Introduction to Hodge Modules and Examples

Introduction to Hodge Modules and Examples Pablo Boixeda October 19, 2016 Definition 1. A filtered D-mod with Q-structure is M = (M, F, K) on X a) K perverse sheaf Q b) regular holonomic D-mod M c) A good filtration F on M with an isomorphism DR(M) = K C Q Remark 1. 1. A Hodge module is a filtered D-mod with Q-structure satisfying certain conditions 2.The categories of MHM(X) consists of (M, W ) where M is a filtered D-mod with Q-structure and W a filtration as such, such that the graded pieces are HM. Axiom 1. 1) M HM(X) is an abelian categories satifying the following rat D b MHM(X) D b cs(x; Q) D b coh (D X) C Q D b cs(x; C) Dmod X 2) MHM(pt) is the category of graded polarizable rational mixed Hodge structures, such that rat is the forgetful functor 3) M MHM(X) admits a weight filtration W satisfying -morphisms preserve W strictly -Grk W M is semisimple -W on pt is the usual weight filtration 4) MHM(X) has a duality functor D For a morphism s : X Y D b MHM(X) has functors f, f!, f, f!, that are sent by rat and Dmod X to the usual such functors. 5) f!, f don t increase weights f!, f don t decrease weight Remark 2. This has the usual weight yoga as in the case of sheaves on schemes over finite fields DR 1

1 Examples 1. Q H pt usual degree (0,0) Hodge structure. Now define for X smooth Q H X [d X] := π X QH pt[d X ] MHM(X) pure and Q H X[d X ] = (O X, F, Q X [d X ]) such that O X k 0 F k O X = 0 else 2. If D a smooth divisor of a smooth variety X of dim n, then if j : U X j Q H U [n] = (O X ( D), F, j Q U [n]) such that O X ((k + 1)D) k 0 F k O X ( D) = 0 else where O X ( D) are the functions with at worst poles along D. In general for D non-smooth we know F k O X ( D) O X ((k + 1)D) Now we present 3 examples of the uses of Hodge modules. The first 2 approach the second example above in the case of D singular 2 Residues and filtered D-mods [S] 2.1 Setting for this approach X projective variety of dim n. O X (1) sufficiently ample line bundle. P = O X (1) linear system of dim(p ) = d X = (p, x) P X x H p }, the incidence variety of dimenson d X =n+d-1. Y = (p, x) P X H p tangent to X at x}, ie points where X p is singular. Note 1. If i : X Q the embedding in projective space given by the ample line bundle. 2. X = P(i θ Q ), where θ Q the cotangent bundle and Y = P(N X Q ), where N X Q is the cotangent bundle of X in Q. 3. φ : X X and ψ : Y X The objective is understanding Rπ Q H X [d X ] where π : X P To do this we introduce the following ingredients N = pr Ω n P X/P ( X ) F k N = pr Ω n P X/P (kx ) = H 0 (X, ω n X (k)) O P (k) k 0 0 else 2

By Bertini X p is generically smooth for p P. Define P sm where this holds. H = R n 1 ev π sm Q = ker(r n 1 π sm Q H n+1 (X, Q)) a variation of Hodge structure on P sm with corresponding D-mod H O. j : P sm P then define Res X /P : N j H O by ω Γ(U, N ) Res X /P (ω)(p) = Res Xp (ω X\ X p ) for p U P sm Note 2. Here the residue map is defined as follows. Ω X (log(e)) Ω X ( E) a quasi-isomorphism for E smooth and define Res(α df f ) = 2πiα Y. Further the induced map on homology is exactly the boundary map of the Gysin sequence. Then define M as the image of N and F k M the image of F k N. Also clearly Res X /P is surjective over P sm so j M = H O and further j F k M = F n k H O if O(1) sufficiently ample. By the Decomposition Theorem we have Rπ Q H X [d X ] = i,j E i,j [ i] where E i,j has strict support on a codim j variety. By Hard Lefschetz for Hodge modules E i,j = E i,j ( i) i X P X U π P we have i i! Q H P X QH P X g g Q H P X distinguished triangle. Further as X smooth Verdier duality gives i! Q H P X = Q H X [ 2]( 1), so we get after applying pr P the LES. ( )... H k 1 π Q H X [d X ]( 1) H k pr P Q H P X[d P X ] H k q Q H U [U]... By base change we get the Knneth formula Lemma 1. H k q Q H U [d U] = 0 k > 0 H k Q H P X[d P X ] = H k+n (X) Q H P [d] Proof. q Q H U [d U] = pr P g Q H U [d U], g Q H U [d U] has D-mod O P X ( X ) so the above has D-mod. Rpr P DR P X/P (O P X ( X )) = pr [O P X ( X )...Ω n P X/P ( X )][n] as these are pr P acyclic by affineness and result follows g q 3

We get H k π Q H X [d X ]( 1) = H n+k+1 (X) Q H P [d] = E k,0 = E k for k > 0 E k = E k,0 = E k,0 = H n+1 k (X)(1 k) Q H P [d] = H n+k 1 (X) Q H P [d] for k < 0 By the semisimplicity we get V n 1 = H n 1 π sm Q H X sm[d] = Vev n 1 H n 1 (X) Q H P sm[d] Further as j H 0 π Q H X [d X ] = H n 1 π sm Q H X sm[d] = V n 1, so j! V n 1 a submodule of H 0 π Q H X [d X ] and as these are the only ones with support on P sm, we get E 0,0 = j! V n 1 = j! Vev n 1 H n 1 (X) Q H P [d] Remark 3. By LES (*) we get, as by Hard Lefschetz some maps are injective And H k q Q H U [d U ] = H n+k (X) H n+k 2 (X)( 1) QH P [d] for k < 0 0 H n 0 (X) Q H P [d] H 0 q Q H U [d U ] j! V n 1 ev R 0 where R = i>0 E 0,i( 1). Now X smooth hypersurface so O P X ((k + 1)X ) k 0 F k O P X ( X ) = 0 else So by the above D-mod of H j q Q H U [d U] is the jth cohomology N j of Further if [pr P O P X ( X )...pr P Ω n P xx/p ( X )][n] F k DR P X/P (O( X )) = [F k Ω 1 P X/P F k+1...ω n P X/P F k+n][n] then Rpr P F k DR P X/P (O P X ( X )) is the filtration on the pushforward, ie has cohomology F k N j Proposition 1. F k N j is the cohomology of E n+k+1 = [pr P F k...pr P Ω n P X/P F k+n][n] provided k 0 or k Z if H q (X, Ω p X (k)) = 0 q > 0 and k > 0 ( ) (ie O(1) sufficiently ample) Proof. Acyclicity of the complex follows by condition or for k 0, as the individual sheaves are E i k = pr P (Ω n+i P X/P F k+i 1O P X ( X )) = H 0 (X, Ω n+i X (k + i)) O P (k + i) k + i 1 0 else 4

Theorem 1. If O(1) ample enough that it satisfies ( ) then H i (P, Ω p P F kn 0 ) = 0 i max(p 1, 0) for p > 0 Proof. Using the stupid filtration we get E i,j 1 = H j (P, Ω p P E i k) H i+j (X, Ω p P E k) As Ω p P Ei k a sum of Ωp P (m) s m 1 and P a projective space, we have E i,j 1 = 0 for j > 0 and clearly E i,0 1 = 0 for i > 0, so H i (X, Ω p P E k ) for i > 0. Using the cohomology filtration E i,j = H i (P, Ω p P F k n 1N j ) H i+j (X, Ω p P E k) so Ω p P F k n 1N j is a bunch of sums of Ω p P for j < 0 by an above remark, so E i,j 2 = 0 unless j = 0 or i = p, so H i (Ω p P F k n 1N 0 ) = E i,0 2 = 0 for i max(p 1, 0) Remark 4. Writing W p k = H0 (P X, Ω p P X/P (kx )) = H 0 (X, Ω p X (k)) H0 (P, O P (k)). Then the same argument shows H 0 (P, F k n 1 N 0 ) = W n k d P xx/p W n 1 k 1 where d P X/P is the map induced by the differential of the de Rham complex 2. Also similarly we get H 0 (P, F k M) W = n k d P X/P W n 1 k 1 +F n+1 k H0 n(x,c) Our objective to understand the cohomology of X is now reduced to computing R Proposition 2. The error term R = E 0,1 (ie E 0,j = 0 j 2) and R = 0 if the vanishing cohomology is non-zero Proof. We compute the characteristic sheaf of M ev R on the projectivized cotangent bundle. By the exact sequence this is the same as for N 0 0 H n 0 (X, C) O P N 0 M ev R 0 Lemma 2. P roj(gr F N 0 ) = ψ Ω n X O Y(n + 1) Proof. P roj(gr F N 0 ) = P roj( k Z H 0 (P, F k N 0 F k 1 N 0 )) By the Theorem above H 1 (P, F k N 0 ) = 0 k 0, so H 0 (P, F k N 0 F k 1 N 0 ) = H0 (P, F k N 0 ) H 0 (P, F k 1 N 0 ) 5

and so by the above we get H 0 (P, F k N 0 F k 1 N 0 ) Wk+n+1 n = d P X/P W n 1 k+n + W k+n n Lemma 3. O X (k n)...φ Ω n X O X (k) is a locally free resolution of ψ Ω n X O Y (k) the differential given by β d P X/P s X β, where s X is the local section defining X P X Proof. Omitted. Koszul type resolution for local complete intersection so for k 0 H 0 (Y, ψ Ω n X O Y(k)) = H 0 (X,φ Ω n 1 X O X (k 1)) Using the SES defining O X in P xx get for k 0 H 0 (X, φ Ω p X O X (k)) = and by the description of the differential above we get H 0 (Y, φ Ω n X O Y (k)) = and so the result follows by comparison H0 (X,φ Ω n X O X (k)) W n k d P xx/p W n 1 k 1 + W n k 1 W p k W p k 1 So we get the characteristic variety of N 0 has multiplicity 1 over Y and some multiplicity over the zero section. So as the characteristic variety of E 0,j j 2 can t be contained in the above, we get E 0,j = 0. Now if R 0 it contains the piece of the characteristic variety not contained in the zero section. So M ev has characteristic variety the zero section, so isa vector bundle and so as P simply connecte its flat sections are a constant sheaf, but H 0 (P sm, Rev n 1 π sm C) = 0, so we would have Rev n 1 π sm C = 0, so R = 0 or M ev = 0 as promised. 3 Hodge ideals [MP] To present the setting for the next example we need the following lemma Lemma 4. F k O X ( D)) O X ((k + 1)D) or equivalently F k n ω X ( D)) ω((k + 1)D) Proof. For SNC we have F k O X ( D) = F k D X O X (D), so if D given locally by x 1...x r = 0, for x i etale coordinates of X, we get F k O X ( D) =< x b1 1...x br r b i 1 b i = r + k > O X ((k + 1)D) For the general case we use a log resolution f : Y X of (X, D). E = (f D) red then f an isomorphism on an open set U whose complement is codimension at least 2 and se we get F k n ω X ( D) j (F k n ω X ( D) U ) = ω X ((k + 1)D) 6

Definition 2. The Hodge ideals are defined by F k n ω X ( D) = ω X ((k + 1)D) I k (D) Example 1. I k (D) =< x α1 1...xαr r 0 α i k, α i = k(r 1) > for D an SNC as in the proof above Note 3. The filtration is defined in the general case as follows. Choosing a log resolution of (X,D) C k n = [f F k n D X Ω 1 Y (log(e)) f F k n+1 D X...ω Y (E) f F k D] [f D X Ω 1 Y (log(e)) f D X...ω Y (E) f D] ω Y ( E) L D Y X which is independent of log resolution. Now we prove some nice properties of these. Using the multiplication of differential operators on the above chain complexes we get F l n ω X ( D) F k D X F l+k n ω X ( D) and so immediately I k 1 (D)O( D) I k (D). In fact it can also be shown I k (D) I k 1 (D) Theorem 2. D reduced effective divisor, then we have I k (D H ) = I k (D)O H for a general element H of a base-point free linear system on X. Further for any smooth divisor H Supp(D) such that D H reduced I k (D H ) I k (D)O H Theorem 3. If dim X=n 2 then the Hodge filtration on O X ( D) is generated at level n 2. More generally k 0 U k open subsets in X whose complement has codimension k + 3 such that O X ( D) Uk is generated at level k Proof. To prove this we ll need the following theorem Theorem 4. For a log resolution as above, the filtration on ω X ( D) is generated at level k if and only if R q f Ω n q Y (log(e)) = 0 for all q > k Proof. We show this by showing for k 0 F k n ω X ( D) F 1 D X = F k n+1 ω X ( D) if and only if R k+1 f Ω n k Y (log(e)) = 0. By definition of the filtration we have, if we define Φ k : C k n f F 1 D X C f k+1 n O X 7

then the result hold iff the map on 0-th cohomology is surjective. Let T be the kernel of Φ k. If R m is the kernel of F m D X F 1 D X F m+1 D X, then we get O X T p = Ω n+p Y (log(e)) f R k+p. So now we use the spectral sequence with the f O X stupid filtration E p,q 1 = R q f T p n R p+q n f T Now using the projection formula and the above formula of T p we get R q f T p n = 0 for p+q > n, using that R p f Ω q Y (log(e)) = 0 for p+q > n, the proof of which we will omitt. Now by the spectral sequence we get immediately R j f T = 0 for j > 0. Now by looking at the definition of C k n and the fact that the multiplication map on differential operators is surjective, we get that Φ k is surjective for k n. But now using the SES with this map as the surjection and taking LES of the pushforward, we get as R 1 f T = 0 that Φ k induces a surjective map on 0th cohomology. In general the only point where Φ k is not surjective is at level k 1, so if we denote B the image, just as above we see that the map on 0th cohomology is still surjective to B. So the original map is surjective if and only if R 0 f B R 0 f C k+1 n is surjective. Now we have the SES and so the result hold if and only if which is the required result. So we need to prove 0 B C k+1 n C k 1 k+1 n [k + 1] 0 0 = R k+1 f C k 1 k+1+n = Rk+1 f Ω n k 1 Y (log(e)) R n f O Y = 0 (1) R n 1 f Ω 1 Y (log(e)) = 0 (2) 1) As f a sequence of blow ups the fibers are n-1 dimensional so result follows. 2) If we let E = D + F, where D is the strict transform and F the reduced exceptional divisor. We can get D smooth, thus 0 Ω 1 Y (log(f )) Ω 1 Y (log(e)) O D 0 as D smooth. It can be shown that R n 1 f Ω 1 Y (log(f )) = 0, so we have to show R n 1 f O D = 0, but in fact as D D a series of blow ups the same argument as for 1) shows the result. We omitt the proof of the second part of the theorem Now we get to the main thm Theorem 5. TFAE 1. D smooth 8

2. the Hodge filtration on O X ( D) is the pole filtration 3. I k (D) = O X k 0 4. I k (D) = O X k n 1 2 Proof. 1) 2) 3) 4) are obvious 4) 1) will follow from the next theorem To state the theorem we make first a definition Definition 3. If W X irreducible closed with ideal I W I (p) W = f O X mult x (f) p x W general} Theorem 6. With the above notation if codim W=r mult W (D) = m where q = min(m 1, (k + 1)m r) I k (D) I (q) W Remark 5. The above theorem follows immediately by taking x D a singular point ie mult x (D) 1 2. We only use the theorem for points, so we will only give the prove in this case. The general case reduces to this one Proof. We will use the following propositions Proposition 3. For an ordinary singularity x of D I k (D) = m (k+1)m n x Proposition 4. h : X T s : T X such that hs = id and D t is reduced t T. Then V q = t T I k (D t ) m q s(t) } is open in T Enough to show the theorem locally. So let g be a local equation for D and x i local coordinates of X. If q = (k + 1)m n ie km < n let A N, be the space of homogeneous polys of degree m in n variables. p : X A N A N projection and s : A N X A N given by s(t) = (x, t). F divisor given by g c u x u c u coordinates of A N V = (y, t) X U y / F or y F and F t is reduced at y} is open as the complement are the points where the fibers are unreduced and is thus closed. Now h : Z = V p 1 (s 1 (V )) s 1 (V ) we can see 0 T and this satisfies the condition of the above proposition. So if I k (D) m q x we get I k (F t V ) m q x generically, but generically x is an ordinary singularity ( the 9

projectivized tangent cone is a generic surface of degree m in P n 1 ) so by the Proposition I k (F t V ) = m q x and thus we get a contradiction. For mk n d = mk n + 1 D in X = X A n inverse image of D via first projection. X embedded via X 0}, then D = D X. The fibers of D are reduced so by a Theorem above I k (D) I k (D )O X and mult (x,0) (D ) = m and mk < n + d by construction. So (k + 1)m n d = m 1, so by the first case I k (D ) m m 1 (x,0) so I k(d) mx m 1 4 Hodge theory and unitary representation of reductive Lie groups [SV] 4.1 Setting Reductive Lie Group G R maximal compact subgroup K R, with complexifications G and K respectively. U R maximal compact subgroup of G The objective is to understand unitary representations of G R ie ones equipped with an invariant positive definite inner product. To do this it is enough to do it for irreducible representations. Further we can associate to each unitary representation a Harish-Chandra modules, ie a U(g)-mod, that is also a K- mod such that the 2 actions of k are the same and st Hom K (W, V ) < for irreducible representations W. This Harish-Chandra module is irreducible and has a positive definite (, ) g R -invariant form. We also know that Z(g) acts via some character and Vogan showed it is enough to understand the ones with character λ h R and λ dominant to understand the unitary representations. Theorem 7. V HC(g, K) λ, for λ as above, has a u R invariant form It has been proven that these forms are strongly related to g R invariant forms 4.2 BB-localization On X = G/B Beilinson-Bernstein proved that you have the following equivalence of categories Theorem 8. Γ : D G/B mod U 0 (g) mod : V M Γ(M) D V U 0(g) 10

Where U 0 (g) = U(g)/ker(χ 0 )U(g) χ 0 the 0-th character of g, where this makes sense as Γ(D) = U 0 (g). Similarly we can define D λ = O X U(g)/ker(χ λ )O X U(g) which are locally isomorphic to the usual sheaf differential operators and we have and analogous equivalence of categories above for λ satisfying < λ, ˇα >/ Z 0 α Φ +. Then for Q locally closed in X such that D λ trivializes on some open containing Q, S an irreducible vector bundle with connection over Q then we define M(Q, λ, S) = H 0 (j S) Mod(D λ ) hol where D λ acts on S via its trivialization. These are called standard modules. Proposition 5. M(Q, λ, S) has a unique irreducible submodule I(Q, λ, S) and every irreducible holonomic D λ -mod can be realize as I(Q, λ, S) for Q, λ, S appropriatley chossen Call a Harish-Chandra sheaves in a similar way as a Harish-Chandra module a K-equivariant sheaf as above, such that the 2 actions of k. Similarly call standard Harish-Chandra sheaves the ones coming from the above construction with Q,S K-equivariant.Now we can get all irreducible Harish-Chandra-mods by this method using Q, S K-equivariant choices on the above construction. It turns out every orbit of K satisfies the trivialization condition so this can be done. And further as K has finitely many orbits, there give all the possible irreducible modules and we get Proposition 6. bijection irred Harrish-Chandra sheaves} standard Harish-Chandra sheaves} 4.3 Mixed Hodge modules and HC-mods As D λ are locally isomorphic to D and MHM is determined locally, so we can define MHM λ. Also in this case we won t have Q-structure. By embedding H into H H(which has an obvious real structure), we can get away with having only a complex structure with 2 filtrations F, F. For S an irreducible D X -mod on a K-orbit Q we give the following filtrations F p S = 0 p < 0 S p 0 W k S = 0 k < dim Q S k dim Q These filtrations (with obvious F choice) makes S HM(Q) λ. And using the same construction as above we get M(Q, λ, S) MHM(X) λ Further as the weight filtration has semisimple subquotients and M(Q, λ, S) contatins a unique irreducible submodules I(Q, λ, S) is the lowest part of the weigth filtration. 11

Theorem 9. Γ : MHM(X) λ U λ (g) mod is exact Now we want to construct inner products of I(Q, λ, S), so by functoriality we can do this by doing this for S. So now assume S comes equipped with a flat hermitian metric <, > Since U R compact we get an essentially unique positive U R -invariant measure dm on X Proposition 7. (σ, τ) = X < σ, τ > dm is u R on Γ(I(Q, λ, S)) Conjecture 1. With the set up as above M = Γ(I(Q, λ, S)) v F p M F p 1 M v 0 ( 1) p c (v, v) ur > 0 p where c = codim X (Q) So this conjecture would give a condition for this to be positive definite according to the Hodge filtration and so we reduce the problem to understanding the Hodge filtration References [1] C. Schnell, An Overview of Morihiko Saito s Theory of Mixed Hodge Modules, 2014. arxiv:1405.3096v1 [2] C. Peters and J. Steenbrink, Mixed Hodge structures, Springer-Verlag, Berlin, 2008 [3] C. Sabbah, Hodge theory, singularities and D-modules, 2007, Lecture notes (CIRM, Luminy, March 2007) [4] C. Schnell, Residues and filtered D-modules, C. Math. Ann. (2012) 354: 727. doi:10.1007/s00208-011-0746-0 [5] M. Mustata and M. Popa Hodge ideals, 20016, arxiv:1605.08088v3 [6] W. Schmid and K. Vilonen, Hodge theory and unitary representations of reductive Lie groups, 2012, arxiv:1206.5547v1 12