Θ αβ = i. Θ αβjk dz j dz k.

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1 Kobayashi-Lübke inequalities for Chern classes of Hermite-Einstein vector bundles and Guggenheimer-Yau-Bogomolov-Miyaoka inequalities for Chern classes of Kähler-Einstein manifolds Let (X, ω) be a compact Kähler manifold, n = dimx, and let E be a holomorphic vector bundle over X, r = rank E. We suppose that E is equipped with a hermitian metric, h and denote by D E,h the Chern connection on (E, h). The Chern curvature form is Θ h (E) = D E,h. In a (local) orthonormal frame (e ) 1 r of E, we write Θ h (E) = (Θ β ) 1,β r where the Θ β are complex valued (1, 1)-forms satisfying the hermitian condition Θ β = Θ β. We denote Θ β = i Θ βjk dz j dz k. 1,β r, 1 j,k n The hermitian symmetry condition can then be read Θ βjk = Θ βkj. If at some point x 0 X the coordinates (z j ) are chosen so that (dz j (x 0 )) is an orthonormal basis of TX,x 0, we define ( ) Tr ω Θ h (E) = Θ βjj C (X, hom(e, E)). j Definition. The hermitian vector bundle (E, h) is said to be Hermite- Einstein with respect to the Kähler metric ω if there is a constant λ > 0 such that Tr ω Θ h (E) = λid E. Recall that the Chern forms c k (E) h are defined by the formula det ( I + tθ h (E) ) = det(δ β + tθ β ) = 1 + t c 1 (E) h t r c r (E) h. This gives in particular the identities c 1 (E) h = Θ, c (E) h = <β Θ Θ ββ Θ β Θ β = 1 Θ Θ ββ Θ β Θ β.,β The trace Tr ω Θ h (E) can be computed by the formula Θ h (E) ωn 1 (n 1)! = Tr ωθ h (E) ωn n!. 1

2 By taking the trace with respect to the indices in E and taking the Hermite- Einstein equation into account, we find c 1 (E) h ωn 1 (n 1)! = λr ωn n!. This implies that the number λ in the definition of Hermite-Einstein metrics is a purely numerical invariant, namely λ = n c 1 (E) ω n 1/ ω n. r X X Kobayashi-Lübke inequality. If E admits a Hermite-Einstein metric h with respect to ω, then [(r 1)c1 (E) h r c (E) h ω n 0 at every point of X. Moreover, the equality holds if and only if Θ h (E) = 1 r c 1(E) h Id E. Observe that the equality holds pointwise already if we have the numerical equality [ (r 1)c1 (E) r c (E) ω n = 0. X If we introduce the (formal) vector bundle Ẽ = E (det E) 1/r (Ẽ normalized vector bundle such that det Ẽ = O), then c1(ẽ) h = 0 and is the Θ h (Ẽ) = ( Θ h (E) 1 r c 1(E) h Id E ) Id (det E) 1/r. By the formula for the chern classes of E L, the Kobayashi-Lübke inequality can be rewritten as c (Ẽ) h ω n 0, with equality if and only if Ẽ is unitary flat. In that case, we say that E is projectively flat. Proof. By the above, (r 1)c 1 (E) h r c (E) h =,β Θ Θ ββ + r Θ β Θ β. Taking the wedge product with ω n /(n )! means taking the trace, i.e. the sum of coefficients of the terms i dz j dz j i dz k dz k for all j < k. For this, we have to look at products of the type (i dz j dz j )(i dz k dz k ) or (i dz j dz k )(i dz k dz j ). This yields [ (r 1)c 1 (E) h r c (E) h ω n = (n )! (Θ jj Θ ββkk Θ jk Θ ββkj ) + r(θ βjj Θ βkk Θ βjk Θ βkj ).

3 The initial factor comes from the fact that the final sum is taken over all unordered indices j, k (terms with j = k cancel). The Hermite-Einstein condition yields j Θ βjj = λ δ β, so we get λ δ δ ββ + rλ δ β δ β Θ jj Θ ββkk + r Θ βjj Θ βkk =,β = r λ + r λ = 0. Hence, using the hermitian symmetry of Θ βjk, we find [ (r 1)c 1 (E) h r c (E) h ω n = = r = r (n )! Θ jk Θ ββjk r Θ βjk β,j,k β,j,k Θ βjk + ( Θ jk Θ ββjk r Θ jk ) j,k,β Θ βjk 1 Θjk Θ ββjk 0. This proves the expected inequality. Moreover, the equality holds if and only if we have Θ βjk = 0 for β, Θ jk = γ jk for all. where γ = i j,k γ jkdz j dz k is a (1, 1)-form (take e.g., γ jk = Θ 11jk ). Hence Θ h (E) = γ Id E. By taking the trace with respect to E in this last equality, we get c 1 (E) h = r γ. Therefore the equality occurs if and only if Θ h (E) = 1 r c 1(E) h Id E. Corollary 1. Let (E, h) be a Hermite-Einstein vector bundle with c 1 (E) = 0 and c (E) = 0. Then E is unitary flat for some hermitian metric h = h e ϕ. Proof. By the assumption c 1 (E) = 0, we can write c 1 (E) h = i π ψ for some global function ψ on X. The equality case of the Kobayashi-Lübke inequality yields Θ h e ψ/r(e) = Θ h (E) 1 i r π ψ Id E = 0. Corollary. Let X be a compact Kähler manifold with c 1 (X) = c (X) = 0. Then X is a finite unramified quotient of a torus. Proof. By the Aubin-Calabi-Yau theorem, X admits a Ricci-flat Kähler metric ω. Since Ricci(ω) = Tr ω Θ ω (T X ), we see that (T X, ω) is a Hermite-Einstein vector bundle, and c 1 (T X ) ω = Ricci(ω) = 0. By the Kobayashi-Lübke inequality, we conclude that (T X, ω) is unitary flat, given by a unitary representation 3

4 π 1 (X) U(n). Let X be the universal covering of X and ω the induced metric. Then (T X, ω) is a trivial vector bundle equipped with a flat metric. Let (ξ 1,...,ξ n ) be an orthonormal parallel frame of T X. Since ξ j = 0, we conclude that dξ j = 0, and it is easy to infer from this that [ξ j, ξ k = 0. The flow of each vector field λj ξ j is defined for all times (this follows from the fact the length of a trajectory is proportional to the time, and ω is complete). Hence we get an action of C n on X, and it follows easily that ( X, ω) (C n, can). Now, π 1 (X) acts by isometries on this C n. The classification of subgroups of affine transformations acting freely (and with compact quotient) shows that π 1 (X) must be a semi-direct product of a finite group of isometries by a group of translations associated to a lattice Λ C n. Hence there is an exact sequence 0 Λ π 1 (X) G 0 where G is a finite group of isometries. It follows that there is a finite unramified covering map C n /Λ X/π 1 (X) X of X by a torus. We now discuss the special case of the tangent bundle T X in case (X, ω) is a compact Kähler-Einstein manifold. The Kähler-Einstein condition means that Ricci(ω) = λω for some real constant λ, i.e., Tr ω Θ ω (T X ) = λid TX. In particular, (T X, ω) is a Hermite-Einstein vector bundle. Here, however, the coefficients (Θ βjk ) 1 n of the curvature tensor Θ ω (T X ) satisfy the additional symmetry relations ( ) Θ βjk = Θ jβk = Θ kjβ = Θ jkβ. These relations follow easily from the identity Θ βjk = ω β / z j z k in normal coordinates, when we apply the Kähler condition ω β / z j = ω jβ / z. It follows that the Chern forms satisfy a slightly stronger inequality than the general inequality valid for Hermite-Einstein bundles. In fact (T X, ω) satisfies a similar inequality where the rank r = n is replaced by n + 1. Guggenheimer-Yau inequality. Let (T X, ω) be a compact n-dimensional Kähler-Einstein manifold, with constant λ IR. If λ = 0, then c (T X ) ω ω n 0. If λ 0, we have the inequality [ nc1 (X) (n + )c (X) (λc 1 (X)) n 0, and the equality also holds pointwise if we replace the Chern classes by the Chern forms c k (T X ) ω. The equality occurs in the following cases: (i) If λ = 0, then (X, ω) is a finite unramified quotient of a torus. (ii) If λ > 0, then (X, ω) (IP n, Fubini Study). (iii) If λ < 0, then (X, ω) (IB n /Γ, Poincaré metric), i.e. X is a compact unramified quotient of the ball in C n. Corollary (Bogomolov-Miyaoka-Yau). Let X be a surface of general type with K X ample. Then there is an inequality c 1 (X) 3 c (X), and the equality occurs if and only if X is a quotient of the ball IB. 4

5 Miyaoka has shown that the inequality holds in fact as soon as X is a surface with general type. Proof. As in the proof of the Kobayashi-Lübke inequality, we find n c 1 (T X ) ω (n + )c (T X ) ω =,β Θ Θ ββ + (n + 1)Θ β Θ β. Taking the wedge product with ω n /(n )!, we get [ n c 1 (T X ) ω (n + ) c (T X ) ω ω n = (n )! (Θ jj Θ ββkk Θ jk Θ ββkj ) + (n + 1)(Θ βjj Θ βkk Θ βjk Θ βkj ). If we had the factor r = n instead of (n + 1) in the right hand side, the terms in jj and kk would cancel (as they did before). Hence we find [ n c 1 (T X ) ω (n + ) c (T X ) ω ω n = = (n )! Θ jk Θ ββkj + Θ βjj Θ βkk (n + 1)Θ βjk Θ βkj Θ jk Θ ββkj (n + 1)Θ βjk Θ βkj. Here, the symmetry relation ( ) was used in order to obtain the equality of the summation of the first two terms. Using also the hermitian symmetry relation, our sum Σ can be rewritten as Σ = Θ jk Θ ββjk (n + 1) Θ βjk + n Θ jk,j,k = Θ jk Θ ββjk (n + 1) Θ βjk, pairwise [ (n + 1) 8 Θ jk + 4 Θ j + 4 Θ jj + Θ j <j [ + n Θ jk + Θ j + Θ jj + Θ j <j = Θ jk Θ ββjk (n + 1) Θ βjk β,j,k (4n + 8) + (n 1) Θ. Θ jk 4 j, pairwise Θ j 4 <j Θ jj All terms are negative except the last one. We try to absorb this term in the 5

6 summations involving the coefficients Θ jj. This gives Σ = Θ jk Θ ββjk (n + 1) Θ βjk β,j k (4n + 8) β,j, pairwise Θ jk 4 Θ j j Θ jj Θ ββjj 4 Θ jj + (n 1) Θ. <j The last line is equal to Θ jj Θ ββjj Θ ββ Θ ββββ β j β 4 Θ ββ + (n 1) Θ <β = Θ jj Θ ββjj β j (n 1) Θ 8 <β Θ ββ + β Θ ββ Θ ββββ + Θ ββ Θ ββββ = β j Therefore we find Σ = Θ jj Θ ββjj β β,j k (4n + 8) β j Θ Θ ββ. Θ jk Θ ββjk (n + 1) Θ jk 4 j Θ jj Θ ββjj β, pairwise Θ j Θ Θ ββ. Θ βjk This proves the expected inequality Σ 0. Moreover, we have Σ = 0 if and only if there is a scalar µ such that Θ ββ = Θ ββ = Θ ββ = µ for β, Θ = µ, and all other coefficients Θ βjk are zero. By taking the trace j Θ jj, we get λ = (n+1)µ. We thus obtain that the hermitian form associated to the curvature tensor is Θ ω (T X )(ξ η), ξ η ω = λ ξ η β ξ n + 1 η β + ξ η β ξ β η ( ),β = λ ( ξ η + ξ, η ) n + 1 for all ξ, η T X. When λ = 0, the curvature tensor vanishes identically and we have already seen that X is a finite unramified quotient of a torus. Assume from 6

7 now on that λ 0. The formula ( ) shows that the curvature tensor is constant and coincides with the curvature tensor of IP n (case λ > 0), or of the ball IB n (case λ < 0), relatively to the canonical metrics on these spaces. By a well-known result from the theory of hermitian symmetric spaces, it follows that (X, ω) is locally isometric to IP n (resp. IB n ). Since the universal covering X is a complete and locally symmetric hermitian manifold, we conclude that X IP n, resp. X IB n. In the case λ > 0, X is a Fano manifold, thus X is simply connected and X = X. The proof is complete. To compute the curvature of IP n and IB n, we use the fact that the canonical metric is ω = i π log(1 + z ) on IP n, resp. ω = i π log(1 z ) on IB n, with respect to the non homogeneous coordinates on IP n. We thus get ω = i ( dz dz dz, ) z, resp. ω = i ( dz dz dz, ) z +. π 1 + z (1 + z ) π 1 z (1 z ) By computing the derivatives Θ βjk = ω β / z j z k at z = 0, we easily see that the curvature is given by ( ) with λ = ±(n + 1). The equality also holds at any other point by the homogeneity of IP n and IB n. Observe that the riemannian exponential map exp : T X,0 X at the origin of C n IP n or IB n is unitary invariant. It follows that the holomorphic part h of the Taylor expansion of exp at 0 is unitary invariant. This invariance forces h to coincide with the identity map in the standard coordinates of IP n and IB n. From this observation, it is not difficult to justify intuitively the local isometry statement used above. In fact let X be a hermitian manifold whose curvature tensor is given by ( ), λ 0. Then ω is proportional to c 1 (T X ) ω and so ω is a Kähler-Einstein metric. Since the Kähler-Einstein equation is (nonlinear) elliptic with real analytic coefficients in terms of any real analytic Kähler form, it follows that ω is real analytic, and so is the exponential map. Fix a point x 0 X and let h : C n T X,x0 X be the holomorphic part of the Taylor expansion of exp at the origin. Then h must provide the holomorphic coordinates we are looking for, i.e. h ω must coincide with the metric of IP n (resp. IB n ) in a neighborhood of the origin. See for example, F. Tricerri et L. Vanhecke, Variétés riemanniennes dont le tenseur de courbure est celui d un espace symétrique riemannien irréductible, C. R. Acad. Sc. Paris, 30 (1986),