Argument shift method and sectional operators: applications to differential geometry

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1 Loughborough University Institutional Repository Argument shift method and sectional operators: applications to differential geometry This item was submitted to Loughborough University's Institutional Repository by the/an author. Citation: BOLSINOV, A.V., Argument shift method and sectional operators: applications to differential geometry. Presented at the 3rd Conference on Finite Dimensional Integrable Systems in Geometry and Mathematical Physics 2015 (FDIS2015), Bedlewo, Poland, 12-17th July. Additional Information: This is a conference paper. Metadata Record: Version: Published Publisher: FDIS Rights: This work is made available according to the conditions of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) licence. Full details of this licence are available at: Please cite the published version.

2 Argument shift method and sectional operators: applications to differential geometry Alexey Bolsinov Loughborough University, UK and Moscow State University FDIS July, Bedlewo, Poland

3 What is it about? Review on joint papers with V.Matveev, V.Kiosak, S.Rosemann, D.Tsonev and A.Konyaev

4 Pre-history Let g be a semisimple Lie algebra, R : g g g a symmetric linear operator. Euler equations on g dx = [x, R(x)] (1) dt are Hamiltonian with H = 1 R(x), x. 2 For which R, are the equations (1) integrable? R : so(n) so(n) is called a sectional operator (with parameters A and B), if [R(X ), A] = [X, B] for all X so(g) (2) where A and B are some fixed symmetric matrices. Theorem (Manakov, Mischenko, Fomenko) Let R satisfy (2). Then (1) can be rewritten as d (X + λa) = [X + λa, R(X ) + λb]; dt Tr(X + λa) k are commuting first integrals of (1); if A is regular, then (1) are completely integrable.

5 Properties of sectional operators 1. A and B commute, moreover, B belongs to the centre of centraliser of A. In particular, B = p(a), where p( ) is some polynomial. 2. R 0 = d dt t=0 p(a + tx ) satisfies (2). If A is regular, then R is unique, otherwise R = R 0 + D where D : so(g) g A = {Y so(g), AY = YA} is arbitrary. 3. if B = 0 = p min(a), then R 0 = d dt t=0 p min(a + tx ) still defines a non-trivial sectional operator whose image is contained in g A. Moreover, if for each eigenvalues of A there are at most 2 Jordan blocks, then the image R 0 coincides with g A. 4. R 0 satisfies the Bianchi identity. 5. Let R satisfy two identities [R(X ), A] = [X, B] and [R(X ), A ] = [X, B ], where A a A + b id. Then R(X ) = k X mod g A. In particular, if A is regular, then R = k id. 6. Let λ 1,..., λ k be the eigenvalues of A. Then p(λ i ) p(λ j ) λ i λ j are eigenvalues of R. Moreover, if A has a nontrivial Jordan λ i -block, then p (λ i ) is an eigenvalue of R.

6 Riemann curvature tensor (quick reminder and new point of view) Let be the Levi-Civita connection of a Riemannian metric g. The Riemann curvature tensor R = (R l ij k) is defined by (formula from a text-book): R(X, Y )Z = X Y Z Y X Z [X,Y ] Z. In other words, R can be understood as a map R : (X, Y ) R(X, Y ) = X Y Y X [X,Y ] End(TM). Algebraic symmetries: R(X, Y ) = R(X, Y ), i.e., R : Λ 2 V gl(v ), V = T xm; g(r(x, Y )Z, W ) = g(r(x, Y )W, Z), i.e. R(X, Y ) so(g); R(X, Y )Z + R(Y, Z)X + R(Z, X )Y = 0 (Bianchi identity); g(r(x, Y )Z, W ) = g(r(z, W )X, Y ). Conclusion: R : so(g) so(g) which is symmetric and satisfying Bianchi.

7 Projectively equivalent metrics g and ḡ are projectively equivalent if they have the same (unparametrised) geodesics. Notation: g Main equation: Let A = Theorem (B., Matveev) proj ḡ. ( det ḡ det g ) 1 n+1 ḡ 1 g. Then g proj ḡ if and only if ua = 1 2 ( u d tr A + (u d tr A) ). Let g proj ḡ. Then the Riemann curvature tensor of g is a sectional operator: [R(X ), A] = [B, X ] for all X so(g), where B = 1 2 ( grad tr A ). Proof. Consider the compatibility condition for the main equation. Theorem (B., Matveev, Kiosak) Let g, ḡ and ĝ be projectively equivalent. Assume that these metrics are linearly independent and g and ĝ are strictly non-proportional, then g, ḡ and ĝ are metrics of constant sectional curvature. Proof. Apply Property 5.

8 New class of holonomy groups in pseudo-riemannian geometry Let M be a smooth manifold endowed with an affine symmetric connection. The holonomy group of is a subgroup Hol( ) GL(T xm) that consists of the linear operators A : T xm T xm being parallel transport transformations along closed loops γ(t) with γ(0) = γ(1) = x. Problem. Given a subgroup H GL(n, R), can it be realised as the holonomy group for an appropriate symmetric connection on M n? Riemannian case and irreducible case: the problem is completely solved (Marcel Berger, D. V. Alekseevskii, R. Bryant, D. Joyce, L. Schwahhöfer, S. Merkulov). Pseudo-Riemannian case: many fundamental results but still open (L. Bérard Bergery, A. Ikemakhen, C. Boubel, D. V. Alekseevskii, T. Leistner, A. Galaev). Theorem (B., Tsonev) For every g-symmetric operator A : V V, its centraliser in SO(g) (the identity connected component of) G A = {Y SO(g) YA = AY } is a holonomy group for a certain (pseudo)-riemannian metric.

9 Classical approach A map R : Λ 2 V gl(v ) is called a formal curvature tensor if it satisfies the Bianchi identity R(u v)w + R(v w)u + R(w u)v = 0 for all u, v, w V. Let h gl(v ) be a Lie subalgebra. Consider the set of all formal curvature tensors R : Λ 2 V gl(v ) such that Im R h: R(h) = {R : Λ 2 V h R(u v)w +R(v w)u +R(w u)v = 0, u, v, w V }. We say that h is a Berger algebra if it is generated as a vector space by the images of the formal curvature tensors R R(h), i.e., h = span{r(u v) R R(h), u, v V }. Berger test: Let be a symmetric affine connection on TM. Then the Lie algebra hol ( ) of its holonomy group Hol ( ) is Berger.

10 Classical approach (with small amendments) A map R : so(g) so(g) is called a formal curvature tensor if it satisfies the Bianchi identity R(u v)w + R(v w)u + R(w u)v = 0 for all u, v, w V, where u v = u g(v) v g(u) so(g). Let h so(g) be a Lie subalgebra. Consider the set of all formal curvature tensors R : so(g) so(g) such that Im R h: R(h) = {R : Λ 2 V h R(u v)w +R(v w)u +R(w u)v = 0, u, v, w V }. We say that h is a Berger algebra if it is generated as a vector space by the images of the formal curvature tensors R R(h), i.e., h = span{r(u v) R R(h), u, v V }. Berger test: Let be a symmetric affine connection on TM. Then the Lie algebra hol ( ) of its holonomy group Hol ( ) is Berger.

11 Step one: Berger test for g A and Magic Formula 1 We have g A = {X so(g) XA = AX } and we need to construct formal curvature tensors R : so(g) so(g) whose images generate g A. Ideally, we want one single formal curvature tensor R such that Im R = g A. Question: How to find R? Answer: Apply Properties 3 and 4, i.e. define a linear mapping R : so(g) so(g) by: R(X ) = d dt t=0 p min(a + tx ), (3) where p min(λ) is the minimal polynomial of A. Conclusion: g A is Berger algebra.

12 Step two: Realisation We need to find an example of g such that hol ( ) = g A. More specifically: For a given operator A : T x0 M T x0 M, we need to find a (pseudo)-riemannian metric g on M and a (1, 1)-tensor field A(x) (with the initial condition A(x 0) = A) such that 1. A(x) = 0; 2. R(x 0) coincides with the formal curvature tensor R formal just defined. The idea is natural: set A(x) = const try to find the desired metric g(x) in the form: i.e., constant + quadratic g ij (x) = g 0 ij + B ij,pq x p x q (4) where B satisfies obvious symmetry relations, namely, B ij,pq = B ji,pq and B ij,pq = B ij,qp.

13 Magic Formula 2 Thus, we need to find B ij,pq with the required properties. Such a tensor can be rewritten in the form B = C α D α, where C α and D α are some symmetric forms. It is more convenient to work with operators rather than forms : B = C α D α B = C α D α, where C α and D α are the g 0-symmetric operators corresponding to C α and D α. Then we can treat B as a linear map Question: How to find B? B : gl(v ) gl(v ) defined by B(X ) = C αxd α, Answer: Amasingly simple B = 1 R( ), i.e. 2 R(X ) = d dt t=0 p min(a + tx ) B = 1 2 d dt t=0 p min(l + t ), More precisely, if p min(λ) = n m=0 amλm is the minimal polynomial of A, then B = 1 2 n m=0 A m 1 j A j. (5) a m m 1 j=0 Conclusion: This B solves the realisation problem.

14 C-projective equivalence and Yano-Obata conjecture A curve γ(t) on a Kähler manifold (M, g, J) is called J-planar, if γ γ = α γ + βj γ where α, β R, and J is the complex structure on M. Two Kähler metrics g and ḡ on a complex manifold (M, J) are called c-projectively equivalent, if they have the same J-planar curves. A vector field ξ on a Kähler manifold is called c-projective, if the flow of ξ preserves J-planar curves. A c-projective vector field is called essential if its flow changes the Levi-Civita connection. Theorem (B., Matveev, Rosemann) Let (M, g, J) be a closed connected Kähler manifold of arbitrary signature which admits an essential c-projective vector field. Then the manifold is isometric to CP n with the Fubini-Study metric.

15 Some ingredients of the proof Statement 1. Let g and ḡ be projectively equivalent Kähler metrics. Then the Riemann curvature tensor of g satisfies [R(X ), A] = [X, B] for all X u(g), ( ) 1 where A = det ḡ 2(n+1) ḡ 1 g and B = (grad tr A) (hermitian operators). det g In other words, R can be considered as a sectional operator for the unitary algebra. Statement 2. If A admits a non-trivial Jordan block, then one of the eigenvalues of R can be explicitly computed from Property 6. Statement 3. Let x(t) be a trajectory of ξ. This eigenvalue λ Spectrum(R), as a function of t, i.e. λ(x(t)) is not bounded. Therefore, M cannot be compact.

16 Thanks for your attention