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The Cotton Tensor and The Ricci Flow / Mantegazza, Carlo; Mongodi, Samuele; Rimoldi, Michele. - In: GEOMETRIC FLOWS. - ISSN 2353-3382. - ELETTRONICO. - 2:1(2017), pp. 49-71. [10.1515/geofl-2017-0001]

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The Cotton Tensor and The Ricci Flow

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Research Article Open Access Carlo Mantegazza, Samuele Mongodi, and Michele Rimoldi*

The Cotton Tensor and the Ricci Flow

https://doi.org/10.1515/geofl-2017-0001

Received April 20, 2017; accepted October 23, 2017

Abstract:We compute the evolution equation of the Cotton and the Bach tensor under the Ricci flow of a Riemannian manifold, with particular attention to the three dimensional case, and we discuss some appli- cations.

Keywords:Ricci flow, Cotton tensor, Bach tensor, Ricci solitons MSC:53C21, 53C25

1 Preliminaries

The Riemann curvature operator of a Riemannian manifold (Mn, g) is defined, as in [6], by Riem(X, Y)Z =YXZXYZ+[X,Y]Z.

In a local coordinate system the components of the (3, 1)–Riemann curvature tensor are given by Rlijk∂xl = Riem ∂xi,∂xj



∂xk and we denote by Rijkl= glmRmijkits (4, 0)–version.

With the previous choice, for the sphere Snwe have Riem(v, w, v, w) = Rabcdvawbvcwd> 0.

In all the paper the Einstein convention of summing over the repeated indices will be adopted.

The Ricci tensor is obtained by the contraction Rik= gjlRijkland R = gikRikwill denote the scalar curva- ture.

We recall the interchange of derivative formula,

2ijωk2jiωk= Rijkpgpqωq, and Schur lemma, which follows by the second Bianchi identity,

2gpqpRqi=iR . They both will be used extensively in the computations that follows.

The so called Weyl tensor is then defined by the following decomposition formula (see [6, Chapter 3, Section K]) in dimension n ≥ 3,

Rijkl= 1

n− 2(Rikgjl− Rilgjk+ Rjlgik− Rjkgil) − R

(n − 1)(n − 2)(gikgjl− gilgjk) + Wijkl. (1.1) The Weyl tensor satisfies all the symmetries of the curvature tensor, moreover, all its traces with the metric are zero, as it can be easily seen by the above formula.

*Corresponding Author: Michele Rimoldi:Dipartimento di Scienze Matematiche "Giuseppe Luigi Lagrange", Politecnico di Torino, Corso Duca degli Abruzzi, 24, Torino, Italy, I-10129, E-mail: [email protected]

Carlo Mantegazza:Dipartimento di Matematica, Università di Napoli "Federico II", Via Cintia, Monte S. Angelo, Napoli, Italy, I-80126, E-mail: [email protected]

Samuele Mongodi:Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci, 32, Milano, Italy, I-20133, E-mail: [email protected]

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In dimension three W is identically zero for every Riemannian manifold. It becomes relevant instead when n≥ 4 since its vanishing is a condition equivalent for (Mn, g) to be locally conformally flat, that is, around every point p Mnthere is a conformal deformationegij = efgijof the original metric g, such that the new metric is flat, namely, the Riemann tensor associated toegis zero in Up(here f : UpR is a smooth function defined in a open neighborhood Upof p).

In dimension n = 3, instead, locally conformally flatness is equivalent to the vanishing of the following Cotton tensor

Cijk=kRijjRik− 1

2(n − 1) kRgijjRgik , (1.2) which expresses the fact that the Schouten tensor

Sij= RijRgij

2(n − 1)

is a Codazzi tensor (see [1, Chapter 16, Section C]), that is, a symmetric bilinear form Tijsuch thatkTij =

iTkj.

By means of the second Bianchi identity, one can easily get (see [1]) that

lWlijk= −n− 3

n− 2Cijk. (1.3)

Hence, when n ≥ 4, if we assume that the manifold is locally conformally flat (that is, W = 0), the Cotton tensor is identically zero also in this case, but this is only a necessary condition.

By direct computation, we can see that the tensor Cijksatisfies the following symmetries

Cijk= −Cikj, Cijk+ Cjki+ Ckij= 0 , (1.4) moreover it is trace–free in any two indices,

gijCijk= gikCijk= gjkCijk= 0 , (1.5) by its skew–symmetry and Schur lemma.

We suppose now that (Mn, g(t)) is a Ricci flow in some time interval, that is, the time–dependent metric g(t) satisfies

∂tgij= −2Rij.

We have then the following evolution equations for the Christoffel symbols, the Ricci tensor and the scalar curvature (see for instance [7]),

∂tΓkij = −gksiRjs− gksjRis+ gkssRij

∂tRij = ∆Rij− 2RklRkijl− 2gpqRipRjq (1.6)

∂tR = ∆R + 2|Ric|2.

All the computations which follow will be done in a fixed local frame, not in a moving frame.

Acknowledgments. The first and second authors were partially supported by the Italian FIRB Ideas “Analysis and Beyond”.

Note1.1. We remark that Huai-Dong Cao also, independently by us, worked out the computation of the evo- lution of the Cotton tensor in dimension three, in an unpublished note.

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2 The Evolution Equation of the Cotton Tensor in 3D

The goal of this section is to compute the evolution equation under the Ricci flow of the Cotton tensor Cijkin dimension three (see [5] for the evolution of the Weyl tensor), the general computation in any dimension is postponed to section 4.

In the special three–dimensional case we have,

Rijkl= Rikgjl− Rilgjk+ Rjlgik− Rjkgil−R

2(gikgjl− gilgjk) , (2.1) Cijk=kRijjRik−1

4 kRgijjRgik , (2.2)

hence, the evolution equations (1.6) become

∂tΓkij= − gksiRjs− gksjRis+ gkssRij

∂tRij= ∆Rij− 6gpqRipRjq+ 3RRij+ 2|Ric|2gij− R2gij

∂tR = ∆R + 2|Ric|2.

From these formulas we can compute the evolution equations of the derivatives of the curvatures assuming, from now on, to be in normal coordinates,

∂tlR = lR + 2l|Ric|2,

∂tsRij = sRij− 6sRipRjp− 6RipsRjp+ 3sRRij+ 3RsRij

+2s|Ric|2gijsR2gij+ (iRsp+sRippRis)Rjp+ (jRsp+sRjppRjs)Rip

= sRij− 5sRipRjp− 5RipsRjp+ 3sRRij+ 3RsRij

+2s|Ric|2gijsR2gij+ (iRsppRis)Rjp+ (jRsppRjs)Rip

= sRij− 5sRipRjp− 5RipsRjp+ 3sRRij+ 3RsRij

+2s|Ric|2gijsR2gij+ CspiRjp+ CspjRip+ Rjp[iRgsppRgis]/4 + Rip[jRgsppRgjs]/4 , where in the last passage we substituted the expression of the Cotton tensor.

We then compute,

∂tCijk =

∂tkRij

∂tjRik

∂t kRgijjRgik/4

= kRij− 5kRipRjp− 5RipkRjp+ 3kRRij+ 3RkRij

+2k|Ric|2gijkR2gij+ CkpiRjp+ CkpjRip

+Rjp[iRgkppRgik]/4 + Rip[jRgkppRgjk]/4

jRik+ 5jRipRkp+ 5RipjRkp− 3jRRik− 3RjRik

−2j|Ric|2gik+jR2gik− CjpiRkp− CjpkRip

−Rkp[iRgjppRgij]/4 − Rip[kRgjppRgkj]/4

+(RijkR − RikjR/2 − kR + 2k|Ric|2gij/4 + jR + 2j|Ric|2gik/4

= kRij− 5kRipRjp− 5RipkRjp+ 3kRRij+ 3RkRij

+3k|Ric|2gij/2 −kR2gij+ CkpiRjp+ CkpjRip

+RjkiR/4 − RjppRgik/4 + RikjR/4 − RippRgjk/4

jRik+ 5jRipRkp+ 5RipjRkp− 3jRRik− 3RjRik

−3j|Ric|2gik/2 +jR2gik− CjpiRkp− CjpkRip

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−RkjiR/4 + RkppRgij/4 − RijkR/4 + RippRgkj/4 +(RijkR − RikjR/2 −k∆Rgij/4 +j∆Rgik/4

= kRij− 5kRipRjp− 5RipkRjp+ 13kRRij/4 + 3RkRij

+3k|Ric|2gij/2 −kR2gij+ CkpiRjp+ CkpjRip− RjppRgik/4

jRik+ 5jRipRkp+ 5RipjRkp− 13jRRik/4 − 3RjRik

−3j|Ric|2gik/2 +jR2gik− CjpiRkp− CjpkRip

+RkppRgij/4 −k∆Rgij/4 +j∆Rgik/4 and

Cijk= ∆kRij− ∆jRik− ∆kRgij/4 + ∆jRgik/4 , hence,

∂tCijk− ∆Cijk = kRijjRik− ∆kRij+ ∆jRik

k∆Rgij/4 +j∆Rgik/4 + ∆kRgij/4 − ∆jRgik/4

−5kRipRjp− 5RipkRjp+ 13kRRij/4 + 3RkRij

+3k|Ric|2gij/2 −kR2gij+ CkpiRjp+ CkpjRip− RjppRgik/4 +5jRipRkp+ 5RipjRkp− 13jRRik/4 − 3RjRik

−3j|Ric|2gik/2 +jR2gik− CjpiRkp− CjpkRip

+RkppRgij/4

Now to proceed, we need the following commutation rules for the derivatives of the Ricci tensor and of the scalar curvature, where we will employ the special form of the Riemann tensor in dimension three given by formula (2.1),

kRij− ∆kRij = 3kllRij3lklRij+3lklRij3llkRij

= −RkppRij+ RkliplRjp+ RkljplRip+3lklRij3llkRij

= −RkppRij+ RikjR/2 + RjkiR/2

−RkpiRjp− RkpjRip+ RlplRjpgik+ RlplRipgjk

−RlilRjk− RljlRik− RjRgik/4 − RiRgjk/4 +RiRjk/2 + RjRik/2 +l RklipRpj+ RkljpRpi



= −RkppRij+ RikjR/2 + RjkiR/2

−RkpiRjp− RkpjRip+ RlplRjpgik+ RlplRipgjk

−RlilRjk− RljlRik− RjRgik/4 − RiRgjk/4 +RiRjk/2 + RjRik/2

+l RikRlj− RilRkj+ RplRpjgik− RpkRpjgil− gikRRlj/2 + gilRRjk/2 +RjkRli− RjlRki+ RplRpigjk− RpkRpigjl− gjkRRli/2 + gjlRRik/2

= −RkppRij+ RikjR/2 + RjkiR/2

−RkpiRjp− RkpjRip+ RlplRjpgik+ RlplRipgjk

−RlilRjk− RljlRik− RjRgik/4 − RiRgjk/4 +RiRjk/2 + RjRik/2

iRpkRpj+iRRjk/2 + gikRpllRpj

−RpkiRpj− gikRjR/4 + RiRjk/2

jRpkRpi+jRRik/2 + gjkRpllRpi

−RpkjRpi− gjkRiR/4 + RjRik/2

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= −RkppRij+ RikjR + RjkiR

−2RkpiRjp− 2RkpjRip+ 2RlplRjpgik+ 2RlplRipgjk

−RlilRjk− RljlRik− RpjiRpk− RpijRpk

−RjRgik/2 − RiRgjk/2 + RiRjk+ RjRik

and

k∆R − ∆kR = RkllppR = −RkppR . Then, getting back to the main computation, we obtain

∂tCijk− ∆Cijk = −RkppRij+ RikjR + RjkiR

−2RkpiRjp− 2RkpjRip+ 2RlplRjpgik+ 2RlplRipgjk

−RlilRjk− RljlRik− RpjiRpk− RpijRpk

−RjRgik/2 − RiRgjk/2 + RiRjk+ RjRik

+RjppRik− RijkR − RkjiR

+2RjpiRkp+ 2RjpkRip− 2RlplRkpgij− 2RlplRipgkj +RlilRkj+ RlklRij+ RpkiRpj+ RpikRpj

+RkRgij/2 + RiRgkj/2 − RiRkj− RkRij

+RkppRgij/4 − RjppRgik/4

−5kRipRjp− 5RipkRjp+ 13kRRij/4 + 3RkRij

+3k|Ric|2gij/2 −kR2gij+ CkpiRjp+ CkpjRip− RjppRgik/4 +5jRipRkp+ 5RipjRkp− 13jRRik/4 − 3RjRik

−3j|Ric|2gik/2 +jR2gik− CjpiRkp− CjpkRip+ RkppRgij/4

= CkpiRjp+ CkpjRip− CjpiRkp− CjpkRip

+[2RlplRjp+ 3RjR/2 − RjppR/2 − 3j|Ric|2/2]gik

+[−2RlplRkp− 3RkR/2 + RkppR/2 + 3k|Ric|2/2]gij

−RkpiRjp+ RjpiRkp− 3kRipRjp− 4RipkRjp+ 9kRRij/4 + 2RkRij

+3jRipRkp+ 4RipjRkp− 9jRRik/4 − 2RjRik

Now, by means of the very definition of the Cotton tensor in dimension three (2.2) and the identities (1.4), we substitute

Ckpj− Cjpk= − Ckjp− Cjpk= Cpkj

lRjp=jRlp+ Cpjl+1

4 lRgpjjRgpl



lRkp=kRlp+ Cpkl+1

4 lRgpkkRgpl



iRjp=jRip+ Cpji+1

4 iRgjpjRgip



iRkp=kRip+ Cpki+1

4 iRgkpkRgip

 in the last expression above, getting

∂tCijk− ∆Cijk = RjpCkpi− RkpCjpi+ RipCpkj

+h

2RlpjRlp+ Cpjl+lRgpj/4 −jRgpl/4 + 3RjR/2 − RjppR/2 − 3j|Ric|2/2i

gik +h

− 2RlpkRlp+ Cpkl+lRgpk/4 −kRgpl/4

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− 3RkR/2 + RkppR/2 + 3k|Ric|2/2i gij

−RkpjRip+ Cpji+iRgjp/4 −jRgip/4 +RjpkRip+ Cpki+iRgkp/4 −kRgip/4

−3kRipRjp− 4RipkRjp+ 9kRRij/4 + 2RkRij

+3jRipRkp+ 4RipjRkp− 9jRRik/4 − 2RjRik

= Rjp Ckpi+ Cpki − Rkp Cjpi+ Cpji + RipCpkj

+2RlpCpjlgik− 2RlpCpklgij

+RjR −j|Ric|2/2gik−RkR −k|Ric|2/2gij

−2kRipRjp− 4RipkRjp+ 2kRRij+ 2RkRij

+2jRipRkp+ 4RipjRkp− 2jRRik− 2RjRik. then, we substitute again

kRjp=pRkj+ Cjpk+1

4 kRgjppRgjk



jRkp=pRjk+ Ckpj+1

4 jRgkppRgkj



kRij=iRkj+ Cjik+1

4 kRgijiRgjk



jRik=iRjk+ Ckij+1

4 jRgikiRgkj , finally obtaining

∂tCijk− ∆Cijk = Rjp Ckpi+ Cpki − Rkp Cjpi+ Cpji + RipCpkj

+2RlpCpjlgik− 2RlpCpklgij

+RjR −j|Ric|2/2gik−RkR −k|Ric|2/2gij

−2kRipRjp− 4RippRkj+ Cjpk+kRgjp/4 −pRgjk/4 +2kRRij+ 2R iRkj+ Cjik+kRgij/4 −iRgjk/4 +2jRipRkp+ 4RippRjk+ Ckpj+jRgkp/4 −pRgkj/4

−2jRRik− 2R iRjk+ Ckij+jRgik/4 −iRgkj/4

= Rjp Ckpi+ Cpki − Rkp Cjpi+ Cpji + RipCpkj

+4Rip Ckpj− Cjpk + 2R Cjik− Ckij

 +2RlpCpjlgik− 2RlpCpklgij

+RjR/2 −j|Ric|2/2gik−RkR/2 −k|Ric|2/2gij

−2kRipRjp+ 2jRipRkp

+kRRijjRRik

= Rjp Ckpi+ Cpki − Rkp Cjpi+ Cpji + 5RipCpkj

+2RCijk+ 2RlpCpjlgik− 2RlpCpklgij

+RjR/2 −j|Ric|2/2gik−RkR/2 −k|Ric|2/2gij +2jRipRkp− 2kRipRjp

+kRRijjRRik, where in the last passage we used again the identities (1.4).

Hence, we can resume this long computation in the following proposition, getting back to a generic coordinate basis.

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Proposition 2.1. During the Ricci flow of a 3–dimensional Riemannian manifold(M3, g(t)), the Cotton tensor satisfies the following evolution equation

t− ∆Cijk = gpqRpj(Ckqi+ Cqki) + 5gpqRipCqkj+ gpqRpk(Cjiq+ Cqij) (2.3) +2RCijk+ 2RqlCqjlgik− 2RqlCqklgij

+1

2k|Ric|2gij−1

2j|Ric|2gik+ R

2jRgik−R 2kRgij

+2gpqRpkjRqi− 2gpqRpjkRqi+ RijkR − RikjR . In particular if the Cotton tensor vanishes identically along the flow we obtain,

0 = k|Ric|2gijj|Ric|2gik+ RjRgik− RkRgij (2.4) +4gpqRpkjRqi− 4gpqRpjkRqi+ 2RijkR − 2RikjR .

Corollary 2.2. If the Cotton tensor vanishes identically along the Ricci flow of a 3–dimensional Riemannian manifold(M3, g(t)), the following tensor

|Ric|2gij− 4RpjRpi+ 3RRij−7 8R2gij is a Codazzi tensor (see [1, Chapter 16, Section C]).

Proof. We compute in an orthonormal basis, 4RpkjRpi− 4 RpjkRpi+ 2RijkR − 2RikjR

= 4j(RpkRpi) − 4k(RpjRpi) − 4RpijRpk+ 4RpikRpj + 2RijkR − 2RikjR

= 4j(RpkRpi) − 4k(RpjRpi) + Rpi(4Cpjk+kRgpjjRgpk) + 2RijkR − 2RikjR

= 4j(RpkRpi) − 4k(RpjRpi) + 3RijkR − 3RikjR

= 4j(RpkRpi) − 4k(RpjRpi) + 3k(RRij) − 3j(RRik) − 3R(kRijjRik)

= 4j(RpkRpi) − 4k(RpjRpi) + 3k(RRij) − 3j(RRik) − 3R(4Cijk+kRgijjRgik)/4

= 4j(RpkRpi) − 4k(RpjRpi) + 3k(RRij) − 3j(RRik) −3

8kR2gij+3

8jR2gik. Hence, we have, by the previous proposition,

0 =k|Ric|2gijj|Ric|2gik+ 4j(RpkRpi) − 4k(RpjRpi) + 3k(RRij) − 3j(RRik) −7

8kR2gij+7

8jR2gik, which is the thesis of the corollary.

Remark2.3. All the traces of the 3–tensor in the LHS of equation (2.4) are zero.

Remark2.4. From the trace–free property (1.5) of the Cotton tensor and the fact that along the Ricci flow there holds

t− ∆gij= 2Rij, we conclude that the following relations have to hold

gij(∂t− ∆)Cijk = −2RijCijk, gik(∂t− ∆)Cijk = −2RikCijk, gjk(∂t− ∆)Cijk = −2RjkCijk= 0 . They are easily verified for formula (2.3).

Corollary 2.5. During the Ricci flow of a 3–dimensional Riemannian manifold(M3, g(t)), the squared norm of the Cotton tensor satisfies the following evolution equation, in an orthonormal basis,

t− ∆

|Cijk|2 = −2|∇Cijk|2− 16CipkCiqkRpq+ 24CipkCkqiRpq+ 4R|Cijk|2 +8CijkRpkjRpi+ 4CijkRijkR .

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Proof.

t− ∆

|Cijk|2 = −2|∇Cijk|2+ 2CijkRipgpqCqjk+ 2CijkRjpgpqCiqk+ 2CijkRkpgpqCijq

+2Cijkh

gpqRpj(Ckqi+ Cqki) + 5gpqRipCqkj+ gpqRpk(Cjiq+ Cqij) +2RCijk+ 2RqlCqjlgik− 2RqlCqklgij

+1

2k|Ric|2gij−1

2j|Ric|2gik+R

2jRgik−R 2kRgij

+2gpqRpkjRqi− 2gpqRpjkRqi+ RijkR − RikjRi

= −2|∇Cijk|2+ 2(Ckij+ Cjki)Ripgpq(Ckqj+ Cjkq) +2CijkRjpgpqCiqk+ 2CikjRkpgpqCiqj

+2Cijkh

2gpqRpj(Ckqi+ Cqki) + 5gpqRipCqkj

i +4R|Cijk|2+ 8gpqCijkRpkjRqi+ 4CijkRijkR

= −2|∇Cijk|2− 16CipkCiqkRpq+ 24CipkCkqiRpq+ 4R|Cijk|2 +8CijkRpkjRpi+ 4CijkRijkR

where in the last line we assumed to be in a orthonormal basis.

3 Three–Dimensional Gradient Ricci Solitons

The structural equation of a gradient Ricci soliton (Mn, g,f) is the following

Rij+ijf = λgij, (3.1)

for some λR.

The soliton is said to be steady, shrinking or expanding according to the fact that the constant λ is zero, positive or negative, respectively.

It follows that in dimension three, for (M3, g,f) there holds

Rij = lRijlf+ 2λRij− 2|Ric|2gij+ R2gij− 3RRij+ 4RisRsj (3.2)

R = lRlf + 2λR − 2|Ric|2 (3.3)

iR = 2Rlilf (3.4)

Cijk = Rlkgij

2 lf −Rljgik

2 lf + Rijkf− Rikjf+Rgik

2 jfRgij

2 kf (3.5)

= kR

4 gijjR 4 gik+

Rij−R 2gij

kf− Rik−R

2gik

jf. In the special case of a steady soliton the first two equations above simplify as follows,

Rij = lRijlf− 2|Ric|2gij+ R2gij− 3RRij+ 4RisRsj

R = lRlf− 2|Ric|2. Remark3.1. We notice that, by relation (3.5), we have

Cijkif = kRjf

4 −jRkf

4 + Rijifkf−R

2jfkf− Rikifjf+R 2kfjf

= jRkf

4 −kRjf

4 ,

where in the last passage we used relation (3.4).

It follows that

Cijkifjf = h∇f,Ri

4 kf|∇f|2 4 kR .

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