Two small goals: Problems of Lie - uni-jena.dematveev/Datei/...Two small goals: Problems of Lie Lie 1882: Problem I: Es wird verlangt, die Form des Bogenelementes einer jeden Fl¨ache

Two small goals: Problems of Lie


Lie 1882:

Lie 1882:


Lie 1882:Problem I: Es wird verlangt, die Form des Bogenelementes einerjeden Flache zu bestimmen, deren geodatische Kurven eine infini-tesimale Transformation gestatten.

Lie 1882:Problem II: Man soll die Form des Bogenelementes einer jedenFlache bestimmen, deren geodatische Kurven mehrere infinitesi-male Transformationen gestatten.






English translation: A vector field is projective w.r.t. a metric, if its flowtakes (unparametrized) geodesics to geodesics.





S. Lie 1882: Describe all 2 dim metrics admitting






◮ Problem I: one projective vector field







◮ Problem II: many projective vector fields







◮ Problem II: many projective vector fields


Both problems are local, in a neighborhood of a generic point,pseudoriemannian metrics are allowed

Examples of Lagrange 1789


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f(X)X


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Radial projection f : S2 → R2


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takes geodesics of the sphere togeodesics of the plane,


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takes geodesics of the sphere togeodesics of the plane, becausegeodesics on sphere/plane are in-tersection of plains containing 0with the sphere/plane.


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Thus, for every Killing vector field on the plane ist pullback is a projectivevector field on the sphere


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Motivation of Lagrange:


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Motivation of Lagrange:

Solution of the 2nd Lie Problem


Theorem (Bryant, Manno, M∼ 2007)


Theorem (Bryant, Manno, M∼ 2007) If a two-dimensional metric gof nonconstant curvature has at least 2 projective vector fields


Theorem (Bryant, Manno, M∼ 2007) If a two-dimensional metric gof nonconstant curvature has at least 2 projective vector fields such thatthey are linear independent at the point p,


Theorem (Bryant, Manno, M∼ 2007) If a two-dimensional metric gof nonconstant curvature has at least 2 projective vector fields such thatthey are linear independent at the point p, then there exist coordinatesx , y in a neighborhood of p such that the metrics are as follows.


Theorem (Bryant, Manno, M∼ 2007) If a two-dimensional metric gof nonconstant curvature has at least 2 projective vector fields such thatthey are linear independent at the point p, then there exist coordinatesx , y in a neighborhood of p such that the metrics are as follows.

1. ǫ1e(b+2) xdx2 + ǫ2beb xdy2, where

b ∈ R \ {−2, 0, 1} and ǫi ∈ {−1, 1}

2. a(

ǫ1e(b+2) xdx2

(eb x+ǫ2)2 + eb xdy2

eb x+ǫ2

)

, where a ∈ R \ {0},b ∈ R \ {−2, 0, 1} and ǫi ∈ {−1, 1}

3. a(

e2 xdx2

x2 + ǫ dy2

x

)

, where a ∈ R \ {0}, and ǫ ∈ {−1, 1}

4. ǫ1e3xdx2 + ǫ2e

xdy2, where ǫi ∈ {−1, 1},

5. a(

e3xdx2

(ex+ǫ2)2 + ǫ1exdy2

(ex+ǫ2)

)

, where a ∈ R \ {0}, ǫi ∈ {−1, 1},

6. a(

dx2

(cx+2x2+ǫ2)2x+ ǫ1

xdy2

cx+2x2+ǫ2

)

, where a > 0, ǫi ∈ {−1, 1}, c ∈ R.

Example for 1st Problem of Lie: infinitesimal homotheties


Def. A vector field is an infinitesimal homothety , if its flow multiplythe metric by a constant (depending on the time).



Example. ∂∂x

= (1, 0) is an infinitesimal homothety for the metriceλx

(E (y)dx2 + F (y)dxdy + G (y)dy2

).



Example. ∂∂x



).

Every infinitesimal homothety is a projective vector field.



Example. ∂∂x



).


Def: Two metrics g and g (on one manifold) are geodesically equivalentif they have the same unparametrized geodesics



Example. ∂∂x



).



Of cause, if v is projective w.r.t. g , then it is projective w.r.t. everygeodesically equivalent g



Example. ∂∂x



).



Of cause, if v is projective w.r.t. g , then it is projective w.r.t. everygeodesically equivalent g

For explicitly given metric g , it is possible (and relatively easy with thehelp of Maple) to describe the space of all geodesically equivalent metricg :Shulikovsky 1954 – Kruglikov 2007 – Bryant-Dunajski-Eastwood 2008

Theorem (M∼ 2008):

Theorem (M∼ 2008): Let v be a projective vector field on (M2, g).Assume the restriction of g to no neighborhood has an infinitesimalhomothety. Then, there exists a coordinate system in a neighborhood ofalmost every point such that certain metric g geodesically equivalent tog is given by

Theorem (M∼ 2008): Let v be a projective vector field on (M2, g).Assume the restriction of g to no neighborhood has an infinitesimalhomothety. Then, there exists a coordinate system in a neighborhood ofalmost every point such that certain metric g geodesically equivalent tog is given by

1. ds2g = (X (x) − Y (y))(X1(x)dx2 + Y1(y)dy2), v = ∂

∂x+ ∂

∂y, sodass

1.1 X (x) = 1x, Y (y) = 1

y, X1(x) = C1 · e−3x

x, Y1(y) = e−3y

y.

1.2 X (x) = tan(x), Y (y) = tan(y), X1(x) = C1 · e−3λx

cos(x) ,

Y1(y) = e−3λy

cos(y) .

1.3 X (x) = C1 · eνx , Y (y) = eνy , X1(x) = e2x , Y1(y) = ±e2y .

2. ds2g = (Y (y) + x)dxdy , v = v1(x , y) ∂

∂x+ v2(y) ∂

∂y, sodass

2.1 Y = e32y ·

√y

y−3 +∫ y

y0e

32ξ ·

√ξ

(ξ−3)2 dξ,

v1 = y−32

(

x +∫ y

y0e

32ξ ·

√ξ

(ξ−3)2 dξ)

, v2 = y2.

2.2 Y = e−32 λ arctan(y) ·

4√

y2+1

y−3λ +∫ y

y0e−

32 λ arctan(ξ) ·

4√

ξ2+1

(ξ−3λ)2 dξ,

v1 = y−3λ2

(

x +∫ y

y0e−

32 λ arctan(ξ) ·

4√

ξ2+1

(ξ−3λ)2 dξ

)

, v2 = y2 + 1.

2.3 Y (y) = yν , v1(x , y) = νx , v2 = y .

It is time to start the talk

Vladimir MatveevJena (Germany)



Quadratic integrals for geodesicsflows and projective vector fields



Quadratic integrals for geodesicsflows and projective vector fields

www.minet.uni-jena.de/∼matveev/

Definition

Definition

A function F : T ∗M → R is called an integral of the geodesicflow of g ,

Definition

A function F : T ∗M → R is called an integral of the geodesicflow of g , if {H, F} = 0,

Definition

A function F : T ∗M → R is called an integral of the geodesicflow of g , if {H, F} = 0, where H := 1

2g ijpipj : T ∗M → R is thekinetic energy corresponding to the metric.

Definition


2g ijpipj : T ∗M → R is thekinetic energy corresponding to the metric.The integral F is quadratic in momenta

Definition


2g ijpipj : T ∗M → R is thekinetic energy corresponding to the metric.The integral F is quadratic in momenta if, in every localcoordinate system (x , y) on M2, it has the form

a(x , y)p2x + b(x , y)pxpy + c(x , y)p2

y .

Definition




y . (1)

The PDE for the function (1) to be an integral for the metricds2

g = f (x , y)dxdy is

Definition




y . (1)



ay = 0 ,f ax + f by + 2fxa + fyb = 0 ,f bx + f cy + fxb + 2fyc = 0 ,

cx = 0 .

(2)

Definition




y . (1)



ay = 0 ,f ax + f by + 2fxa + fyb = 0 ,f bx + f cy + fxb + 2fyc = 0 ,

cx = 0 .

(2)

We see that the system is overdetermined and of finite type.

Why to study quadratic integrals ?


◮ Because they are classical


◮ Because they are classical (Jacobi 1839, Liouville, Darboux,Eisenhart, ... )

◮ Because they are useful in physics



◮ Because they are useful in physics (Wittaker, Birkhoff,...)



◮ Because they are useful in physics (Wittaker, Birkhoff,...)(conservative quantities, separation of variables)




.

◮ Because they are useful for description ofmetrics with the same geodesics:




.

◮ Because they are useful for description ofmetrics with the same geodesics:Theorem (Dini,Darboux < −−−−−− > Topalov, M∼ 1998)g ∼ g




.

◮ Because they are useful for description ofmetrics with the same geodesics:Theorem (Dini,Darboux < −−−−−− > Topalov, M∼ 1998)g ∼ g iff the function

I : TM2 → R,




.


I : TM2 → R, I (ξ) := g(ξ, ξ)

(det(g)

det(g)

)2/3




.


I : TM2 → R, I (ξ) := g(ξ, ξ)

(det(g)

det(g)

)2/3

is an integral of the geodesic flow of g .




.


I : TM2 → R, I (ξ) := g(ξ, ξ)

(det(g)

det(g)

)2/3

is an integral of the geodesic flow of g . We identify TM and T ∗Mby g .

What people studied about quadratic integrals?


◮ Local description/classification.



Theorem In a neighborhood of almost every point there existcoordinates x , y such that the metric and the integral are




Liouville case Complex-Liouville case Jordan-block case

(X (x) − Y (y))(dx2 ± dy2) ℑ(h)dxdy (1 + xY ′(y)) dxdyX (x)p2

y±Y (y)p2x

X (x)−Y (y) p2x − p2

y + 2ℜ(h)ℑ(h)pxpy p2

x − 2 Y (y)1+xY ′(y)pxpy






y±Y (y)p2x

X (x)−Y (y) p2x − p2



Dini, Liouville 1869






y±Y (y)p2x

X (x)−Y (y) p2x − p2



Dini, Liouville 1869 Its complexification






y±Y (y)p2x

X (x)−Y (y) p2x − p2



Dini, Liouville 1869 Its complexification Bolsinov, Pucacco, M∼08

where ℜ(h) and ℑ(h) are the real and imaginary parts of aholomorphic function h of the variable z := x + i · y .






y±Y (y)p2x

X (x)−Y (y) p2x − p2





◮ Superintegrability (when there exist many quadratic integrals)






y±Y (y)p2x

X (x)−Y (y) p2x − p2





◮ Superintegrability (when there exist many quadratic integrals)(Koenigs 1896,






y±Y (y)p2x

X (x)−Y (y) p2x − p2





◮ Superintegrability (when there exist many quadratic integrals)(Koenigs 1896, Winternitz 1969






y±Y (y)p2x

X (x)−Y (y) p2x − p2





◮ Superintegrability (when there exist many quadratic integrals)(Koenigs 1896, Winternitz 1969 , Miller, Kalnins, Kress(nowdays)).






y±Y (y)p2x

X (x)−Y (y) p2x − p2











y±Y (y)p2x

X (x)−Y (y) p2x − p2






Theorem (Koenigs 1896, Lie 1882 < −−− > Bryant, Manno, M∼07)

The space of quadratics inte-grals is ≥ 4-dimensional

⇐⇒ the space of projective vectorfields is ≥ 3-dimensional






y±Y (y)p2x

X (x)−Y (y) p2x − p2






Theorem (Koenigs 1896, Lie 1882 < −−− > Bryant, Manno, M∼07)

The space of quadratics inte-grals is ≥ 4-dimensional

⇐⇒ the space of projective vectorfields is ≥ 3-dimensional

We will use it for Lie Problems

PDE for “g is geodesically equivalent to a given g“


Is nonlinear


Is nonlinear . One can linearize it with the help of quadraticintegrals.



◮ First observation:




If g ∼ g then I (ξ) := g(ξ, ξ)(

det(g)det(g)

)2/3





det(g)det(g)

)2/3is an integral





det(g)det(g)

)2/3is an integral

◮ Second observation:





det(g)det(g)

)2/3is an integral

◮ Second observation: I is integral





det(g)det(g)

)2/3is an integral

◮ Second observation: I is integral ⇐⇒ {I , H}g = 0





det(g)det(g)

)2/3is an integral

◮ Second observation: I is integral ⇐⇒ {I , H}g = 0 ⇐⇒{(

det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0





det(g)det(g)

)2/3is an integral


det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0

The last equation is linear in g/det(g)2/3.





det(g)det(g)

)2/3is an integral


det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0


Moreover, the equation is projectively invariant





det(g)det(g)

)2/3is an integral


det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0


Moreover, the equation is projectively invariant : ifwe replace g by any other geodesically equivalentmetric g





det(g)det(g)

)2/3is an integral


det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0


Moreover, the equation is projectively invariant : ifwe replace g by any other geodesically equivalentmetric g , the equation will not be changed





det(g)det(g)

)2/3is an integral


det(g)det(g)

)2/3g(ξ, ξ), H(ξ)}g = 0


Moreover, the equation is projectively invariant : ifwe replace g by any other geodesically equivalentmetric g , the equation will not be changed .

The linear equation on a := g/det(g)2/3 (R. Liouville1889)


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.

Important observation:


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.

Important observation: The system of PDE is projectively invariant: itdoes not depend on the connection within connections with the samegeodesics.


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.


PDE-background of the observation:


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.


PDE-background of the observation: Ki determine unparameterizedgeodesics.


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.



Geometric background of observation:


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.



Geometric background of observation: The integrals for g allow toconstruct integrals for g , if g ∼ g ,


∂a11

∂x+ 2K0a12 − 2/3K1a11 = 0

2 ∂a12

∂x+ ∂a11

∂y+ 2K0a22 + 2/3K1a12 − 4/3K2a11 = 0

∂a22

∂x+ 2 ∂a12

∂y+ 4/3K1a22 − 2/3K2a12 − 2K3a11 = 0

∂a22

∂y+ 2/3K2a22 − 2K3a12 = 0

where K0 = −Γ211, K1 := Γ1

11 − 2Γ212, K2 = −Γ2

22 + 2Γ112, K3 = Γ1

22.



Geometric background of observation: The integrals for g allow toconstruct integrals for g , if g ∼ g , because g and g have the samegeodesics.

How we solved Lie Problems


Let A be the space of all solutions of the above system (for agiven metric g).


Let A be the space of all solutions of the above system (for agiven metric g). It is a linear vector space


Let A be the space of all solutions of the above system (for agiven metric g). It is a linear vector space . If dim(A) ≥ 4, thenthe metric admits 3 projective vector fields.


Let A be the space of all solutions of the above system (for agiven metric g). It is a linear vector space . If dim(A) ≥ 4, thenthe metric admits 3 projective vector fields. If dim(A) = 1, allprojective vector fields are infinitesimal homotheties.


Let A be the space of all solutions of the above system (for agiven metric g). It is a linear vector space . If dim(A) ≥ 4, thenthe metric admits 3 projective vector fields. If dim(A) = 1, allprojective vector fields are infinitesimal homotheties.We assume dim(A) = 2 or dim(A) = 3.



Let v be a projective vector field



Let v be a projective vector field .Since the system is projectively invariant



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A . Thus, Lv : A → A is a linear map



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A . Thus, Lv : A → A is a linear map . Sincedim(A) = 2 or 3



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A . Thus, Lv : A → A is a linear map . Sincedim(A) = 2 or 3 , there exists a two-dimensional subspace Ainvariant w.r.t. Lv



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A . Thus, Lv : A → A is a linear map . Sincedim(A) = 2 or 3 , there exists a two-dimensional subspace Ainvariant w.r.t. Lv .In a basis σ1, σ2 ∈ A



Let v be a projective vector field .Since the system is projectively invariant , for every a ∈ A , its Liederivative Lva ∈ A . Thus, Lv : A → A is a linear map . Sincedim(A) = 2 or 3 , there exists a two-dimensional subspace Ainvariant w.r.t. Lv .In a basis σ1, σ2 ∈ A ,Lv

�σ1σ2

�




�σ1σ2

�= B

�σ1σ2

�,




�σ1σ2

�= B

�σ1σ2

�,

where B is a 2 × 2 matrix




�σ1σ2

�= B

�σ1σ2

�,

where B is a 2 × 2 matrix .By choice of basis we can makeB




�σ1σ2

�= B

�σ1σ2

�,

where B is a 2 × 2 matrix .By choice of basis we can makeB =

�α

β

�,




�σ1σ2

�= B

�σ1σ2

�,


�α

β

�, or

�α 1

α

�,




�σ1σ2

�= B

�σ1σ2

�,


�α

β

�, or

�α 1

α

�, or

�α β

−β α

�.

All together

◮ There are 3 cases for the matrix B of Lv : A → A.

All together


◮ For a fixed matrix B, the condition Lv

�σ1σ2

�

All together



�σ1σ2

�= B

�σ1σ2

�

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.

◮ There are 3 cases for the normal form of the pair( g︸︷︷︸

metric

, F︸︷︷︸

quadratic integral

) (which is essentially the same as (σ1, σ2)):

Liouville, complex-liouville und Jordan-block cases.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral



◮ all together we have 9 cases to consider.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral




◮ In every case the data in the normal form of (σ1, σ2), i.e., thefunctions X (x),Y (y) for Liouville case, h for complex-liouville case,Y (y) for Jordan-block case, have at most two first derivatives.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral





◮ Thus, the equation Lv

�σ1σ2

�

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral






�σ1σ2

�= B

�σ1σ2

�

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral






�σ1σ2

�= B

�σ1σ2

�contains 6 PDE and 6

highest derivatives of the unknown functions. i.e., is a Frobeniussystem.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral






�σ1σ2

�= B

�σ1σ2



◮ Such systems can be solved by hands. We did it and solved the LieProblems.

All together



�σ1σ2

�= B

�σ1σ2

�is a system

of 6 PDE.


metric

, F︸︷︷︸

quadratic integral






�σ1σ2

�= B

�σ1σ2



◮ Such systems can be solved by hands. We did it and solved the LieProblems.

If somebody did not manage to follow

Solution of Lie problem is based on



1. restriction on the dimension of the space of solutions ofLiouville system due to connection with quadratic integrals




2. projective invariance of the Liouville system of equations dueto connection with quadratic integrals





3. Jordan normal forms for matrices






These allowed to simplify the equations: insteadof 4 equations of the second order whichimmediately come out,






These allowed to simplify the equations: insteadof 4 equations of the second order whichimmediately come out, we construct 6equations of the first order.






These allowed to simplify the equations: insteadof 4 equations of the second order whichimmediately come out, we construct 6equations of the first order. The price we paid:there are 9 different systems







4. solving 9 Frobenius systems.







4. solving 9 Frobenius systems.

Global questions: Schouten 1924: List all complete metricsadmitting complete projective vector field


I proved Lichnerowicz-Obata-Solodovnikov Conjecture (50th):


I proved Lichnerowicz-Obata-Solodovnikov Conjecture (50th): Let acomplete Riemannian manifold (of dim ≥ 2) admit a complete projectivevector field.


I proved Lichnerowicz-Obata-Solodovnikov Conjecture (50th): Let acomplete Riemannian manifold (of dim ≥ 2) admit a complete projectivevector field. Then, the manifold is covered by the round sphere, or thevector field is affine.





It is hard to relax the assumptions







History of L-O-S conjecture:









France(Lichnerowicz)






Japan(Yano, Obata, Tanno)







Soviet Union(Raschewskii)








Couty (1961) provedthe conjecture assu-ming that g is Einsteinor Kahler









Yamauchi (1974) pro-ved the conjecture as-suming that the scalarcurvature is constant










Solodovnikov (1956)proved the conjecture










Solodovnikov (1956)proved the conjectureassuming that all ob-jects are real analytic










Solodovnikov (1956)proved the conjectureassuming that all ob-jects are real analytic










Solodovnikov (1956)proved the conjectureassuming that all ob-jects are real analyticand that n ≥ 3.

Proof of Lichnerowich conjecture



(Lichnerowicz Conjecture: Among closed Riemannian manifolds, onlythe round sphere admits essential projective vector fields)


(Lichnerowicz Conjecture: Among closed Riemannian manifolds, onlythe round sphere admits essential projective vector fields)

The multidimensional version of Liouville equations


Sinjukov 1962/


Sinjukov 1962/ Mikes, Berezovski 1989 /


Sinjukov 1962/ Mikes, Berezovski 1989 / Bolsinov, M∼ 2003.


Sinjukov 1962/ Mikes, Berezovski 1989 / Bolsinov, M∼ 2003.Theorem (Eastwood, M∼ 2007) g is geodesically equivalent to aconnection Γi

jk iff σab := g ab · det(g)1/(n+1) is a solution of




Tracefreepartof(∇aσ

bc)

= 0.





bc)

= 0.

Here σab := g ab · det(g)1/(n+1) should be understood as an element ofS2M ⊗ (Λn)

2/(n+1)M.





bc)

= 0.


2/(n+1)M. In particular,

∇aσbc =

∂

∂xaσbc + Γb

adσdc + Γcdaσ

bd

︸︷︷︸

Usual covariant derivative

− 2

n + 1Γd

da σbc

︸︷︷︸

addition coming from volume form





bc)

= 0.



∇aσbc =

∂

∂xaσbc + Γb

adσdc + Γcdaσ

bd

︸︷︷︸


− 2

n + 1Γd

da σbc

︸︷︷︸


Advantages:





bc)

= 0.



∇aσbc =

∂

∂xaσbc + Γb

adσdc + Γcdaσ

bd

︸︷︷︸


− 2

n + 1Γd

da σbc

︸︷︷︸


Advantages:

1. It is a linear PDE-system of the first order.





bc)

= 0.



∇aσbc =

∂

∂xaσbc + Γb

adσdc + Γcdaσ

bd

︸︷︷︸


− 2

n + 1Γd

da σbc

︸︷︷︸


Advantages:


2. The system does not depend on the choice of Γ within a projectiveclass.





bc)

= 0.



∇aσbc =

∂

∂xaσbc + Γb

adσdc + Γcdaσ

bd

︸︷︷︸


− 2

n + 1Γd

da σbc

︸︷︷︸


Advantages:


2. The system does not depend on the choice of Γ within a projectiveclass. (short tensor calculations)

3. For dim2 it is the Liouville system we used to solve the Lie problems.

Proof of Lichnerowicz conjecture if dim(A) = 2

Denote by A the space of solutions.


Denote by A the space of solutions.Assume first A is two-dimenional.


Denote by A the space of solutions.Assume first A is two-dimenional.Every projective vector field v does not change theLiouville-Sinjukov-Eastwood-Matveev equations.


Denote by A the space of solutions.Assume first A is two-dimenional.Every projective vector field v does not change theLiouville-Sinjukov-Eastwood-Matveev equations.Then,


Denote by A the space of solutions.Assume first A is two-dimenional.Every projective vector field v does not change theLiouville-Sinjukov-Eastwood-Matveev equations.Then, for every solution σ ∈ A,


Denote by A the space of solutions.Assume first A is two-dimenional.Every projective vector field v does not change theLiouville-Sinjukov-Eastwood-Matveev equations.Then, for every solution σ ∈ A, we have: Lvσ is also a soluiton. Thus,Lv is a linear mapping :A → A.



Then, for the appropiate choice of the basis σ1, σ2 we get



Then, for the appropiate choice of the basis σ1, σ2 we getLv

�σ1σ2

�=




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2

�




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2

�depending on the eigenvalues of Lv .

In the left case




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds.




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds. In thesecond case




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds. In thesecond case the determinant of σ vanishes at some point




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds. In thesecond case the determinant of σ vanishes at some point (becausecos(t) vanishes for some t)




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds. In thesecond case the determinant of σ vanishes at some point (becausecos(t) vanishes for some t) implying that g = σ · det(σ) vanishessomewhere.




�σ1σ2

�=

�α ∗

0 β

��σ1σ2

�, or Lv

�σ1σ2

�=

�α β

−β α

��σ1σ2


In the left case there exists a solution s.t. Lvσ = ασ implying v ishomothety vector for g which is impossible for closed manifolds. In thesecond case the determinant of σ vanishes at some point (becausecos(t) vanishes for some t) implying that g = σ · det(σ) vanishessomewhere. Lichnerowich conjecture is proved under assumption thatthe space of solutions is two-dimenional

What happens if the space of solutions is more thantwo-dimensional?


Theorem (Matveev 2004) Assume Γ is the Levi-Civita connection of aRiemannian metric on a closed or complete manifold.


Theorem (Matveev 2004) Assume Γ is the Levi-Civita connection of aRiemannian metric on a closed or complete manifold. If dim(A) ≥ 3,then g has constant curvature.


Theorem (Matveev 2004) Assume Γ is the Levi-Civita connection of aRiemannian metric on a closed or complete manifold. If dim(A) ≥ 3,then g has constant curvature.

Initial proof is complicated.

Explanation using newer results


Theorem (Bolsinov, Kiosak, M∼ 2008)


Theorem (Bolsinov, Kiosak, M∼ 2008) Let dim(A) ≥ 3 anddim(M) ≥ 3


Theorem (Bolsinov, Kiosak, M∼ 2008) Let dim(A) ≥ 3 anddim(M) ≥ 3 . Then, for every solution σ its tracef := σijgij/det(g)1/(n+1) satisfy the equation


Theorem (Bolsinov, Kiosak, M∼ 2008) Let dim(A) ≥ 3 anddim(M) ≥ 3 . Then, for every solution σ its tracef := σijgij/det(g)1/(n+1) satisfy the equation (for a certain constant K).

f,ijk = K · (2fkgij + figjk + fjgik). (3)




Method of Proof: We prolonged two times theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equations




Method of Proof: We prolonged two times theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equations .





(3) is a famous equation: it appears everywhere in local differentialgeometry:






◮ Solodovnokov 1956: in projective transformations







◮ deVries 1953: in conformal transformations








◮ Mikes-Kiosak 2003: in Einstein manifolds









◮ Obata – Tanno 1970: eigenfunction of Laplace equation on theround sphere corresponding to second eigenvalue satisfy thisequation.









◮ Obata – Tanno 1970: eigenfunction of Laplace equation on theround sphere corresponding to second eigenvalue satisfy thisequation.

Theorem of Tanno 1970 finishes the proof:


Theorem (Tanno 1970 for positive K ; Kiosak 2003 < −− >M∼ 2004 for all K)


Theorem (Tanno 1970 for positive K ; Kiosak 2003 < −− >M∼ 2004 for all K) Let (Mn≥3, g) be a closed Riemannianmanifold


Theorem (Tanno 1970 for positive K ; Kiosak 2003 < −− >M∼ 2004 for all K) Let (Mn≥3, g) be a closed Riemannianmanifold such that there exists a nontrivial solution of theequation (3):

f,ijk = K · (2fkgij + figjk + fjgik).




Then, g has constant positive curvature.




Then, g has constant positive curvature.

Lichnerowich conjecture in the pseudo-Riemannian case:Work in Progress:

Conjecture


Conjecture M∼ 2008:


Conjecture M∼ 2008: A complete pseudoriemannian manifold ofdim ≥ 3 does not admit a complete essential projective vectorfield.



Joint project with Bolsinov and Kiosak:

Scheme of the possible proof/way to find counterexample.





◮ Let A be the space of solutions of theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equation.





◮ Let A be the space of solutions of theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equation. If A istwo-dimensional, then, as in the solution of S. Lie problem





◮ Let A be the space of solutions of theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equation. If A istwo-dimensional, then, as in the solution of S. Lie problem ,one can describe all such metrics





◮ Let A be the space of solutions of theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equation. If A istwo-dimensional, then, as in the solution of S. Lie problem ,one can describe all such metrics – one need to understandwhether they could be complete.





◮ Let A be the space of solutions of theLiouville-Sinjukov-Bolsinov-Eastwood-M∼ equation. If A istwo-dimensional, then, as in the solution of S. Lie problem ,one can describe all such metrics – one need to understandwhether they could be complete.

◮ If dim(A) ≥ 3

◮ If dim(A) ≥ 3 , then we are done because of the following result

◮ If dim(A) ≥ 3 , then we are done because of the following result :

Theorem: (Kiosak, M∼ April 2008)


Theorem: (Kiosak, M∼ April 2008) Let (Mn≥3, g) be a completepseudoriemannian manifold


Theorem: (Kiosak, M∼ April 2008) Let (Mn≥3, g) be a completepseudoriemannian manifold such that there exists a nontrivialsolution of the equation (3)


Theorem: (Kiosak, M∼ April 2008) Let (Mn≥3, g) be a completepseudoriemannian manifold such that there exists a nontrivialsolution of the equation (3) :







Then, every g geodesically equivalent to g




Then, every g geodesically equivalent to g has the same Levi-Civitaconnection with g.





◮ Corollary:





◮ Corollary: Complete Einstein manifolds are geodesically rigid:





◮ Corollary: Complete Einstein manifolds are geodesically rigid:Let (M, g) is a complete pseudoriemannian Einstein manifold.






Then, every complete g geodesically equivalent to g






Then, every complete g geodesically equivalent to g has the sameLevi-Civita connection with g.













Thanks a lot!!!

Documents

Two small goals: Problems of Lie - uni-jena.dematveev/Datei/...Two small goals: Problems of Lie Lie 1882: Problem I: Es wird verlangt, die Form des Bogenelementes einer jeden Fl¨ache