HODGE GROUPS OF K3 SURFACES

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1 HODGE GROUPS OF K3 SURFACES GREGORIO BALDI ABSTRACT. First exposé for the study group about K3s. We present the homonymous paper of Zarhin. Some other recent applications and examples are also presented. Sections 2.5, 2.7 and 3 were not discussed during the talk. First draft (the missing proofs will be added soon), comments are very welcome. CONTENTS Notation 1 1. Preliminaries 1 2. Endomorphism algebra and Hodge group 2 3. Applications (to be expanded!) 6 4. Algebraicity of invariant cycles in families: finite vs infinite monodromy 7 5. Appendix: some facts from Hodge II 9 NOTATION Let X/C be a smooth projective variety, we denote with H 2 (X) is the image of H 2 (X, Z) into H 2 (X, Q), A(X) = H 2 (X) H 1,1 (X) and A(X) Q = H 2 (X, Q) H 1,1. A(X) is a lattice in A(X) Q, the lattice of algebraic cycles. The transcendental part (integral/rational), denoted with T, of an Hodge structure of K3 type can be also defined as the minimal (primitive in the integral case) sub-hodge structure of K3 type; it is irreducible: let T T a sub- Hodge structure, it it is pure of type (1,1), then its orthogonal complement in T is of K3 type and smaller than T, if it is not of type (1,1), then T 2,0 T C, and by minimality and primitivity of T, we have T = T. In any case if X is a K3, T (X) is a polarizable irreducible Hodge structure. 1. PRELIMINARIES Let V be a finite dimensional Q-vector space. A Hodge structure of weight n is a decomposition of its complexification V C = V p,q such that the conjugation swaps the degrees. Equivalently it is a representation of the Deligne torus: h : S GL(V R )/R such that h Gm,R : t t n. With this interpretation we have natural notions of duals, Hodge sub-structure, homomorphism of Hodge structures, tensor product, direct sum, irreducibility. Notice that if V is irreducible (i.e. it does not contain non-trivial Hodge substructures) then End Hdg (V ) is a (finitely generated) division algebra, i.e. every non-zero endomorphism is an iso. Definition 1.1 (Polarization). A polarization of (V, h, n) is a bilinear map of Hodge structures ϕ : V V Q( n) (i.e. ϕ R (h(z)v, h(z)w) = (zz) n ϕ R (v, w) such that defines a positive definite symmetric form on V R. ϕ(v, h(i)w) Remark. Since h(i) 2 = ( 1) k, ϕ is symmetric if the weight is even, alternating if it is odd. Remark. The category of Q-HS is abelian, the category of polarized Q-HS is semisimple. 1

2 2 GREGORIO BALDI Definition 1.2 (Hodge structure of K3 type). A Hodge structure of weight two, V, is of K3 type if dim C V 2,0 = 1 and V p,q = 0 for p q > 2. Examples. The H 2 of a K3 gives an integral and a rational polarizable Hodge structure. The transcendental lattice of a PHS of K3 type is a irreducible PHS of K3 type. Definition 1.3 (Mumford-Tate and Hodge groups). We define the Mumford-Tate group of a Hodge structure (V, h) as the smallest Q-algebraic subgroup H of GL(V Q ) such that h factorizes through H R. Notice that, since S is connected, the Mumford-Tate group is always connected. The smallest rational group containing h(u(r)) is called the Hodge group of (V, h), where U is the kernel of the norm map S G m,r. Remark. MT = Hdg holds iff V has weight zero. Examples. MT(Q(i)) = 0 if i = 0, G m /Q otherwise Around the Mumford-Tate conjecture. Let k be a number field, X/k a smooth projective variety, and denote with V Q the group H 2n (X, Q)(n). We have an Hodge structure on V Q and a Galois action on V l (X) = V Q Q l or, in terms or representations, two arrows: ρ l : G k GL(V l (X)) h : S GL(V Q R). We define the l-adic monodromy group G l of X as the Zariski closure of the image of ρ l in GL(V l (X)) Conjecture 1.1 (Mumford-Tate conjecture). Via the canonical identification V l (X) = V Q Q l we have G 0 l = MT Q l. Remark. For motives of abelian type (the H 1 of an abelian varieties, the H 2 (X, Q)(1) for X a K3... ), it is known that the image of the Galois is contained in the Mumford-Tate. Moreover Bogomolov proved that Im(ρ l ) is open in G l (Q l ), in particular the Mumford-Tate conjecture is equivalent to the following statement: the image of ρ l is an open subgroup of MT(Z l ). 2. ENDOMORPHISM ALGEBRA AND HODGE GROUP 2.1. Rational irreducible PHS of K3 type. The aim of this talk is to present the following theorem. Theorem 2.1 (Zarhin). Let (T, ψ) be a polarised irreducible Q-HS of K3-type. (1) The endomorphism algebra E of T is a (commutative) number field. (2) The field E is totally real or a CM field. (3) Let E 0 be the maximal totally real subfield of E and ψ be the unique E-bilinear (resp- hermitian) form such that ψ = tr E/Q ψ if E is totally real (resp. CM). The Hodge group of T is as big as possible, i.e. { (2.1.1) Hdg(T ) Res E/Q SO E (T, = ψ), if E is TR, Res E0/Q U E0 (T, ψ), if E is CM. U E0 is the unitary group over E 0 associated to the hermitian form ψ. (4) If E is totally real, then dim E T 3. (5) If E is CM, then there exists a Hodge isometry η such that E = Q(η). To be more precise about ψ. We define ψ : T T E, x, y α by the formula ψ(ex, y) = tr E/Q (eα), for all e E. This makes sense: the existence and uniqueness follow from the non degeneracy of the pairing tr E/Q : E E Q. One sees that ψ is non degenerate, and ψ(ex, y) = e ψ(x, y), ψ(x, y) = ψ(y, x).

3 HODGE GROUPS OF K3 SURFACES 3 So if E is TR this means that ψ is a non-degenerate symmetric E-bilinear form (obtained from ψ, a Q-bilinear map). If E is an imaginary quadratic extension of the totally real field E 0, then ψ is a non-degenerate Hermitian E-sesquilinear form w.r.t. the complex conjugation. Since Hdg commutes with the action of E and preserves ψ, then Hdg preserves ψ; this explains why the Hodge group is as big as possible Key observation. The importance of the condition dim T 2,0 = 1 is encoded in the following. Consider the map of Q-algebras: ɛ : E = End Hdg (T ) End C T 2,0 = C, a a T 2,0 = ɛ(a) Id. We claim that ɛ is injective: Let a E such that a = 0 on T 2,0, since a is an endomorphism of Hodge structure, its kernel is a sub-hodge structure, non trivial in degree (2, 0); it follows that ker(a) = T, by minimality. It follows that E is a field Isometries of the transcendental lattice. Hodge isometries of the transcendental lattice of a K3 are easily described. Warning: given X a K3 we have T such that H 2 (X, Q) = NS(X) Q T (X) called the rational lattice of transcendental cycles, and T (X) = H 2 (X, Z) T the lattice of transcendental cycles. Theorem 2.2. Let T (X) be the transcendental lattice of a complex K3 surface, and a : T (X) T (X) be a Hodge isometry. The following are true. a) There exists n > 0 such that a n = Id. b) The group of all Hodge isometries is finite cyclic. c) If T (X) has odd rank, then the only Hodge isometries are ± Id. Remark. With c) on can prove the following result. Proposition 2.3. Let X be a complex K3 with Pic(X) = Z H, then { Aut(X) Id, if H 2 > 2, = ± Id, if H 2 = 2. Sketch of the proof: Let f Aut(X), we have f = ± Id on T (X) and f = 1 on NS(X). Recall that f = Id on H 2 (X, Z) implies f = Id... Proof of a). Define V (X) := T (X) R (H 2,0 H 0,2 ). The key point is the following: the restricted intersection form is positive definitive on V (X), and negative definite on its orthogonal complement. Since a is an isometry, then it preserves the orthogonal decomposition T (X) R = V (X) V (X), in particular all eigenvalues of a have absolute value one. But a has integral coefficient, hence its eigenvalues are all algebraic integers (the conjugate of an eigenvalue is still an eigenvalue); in particular roots of unity, hence a n = Id for some n > 0. Proof of b). The group of Hodge isometries is a discrete (thanks to a)) subgroup of O(V (X)) O(V (X) ), which is compact. Moreover the map ɛ realizes this group as a finite subgroup of C, hence it is cyclic. Proof of c). Let λ be an eigenvalue of a, then λ 1 = λ, thus they are ±1 or in pairs. The oddness assumption implies that 1 or 1 occurs as eigenvalue of an eigenvector β (we chose β T (X) such that a(β) = ±β). As above a = ξ m Id on H 2,0 (X), and if a Id, then m > 2. The contradiction is established, by construction of T (X), if we show that β H 1,1 (X, Z), i.e. β is orthogonal to H 2,0. To do so notice that β, v = a(β), a(v) = ±β, ξ m v β, v = 0 v H 2,0.

4 4 GREGORIO BALDI 2.4. Two examples. A K3 X is said to have CM if Hdg(T (X)) is commutative dim E T = 1 that E is a CM field. Proof of. Suppose dim E T = 1, i.e. [E : Q] = dim Q T, and write T Q C = T ρ ( = C dim Q T ) where the sum runs over all the [E : Q] embeddings ρ : E C and T ρ is the complex subspace on which the elements α E act via multiplication by ρ(α) (one of those ρ is our ɛ!). It follows that every T ρ has dimension one. Suppose K were totally real, so E = Q(α) with ɛ(α) R, then T 2,0 and T 0,2 are contained in T ɛ, contradiction. Definition 2.4. For future reference we define And hdg ɛ for its lie algebra, which is contained in hdg R. T ɛ = {x T R such that ax = ɛ(a)x for a E}. Remark. We will see that a K3 with CM is defined over a number field (cf. abelian varieties). Examples (Complex multiplication). Any complex K3 with maximal Picard rank is CM: SO(2) is commutative! Remark. One can prove that any totally real field can be realized as the endomorphism algebra of a Q-PHS of K3 type, and the m := dim E T 3 can be prescribed. Example: If m [E : Q] 10, then E is realized from the Q-lattice of transcendental classes of the general member of a m 2-dimensional family of K3 surfaces. See vg lemma 3.2, Proposition 3.3 and Example 3.4 Examples (Real multiplication). Van Geemen proved that there is a one parameter family inside four dimensional family of double plains ramified over the union of six lines (defined over Q( d) with d = a 2 + b 2 and odd, rank 16); not explicit equation. But it is also possible to find examples defined over the rational numbers (with rank 16) and explicit equations (see thm 5.12 in Examples of K3 surfaces with real multiplication, Elsenhans, Jahnel): Namely consider the one parameter family of K3s over Q: Infinitely many of them have real multiplication by Q( 2) and Picard rank 16 (for t with v 17 (1 t) > 0 and v 23 (1 t) > 0). To understand whether or not a K3 has real multiplication the following can be useful: Real multiplication by Q( d) implies X p (F p ) = 1 mod p for all primes of good reduction p that are inert under the field extension Q( d)/q, see cor 4.13 of the paper (cf. with elliptic curves: Deuring criterion) Kostant s theorem. Theorem 2.5. Let W be a m-dimensional R-vector space, ϕ : W W R a non-degenerate symmetric bilinear form, g End(W ) an irreducible Lie algebra, which preserves ϕ. Assume that g contains a semisimple endomorphism f of rank 2. One of the following statements is true. i) End g (W ) = R and g = so(w, ϕ). In particular dim R g = m2 m 2. ii) End g (W ) = C, so we can consider W as a C-vector space, and define a form by the formula ϕ : W W C, x, y a ϕ(bx, y) = tr C/R (ba) = ba + ba, for all b C. It is a non-degenerate hermitian form and g = u(w, ϕ ). In particular dim R g = m2 4. Proof. will be added soon Remark. Let G/k be a linear algebraic group, and H 1 H 2 two closed connected subgroups of G. If dim Lie(H 1 ) = dim Lie(H 2 ), then H 1 = H 2. PF: smooth (algebraic groups are smooth in char zero) and connected imply irreducible, and a closed subset of an irreducible finite dimensional topological space Y X with dim Y = dim X is such that Y = X.

5 HODGE GROUPS OF K3 SURFACES How to apply it. Recall from section 2.2 we have a distinguished embedding ɛ : E 0 R. T ɛ is an R-vector space with a non degenerate bilinear form F ɛ : T ɛ T ɛ R, an irreducible Lie algebra hdg ɛ of End R (T ɛ ), with the following conditions: (1) dim R (T ɛ ) = dim E0 T, dim R hdg ɛ = dim E0 hdg; (2) End hdgɛ (T ɛ ) = R if E = E 0, C otherwise; (3) hdg ɛ contains a semisimple endomorphism f H of rank 2 defined as the restriction of the generator of the Lie algebra associated to h(u 1 ). It acts via the following rules: f H x = 0, x T 1,1 f H x = (2i)x for x T 2,0, f H x = ( 2i)x for x T 0,2 ; its non-zero eigenvalues are non real: they are 2i and 2i with multiplicities h 2, Proof of TR or CM. Write [E : Q] = r + 2s and the real and complex embeddings Consider T as a K-module and write {ρ i : E R C} i, {σ j, σ j : E C} j. dim Q T = dim E T [E : Q], tr T/Q (a E) = dim E T tr E/Q (a), tr E/Q (a E) = ρ i (a) + 2 Re(σ j (a)). Define, = ϕ(, ), it s a non-degenerate symmetric bilinear form on T of signature (2, m) such that its real extension is positive definite on (T 2,0 T 0,2 ) T R and negative definite on the orthogonal T 1,1 T R. We get an involution E E, a a, by a, =, a Properties: a E, i.e. with a also a preserves the Hodge structure. PF: let w T 1,1, then a (w) is orthogonal to T 2,0 T 0,2... is an automorphism of E: (ab) = a b ; a a = 1 iff a is an isometry Let E 0 be the fixed field of, we have [E : E 0 ] 2 For a E define ξ a = a a = aa K, it is self adjoint and notice i) ξ a v, v = av, av > 0 for all v T R 0 (up to setting. =. on T 1,1 T R ). In particular all eigenvalues of ξ a are positive, so tr T/Q (ξ a ), tr E/Q (ξ a ) > 0; ii) ξ a = a 2 for a E 0 ; iii) E 0 is a totally real field. Recall that if F is a number field and tr F/Q (a 2 ) > 0 for all a F, then F is TR. PF: suppose there exists a complex embedding σ s, and find an element a F close to 0 via every embedding different from σ s and σ s (a) close to i (to do this use F Q R = R r C s and Q is dense in R)... To end the proof we have to show that either E 0 = E or, if not, then E 0 /E is a purely imaginary quadratic extension. If E 0 E, then it is an extension of degree two, and write E = E 0 ( α), α E 0. Fix a real embedding E 0 R, and heading for a contradiction suppose that α > 0. We get two real embeddings of E: ρ 1 : E R, α α and ρ 2 = ρ 1 () : E R, α α As above one can choose a E such that ρ 1 (a) is close to 1, ρ 2 (a) 1 and it is close to zero w.r.t. every other embedding (real or complex). Here is the contradiction: 0 < tr E/Q (ξ a ) = tr E/Q (aa ) ρ 1 (aa ) + ρ 2 (aa ) = ρ 1 (a)ρ 2 (a) + ρ 2 (a)ρ 1 (a) 2. Theorem 2.1 (2) is finally proved.

6 6 GREGORIO BALDI 2.7. Real and complex points of the Hodge group in the TR case. Let T be a irreducible Q-PHS such that E is TR, and n = [E : Q], the theorem implies Hdg(V ) = SO(T, ψ) and one can prove: SO(T, ψ)(r) = SO(2, m 2) SO(m, R) n 1, SO(T, ψ)(c) = SO(m, C) n. Moreover the representation of these algebraic groups are the direct sum of the standard representations. See vg Thm Proof of dim E T 3 in the TR case. Thanks to the first proof of section 2.4, we are left to prove that dim E T 2. Write n = [E : Q], and suppose 2 = dim E T, by the main theorem we have Hdg(T )(C) = SO(2, C) n. Notice that SO(2, C) = C and its standard representation in C 2 is given by z diag(z, z 1 ); with this action one computes End C (C 2 ): is made by the diagonal matrices. Hence End Hdg (T ) C = (End C (C 2 )) n = C 2n But E = End Hdg (T ), so 2n = n is the contradiction we were heading for. 3. APPLICATIONS (TO BE EXPANDED!) The applications named here need also the Tate conjecture and the Mumford-Tate conjecture (now theorems) Picard rank (Charles). X a K3 defined over a number field k, T, E defined as in the previous section. Denote with ρ the geometric Picard rank of X, if p is a place of good reduction, ρ p is the geometric Picard rank of X p. One always has ρ ρ p = 2n Theorem 3.1. If E is CM or dim E T space is even, then there exist infinitely many places p of good reduction such that ρ = ρ p (with density one if one allows finite extensions of k). If E is TR and dim E T space is odd, then ρ + [E : Q] ρ p ; and the equality holds for infinitely many primes. Corollary 3.2. Let X be either a K3 surface of Picard rank 2 with E a totally real field of degree 4, or a K3 surface of Picard rank 4 with E a totally real field of even degree. Then X contains infinitely many rational curves Mumford-Tate conjecture for the product of an abelian surface and a K3. Commelin: the MT conjecture is true for H 2 (X A)(1) X a K3, A an abelian surface. Combine Albert classification and Zarhin s theorem. Easiest case: the two endomorphism algebras are not isomorphic, then proceed case by case An adelic open image theorem for K3s. One of the next lectures will be about the following result. Theorem 3.3 (Cadoret-Moonen). Let X be a K3 surface defined over a number field k, and consider the Galois representation ρ X : G k GL(H 2 (X(C), Ẑ)(1)). Then the image of ρ X is an open subgroup of the A f -points of the Mumford-Tate group associated to H 2 (X(C), Z)(1). A necessary condition to obtain the adelic version of the Mumford-Tate conjecture is that the Hodge-structure associated to H 2 (X, Ẑ) is Hodge-maximal (and the classical MT conjecture for K3s). The proof of this fact is a nice application of Theorem 2.1

7 HODGE GROUPS OF K3 SURFACES Hodge-maximality. About the adelic points of a map between algebraic groups we have the following results. Proposition 3.4. Let G, H be connected algebraic k-groups, π : G H/k a surjective morphism, and write F := ker π. (1) If F is finite, then for almost all v Ω k we have δ kv (H Ov ) = H 1 (kv nr /k v, F ), where δ kv : H Kv H 1 (k v, F )is the coboundary map; (2) If F is connected, then π(g Ov ) = H Ov for almost all v, i.e. π is surjective at the level of the adelic points. Definition 3.5. Let V Q be a rational Hodge structure given by h : S GL(V R ) and M = MT(h) GL(V Q ) its Mumford-Tate group. V is Hodge-maximal is there is no non-trivial isogeny of connected k-groups M M such that h can be factorised as h : S M R M R. Proposition 3.6. Let (V, ψ) be a polarised Q-HS of type ( 1, 1)+(0, 0)+(1, 1) with Hodge numbers 1 n 1, for some n. Then V is hodge maximal. Remark. It is necessary to work with HS of weight 0! So we are working with the twisted H 2 of a K3: H 2 (X, Q)(1). Proof. We may assume (V, h) to be irreducible (i.e. that there are no non-zero Hodge classes in V ): Hodgemaximality depends only on the MT and on h, and they do not change if we add copies of Q(0)s. We treat the TR and CM cases separately. Suppose E is totally real, we know that n := dim E V 3. If n is odd, the Mumford-Tate group is simply connected, and we are done. If the dimension is even... Suppose E is CM, ALGEBRAICITY OF INVARIANT CYCLES IN FAMILIES: FINITE VS INFINITE MONODROMY We start with a lemma. Lemma 4.1. Let Γ be a subgroup of Aut(V ), and G 0 the connected component of its Zariski closure in Aut(V C ). Assume that G 0 is connected and semisimple group (hence contained in SL(V C )), and that Γ contains a subgroup of finite index lying in MT(Q). Then Γ contains a subgroup of finite index lying in Hdg(Q). Proof. Γ SL(V C ) MT(C) is a subgroup of finite index in Γ, Hdg(C) is of finite index in SL(V C ) MT(C), so Γ SL(V C ) MT(C) Hdg(C) works. We can now prove the following dichotomy. Theorem 4.2. Let X be a complex variety with h 2,0 = 1, and Γ a subgroup of Aut H 2 (X, Q) satisfying the following conditions. C1. H 2 (X, Q) Γ Q R is an S invariant subspace of H 2 (X, R); C2. there exists a subgroup of finite index Γ Γ such that Γ MT(Q); C3. The connected component of the Zariski closure of Γ in Aut H 2 (X, Q) is semisimple. Then one of the following is true: D1. Γ is finite; D2. H 2 (X, Q) Γ H 2 (X, Q) Hdg = A(X) Q, and if Γ lies in the automorphisms of H 2 (X, Z), then the previous inclusion holds at the integral level. D2. is a geometric analogue of the Tate conjecture. Proof. Thanks to C1, H 2 (X, Q) Γ is a Hdg-submodule of the H 2. The decomposition H 2 (X, Q) = H 2 (X, Q) Hdg T and the simplicity of T (X), established in the second lemma, imply that either H 2 (X, Q) Γ H 2 (X, Q) Hdg = A(X) Q or H 2 (X, Q) Γ T (X).

8 8 GREGORIO BALDI In the first case we are done. In the second, notice that, thanks to the first lemma, there exists a subgroup of Γ, contained in Hdg(Q) with finite index in Γ. So Γ := Γ Hdg(Q) is a subgroup of finite index in Γ. To prove the theorem we show that Γ = 1. We have T (X) H 2 (X, Q) Γ H 2 (X, Q) Γ, and H 2 (X, Q) Hdg H 2 (X, Q) Γ. This implies H 2 (X, Q) = H 2 (X, Q) Hdg H 2 (X, Q) Γ H 2 (X, Q). This means exactly Γ = Interlude: families of surfaces. In the next two subsections we work with a family of surfaces, namely a smooth projective morphism of relative dimension two f : X S/C where X and S are connected schemes of finite type, and each geometric fiber X s is connected (hence they are irreducible smooth projective surfaces). We may factorise f as X P n S S, and fix a polarization L, i.e. an invertible O X -module L which is the inverse image of O(1) P n. Notice that L is relative ample and each L s is a very ample invertible O Xs -module. Remark. Euler characteristic is constant in family, and also the hodge numbers are constant. Notation: π(s, s) Aut H 2 (X s ) Aut H 2 (X s, Q) and Γ s its image, global monodromy of the family (topological π 1 ). Everything behaves well, if you chose another point t Application to family of surfaces with P g = 1. Proposition 4.3. Outside of a countable union of analytic subvarieties of S,the Mumford-Tate group MT(X s ) is locally constant and contains a finite-index subgroup of the algebraic monodromy group. Proof. Proof will be added, see Deligne, Weil pour K3s prop 7.5 Theorem 4.4. Assume h 2,0 (X s ) = 1 and Γ s is infinite, then and rationally. H 2 (X s ) π(s,s) = H 2 (X s ) Γs A s Proof. We can apply Theorem 4.2: C2 is the result named above, C1 and C3 are always true for such a family (see the appendix) Application to families of K3s. The main claim of this section is the following result. Theorem 4.5. Let X S be a non-isotrivial family of K3 surfaces, i.e. there is no étale covering S S such that X S T is isomorphic to the constant family Y S for Y = Y t. Then for all s S H 2 (X s, Q) π(s,s) A s,q and H 2 (X s ) π1(s,s) A s. In virtue of Theorem 4.4, and the alternative of Theorem 4.2, it is enough to prove the following. Theorem 4.6. Let X S be a family of K3 surfaces, the following are equivalent. The global monodromy group is finite; The family is isotrivial.

9 HODGE GROUPS OF K3 SURFACES 9 Remark (Isotriviality vs isomorphic fibers). Let X S be a family of curves (of genus g 2). Fix a level structure on each fiber, i.e. a symplectic isomorphism ϕ s : Z/nZ 2g = H 1 (X s, Z/nZ). Let S S be a finite étale cover such that the local system H 1 (X s, Z/nZ) s is trivial and consider the period map S M g,n. If all the fibers X s are isomorphic then S M g,n is constant, but since M g,n is a fine moduli space the pullback of the family X S along S S must be trivial. If the family is isotrival then the monodromy, after a finite étale base change, is trivial. For the converse we may assume the monodromy to be trivial; the strategy is to show that all the fibers are isomorphic (thanks to the Torelli theorem), and to emulate the argument of the remark to conclude that the family is constant (using the existence of a fine moduli space of K3s, fixing a polarization and a level structure). Proof. Let s, t S, the natural map H 2 (X s, Q) = H 2 (X s, Q) Γs = H 2 (X s, Q) Γt = H 2 (X s, Q) is an isomorphism of Q-HS, which gives also an isometry H 2 (X s ) = H 2 (X t ). This, together with the initial choice of a polarization L for X, gives an isomorphism of polarized K3: As in the remark this is enough to conclude. Torelli: (X s, l s ) = (X t, l t ). 5. APPENDIX: SOME FACTS FROM HODGE II Let f : X S as in the text. We collect here some facts, see Deligne Hodge II, Section 4. Fixed part theorem. The Q-subspace H 2 (X s, Q) π1(s,s) = H 2 (X s, Q) Γs is a Q-Hodge substructure of H 2 (X s, Q), and does not depend on s. The s independence means the following: s, t S then H 2 (X s, Q) Γs Q R = H 2 (X s, Q) Γt Q R compatibly with the action of S. Semi-simplicity theorem. If R 2 f Q is a local system on S, then the connected component of the Zariski closure of the image of π 1 (S, s) in Aut((R n f C) s ) is semi-simple. The main point is that the Leray ss degenerates, i.e. E = E 2. E p,q 2 = H p (S, R q f Q) H p+q (X, Q)