓijk + gpq

(Rr

ijpRℓrqk − 2Rr

pikRℓjqr + 2Rℓ

pirRrjqk

)−Rp

iRℓpjk −Rp

jRℓipk −Rp

kRℓijp +Rℓ

pRpijk.

By applying the second Bianchi identity and commuting covariant derivatives, we compute

∆Rℓijk = gpq∇p∇qR

ℓijk = gpq∇p

(−∇iR

ℓjqk −∇jR

ℓqik

)= gpq

−∇i∇pR

ℓjqk +Rr

pijRℓrqk +Rr

piqRℓjrk +Rr

pikRℓjqr −Rℓ

pirRrjqk

+∇j∇pRℓqik −Rr

pjiRℓrqk −Rr

pjqRℓirk −Rr

pjkRℓiqr +Rℓ

pjrRriqk

and then apply the second Bianchi identity again to obtain

gpq∇pRℓjqk = gpqgℓm (−∇kRjqmp −∇mRjqpk) = ∇kR

ℓj −∇ℓRjk.

This lets us rewrite the formula for ∆Rℓijk given above as

∆Rℓijk = −∇i∇kR

ℓj +∇i∇ℓRjk +∇j∇kR

ℓi −∇j∇ℓRik

+ gpq

Rr

pijRℓrqk +Rr

piqRℓjrk +Rr

pikRℓjqr −Rℓ

pirRrjqk

−RrpjiR

ℓrqk −Rr

pjqRℓirk −Rr

pjkRℓiqr +Rℓ

pjrRriqk

.

Now the first Bianchi identity shows that

RrpijR

ℓrqk −Rr

pjiRℓrqk = −Rr

ijpRℓrqk,

which implies that ∆Rℓijk may be written as

∆Rℓijk = −∇i∇kR

ℓj +∇i∇ℓRjk +∇j∇kR

ℓi (9a)

−∇j∇ℓRik −RriR

ℓjrk +Rr

jRℓirk (9b)

+ gpq

−Rr

ijpRℓrqk +Rr

pikRℓjqr +Rℓ

pjrRriqk

−RℓpirR

rjqk −Rr

pjkRℓiqr

. (9c)

3

On the other hand, rewriting the commutator term gℓp (∇j∇i −∇i∇j)Rkp yields

∂

∂tRℓ

ijk = −∇i∇kRℓj +∇i∇ℓRjk +∇j∇kR

ℓi −∇j∇ℓRik (10a)

+ gℓp(Rq

ijkRqp +RqijpRkq

). (10b)

The reaction-diffusion equation for the Riemann curvature tensor now follows from comparing termsin formulas (9) and (10).

2 Uhlenbeck’s trick and the evolution of the Einstein tensor

Reference: §2 of [4man].The main goal is to prove (16). One can skip this section and simply first use (16) and prove it

later.Uhlenbeck’s trick (more concrete version for 3D; the n ≥ 4 case is essentially the same). Let

(M3, g(t)), t ∈ [0, T ), be a solution to the Ricci flow on a closed 3-manifold. By a standard result,M is parallelizable and so there exists a global orthonormal frame field {e0a}3a=1 for g(0) on M. Foreach x ∈ M let ea(x, t) be the solution to the ODE:

d

dtea(x, t) = Rcg(t)(ea(x, t)),

ea(x, 0) = e0a,

where we consider Rcg(t) : TxM → TxM as a (1, 1)-tensor. Then

d

dtg(t) (ea(t), eb(t)) = 0.

Hence {ea(t)}3a=1 is a global orthonormal frame field for g(t) on M for all t ∈ [0, T ).Let E = M× R3 be the trivial (flat) bundle with the Euclidean metric h on the fibers. Then we

have bundle isometriesι(t) : (E, h) → (TM, g(t))

defined by

ι(t) (x,v) =3∑

a=1

vaea(x, t),

where v = (v1, v2, v3). Note that ι(t)∗ (ea(x, t)) = ea, where ea is the standard Euclidean basis elementwith 1 in the a-th slot and 0 in the other slots. Of course, g(t)(ea(t), eb(t)) = h(ea, eb) = δab.

Exercise 5 For all dimensions, let ι0 : E → TM be a bundle isomorphism and let h = ι∗0(g(0)).Define bundle isomorphisms ι(t) : E → TM to solve the ODE:

d

dtι(t) = Rcg(t) ◦ι(t),

ι(0) = ι0.

Show that ι(t)∗(g(t)) is independent of t and hence is equal to h.

4

In the following, one can skip the evolution of Rabcd and instead just read the evolution of Eab.n ≥ 2. Now Rmg(t) is a section of the vector bundle Λ2T ∗M⊗S Λ

2T ∗M and the isomorphism ι(t)induces a pull back of this to Λ2E∗ ⊗S Λ2E∗, where the metric induced by g(t) is pulled back to themetric induced by h. Let R denote the pull-back of Rmg(t) by ι(t). We have

Rabcd = R(ea, eb, ec, ed)

= Rmg(t)(ea(t), eb(t), ec(t), ed(t)).

For the pulled back version of (7), we have the simplification:(∂

∂t−∆

)Rabcd = 2(Babcd − Babdc + Bacbd − Badbc), (11)

where Babcd = −RaebfRcedf .

Exercise 6 Prove (11).

With this, we can rewrite the evolution of R in a nice form. Define the square of R by

R2 = R ◦R : Λ2E∗ → Λ2E∗.

Define a Lie algebra structure on Λ2E∗ by

[U, V ]ab +∑c

(UacVcb − VacUcb)

for U, V ∈ Λ2E∗. Then Λ2E∗ ∼= so (n). Choose a basis {φα} of Λ2E∗ and let Cαβγ denote the structure

constants defined by[φα, φβ

]+∑

γ Cαβγ φγ. We define the Lie algebra square R# : Λ2E∗ → Λ2E∗ by

R#αβ = Cγδ

α Cεζβ RγεRδζ . (12)

Exercise 7 Show that∂

∂tR = ∆R+R2 +R#. (13)

When n = 3, we may express R2 +R# in the following simple way. Diagonalize R, i.e., choosean orthonormal basis {φα} of Λ2E∗ so that

(R(φα, φβ)

)=

λ 0 00 µ 00 0 ν

.

Then the matrix((R2 +R#)(φα, φβ)

)is also diagonal and

((R2 +R#)(φα, φβ)

)=

λ2 + µν 0 00 µ2 + λν 00 0 ν2 + λµ

. (14)

Here is an alternate (equivalent) way to obtain the 3D formula (14). With respect to an evolvingorthonormal frame {ea(t)} we have ∂

∂tgab = 0 and

∂

∂tRab = ∆Rab + 2RcabdRcd.

5

Hence Eab = Rgab − 2Rab (negative of the Einstein tensor) satisfies

∂

∂tEab = ∆Eab − 4RcabdRcd + 2 |Rc|2 gab.

Now assume n = 3. Then the eigenvalues of Eab and Rm are the same (twice the principal sectionalcurvatures). Moreover,

Rcabd = Rcdgab +Rabgcd −Rcbgad −Radgcb −1

2R (gcdgab − gcbgad)

Thus

∂

∂tEab = ∆Eab + 8RcbRac − 6RabR + 2

(R2 − |Rc|2

)gab

= ∆Eab +Qab,

where Qab = 2EcbEac − EabR +(R2 − 2 |Rc|2

)gab.

Let λ, µ, ν be the eigenvalues of Eab. Then λ+ µ+ ν = R. Since |Eab|2 = 4 |Rc|2 −R2 we have

R2 − 2 |Rc|2 = −1

2|Eab|2 +

1

2R2

= λµ+ λν + µν.

So

Q11 = 2E11E11 − E11R +(R2 − 2 |Rc|2

)g11

= 2λ2 − λ (λ+ µ+ ν) + (λµ+ λν + µν)

= λ2 + µν.

Similarly,

Q22 = µ2 + λν,

Q33 = ν2 + λµ.

This derives the evolution equation for Rm in dimension 3.For each x ∈ M, we may consider the ODE obtained from (13) by dropping the Laplacian term,

i.e., the ODE for R(t) : Λ2E∗x → Λ2E∗

x defined by

d

dtR = R2 +R#. (15)

Denoting the eigenvalues of R in the same way as for R, the ODE is (this also follows from theevolution of Eab):

dλ

dt= λ2 + µν,

dµ

dt= µ2 + λν,

dν

dt= ν2 + λµ.

Reference: §3 of [4man].The following is Hamilton’s maximum principle for systems. Let

V = Λ2E∗ ⊗S Λ2E∗.

6

Theorem 8 Let g (t), t ∈ [0, T ), be a solution to the Ricci flow on a closed manifold Mn. Let K ⊂ Vbe a subset which is invariant under parallel translation and whose intersection Kx + K ∩ Vx witheach fiber is closed and convex.Suppose the ode (15) has the property that for any R (0) ∈ K, we have R (t) ∈ K for all t ∈ [0, T ).If R (0) ∈ K, then R (t) ∈ K for all t ∈ [0, T ).

Let Rx ∈ Vx. For any path γ : [0, 1] → M with γ(0) = x, let R(u) be the parallel translation of Rx

along γ. Let vx ∈ Λ2E∗x be an eigenvector of Rx with eigenvalue λ. Let v(u) be the parallel translation

of vx along γ. Then R(u) (v(u)) is a parallel section of Λ2E∗ with R(0) (v(0)) = Rx(vx) = λvx. Bythe uniqueness of parallel translation, we have R(u) (v(u)) = λv(u) for u ∈ [0, 1]. So the eigenvaluesof an element R ∈ V are preserved under parallel translation. So if a subset K of V is defined byinequalities regarding eigenvalues of R, then K is invariant under parallel translation.

Note also that for any k ≥ 1, the minimum over all k-planes P in Λ2E∗ of trace(R|P ) is a concavefunction since for each P the function R 7→ trace(R|P ) is linear. The set of all R ∈ Vx where aconcave function of R is at least a constant c is a convex subset of Vx.

Sets defined by weak inequalities, i.e., ≥ or ≤, involving continuous functions of the eigenvaluesof R are closed.

For an exposition of the following, see chapter 3 of [4].

Exercise 9 Show that for any constant C0 ∈ R, if Rg(0) ≥ C0, then Rg(t) ≥ C0 for all t ∈ [0, T ). Thisis true for all n ≥ 3, but just prove this for n = 3. Hint: Let

K = {R ∈ V : λ(R) + µ(R) + ν(R) ≥ C0}

and show that ddt(λ+ µ+ ν) ≥ 0.

Answer : The set K is preserved by the ODE because

d

dt(λ+ µ+ ν) =

1

2

((λ+ µ)2 + (λ+ ν)2 + (µ+ ν)2

)≥ 0.

Exercise 10 n = 3. Show that if Rcg(0) ≥ 0, then Rcg(t) ≥ 0 for all t ∈ [0, T ). Hint: Order theeigenvalues by λ ≥ µ ≥ ν. Let

K = {R ∈ V : µ(R) + ν(R) ≥ 0} .

Note that µ+ ν is a concave function and

(µ+ ν) (R) = min{R(φ1, φ1

)+R

(φ2, φ2

):{φ1, φ2

}orthonormal

}.

This is the same as the minimum over all 2-planes P in Λ2E∗ of trace(R|P ).

Answer : The set K is preserved by the ODE because

d

dt(µ+ ν) = λ2 + µ2 + (µ+ ν)λ ≥ 0

whenever µ+ ν ≥ 0 (since λ ≥ µ ≥ 0).

Exercise 11 Suppose that M3 is closed and (M, g(0)) has positive Ricci curvature.

7

1. Show that there exists ε > 0 such that Rcg(0) ≥ εRg(0)g(0) on M. Show also that there existsC ≥ 1/2 such that λ

(Rmg(0)

)≤ C

(µ(Rmg(0)

)+ ν

(Rmg(0)

)).

2. Show that there exists ε > 0 such that Rcg(t) ≥ εRg(t)g(t) on M for all t ∈ [0, T ). Hint: GivenC ≥ 1/2, let

K = {R ∈ V : λ (R) ≤ C (µ (R) + ν (R))} ,where λ ≥ µ ≥ ν. Show that if λ−C (µ+ ν) = 0, then d

dt[λ− C (µ+ ν)] ≤ 0. (Note that λ (R)

for R ∈ Vx is the maximum of R(v, v) over all unit vectors v ∈ Λ2E∗x and hence is a convex

function of R.)

Answer : (1) is easy.(2)

d

dt(λ− C (µ+ ν)) = λ (λ− C (µ+ ν))− C

(µ2 − 1

Cµν + ν2

).

Hence, if λ− C (µ+ ν) = 0 and C ≥ 1/2 (i.e., 1C≤ 2), then

d

dt(λ− C (µ+ ν)) ≤ 0.

Exercise 12 Given C0 > 0, C1 ≥ 1/2, C2 < ∞ and δ > 0, let

K =

R ∈ V :λ(R)−ν(R)−C2 (λ(R)+µ(R)+ν(R))1−δ≤ 0,

λ (R) ≤ C1 (µ (R) + ν (R)) ,λ(R)+µ(R)+ν(R) ≥ C0

.

Show thatd

dtln

(λ− ν

(λ+ µ+ ν)1−δ

)≤ δ (λ+ ν − µ)− (1− δ)

µ2

λ+ µ+ ν.

Show that for δ > 0 small enough (e.g., δ1−δ

≤ 112C2

1works), we have

d

dtln

(λ− ν

(λ+ µ+ ν)1−δ

)≤ 0.

Conclude that there exist constants C < ∞ and δ > 0 such that∣∣∣∣Rc−1

3Rg

∣∣∣∣ ≤ CR1−δ. (16)

Note that (λ− ν) (R)− C ((λ+ µ+ ν) (R))1−δ is a convex function of R.

Answer : λ+ µ+ ν ≥ C0 and λ ≤ C1 (µ+ ν) are preserved. We compute

d

dtln

(λ− ν

(λ+ µ+ ν)1−δ

)

= δ (λ+ ν − µ)− (1− δ)(µ+ ν)µ+ (µ− ν)λ+ µ2

λ+ µ+ ν

≤ δ (λ+ ν − µ)− (1− δ)µ2

λ+ µ+ ν.

8

Note thatµ2

λ+ µ+ ν≥ 1

6

(µ+ ν)µ

λ≥ 1

6C1

µ

since λ+ µ ≤ 2λ ≤ 2C1 (µ+ ν) , and we also have

λ+ ν − µ ≤ λ ≤ C1 (µ+ ν) ≤ 2C1µ.

Hence, choosing δ > 0 small enough so that δ1−δ

≤ 112C2

1, we have

d

dtlog

(λ− ν

(λ+ µ+ ν)1−δ

)≤ 0.

Since λ− ν ≥∣∣Rc−1

3Rg∣∣ , we obtain (16).

3 Entropy monotonicity and no local collapsing

Reference: §3, §4, and §9 of [GP1]. (Exposition: chapter 6 of [3].)

∂

∂tgij = −2Rij, (17a)

∂f

∂t= −R−∆f + |∇f |2 + n

2τ, (17b)

dτ

dt= −1, (17c)

v =(τ(R + 2∆f − |∇f |2

)+ f − n

)τ−

n2 e−f .

W(g, f, τ) =

∫M

(τ(R + |∇f |2

)+ f − n

)τ−

n2 e−fdµ =

∫M

vdµ.

�∗ = − ∂∂t−∆+R.

−�∗(τ−n2 e−f ) =

(n

2τ− ∂f

∂t−∆f + |∇f |2 −R

)τ−

n2 e−f = 0.

−�∗v = 2τ

∣∣∣∣Rc+∇2f − 1

2τg

∣∣∣∣2 τ−n2 e−f . (18)

(18) implies

d

dtW(g(t), f(t), τ(t)) =

∫M

(∂

∂tv −Rv

)dµ

=

∫M

(−�∗v) dµ

= 2τ

∫M

∣∣∣∣Rc+∇2f − 1

2τg

∣∣∣∣2 τ−n2 e−fdµ.

9

Proof of (18). (∂

∂t+∆−R

)R = 2∆R + 2 |Rc|2 −R2

(∂

∂t+∆−R

)(∆f) = ∆

(∂f

∂t

)+ 2Rc ·∇2f + (∆−R) (∆f)

= −∆R +∆|∇f |2 + 2Rc ·∇2f −R∆f(∂

∂t+∆−R

)|∇f |2 = 2Rc(∇f,∇f) + 2∇f · ∇

(−R−∆f + |∇f |2

)+ (∆−R) |∇f |2

So

−�∗ (R + 2∆f − |∇f |2)= 2 |Rc|2 −R2 +∆|∇f |2 + 4Rc ·∇2f − 2R∆f

− 2Rc(∇f,∇f) + 2∇f · ∇(R +∆f − |∇f |2

)+R|∇f |2.

Using∆ |∇f |2 = 2 |∇∇f |2 + 2Rc(∇f,∇f) + 2∇f · ∇∆f,

this simplifies to

−�∗ (R + 2∆f − |∇f |2)= 2

∣∣Rc+∇2f∣∣2 + 2∇f · ∇

(R + 2∆f − |∇f |2

)−R

(R + 2∆f − |∇f |2

).

By −�∗(τ−n2 e−f ) = 0 and

−�∗(ab) = −b�∗a− a�∗b+ 2∇a · ∇b+Rab,

we have−�∗ ((R + 2∆f − |∇f |2

)τ−

n2 e−f

)= 2τ−

n2 e−f

∣∣Rc+∇2f∣∣2 .

By −�∗(τ−n2 e−f ) = 0 again and −�∗f = −Rf −R + |∇f |2 + n

2τ,

−�∗ (τ−n2 e−ff

)= τ−

n2 e−f (−�∗f) + 2∇

(τ−

n2 e−f

)· ∇f +Rτ−

n2 e−ff

= τ−n2 e−f

(−R− |∇f |2 + n

2τ

).

Finally,

−�∗ ((τ (R + 2∆f − |∇f |2)+ f − n

)τ−

n2 e−f

)= 2τ 1−

n2 e−f

∣∣Rc+∇2f∣∣2 − (R + 2∆f − |∇f |2

)τ−

n2 e−f

+ τ−n2 e−f

(−R− |∇f |2 + n

2τ

)= 2τ

∣∣∣∣Rc+∇2f − 1

2τg

∣∣∣∣2 τ−n2 e−f .

Logarithmic Sobolev inequality. Let (Mn, g) be a closed Riemannian manifold. Then for any a > 0there exists C(a, g) such that for any φ satisfying

∫φ2 = 1 we have∫

φ2 lnφ ≤ a

∫|∇φ|2 + C(a, g).

10

Entropy lower bound. Let w = τ−n4 e−

f2 , so that

∫w2 = 1 by assumption. Taking a = 2τ , we have

W (g, f, τ) =

∫M

(τ(Rw2 + 4 |∇w|2

)−(2 lnw +

n

2ln τ + n

)w2)dµ (19)

≥ τRmin −n

2ln τ − n− 2C(2τ, g)Vol(g).

µ(g, τ) = inf∫τ−

n2 e−f=1

W(g, f, τ),

ν(g) = infτ>0

µ(g, τ).

Scaling:

µ(g, τ) = µ (cg, cτ) ,

ν(g) = ν(cg).

By (19) we have µ(g, τ) > −∞.Standard elliptic theory implies the existence of a minimizer f of W(g, · , τ), and each such mini-

mizer satisfiesτ(R + 2∆f − |∇f |2

)+ f − n = µ (g, τ) .

Under the Ricci flow, if t1 ≤ t2, then

µ (g (t2) , τ (t2)) ≥ µ (g (t1) , τ (t1)) .

In particular,µ(g (t) , r2

)≥ µ

(g (0) , r2 + t

).

Nonrigorous proof of no local collapsing. Suppose R ≤ r−2 in B = Btr(x), where t ≥ T

2and r ≤ ρ.

Let f(t) = cχB. Then∫M r−ne−fdµ = 1 implies c = ln VolB

rn. By entropy monotonicity and since

T2< r2 + t ≤ ρ2 + T , we have (ignoring the |∇f |2 term)

lnκ + minτ∈[T

2,ρ2+T ]

µ (g (0) , τ)

≤ W(g (0) , f(0), r2 + t

)≤ W

(g (t) , f(t), r2

)≤ r2 max

BR + c− n

≤ lnVolB

rn.

See the aforementioned references for the actual proof.

4 Three-manifolds with positive curvature

Short time existence ([3man] (do not read), [DeTurck] (read)):

Theorem 13 If Mn is a closed Riemannian manifold and if g0 is a C∞ Riemannian metric, thenthere exists a unique smooth solution g (t) to the Ricci flow defined on some time interval [0, δ), δ > 0,with g (0) = g0.

11

The normalized Ricci flow is∂

∂tgij = −2Rij +

2

nrgij, (20)

where r = Vol (g)−1 ∫M Rdµ is the average scalar curvature.

Exercise 14 Show that ddtVol (g (t)) = 0 under the normalized Ricci flow.

Reference: [7], [2].

Theorem 15 If (M2, g0) is a closed Riemannian surface, there exists a unique solution g (t) of thenormalized Ricci flow (when n = 2, 2Rij = Rgij)

∂

∂tg = (r −R) g,

g (0) = g0.

The solution exists for all time. As t → ∞, the metrics g (t) converge uniformly in any Ck-norm toa smooth metric g∞ of constant curvature.

Reference: [3man].

Theorem 16 Let (M3, g0) be a closed Riemannian 3-manifold with positive Ricci curvature. Thenthere exists a unique solution g (t) of the normalized Ricci flow with g (0) = g0 for all t ≥ 0. Further-more, as t → ∞, the metrics g(t) converge exponentially fast in every Ck-norm to a C∞ metric g∞with constant positive sectional curvature. Hence M is diffeomorphic to a spherical space form.

Hamilton’s Cheeger–Gromov compactness theorem for solutions to the Ricci flow is discussed insection 5 below.

Hamilton’s point selection for obtaining singularity models. Reference: §16 of [8]Curvature blows up at least at the Type I rate. If the singular time T is finite, then

K (t) = maxx∈M

|Rm| (x, t) ≥ 1

8 (T − t)for all t ∈ [0, T ). (21)

Since ∂∂t|Rm|2 ≤ ∆ |Rm|2 + 16 |Rm|3, by the maximum principle, K (t) satisfies dK

dt≤ 16K

32 . The

result follows from limt→T K (t)−12 = 0.

Finite time singularity : Since ∂R∂t

= ∆R+ 2 |Rc|2 ≥ ∆R+ 23R2, Rmin(t) ≥

1

Rmin(0)−1 − 23t. Hence

a singularity forms at time T ≤ 32Rmin(0)

−1.Elementary point selection. We describe how to choose (xi, ti) so that the rescaled solutions

(M, gi (t) , xi), where

gi (t) = Kig(K−1

i t+ ti), Ki = |Rm| (xi, ti) , t ∈ [−Kiti, Ki (T − ti))

converge.

12

Type I. Let ti → T . By (21), there exist xi such that (T − ti)Ki ≥ 18. Since |Rm| (x, t) ≤ C

T−t, we

have

|Rm gi| (x, t) = K−1i |Rm g|

(x,K−1

i t+ ti)

≤ CK−1i

(T − ti)−K−1i t

≤ C18− t

for t ∈ [−Kiti,18). By this curvature bound, by no local collapsing, and by Hamilton’s Cheeger–

Gromov compactness theorem, (M, gi (t) , xi) converges to an ancient solution (M3∞, g∞(t), x∞), t ∈

(−∞, 18), satisfying |Rm g∞| (x, t) ≤ C

18−t.

Type II. Let Ti → T and choose (xi, ti) ∈ M× [0, Ti] so that

(Ti − ti)Ki = maxM×[0,Ti]

(Ti − t) |Rm| (x, t) ,

which → ∞ by the Type II assumption. Since |Rm| (x, t) ≤ (Ti−ti)Ki

Ti−t, we have

|Rm gi| (x, t) = K−1i |Rm g|

(x,K−1

i t+ ti)

≤ (Ti − ti)Ki

(Ti − ti)Ki − t

on M × [−tiKi, (Ti − ti)Ki). Since (Ti − ti)Ki → ∞, we have that (M, gi (t) , xi) converges to aneternal solution (M3

∞, g∞(t), x∞), t ∈ (−∞,∞), satisfying |Rm g∞ | (x, t) ≤ 1.In either the Type I or Type II case, the limit is a complete ancient solution with Rc g∞ ≥ 0

and |Rm g∞| (x∞, 0) = 1, in particular, Rg∞ (x∞, 0) > 0. Let U be the open subset of M∞ on whichRg∞ > 0. For the original solution g(t) there exists δ > 0 such that∣∣Rc−1

3Rg∣∣

R≤ δ−1R−δ. (22)

Since Ki → ∞, on U × (−∞, 0] we have|Rcg∞ − 1

3Rg∞g∞|

Rg∞≡ 0, which implies Rg∞ is constant by the

contracted second Bianchi identity. (For the rescaled sequence of solutions, (22) says|Rcgi −

13Rgigi|

Rgi≤

K−δi δ−1R−δ

gi, which limits to zero in U as i → ∞ since Rgi → Rg∞ > 0 in U .) This implies that

U = M∞. This implies that (M∞, g∞(0)) satisfies Rcg∞(0) =13Rg∞(0)g∞(0). Hence M∞ is compact,

which implies that it is diffeomorphic to M.

5 Cheeger–Gromov compactness theorem

Reference: [9]. (Exposition: chapters 3 and 4 of [3].)

Definition 17 Let K ⊂ M be an open set whose closure K is compact and let {gi} and g∞ beRiemannian metrics on K. Given k ≥ 0, we say that gi converges to g∞ in Ck on K if for everyε > 0 there exists i0 such that for i ≥ i0,

sup0≤α≤k

supx∈K

|∇α (gi − g∞)|g < ε,

13

where the covariant derivative ∇ is with respect to a fixed complete Riemannian metric g on K. Thisdefinition is independent of the choice of g.

Definition 18 Suppose {Ui} is an exhaustion of a smooth manifold Mn by open sets and gi areRiemannian metrics on Ui. We say that (Ui, gi) converges to (M, g∞) in C∞ on compact setsin M if for any subset K of M with compact closure and for each k > 0, {gi|K}i≥i0

converges to g∞|Kin Ck on K, where i0 is such that K ⊂ Ui for i ≥ i0.

C∞ pointed Cheeger–Gromov convergence:

Definition 19 A sequence {(Mni , gi (t) , xi)} , t ∈ (α, ω) , of complete pointed solutions to the Ricci

flow converges to a complete pointed solution to the Ricci flow (Mn∞, g∞ (t) , x∞) , t ∈ (α, ω) , if there

exist :

1. an exhaustion {Ui} of M∞ by open sets with x∞ ∈ Ui, and

2. a sequence of diffeomorphisms Φi : Ui → Vi = Φi (Ui) ⊂ Mi with Φi (x∞) = xi

such that(Ui × (α, ω) ,Φ∗

i

[gi (t)|Vi

]+ dt2

)converges to (M∞×(α, ω),g∞ (t)+dt2) in C∞ on compact

sets in M∞ × (α, ω) .

Hamilton’s Cheeger–Gromov compactness theorem for solutions to the Ricci flow:

Theorem 20 Let {(Mni , gi (t) , xi)} , t ∈ (α, ω) ∋ 0, be a sequence of complete pointed solutions to

the Ricci flow where there exists a constant C0 such that :

1.∣∣Rm gi(t)

∣∣ ≤ C0 on Mi × (α, ω),

2. Volgi(0)Bgi(0)

1 (xi) ≥ C−10 .

Then there exists a subsequence such that {(Mni , gi (t) , xi)} converges to a complete pointed solution

to the Ricci flow (Mn∞, g∞ (t) , x∞) , t ∈ (α, ω) , as i → ∞.

Example. Let (R× S1 (2) , h) denote the flat cylinder, where h = dx2 + dθ2 and θ ∈ S1 (2) =R/4πZ. The Rosenau solution is the solution g (t) = u (t)h to the Ricci flow defined for t < 0 by

u (x, t) =sinh (−t)

coshx+ cosh t. (23)

The metrics g (t) extend to smooth metrics on S2 by adding the north pole N and south pole S atthe two ends of R× S1 (2). From

Rg(t) = −∆h log u

u=

cosh t coshx+ 1

sinh (−t) (coshx+ cosh t),

one checks that ∂∂tgij = −Rgij = −2Rij.

Observe that for x ∈ R fixed,

limt→−∞

u (x, t) = limt→−∞

sinh (−t)

coshx+ cosh t= 1.

14

Hence, for any fixed p ∈ S2 − {N,S} (which corresponds to (x, θ) ∈ R × S1 (2)), we have that(S2, g(t), p) converges to (R× S1 (2) , h) as t → −∞.

On the other hand, we have

u (x+ t, t) = (− coshx coth t− sinhx− coth t)−1 ,

so thatlim

t→−∞u (x+ t, t) = (cosh x− sinhx+ 1)−1 =

(e−x + 1

)−1.

Let ϕt : R×S1 (2) → R×S1 (2) (ϕt extends to a map from S2 to itself which preserves both N and S)be defined by ϕt (x, θ) = (x+ t, θ) . Then (ϕ∗

tg) (x, t) = u (x+ t, t)h. Hence

limt→−∞

(ϕ∗t g) (x, t) =

(e−x + 1

)−1h (x) . (24)

Changing variables by x̃ = x/2 and θ̃ = θ/2, we have(e−x + 1

)−1h (x) = 4

(e−2x̃ + 1

)−1(dx̃2 + dθ̃2

), (25)

where x̃ ∈ R and θ̃ ∈ S1 (1) . We now show that this is 4 times the cigar metric. Recall

gcig =dx2 + dy2

1 + x2 + y2= ds2 + tanh2 s dθ̃2.

Let x̃ = ln(sinh s). Then dx̃ = coth s ds, so that ds2 = tanh2 s dx̃2. ex̃ = sinh s. So e−2x̃+1 = coth2 s.

Thus gcig =(e−2x̃ + 1

)−1(dx̃2 + dθ̃2

).

6 Compact singularity models

Reference: [13]. (Exposition: chapter 17, §1 of [3].)Energy. Define

λ(g) = inf∫e−fdµ=1

F(g, f),

where F(g, f) =∫ (

R + |∇f |2)e−fdµ. Scaling property: λ(cg) = c−1λ(g). It is easy to see

Rmin ≤ λ(g) ≤ Ravg,

where the upper bound follows from taking f = lnVol(g).Standard elliptic theory implies that there exists a C∞ minimizer f0 of F(g, f) under the constraint∫

e−fdµ = 1. Another standard argument implies that the minimizer is unique. The Euler–Lagrangeequation is:

2∆f0 − |∇f0|2 +R = λ(g).

Let g(t), t ∈ I (I is an interval), be a solution to the Ricci flow and let λ(t) = λ(g(t)). Given t0,let f0 be the minimizer for λ(t0). Solve ∂f

∂t= −R − ∆f + |∇f |2 for t ≤ t0 with f(t0) = f0. Since

15

λ(t) ≤ F(g(t), f(t)) + F(t) for t ≤ t0 and λ(t0) = F(t0), we have in the sense of the lim inf ofbackward difference quotients:

d

dt−λ(t)|t=t0 ≥

d

dtF(t)|t=t0

= 2

∫ ∣∣Rc+∇2f∣∣2 e−fdµ

≥ 2

n

∫(R +∆f)2 e−fdµ

≤ 2

n

(∫(R +∆f) e−fdµ

)2

=2

nF(t0)

2

=2

nλ(t0)

2

since∫e−fdµ = 1.

No breathers theorem: Suppose g(t) is such that there exist t1 < t2 and a diffeomorphism φ :M → M such that g(t2) = φ∗g(t1). g(t) is called a breather solution. Then λ(g(t2)) = λ(g(t1)). Letf2 be a minimizer for F(g(t2), · ). Solve ∂f

∂t= −R−∆f + |∇f |2 for t ∈ [t1, t2] with f(t2) = f2. Then

for t ∈ [t1, t2] we have∫e−f(t)dµg(t) = 1 and

λ(g(t2)) = F(g(t2), f2) ≥ F(g(t), f(t)) ≥ F(g(t1), f1) ≥ λ(g(t1)).

Since λ(g(t2)) = λ(g(t1)), we conclude that F(g(t), f(t)) = λ(g(t)) is constant for t ∈ [t1, t2]. Hencef(t) is a minimizer for F(g(t), · ) and d

dtF(g(t), f(t)) = 0, which implies Rcg(t) +∇2

g(t)f(t) = 0 for

t ∈ [t1, t2]. Thus g(t) is a steady GRS.

Entropy. Recall the entropy:

W(g, f, τ) =

∫ (τ(R + |∇f |2

)+ f − n

)τ−

n2 e−fdµ.

µ(g, τ) = inf∫τ−

n2 e−f=1

W(g, f, τ).

Let f̄ = f + n2ln τ , so that e−f̄ = τ−

n2 e−f and the constraint for W is equivalent to

∫e−f̄ = 1. Then,

under the constraint:

µ(g, τ) ≤ W(g, f, τ) = τ

∫(R + |∇f̄ |2)e−f̄dµ+

∫ (f̄ − n

2ln τ)e−f̄dµ− n. (26)

So taking f so that f̄ is the minimizer for λ(g), we obtain

µ(g, τ) ≤ τλ(g) + e−1Vol(g)− n

2ln τ − n

since xe−x ≤ e−1.Suppose λ(g) ≤ 0. Then

µ(g, τ) ≤ e−1Vol(g)− n

2ln τ − n.

16

In particular, limτ→∞ µ(g, τ) = −∞, so that ν(g) = −∞. Now, by scale-invariance, for any c > 0 wehave λ(cg) = c−1λ(g) ≤ 0 and

µ(g, τ) = µ(cg, cτ) ≤ e−1Vol(cg)− n

2ln(cτ)− n.

Taking c = Vol(g)−2n , we have Vol(cg) = 1, so that

µ(g, τ) ≤ ln(τ−

n2 Vol(g)

)+ e−1 − n ≤ lnVol(τ−1g).

Let g(t), t ∈ [0, T ), T < ∞, be a solution to the Ricci flow. Suppose that λ(g(t)) ≤ 0 for all t.Then taking τ = 1, we have

lnVol(g(t)) ≥ µ(g(t), 1) ≥ µ(g(0), 1 + t) ≥ −C = minτ∈[1,1+T ]

µ(g(0), τ) > −∞.

In particular, we cannot have limt→T Vol(g(t)) = 0.Suppose g(t), t ∈ [0, T ), T < ∞, is a singular solution to the Ricci flow with a compact singularity

model. Then limt→T Vol(g(t)) = 0, so that λ(g(t0)) > 0 for some t0. This also implies λ(g(t)) ≥λ(g(t0)) > 0 for t ≥ t0.

By (26), we have for any f satisfying the constraint that

W(g, f, τ) = τF(g, f̄) +

∫f̄ e−f̄dµ− n

2ln τ − n.

By the logarithmic Sobolev inequality (see Notes 3) with a = 2 and since∫e−f̄dµ = 1 (let φ = e−

f̄2 ),

we have ∫f̄ e−f̄dµ ≥ −

∫|∇f̄ |2e−f̄dµ− 2C(2, g)

≥ F(g, f̄) +Rmin − 2C(2, g).

Hence, for τ > 1 we have

µ(g, τ) = inffW(g, f, τ) ≥ (τ − 1)λ(g) +Rmin − 2C(2, g)− n

2ln τ − n.

Thus, if λ(g) > 0, then limτ→∞ µ(g, τ) = ∞.Claim. µ(g, τ) < 0 for τ sufficiently small.Proof of the claim. By the short time existence theorem, there exists δ > 0 and a unique solution

g(t), t ∈ [0, δ], to the Ricci flow with g(0) = g. Let u = τ−n2 e−f , where τ(t) = δ − t, be a solution

to �∗u = 0 with u = (4π)n2 δx0 for any x0 ∈ M, where δx0 is the Dirac delta function centered at

x0. Then limt→δ W(g(t), f(t), τ(t)) = 0. By entropy monotonicity, we have W(g(t), f(t), τ(t)) ≤ 0 fort ∈ [0, δ]. If W(g(0), f(0), τ(0)) = 0, then we are on a shrinking GRS with scale 0 at time δ, whereg(δ) is smooth. So Rm ≡ 0, which contradicts M being compact.

Claim. limτ→0 µ(g, τ) = 0. (We prove this later.)Since τ 7→ µ(g, τ) is continuous, we have

ν(g) = infτ>0

µ(g, τ) ∈ (−∞, 0).

Now since t 7→ ν(g(t)) is nondecreasing, we have limt→T ν(g(t)) + νT ∈ (−∞, 0] exists.

17

Theorem 21 (Zhenlei Zhang) Any compact singularity model is a shrinking GRS.

Proof. Suppose g(t), t ∈ [0, T ), T < ∞, is a singular solution to the Ricci flow with a compactsingularity model (Mn

∞, g∞(t)). Then there exist (xi, ti) with Ki = |Rm |(xi, ti) → ∞ such that(Mn

i , gi(t)) converges to (Mn∞, g∞(t)) and Mn

∞ is diffeomorphic to Mn. Hence λ(gi(t)) → λ(g∞(t)).Since λ(gi(t)) > 0, this implies λ(g∞(t)) ≥ 0.

Suppose there exists t0 such that λ(g∞(t)) = 0. Then λ(g∞(t)) = 0 for t ≤ t0. This implies g∞(t),t ≤ t0, is a steady GRS. Since Mn

∞ is compact, this implies Rcg∞(t) = 0. However, this contradictsno local collapsing: Let r be any positive real number. Since Rg∞(t) = 0 on Mn

∞, we have for any

x0 ∈ Mn∞ that Rg∞(t) ≤ r−2 in B = B

g∞(t)r (x0). Hence

Vol(Mn∞, g∞(t)) ≥ Vol(B, g∞(t)) ≥ κrn.

Since r is arbitrary, we have a contradiction.Finally, since after pullback gi(t) converges to g∞(t) pointwise in C∞, we have

ν(g∞(t)) = limi→∞

ν(gi(t)) = limi→∞

ν(g(K−1i t+ ti)) = νT

since for each t we have K−1i t + ti → T . Since ν(g∞(t)) is constant, we conclude that g∞(t) is a

shrinking GRS. This also implies that Mn admits a shrinking GRS structure. �

7 The proof of the strong maximum principle

Reference: §8 and §9 of [4man]. (Exposition: chapter 12 of [3].)We assume the following result:

Strong maximum principle (for the scalar heat equation). Let (Mn, g(t)), t ∈ I, be a family ofRiemannian metrics on a connected manifold, not necessarily complete. Suppose that u : M× I → Rsatisfies ∂u

∂t≥ ∆u and u ≥ 0. If u (x0, t0) = 0, then u(x, t) = 0 for all x and t ≤ t0.

Since ∂R∂t

≥ ∆R under the Ricci flow, if R ≥ 0 on M× I and R (x0, t0) = 0, then R(x, t) = 0 for

all x and t ≤ t0. Since∂R∂t

= ∆R + 2 |Rc|2, this implies Rc(x, t) = 0 for all x and t ≤ t0.

Strong maximum principle for Rm (statement). Let (Mn, g(t)), t ∈ [0, T ) be a solution to theRicci flow with nonnegative curvature operator on a connected manifold. Then for each t ∈ (0, T )the set image

(Rmg(t)

)⊂ Λ2T ∗M is a smooth subbundle which is invariant under parallel translation

(in space). Moreover, there exists a finite sequence of times 0 = t0 < t1 < t2 < · · · < tk < T suchthat image (Rm) (x, t) is a Lie subalgebra of Λ2T ∗

xM ∼= so (n) independent of time for t ∈ (ti−1, ti],1 ≤ i ≤ k, and for t ∈ (tk, T ). Furthermore, image(Rmg(0)) ⊂ image(Rmg(t1)) and image(Rmg(ti)) is aproper subalgebra of image(Rmg(tj)) for 1 ≤ i < j ≤ k and image(Rmg(tk)) is a proper subalgebra ofimage(Rmg(t)) for t > tk. If M is closed, then we may take k = 0.

Strong maximum principle for Rm (applications). We have the following immediate consequences:

1. If Rm = 0 at (x̄, t̄), then Rm (x, t) = 0 for all x ∈ M and t ≤ t̄.

2. If Rm > 0 at (x̄, 0), then Rm (x, t) > 0 for all x ∈ M and t > 0.

For n = 3 any subalgebra of so (3) besides 0 and so (3) is isomorphic to so (2).Let (M3, g(t)), t ∈ (α, ω), be a complete solution to the Ricci flow with bounded nonnegative

sectional curvature.

18

1. Suppose Rm = 0 at (x′, t′). Then the strong maximum principle implies Rm (x, t) = 0 for allx ∈ M and t ≤ t′. By the (forward) uniqueness theorem of Hamilton (compact case) and Chen andZhu (noncompact case), we have Rm = 0 on M× (α, ω).

2. Suppose image (Rm) (x′, t′) ∼= so (2). Then image (Rm) (x, t′) ∼= so (2) for x ∈ M is invari-ant under parallel translation. Lift the metric (M, g(t′)) to the universal cover (M̃, g̃(t′)). Then(M̃, g̃(t′)) = (N 2, h′) × R, where h′ has bounded curvature. By forward existence and uniqueness,we have for t ∈ [t′, ω) that (M̃, g̃(t)) is the product of R and some solution (N 2, h(t)). We cannothave Rm > 0 at any point at any time t < t′ since that would imply that Rm > 0 on M × [t′, ω),a contradiction. By (1), we also cannot have image (Rm) = 0 at any (x, t) ∈ M × (α, ω). Henceimage (Rm) (x, t) ∼= so (2) for all (x, t) ∈ M × (α, ω) is invariant under parallel translation andindependent of time. Hence (M̃, g̃(t)) is the product of R and a solution (N 2, h(t)) for t ∈ (α, ω).

3. Suppose Rm(x′, t′) > 0. Then Rm > 0 on M× (α, ω) for otherwise we would be in case (1) or(2), a contradiction.

To summarize, on all of M× (α, ω) either Rm = 0, Rm > 0, or the universal cover splits as theproduct of R and solution on a surface with positive curvature.

Strong maximum principle for Rm (proof in dimension 3). Let n = 3. Given a self-adjoint linearmap R : Λ2T ∗

xM → Λ2T ∗xM, let {λi (R)} denote the eigenvalues of R in nondecreasing order. Given

1 ≤ k ≤ 3, defineϕk (R) = λ1 (R) + · · ·+ λk (R) = inf

dimS=ktrace(R|S),

where the infimum is taken over all k-dimensional subspaces S of Λ2T ∗xM. In each fiber ϕk is a

concave function of R. Define ϕk : M× [0, T ) → R by

ϕk (x, t) = ϕk (Rm (x, t)) .

Lemma. If ϕk (x0, t0) > 0 for some k and x0 ∈ M, then ϕk > 0 on M× (t0, T ).

Proof of the lemma. Choose any smooth function h0 : M → R with 0 ≤ h0 ≤ ϕk and h0(x0) > 0.Let h : [t0, T ) → R be the solution to ∂h

∂t= ∆g(t)h with h(t0) = h0. By the strong maximum principle,

h > 0 on M× (t0, T ). So the lemma follows from:Claim. ϕk(x, t) ≥ h(x, t) on M× [t0, T ).Proof of the claim. Consider the system of PDE

∂

∂t(Rm, h) = ∆(Rm, h) + (Q, 0) .

Let K = {(R,h) |ϕk (R) ≥ h}. This a closed convex set which is invariant under parallel translation.The associated ODE is

d

dt(R,h) = (Q(R), 0),

where Q(R) = R2 +R#. Since Q(R) ≥ 0 when R ≥ 0, we have that K is preserved by the ODE.Hence K is preserved by the PDE. �

The null space of Rm at (x, t) is

null (Rm (x, t)) ={V ∈ Λ2T ∗

xM : Rm (x, t) (V,W ) = 0 for all W ∈ Λ2T ∗xM

}.

Since Rm ≥ 0,null (Rm (x, t)) =

{V ∈ Λ2T ∗

xM : Rm (x, t) (V, V ) = 0}.

19

By the lemma, there exist times 0 = t0 < t1 < t2 < · · · < tk < tk+1 = T such that dimnull (Rm (x, t))is constant on M× (ti−1, ti], 1 ≤ i ≤ k, and on M× (tk, T ).

Now suppose Rm11 (x0, t0) = 0, i.e., λ = 0. Then Q11 = 0, i.e., λ2 + µν = 0, so that µν = 0. ThusRm (x0, t0) = 0 or image(Rm (x0, t0)) is 1-dimensional.

If ϕ3 (x0, t0) = 0 somewhere, then ϕ3 = 0 on M× (α, t0]. By this and forward uniqueness, we haveRm = 0 on M× (α, ω).

So we may assume ϕ3 > 0 on M×(α, ω). For each (x0, t0) we have ϕ1 (x0, t0) > 0 or ϕ1 (x0, t0) > 0,which is equivalent to ϕ2 (x0, t0) = 0.

If ϕ1 > 0 on M× (α, ω), then Rm > 0 on M× (α, ω).Otherwise ϕ2 (x0, t0) = 0 somewhere, which implies ϕ2 = 0 on M×(α, t0]. This implies that either

ϕ2 = 0 on M× (α, ω) or there exists t1 ∈ (α, ω) such that ϕ2 = 0 on M× (α, t1] and ϕ1 > 0 (Rm > 0)on M× (t1, ω).

Claim. null(Rm) is invariant under parallel translation a subset of null(Q).

Proof of the claim. Let V (x, t) be a smooth local section of the subbundle null (Rm (x, t)) definedon an open subset of M× (α, ω). Using ∆ (Rm (V, V )) = 0 and Rm (∆V, V ) = 0, we compute that

0 =∂ Rm

∂t(V, V ) + 2Rm

(∂V

∂t, V

)(27)

= (∆Rm+Q) (V, V )

= −2gij Rm(∇iV,∇jV )− 4gij (∇i Rm) (∇jV, V ) +Q (V, V ) .

Since0 = ∇ (Rm (V )) = (∇Rm) (V ) + Rm (∇V ) ,

we have0 = gij (∇iRm) (∇jV, V ) + gij Rm(∇iV,∇jV ) .

Hence (27) yields2gij Rm(∇iV,∇jV ) +Q (V, V ) = 0.

Since both terms in the above equation are nonnegative, we conclude thatQ (V, V ) = 0 and gij Rm(∇iV,∇jV ) =0. In particular, null (Rm) ⊂ null (Q) and Rm (∇Y V,∇Y V ) = 0 for all Y , that is,

∇Y V (x, t) ∈ null (Rm (x, t)) for all Y. (28)

Now prove (28) implies that null (Rm (x, t)) is invariant under parallel translation. Let p =dimnull (Rm (x, t)), which is independent of x. Given (x0, t0), let {Wi(x)}pi=1 be a basis of smoothsections of null (Rm (x, t0)) defined in a neighborhood U of x0 in M and let γ : [0, L] → U be anypath with γ(0) = x0. Since ∇γ′(s)Wi ∈ null (Rm (γ(s), t0)), there exist functions Aj

i (s) such that

∇γ′(s)Wi =

p∑j=1

Aji (s)Wj.

Let V0 ∈ null (Rm (x0, t0)). Then V0 =∑p

i=1 fi0Wi (x0). Let V (s) =

∑pi=1 f

i(s)Wi(γ(s)). The IVP:

0 = ∇γ′(f iWi

)=

p∑i=1

(df i

dsWi + f i∇γ′Wi

), f i (0) = f i

0,

20

is equivalent to IVP for the system of linear ODE:

df i

ds+ f j

p∑j=1

Aij(s) = 0, f i (0) = f i

0,

which has a unique solution. Hence the parallel translate V (s) of V0 along γ is in null (Rm (γ(s), t0)).Finally we prove that null (Rm (x, t)) is constant in time onM×(ti−1, ti), 1 ≤ i ≤ k+1. Let V (x, t)

be a smooth local section of null (Rm (x, t)) defined in a space-time neighborhood in M× (ti−1, ti).Claim. V (x, t) ∈ null (∆Rm(x, t)).Proof of the claim. Since Rm (∇jV ) = 0, we have

0 = gij∇i (Rm (∇jV )) (29)

= gij (∇i Rm) (∇jV ) + Rm (∆V ) .

Since ∇eiV ∈ null(Rm) for any vector field ei, we have ∇ei (∇eiV ) ∈ null(Rm) and ∇∇eieiV ∈

null(Rm). Thus ∆V (x, t) ∈ null (Rm (x, t)), which implies

gij (∇i Rm) (∇jV ) = 0.

Using the above, we compute

0 = ∆ (Rm (V ))

= (∆Rm) (V ) + Rm (∆V ) + 2gij (∇iRm) (∇jV )

= (∆Rm) (V ) ,

which proves the claim.Now, since V ∈ null (Q (Rm)), we have

0 =∂

∂t(Rm (V ))

= (∆Rm+Q (Rm)) (V ) + Rm

(∂V

∂t

)= Rm

(∂V

∂t

).

Hence ∂V∂t

∈ null (Rm (t)). This implies null (Rm (t)) is constant in time.

8 Appendix: Review of Riemannian geometry.

Let (Mn, g) be a Riemannian metric and let ∇ denote the Levi-Civita connection. Given localcoordinates {xi}ni=1, the Christoffel symbols are the components of ∇: ∇ ∂

∂xi

∂∂xj = Γk

ij∂

∂xk .

Lemma 22

Γkij =

1

2gkℓ(

∂

∂xigjℓ +

∂

∂xjgiℓ −

∂

∂xℓgij

). (30)

21

Proof. For any vector fields X, Y and Z that

2g (∇XY, Z) = X (g (Y, Z)) + Y (g (X,Z))− Z (g (X, Y )) (31)

+ g ([X,Y ] , Z)− g ([X,Z] , Y )− g ([Y, Z] , X) .

Applying this to X = ∂∂xi , Y = ∂

∂xj , Z = ∂∂xℓ , multiplying bygkℓ, and using

[∂∂xi ,

∂∂xj

]= 0, we obtain

equation (30).Let α be a covariant r-tensor and X be a vector field. Lie derivative of α with respect to X is

denoted by LXα. Formula:

(LXα) (Y1, . . . , Yr) = (∇Xα) (Y1, . . . , Yr) (32)

+n∑

i=1

α (Y1, . . . , Yi−1,∇YiX,Yi+1, . . . , Yr) .

Since ∇Xg = 0, the Lie derivative of the metric is given by

(LXg) (Y1, Y2) = g (∇Y1X, Y2) + g (Y1,∇Y2X) . (33)

Lemma 23 If α is a covariant 2-tensor, then

(LXα)ij = ∇Xαij + gkℓ (∇iXkαℓj +∇jXkαiℓ)

and in particular(LXg)ij = ∇iXj +∇jXi,

so that if f is a function, then (Lgradg fg

)ij= 2∇i∇jf. (34)

Proof. Apply (32) with Y1 = ∂∂xi , Y2 = ∂

∂xj . To obtain equation (34), let Xi = ∇if . Note that∇i∇jf = ∇2f

(∂∂xi ,

∂∂xj

)and ∇2f (X, Y ) = X (Y f)− (∇XY ) f .

Let Rijkℓ = gℓmRmijk be the components of the Riemann curvature 4-tensor. The first and second

Bianchi identities are

Rijkℓ +Rjkiℓ +Rkijℓ = 0, (35)

∇iRjkℓm +∇jRkiℓm +∇kRijℓm = 0. (36)

Lemma 24 Let Rij be the Ricci tensor and let R be the scalar curvature. We have the contractedsecond Bianchi identity :

2gij∇iRjk = ∇kR. (37)

If g is an Einstein metric, i.e., Rij =1nRgij, and n ≥ 3, then R is a constant.

Proof. Multiplying (36) by gim, i.e., tracing, we obtain

0 = gim (∇iRjkℓm +∇jRkiℓm +∇kRijℓm)

= gim∇iRjkℓm −∇jRkℓ +∇kRjℓ.

Tracing again, i.e., multiplying by gkℓ, we obtain

gim∇iRjm −∇jR + gkℓ∇kRjℓ = 0,

22

which yields (37).Taking the divergence of Rjk =

1nRgjk yields

1

2∇kR = gij∇iRjk = gij∇i

(1

nRgjk

)=

1

n∇kR

and the result follows from n ̸= 2.The Ricci identities:

(∇i∇j −∇j∇i)αk1···kr = −r∑

ℓ=1

Rmijkℓ

αk1···kℓ−1mkℓ+1···kr . (38)

As special cases of (38), we have:

Lemma 25 1. If α is a 1-form, then

(∇i∇j −∇j∇i)αk = −Rℓijkαℓ.

2. If β is a 2-tensor, then

∇i∇jβkℓ −∇j∇iβkℓ = −Rpijk βpℓ −Rp

ijℓ βkp. (39)

Acting on functions, the Laplacian is

∆f = div∇f = trace g(∇2f) = gij∇i∇jf = gij(

∂2f

∂xi∂xj− Γk

ij

∂f

∂xk

).

One can show that

∆f =1√|g|

n∑i,j=1

∂

∂xi

(√|g|gij ∂f

∂xj

), (40)

where |g| + det (gij).

Lemma 26 1.∆∇if = ∇i∆f +Rij∇jf. (41)

2.∆ |∇f |2 = 2 |∇∇f |2 + 2Rij∇if∇jf + 2∇if∇i (∆f) . (42)

3. If Rc ≥ 0, ∆f ≡ 0, and |∇f | = 1, then ∇2f = ∇(∇f) = 0, i.e., ∇f is parallel, andRc (∇f,∇f) = 0.

Proof. (1) This follows from

∆∇if = ∇j∇i∇jf = ∇i∇j∇jf −Rjijk∇kf,

where we used the Ricci identities.(2) We compute

∆ |∇f |2 = ∇i∇i |∇jf |2 = 2∇i (∇i∇jf∇jf)

= 2 |∇i∇jf |2 + 2∆∇jf∇jf

23

and obtain part (2) from part (1).(3) If Rc ≥ 0 and ∆f = 0, then part (2) implies

∆ |∇f |2 = 2∣∣∇2f

∣∣2 + 2Rc (∇f,∇f) ≥ 2 |∇∇f |2 .

Hence, if also |∇f | = 0, then |∇2f | = 0 and Rc (∇f,∇f) = 0.Divergence theorem:

Lemma 27 Let (Mn, g) be a compact oriented Riemannian manifold. Let dµ denote the volume formof (M, g), let dσ denote the area form of ∂M, and let ν denote the unit outward normal to ∂M. IfX is a vector field, then ∫

Mdiv (X) dµ =

∫∂M

⟨X, ν⟩ dσ, (43)

where div (X) = ∇iXi. In particular, if M is closed, then

∫M div (X) dµ = 0.

Proof. Define the (n− 1)-form α byα + ιX(dµ).

Using d2 = 0 and LX(dµ) = (d ◦ ιX + ιX ◦ d) (dµ), we compute

dα = d ◦ ιX(dµ) = (d ◦ ιX + ιX ◦ d) (dµ) = LX (dµ) = div(X)dµ,

where to obtain the last equality, we may compute in an orthonormal frame e1, . . . , en :

LX (dµ) (e1, . . . , en) =n∑

i=1

dµ (e1, . . . ,∇eiX, . . . , en)

= div(X)dµ (e1, . . . , en) .

Now Stokes’s theorem implies that∫M

div(X)dµ =

∫M

dα =

∫∂M

α =

∫∂M

ιX(dµ) =

∫∂M

⟨X, ν⟩ dσ.

To verify the last equality, we used

(ιX(dµ)) (e2, . . . , en) = n (dµ) (X, e2, . . . , en)

= n ⟨X, ν⟩ (dµ) (e1, e2, . . . , en)

=1

(n− 1)!⟨X, ν⟩ .

Lemma 28 1. On a closed manifold,∫M ∆udµ = 0.

2. On a compact manifold,∫M

(u∆v − v∆u) dµ =

∫∂M

(u∂v

∂ν− v

∂u

∂ν

)dσ.

In particular, on a closed manifold∫M

u∆vdµ =

∫M

v∆udµ.

24

3. If f is a function and X is a 1-form, then∫M

f div (X) dµ = −∫M

⟨∇f,X⟩ dµ+

∫∂M

f ⟨X, ν⟩ dσ.

Proof. (1) This follows from the divergence theorem and ∆u = div(∇u).(2) This follows from the divergence theorem applied to X = u∇v − v∇u which has div(X) =

u∆v − v∆u.(3) This follows from the divergence theorem applied to the vector field fX.An easy calculation using coordinates where gij = δij at a point shows that for any 2-tensor aij,

|aij|2g ≥1

n

(gijaij

)2.

More generally, for a p-tensor a, p ≥ 2,∣∣ak1···kp∣∣2 ≥ 1

n

∣∣gijaijk3···kp∣∣2 .Let Rc denote the Ricci tensor.

Lemma 29 Show that on a closed manifold∫M

∣∣∇2f∣∣2 dµ+

∫Mn

Rc (∇f,∇f) dµ =

∫M

(∆f)2 dµ. (44)

Since |∇2f |2 ≥ 1n(∆f)2 , this implies∫

MRc (∇f,∇f) dµ ≤ n− 1

n

∫M

(∆f)2 dµ. (45)

Proof. This follows from (42) since integration by parts yields∫M ∇f ·∇∆fdµ = −

∫M (∆f)2 dµ.

Lichnerowicz inequality:

Lemma 30 Suppose f is an eigenfunction of the Laplacian with eigenvalue λ > 0:

∆f + λf = 0.

If Rc ≥ (n− 1)K, where K > 0 is a constant, then

λ ≥ nK. (46)

Proof. If ∆u+ λu = 0, then

1

2∆ |∇u|2 = |∇∇u|2 + ⟨∇∆u,∇u⟩+Rc (∇u,∇u)

≥ 1

n(∆u)2 − λ |∇u|2 + (n− 1)K |∇u|2

=1

nλ2u2 + ((n− 1)K − λ) |∇u|2 .

Integrating this, we have

0 ≥ 1

nλ2

∫M

u2dµ+ ((n− 1)K − λ)

∫M

|∇u|2 dµ.

Dividing by∫Mu2dµ and using the fact that λ =

∫M|∇u|2 dµ

/∫Mu2dµ implies

0 ≥ 1

nλ2 + ((n− 1)K − λ)λ.

25

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