Hermitian-Yang-Mills approach to the conjecture of Griffiths on the - PDF document

Hermitian-Yang-Mills approach to the conjecture of Griffiths on the positivity of ample vector bundles Jean-Pierre Demailly Institut Fourier, Universit´ e Grenoble Alpes & Acad´ emie des Sciences de Paris Virtual Conference on Several Complex Variables organized by Shiferaw Berhanu and Ming Xiao August 20, 2020, 09:00 EDT J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 1/21 Ample vector bundles Let X be a projective n -dimensional manifold and E → X a holomorphic vector bundle of rank r ≥ 1. Ample vector bundles E → X is said to be ample in the sense of Hartshorne if the associated ⇒ ∃ C ∞ line bundle O P ( E ) (1) on P ( E ) is ample, i.e. by Kodaira ⇐ hermitian metric on O P ( E ) (1) with positive curvature. This is equivalent to the existence of a a strongly pseudoconvex tubular neighborhood U of the 0-section in E ∗ , i.e. of a negatively curved Finsler metric on E ∗ . Geometric interpretation: U can be taken S 1 invariant U � � | λ | =1 λ U J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 2/21

Chern curvature tensor and positivity concepts Chern curvature tensor This is Θ E , h = i ∇ 2 E , h ∈ C ∞ (Λ 1 , 1 T ∗ X ⊗ Hom( E , E )), written � c jk λµ dz j ∧ dz k ⊗ e ∗ Θ E , h = i λ ⊗ e µ 1 ≤ j , k ≤ n , 1 ≤ λ,µ ≤ r in terms of an orthonormal frame ( e λ ) 1 ≤ λ ≤ r of E . Griffiths and Nakano positivity One looks at the associated quadratic form on S = T X ⊗ E � � Θ E , h ( ξ ⊗ v ) := � Θ E , h ( ξ, ξ ) · v , v � h = c jk λµ ξ j ξ k v λ v µ . 1 ≤ j , k ≤ n , 1 ≤ λ,µ ≤ r Then E is said to be: Griffiths positive (Griffiths 1969) if at any point z ∈ X � Θ E , h ( ξ ⊗ v ) > 0 , ∀ 0 � = ξ ∈ T X , z , ∀ 0 � = v ∈ E z Nakano positive (Nakano 1955) if at any point z ∈ X � � Θ E , h ( τ ) = c jk λµ τ j ,λ τ k ,µ > 0 , ∀ 0 � = τ ∈ T X , z ⊗ E z . 1 ≤ j , k ≤ n , 1 ≤ λ,µ ≤ r J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 3/21 Geometric interpretation of Griffiths positivity Griffiths positivity of E is equivalent to the existence of a strongly pseudoconvex neighborhood U ′ of the 0-section in E ∗ whose fibers are (varying) hermitian balls. (Nakano > 0 is more restrictive than strict pseudoconvexity of U ′ .) Easy and well known facts E Nakano positive ⇒ E Griffiths positive ⇒ E ample. In fact, E Griffiths positive ⇒ O P ( E ) (1) positive. J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 4/21

Dual Nakano positivity – a conjecture Curvature tensor of the dual bundle E ∗ � λ ) ∗ ⊗ e ∗ Θ E ∗ , h = − T Θ E , h = − c jk µλ dz j ∧ dz k ⊗ ( e ∗ µ . 1 ≤ j , k ≤ n , 1 ≤ λ,µ ≤ r Dual Nakano positivity One requires � − � ∀ 0 � = τ ∈ T X , z ⊗ E ∗ Θ E ∗ , h ( τ ) = c jk µλ τ j λ τ k µ > 0 , z . 1 ≤ j , k ≤ n , 1 ≤ λ,µ ≤ r Dual Nakano positivity is clearly stronger than Griffiths positivity. Also, it is better behaved than Nakano positivity, e.g. E dual Nakano positive ⇒ any quotient Q = E / S is also dual Nakano positive. (Very speculative) conjecture Is it true that E ample ⇒ E dual Nakano positive ? J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 5/21 Geometric interpretation of the conjecture Basic question. Is there a (geometric, analytic) procedure that turns the strictly pseudoconvex neighborhood U into another strictly pseudoconvex U ′ that would be a ball bundle ? Answer is yes if n = dim X = 1 (Umemura, Campana-Flenner) !! J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 6/21

Brief discussion around this positivity conjecture If true, Griffiths conjecture would follow: E ample ⇔ E dual Nakano positive ⇔ E Griffiths positive. Remark E ample �⇒ E Nakano positive, in fact E Griffiths positive �⇒ E Nakano positive. For instance, T P n is easy shown to be ample and Griffiths positive for the Fubini-Study metric, but it is not Nakano positive. Otherwise the Nakano vanishing theorem would then yield H n − 1 , n − 1 ( P n , C ) = H n − 1 ( P n , Ω n − 1 P n ) = H n − 1 ( P n , K P n ⊗ T P n ) = 0 !!! Let us mention here that there are already known subtle relations between ampleness, Griffiths and Nakano positivity are known to hold – for instance, B. Berndtsson has proved that the ampleness of E implies the Nakano positivity of S m E ⊗ det E for every m ∈ N . J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 7/21 “Total” determinant of the curvature tensor If the Chern curvature tensor Θ E , h is dual Nakano positive, then one can introduce the ( n × r )-dimensional determinant of the corresponding Hermitian quadratic form on T X ⊗ E ∗ det T X ⊗ E ∗ ( T Θ E , h ) 1 / r := det( c jk µλ ) 1 / r ( j ,λ ) , ( k ,µ ) idz 1 ∧ dz 1 ∧ ... ∧ idz n ∧ dz n . This ( n , n )-form does not depend on the choice of coordinates ( z j ) on X , nor on the choice of the orthonormal frame ( e λ ) on E . Basic idea Assigning a “matrix Monge-Amp` ere equation” det T X ⊗ E ∗ ( T Θ E , h ) 1 / r = f > 0 where f is a positive ( n , n )-form, may enforce the dual Nakano positivity of Θ E , h if that assignment is combined with a continuity technique from an initial starting point where positivity is known. J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 8/21

Continuity method (case of rank 1) For r = 1 and h = h 0 e − ϕ , we have T Θ E , h = Θ E , h = − i ∂∂ log h = ω 0 + i ∂∂ϕ, and the equation reduces to a standard Monge-Amp` ere equation (Θ E , h ) n = ( ω 0 + i ∂∂ϕ ) n = f . ( ∗ ) If f is given and independent of h , Yau’s theorem guarantees the existence of a unique solution θ = Θ E , h > 0, provided E is an ample line � X f = c 1 ( E ) n . bundle and When the right hand side f = f t of ( ∗ ) varies smoothly with respect to some parameter t ∈ [0 , 1], one then gets a smoothly varying solution Θ E , h t = ω 0 + i ∂∂ϕ t > 0 , and the positivity of Θ E , h 0 forces the positivity of Θ E , h t for all t . J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 9/21 Undeterminacy of the equation Assuming E to be ample of rank r > 1, the equation det T X ⊗ E ∗ ( T Θ E , h ) 1 / r = f > 0 ( ∗∗ ) becomes underdetermined, as the real rank of the space of hermitian matrices h = ( h λµ ) on E is equal to r 2 , while ( ∗∗ ) provides only 1 scalar equation. (Solutions might still exist, but lack uniqueness and a priori bounds.) Conclusion In order to recover a well determined system of equations, one needs an additional “matrix equation” of rank ( r 2 − 1). Observation 1 (from the Donaldson-Uhlenbeck-Yau theorem) Take a Hermitian metric η 0 on det E so that ω 0 := Θ det E ,η 0 > 0. If E is ω 0 -polystable, ∃ h Hermitian metric h on E such that ω n − 1 ∧ Θ E , h = 1 0 ⊗ Id E (Hermite-Einstein equation, slope 1 r ω n r ). 0 J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 10/21

Resulting trace free condition Observation 2 The trace part of the above Hermite-Einstein equation is “automatic”, hence the equation is equivalent to the trace free condition ω n − 1 ∧ Θ ◦ E , h = 0 , 0 when decomposing any endomorphism u ∈ Herm ( E , E ) as u = u ◦ + 1 r Tr( u ) Id E ∈ Herm ◦ ( E , E ) ⊕ R Id E , tr ( u ◦ ) = 0 . Observation 3 The trace free condition is a matrix equation of rank ( r 2 − 1) !!! Remark In case dim X = n = 1, the trace free condition means that E is projectively flat, and the Umemura proof of the Griffiths conjecture proceeds exactly in that way, using the fact that the graded pieces of the Harder-Narasimhan filtration are projectively flat. J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 11/21 Towards a “cushioned” Hermite-Einstein equation In general, one cannot expect E to be ω 0 -polystable, but Uhlenbeck-Yau have shown that there always exists a smooth solution q ε to a certain “cushioned” Hermite-Einstein equation. To make things more precise, let Herm ( E ) be the space of Hermitian (non necessarily positive) forms on E . Given a reference Hermitian metric H 0 > 0, let Herm H 0 ( E , E ) be the space of H 0 -Hermitian endomorphisms u ∈ Hom( E , E ); denote by Herm ( E ) ≃ → Herm H 0 ( E , E ) , q �→ � q s.t. q ( v , w ) = � � q ( v ) , w � H 0 the natural isomorphism. Let also � � Herm ◦ H 0 ( E , E ) = q ∈ Herm H 0 ( E , E ) ; tr ( q ) = 0 be the subspace of “trace free” Hermitian endomorphisms. In the sequel, we fix H 0 on E such that Θ det E , det H 0 = ω 0 > 0. J.-P. Demailly, Virtual Conference on SCV, August 20, 2020 Griffiths conjecture on the positivity of vector bundles 12/21