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., 2015. 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: https://dspace.lboro.ac.uk/2134/19946 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: https://creativecommons.org/licenses/by-nc-nd/4.0/ Please cite the published version.
Argument shift method and sectional operators: applications to differential geometry Alexey Bolsinov Loughborough University, UK and Moscow State University FDIS 2015 13 17 July, Bedlewo, Poland
What is it about? Review on joint papers with V.Matveev, V.Kiosak, S.Rosemann, D.Tsonev and A.Konyaev
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
Thanks for your attention