How to recognize a conformally Kähler metric

How to recognize a conformally Kähler metric Maciej Dunajski Department of Applied Mathematics and Theoretical Physics University of Cambridge MD, Paul Tod arxiv:0901.2261, Mathematical Proceedings of the Cambridge Philosophical Society (2010). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 1 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that ĝ = Ω 2 g is flat? Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that ĝ = Ω 2 g is flat? Answer: Need Weyl = 0. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that ĝ = Ω 2 g is flat? Answer: Need Weyl = 0. Curvature decomposition: Riemann=Weyl+Ricci+scalar. ĝ = Ω 2 g is Einstein? Answer:??? (but lots of neccessary conditions are known: e.g vanishing of the Bach tensor). ĝ = Ω 2 g is Kähler? 1 J : T M T M is a complex structure: J 2 = Id and [T (1,0), T (1,0) ] T (1,0), where T (1,0) = {X T M C, J(X) = ix}. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that ĝ = Ω 2 g is flat? Answer: Need Weyl = 0. Curvature decomposition: Riemann=Weyl+Ricci+scalar. ĝ = Ω 2 g is Einstein? Answer:??? (but lots of neccessary conditions are known: e.g vanishing of the Bach tensor). ĝ = Ω 2 g is Kähler? 1 J : T M T M is a complex structure: J 2 = Id and [T (1,0), T (1,0) ] T (1,0), where T (1,0) = {X T M C, J(X) = ix}. 2 ĝ(x, Y ) = ĝ(jx, JY ) for any vector fields X, Y. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Three can you find questions Let (M, g) be a Riemannian four-manifold. Can you find a non zero function Ω : M R such that ĝ = Ω 2 g is flat? Answer: Need Weyl = 0. Curvature decomposition: Riemann=Weyl+Ricci+scalar. ĝ = Ω 2 g is Einstein? Answer:??? (but lots of neccessary conditions are known: e.g vanishing of the Bach tensor). ĝ = Ω 2 g is Kähler? 1 J : T M T M is a complex structure: J 2 = Id and [T (1,0), T (1,0) ] T (1,0), where T (1,0) = {X T M C, J(X) = ix}. 2 ĝ(x, Y ) = ĝ(jx, JY ) for any vector fields X, Y. 3 Fundamental two form Σ(X, Y ) = ĝ(jx, Y ) is closed. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 2 / 13

Degrees of freedom Always? No - lets count: Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 3 / 13

Degrees of freedom Always? No - lets count: A metric in 2n dimensions: n(2n + 1) arbitrary functions of 2n variables. Diffeormorphisms + conf. rescalling: 2n 2 n 1. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 3 / 13

Degrees of freedom Always? No - lets count: A metric in 2n dimensions: n(2n + 1) arbitrary functions of 2n variables. Diffeormorphisms + conf. rescalling: 2n 2 n 1. The general Kähler metric can be locally described by the Kähler potential: there exists a function K : M R and a holomorphic coordinate system (z 1,..., z n ) such that g = 2 K z j z k dzj dz k. The difference between the number of arbitrary functions is 2n 2 n 2, which is positive if n > 1 (every metric in 2D is Kähler). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 3 / 13

Degrees of freedom Always? No - lets count: A metric in 2n dimensions: n(2n + 1) arbitrary functions of 2n variables. Diffeormorphisms + conf. rescalling: 2n 2 n 1. The general Kähler metric can be locally described by the Kähler potential: there exists a function K : M R and a holomorphic coordinate system (z 1,..., z n ) such that g = 2 K z j z k dzj dz k. The difference between the number of arbitrary functions is 2n 2 n 2, which is positive if n > 1 (every metric in 2D is Kähler). Obstructions should be given by conformally invariant tensors. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 3 / 13

Summary of results in four dimensions One to one correspondence between Kähler metrics in the conformal class of g and parallel sections of a certain (canonical) connection D on a rank ten vector bundle E = Λ 2 +(M) Λ 1 (M) Λ 2 (M). If the self dual (SD) part of Weyl tensor C + of g is non zero we find the necessary and sufficient conditions : C + spinor must be of algebraic type D and T = 0, dv = 0, where T Γ(S Sym 5 (S )), V Γ(Λ 1 (M)) depend on C + and C +. If C + = 0 we get some necessary conditions from the holonomy of the curvature of D. E.g. A metric with C + = 0 is conformal to Einstein AND conformal to Kähler if and only if it admits an isometry. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 4 / 13

Summary of results in four dimensions One to one correspondence between Kähler metrics in the conformal class of g and parallel sections of a certain (canonical) connection D on a rank ten vector bundle E = Λ 2 +(M) Λ 1 (M) Λ 2 (M). If the self dual (SD) part of Weyl tensor C + of g is non zero we find the necessary and sufficient conditions : C + spinor must be of algebraic type D and T = 0, dv = 0, where T Γ(S Sym 5 (S )), V Γ(Λ 1 (M)) depend on C + and C +. If C + = 0 we get some necessary conditions from the holonomy of the curvature of D. E.g. A metric with C + = 0 is conformal to Einstein AND conformal to Kähler if and only if it admits an isometry. Link with the metrisability problem (Bryant, MD, Eastwood. J. Diff. Geom. 2009): A projective structure on a surface U is metrisable if and only if the induced (2, 2) conformal structure on M = T U admits a Kähler metric or a para Kähler metric. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 4 / 13

self duality in four dimensions Oriented Riemannian four-manifold (M, g) : Λ 2 Λ 2, 2 = Id, Λ 2 = Λ 2 + Λ 2. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 5 / 13

self duality in four dimensions Oriented Riemannian four-manifold (M, g) : Λ 2 Λ 2, 2 = Id, Λ 2 = Λ 2 + Λ 2. Riemann tensor gives rise to R : Λ 2 Λ 2. C + + R 12 φ R = φ C + R 12. C ± = SD/ASD Weyl tensors, φ = trace-free Ricci curvature, R =scalar curvature. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 5 / 13

spinors in four dimensions C T M = S S, where (S, ε), (S, ε ) are rank two complex symplectic vector bundles (spin bundles) over M. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 6 / 13

spinors in four dimensions C T M = S S, where (S, ε), (S, ε ) are rank two complex symplectic vector bundles (spin bundles) over M. g(v 1 w 1, v 2 w 2 ) = ε(v 1, v 2 )ε (w 1, w 2 ), where v 1, v 2 Γ(S), w 1, w 2 Γ(S ). Two component spinor notation (love it or hate it): µ Γ(S), µ = µ A. Spinor indices A, B, C, = 0, 1. µ A = ε AB µ B, µ A = µ B ε BA. Metric g ab = ε AB ε A B. Spinors and self duality. Σ Λ 2 (M), Σ ab = Σ [ab]. Σ AA BB = ω ABε A B + ω A B ε AB, where ω AB = ω (AB) and ω A B = ω (A B ). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 6 / 13

spinors in four dimensions C T M = S S, where (S, ε), (S, ε ) are rank two complex symplectic vector bundles (spin bundles) over M. g(v 1 w 1, v 2 w 2 ) = ε(v 1, v 2 )ε (w 1, w 2 ), where v 1, v 2 Γ(S), w 1, w 2 Γ(S ). Two component spinor notation (love it or hate it): µ Γ(S), µ = µ A. Spinor indices A, B, C, = 0, 1. µ A = ε AB µ B, µ A = µ B ε BA. Metric g ab = ε AB ε A B. Spinors and self duality. Σ Λ 2 (M), Σ ab = Σ [ab]. Σ AA BB = ω ABε A B + ω A B ε AB, where ω AB = ω (AB) and ω A B = ω (A B ). Spinor curvature decomposition R abcd = ψ ABCD ε A B ε C D + ψ A B C D ε ABε CD +φ ABC D ε A B ε CD + φ A B CDε AB ε C D + R 12 (ε ACε BD ε A C ε B D ε ADε BC ε A D ε B C ). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 6 / 13

Twistor equation J 2 = Id 2 Σ is SD or ASD. Make a choice: Σ = ω ε, ω Γ(S S ) (self dual). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 7 / 13

Twistor equation J 2 = Id 2 Σ is SD or ASD. Make a choice: Σ = ω ε, ω Γ(S S ) (self dual). Conformal rescallings g Ω 2 g, Σ Ω 3 Σ, so ω Ω 2 ω. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 7 / 13

Twistor equation J 2 = Id 2 Σ is SD or ASD. Make a choice: Conformal rescallings Σ = ω ε, ω Γ(S S ) (self dual). g Ω 2 g, Σ Ω 3 Σ, so ω Ω 2 ω. Lemma. The metric g is conformal to a Kähler metric if and only if there exists a real, symmetric spinor field ω Γ(S S ) satisfying A(A ω B C ) = 0, ( ), and such that ω 2 0. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 7 / 13

Twistor equation J 2 = Id 2 Σ is SD or ASD. Make a choice: Conformal rescallings Σ = ω ε, ω Γ(S S ) (self dual). g Ω 2 g, Σ Ω 3 Σ, so ω Ω 2 ω. Lemma. The metric g is conformal to a Kähler metric if and only if there exists a real, symmetric spinor field ω Γ(S S ) satisfying and such that ω 2 0. A(A ω B C ) = 0, ( ), ( ω = dω) Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 7 / 13

Twistor equation J 2 = Id 2 Σ is SD or ASD. Make a choice: Conformal rescallings Σ = ω ε, ω Γ(S S ) (self dual). g Ω 2 g, Σ Ω 3 Σ, so ω Ω 2 ω. Lemma. The metric g is conformal to a Kähler metric if and only if there exists a real, symmetric spinor field ω Γ(S S ) satisfying and such that ω 2 0. A(A ω B C ) = 0, ( ), ( ω = dω) ( ) is the (conformally invariant) twistor equation. Idea: prolong it, look for integrability conditions. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 7 / 13

Prolongation of A(A ω B C ) = 0 Drop symmetrisation: AA ω B C ε A B K C A ε A C K B A = 0 for some K Λ 1 (M). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 8 / 13

Prolongation of A(A ω B C ) = 0 Drop symmetrisation: AA ω B C ε A B K C A ε A C K B A = 0 for some K Λ 1 (M). Differentiate and commute derivatives: ψ E (A B C ω D )E = 0 and AA K BB + P ABA C C ωb ε A B ρ AB = 0 (where P ab = (1/2)R ab (1/12)Rg ab ) for some ρ Λ 2 (M). Differentiate and commute derivatives: AA ρ BC ωa E E D ψ ABCD + KA D ψ ABCD 2P A E A(BKC) E = 0. Now the system is closed: All derivatives of unknowns have been determined. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 8 / 13

Prolongation of A(A ω B C ) = 0 Drop symmetrisation: AA ω B C ε A B K C A ε A C K B A = 0 for some K Λ 1 (M). Differentiate and commute derivatives: ψ E (A B C ω D )E = 0 and AA K BB + P ABA C C ωb ε A B ρ AB = 0 (where P ab = (1/2)R ab (1/12)Rg ab ) for some ρ Λ 2 (M). Differentiate and commute derivatives: AA ρ BC ωa E E D ψ ABCD + KA D ψ ABCD 2P A E A(BKC) E = 0. Now the system is closed: All derivatives of unknowns have been determined. Geometric interpretation Ψ = (ω, K, ρ) is a section of a rank 10 vector bundle E = Λ 2 +(M) Λ 1 (M) Λ 2 (M) which is parallel with respect to a connection D determined by the blue equations. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 8 / 13

Example: Local vs. Global obstructions Compact hyperbolic four manifold (M, g). Weyl= 0, R = 1. All local obstructions vanish. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 9 / 13

Example: Local vs. Global obstructions Compact hyperbolic four manifold (M, g). Weyl= 0, R = 1. All local obstructions vanish. Assume globally defined non-degenerate ω satisfies the twistor eq. ω 0 so ω defines a Killing vector K a. Killing identity K a + R ab K b = 0, where R ab = g ab /4. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 9 / 13

Example: Local vs. Global obstructions Compact hyperbolic four manifold (M, g). Weyl= 0, R = 1. All local obstructions vanish. Assume globally defined non-degenerate ω satisfies the twistor eq. ω 0 so ω defines a Killing vector K a. Killing identity K a + R ab K b = 0, where R ab = g ab /4. Contract with K a, integrate by parts K 2 gd 4 x = 1 4 M M K 2 gd 4 x. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 9 / 13

Example: Local vs. Global obstructions Compact hyperbolic four manifold (M, g). Weyl= 0, R = 1. All local obstructions vanish. Assume globally defined non-degenerate ω satisfies the twistor eq. ω 0 so ω defines a Killing vector K a. Killing identity K a + R ab K b = 0, where R ab = g ab /4. Contract with K a, integrate by parts K 2 gd 4 x = 1 4 M M K 2 gd 4 x. Therefore K a = 0 and our assumption was wrong. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 9 / 13

Generic case C + 0 Recall ψ E (A B C ω D )E = 0 ( ). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 10 / 13

Generic case C + 0 Recall ψ E (A B C ω D )E = 0 ( ). Find that C + is of type D i.e. ψ A B C D = ±ω (A B ω C D ). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 10 / 13

Generic case C + 0 Recall ψ E (A B C ω D )E = 0 ( ). Find that C + is of type D i.e. ψ A B C D = ±ω (A B ω C D ). Differentiate ( ), impose the twistor equation. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 10 / 13

Generic case C + 0 Recall ψ E (A B C ω D )E = 0 ( ). Find that C + is of type D i.e. ψ A B C D = ±ω (A B ω C D ). Differentiate ( ), impose the twistor equation. Theorem. Let (M, g) be a 4 manifold such that the self dual part of the conformal curvature is non zero. Then there exists a Kähler metric in [g] if and only if C + is of type D and A(A ψ B C D E ) V A(A ψ B C D E ) = 0, [a V b] = 0, where V AA = 1 ψ 2 ( 1 6 AA ψ 2 + 4 3 ψb C D E AB ψ C D E A ). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 10 / 13

Anti self dual case Theorem. Parallel sections Ψ of D on a rank 10 vector bundle E M correspond to Kähler metrics in a conformal class. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 11 / 13

Anti self dual case Theorem. Parallel sections Ψ of D on a rank 10 vector bundle E M correspond to Kähler metrics in a conformal class. Integrability conditions: FΨ = 0 where F = [D, D] (there are some indices, but lets not write them down). If F = 0 then g is conformally flat. Otherwise differentiate: (DF)Ψ = 0, (DDF)Ψ = 0,... After K steps F K Ψ = 0, where F K is a matrix of linear blue eqn. Stop when rank (F K ) = rank (F K+1 ). The space of parallel sections has dimension (10 rank(f K )). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 11 / 13

Anti self dual case Theorem. Parallel sections Ψ of D on a rank 10 vector bundle E M correspond to Kähler metrics in a conformal class. Integrability conditions: FΨ = 0 where F = [D, D] (there are some indices, but lets not write them down). If F = 0 then g is conformally flat. Otherwise differentiate: (DF)Ψ = 0, (DDF)Ψ = 0,... After K steps F K Ψ = 0, where F K is a matrix of linear blue eqn. Stop when rank (F K ) = rank (F K+1 ). The space of parallel sections has dimension (10 rank(f K )). Theorem. An anti-self dual Einstein metric g with Λ 0 is conformal to a Kähler metric iff g admits a Killing vector. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 11 / 13

Anti self dual case Theorem. Parallel sections Ψ of D on a rank 10 vector bundle E M correspond to Kähler metrics in a conformal class. Integrability conditions: FΨ = 0 where F = [D, D] (there are some indices, but lets not write them down). If F = 0 then g is conformally flat. Otherwise differentiate: (DF)Ψ = 0, (DDF)Ψ = 0,... After K steps F K Ψ = 0, where F K is a matrix of linear blue eqn. Stop when rank (F K ) = rank (F K+1 ). The space of parallel sections has dimension (10 rank(f K )). Theorem. An anti-self dual Einstein metric g with Λ 0 is conformal to a Kähler metric iff g admits a Killing vector. Examples of conf. classes with more than one (local) Kähler metrics: Fubini-Study metric on CP 2 with reversed orientation. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 11 / 13

Metrisability in 2D A projective structure on an open set U R 2 is an equivalence class of torsion free connections [Γ]. Two connections Γ and ˆΓ are equivalent if they share the same unparametrised geodesics. ˆΓ k ij = Γ k ij + δ i k ω j + δ j k ω i, i, j, k = 1, 2. for some one form ω = ω i dx i. Projective connection (T. Y. Thomas) Π k ij = Γk ij 1 3 Γl li δk j 1 3 Γl lj δk i. Geodesics of a generic projective structures are not unparametrised geodesics of any metrics. The necessary and sufficient conditions for metrisability have recently been found (Bryant, MD, Eastwood, 09). Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 12 / 13

Metrisability in 2D A projective structure on an open set U R 2 is an equivalence class of torsion free connections [Γ]. Two connections Γ and ˆΓ are equivalent if they share the same unparametrised geodesics. ˆΓ k ij = Γ k ij + δ i k ω j + δ j k ω i, i, j, k = 1, 2. for some one form ω = ω i dx i. Projective connection (T. Y. Thomas) Π k ij = Γk ij 1 3 Γl li δk j 1 3 Γl lj δk i. Geodesics of a generic projective structures are not unparametrised geodesics of any metrics. The necessary and sufficient conditions for metrisability have recently been found (Bryant, MD, Eastwood, 09). Construct a signature signature (2, 2) metric on T U g = dz i dx i Π k ij(x) z k dx i dx j, i, j, k = 1, 2. Walker (1953), Yano Ishihara,.... Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 12 / 13

Metrisability in 2D A projective structure on an open set U R 2 is an equivalence class of torsion free connections [Γ]. Two connections Γ and ˆΓ are equivalent if they share the same unparametrised geodesics. ˆΓ k ij = Γ k ij + δ i k ω j + δ j k ω i, i, j, k = 1, 2. for some one form ω = ω i dx i. Projective connection (T. Y. Thomas) Π k ij = Γk ij 1 3 Γl li δk j 1 3 Γl lj δk i. Geodesics of a generic projective structures are not unparametrised geodesics of any metrics. The necessary and sufficient conditions for metrisability have recently been found (Bryant, MD, Eastwood, 09). Construct a signature signature (2, 2) metric on T U g = dz i dx i Π k ij(x) z k dx i dx j, i, j, k = 1, 2. Walker (1953), Yano Ishihara,.... Theorem.The metric g is conformal to (para) Kähler iff the projective structure is metrisable. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 12 / 13

Summary and Outlook Conformal to Kähler in dimensions four. Overdetermined system of linear PDEs. Necessary and sufficient conditions in the generic case. Prolongation bundle of rank 10 in the anti self dual case. Take this number, type it into Google,.... Twistor theory. Preferred section of κ Z 1/2 of Z M. Prolongation bundle E with connection over M Holomorphic vector bundle (no connection!) E over Z. Find E = J 2 (κ Z 1/2 ). MD (2009) Solitons, Instantons and Twistors. OUP. Conformal to Kähler in higher dimensions (work in progress MD, Rod Gover). The spinor/self duality methods fail. Use tractor formalism, conformal Killing Yano obstructions (Semmelmann, arxiv:math/0206117),.... Non linear connections? Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 13 / 13

Summary and Outlook Conformal to Kähler in dimensions four. Overdetermined system of linear PDEs. Necessary and sufficient conditions in the generic case. Prolongation bundle of rank 10 in the anti self dual case. Take this number, type it into Google,.... Twistor theory. Preferred section of κ Z 1/2 of Z M. Prolongation bundle E with connection over M Holomorphic vector bundle (no connection!) E over Z. Find E = J 2 (κ Z 1/2 ). MD (2009) Solitons, Instantons and Twistors. OUP. Conformal to Kähler in higher dimensions (work in progress MD, Rod Gover). The spinor/self duality methods fail. Use tractor formalism, conformal Killing Yano obstructions (Semmelmann, arxiv:math/0206117),.... Non linear connections? General approach to overdetermined systems: prolong, construct connection, restrict its holonomy. Applicable to other can you find problems. Dunajski (DAMTP, Cambridge) Conformal to Kahler April 2011 13 / 13