These are my live-TeXed notes for the course Mathematics 262y: Perverse Sheaves in Representation Theory taught by Carl Mautner at Harvard (Fall 2011). I eventually put some effort and editted the part before the Verdier duality during the winter. Please let me know if you notice any errors or have any comments!

## The Lefschetz hyperplane theorem

Theorem 1 (Lefschetz Hyperplane Theorem) Let be a projective variety and be the intersection of with a hyperplane such that is smooth, then the map induced by the inclusion is an isomorphism for and an injection for .
Example 1 Let be a hypersurface of degree . Then for some hyperplane , where is the Veronese embedding. Therefore by the Lefschetz hyperplane thoerem, for , we have Moreover, when is smooth, we have the same for by Poincare duality. If we replace by a field, this holds except the middle degree .

If is smooth, by Poincare duality we also have a Gysin homomorphism . The composition is given by , where is the first Chern class of the hyperplane section.

Theorem 2 (Hard Lefschetz) Let be a smooth projective variety of dimension . Then for any embedding and , the map is an isomorphism.

The Hard Lefschetz can be generalized to any Kähler manifold. Let be a complex manifold. One can show that any can be endowed wit a Hermitian metric on . Write , where is a Riemann metric and is an anti-symmetric 2-form.

Definition 1 A Kähler manifold is a complex manifold with a Hermitian metric where .
Example 2 Any complex smooth projective variety is Kähler. The idea is to pull back the Hermitian metric from .
Theorem 3 (Kahler version) Let be a Kähler manifold of dimension with Kähler class . Then for , is an isomorphism.
Remark 1 We are going to sketch the proof of this Kähler vesion of the Hard Lefschetz theorem using Hodge theory in the following several sections.

## Hodge theory on Riemannian manifolds

Let be a real vector space of dimension . Let be an inner product. Let be the exterior algebra of . Then there is an induced inner product on such that for .

Definition 2 An orientation of is a choice of a vector of length 1. Define an operator such that is characterized by . One can check that .

Let be a Riemannian manifold. Then is an inner product on , which extends to a smooth inner product on . Let be the smooth -forms.

Definition 3 An orientation on is a choice of a top form on with norm 1 in each fiber.
Definition 4 If is oriented, we define the Hodge star operator using the above local construction pointwise.
Definition 5 For , we define the inner product on by
Definition 6 Define by . Using Stokes' theorem, one can show that .
Definition 7 Define the Laplacian by . The kernel of is called the space of harmonic forms.
Proposition 1 is harmonic if and only if and .
Definition 8 A direct check using definition.
Theorem 4 (Hodge) Let be a compact oriented Riemannian manifold. Then there is an isomorphism (depending on ) .

The proof of the Hodge theorem involves the analysis of elliptic operators in order to construct Green's operator, which we shall not get into here.

Corollary 1 (Poincare Duality with real coefficients) The pairing is non-degenerated.
Proof Let be a harmonic form, then is also harmonic by Proposition 1 and So the pairing is non-degenerated. ¡õ

## Complex manifolds and the -Hodge Theorem

Definition 9 Let be a complex manifold of dimension . We denote the real tangent bundle (think of as real manifold). It is a real vector bundle of rank with . The complex tangent bundle is defined to be , namely .

A map induces the tangent map . If the is holomorphic, then the tangent map (or, its Jacobian matrix) has very restricted form, since the subspaces and are preserved under holomorphic change of variables.

Definition 10 Define the holomorphic tangent bundle to be the holomorphic subbundle of with . Similarly define be the anti-holomorphic tangent bundle with .

We have -isomorphisms between and , and also .

Definition 11 Using the dual basis , we have similar notions of real, complex, holomorphic and anti-holomorphic cotangent bundles.
Remark 2 Every complex manifold admits a canonical orientation.
Definition 12 The space of complex-valued smooth -forms is defined to be .

For , nonnegative integers, we denote . Then

Definition 13 A -form is an element of , which can be locally written as .

We have . Let be the projection. Define the Dolbeault operators by and similarly . Then and . We denote the corresponding Dolbeault cohomology groups of and by and .

Let be a Hermitian form, then is a positive definite quadratic form. Let be a complex manifold with the Hermitian metric , then we can define and on using . We can extend them -linearly to . Then and is a linear isometry. We can also extend the pairing on to by

Definition 14 Define and . Again using Stokes' theorem, one can show that and are adjoint pairs.
Definition 15 Define and . Define . Define similarly for .
Theorem 5 (-Hodge Theorem) Let be a Hermitian manifold. Then there are isomorphisms of finite dimensional vector spaces

## The Hard Lefschetz theorem on Kähler manifolds

Definition 16 A Hermitian manifold is Kähler if the real 2-form is closed, i.e., .
Definition 17 Let be a Kähler manifold. Define the Lefschetz operator by . As , this induces a map . Define to be the adjoint to . We have since
Lemma 1 on -forms.
Corollary 2 The actions of and form an action of on . This implies , hence .

By Corollary 2, to prove the Kähler version of the Hard Lefschetz Theorem, it suffices to show that sends harmonic forms to harmonic forms.

Theorem 6 (Kahler identities)
1. , .
2. .
3. .
Proof For (a), see Griffiths-Harris. For (b), use (a). For (c), using the definition of and (b), it suffices to check that , which can be shown using (a). ¡õ
Corollary 3 . .
Corollary 4 commutes with .

As commutes with , we now know that sends harmonic forms to harmonic forms, which implies the Hard Lefschetz Theorem.

## The Lefschetz decomposition and Hodge-Riemann bilinear relations

Definition 18 For , we define to be , called the space of primitive -forms. We define to be , called the space of primitive cohomology classes.

The Hard Lefschetz Theorem then has the following easy consequence.

Corollary 5 (Lefschetz Decomposition) Any can be written uniquely as a sum , where . For , we have the Lefschetz decomposition
Remark 3 The same is true for bihomogeneous elements of .
Lemma 2 If , then , where .

We omit the proof of the above useful fact.

Let be the Poincare duality pairing. Define the intersection form on using , Then is symmetric when is even and skew-symmetric if is odd. Define the Hermitian form on by

g

Lemma 3 The Lefschetz decomposition on is an orthogonal decomposition, i.e., for .

Proof as . ¡õ
Theorem 7 (Hodge-Riemann bilinear relation) The decomposition is orthogonal with respect to . Moreover, is positive definite on .
Proof is a -form in . It is non-zero only when and , hence , which implies the orthogonality. Let , then there exists primitive and harmonic with . So is also primitive and harmonic. Then by Lemma 2, The positive definiteness then follows. ¡õ
Remark 4 Using the Hodge-Riemann bilinear relations, one can show the Hodge index theorem: suppose is even, then the signature of the intersection form on is equal to , where are the Hodge numbers.

## Cohomology of sheaves and categories

We construct a sequence of vector bundles which is an injective ?? resolution of the trivial bundle. The global sections form a complex and its cohomology is the de Rham cohomology. More generally, we would like to replace the vector bundles by any sheaves. In abstract language, we would like to define a new category of sheaves such that

1. Any object in should be identified with all its resolutions.
2. Functors should only be applied to special representatives in the isomorphism class of an object to obtain its cohomology.
Definition 19 A morphism of of complexes in an abelian category is called a quasi-isomorphism if the induced morphism on the cohomology is an isomorphism.
Example 3 Let be a resolution. Then the morphism is an quasi-isomorphism.
Definition 20 Let be an abelian category and be the category of complexes in . There is a category and a functor (called the derived category of )such that
1. If is a quasi-isomorphism, then is an isomorphism.
2. For any functor satisfying (a), there is a unique functor such that .

Let be a category and be a class of morphisms in . We can construct a functor such that for any , is an isomorphism, where has the same object as with morphisms in formally inverted. However, this construction loses the additive structure. To solve this problem, we shall only do this construction for a localizing system.

Definition 21 Let be a class of morphisms in . We say that is a localizing system (or multiplicative system) for the category if
1. is closed under composition: for any and for any , we have (if they are composable).
2. Extension condition: for any , , we have the following commutative diagram and a similar commutative diagram with arrows inverted.
3. For , there exists such that if and only if there exists such that .

Unfortunately, the class of quasi-isomorphisms does not form a localizing system for . In order to construct the derived category, we need to pass to the homotopy category . It turns out that the quasi-isomorphisms form a localizing system for and the category is the desired derived category .

Remark 5 The construction and properties of homotopy category, derived category and derived functor are discussed in class, which are not reproduced here. For this part of material, see Haiman's notes.

## Aside on spectral sequences

Definition 22 Let . A decreasing filtration is a sequence Let be a complex, a decreasing filtration is a double sequence such that and . The filtration on induces a filtration on given by .
Theorem 8 (Spectral sequence of a filtered complex) Let be a filtered complex as above such that for any , there exists , . Then there exists complexes , where such that
1. and is the map induced from .
2. .
3. For a fixed and , .
Proof Let . Note that and . We can define . Then we define and . One can check that .
1. , . So .
2. By construction.
3. For , . So and . Therefore and This completes the proof. ¡õ
Example 4 Let be a double complex and be the total complex. Define . The associated spectral sequence for this filtration is and .

## Ordinary cohomology = Sheaf cohomology of the constant sheaf

Lemma 4 If is a contractible space, then .
Proof Cf. Corollary 2.7.7 (iii) of Kashiwara M., Schapira P, Sheaves on manifolds. ¡õ
Proposition 2 Let be a sheaf on , be an open cover of , such that for any , , . Then .
Remark 6 This justifies writing .
Proof To define the Cech cohomology , we form with So we get a complex of sheaves on by restrictions. The sequence is a resolution, hence is quasi-isomorphic to . Let be a flabby resolution of . Then we can construct with the Cech complex of the 's a double complex . Consider the spectral sequence from the filtration of the global sections. As is flabby and the columns are resolutions, we get And Using the other filtration we know that . ¡õ
Remark 7 If is flabby, then so is .
Corollary 6 If is a CW-complex, then .
Proof Choose a cover of such that is contractible for any and apply the previous proposition. ¡õ
Proposition 3 Let be a continuous map, then is the sheafification of the presheaf
Proof Let be an injective resolution. Then is the sheafification of . ¡õ

So gives a good notion of cohomology in families. For example, to compute the cohomology of , we can compute for any map .

## Degeneracy of the Leray spectral sequences

Let be abelian categories and be a left exact functor. Suppose has a class of -acyclic objects.

Theorem 9 For any , there exists a spectral sequence .

More generally, let , where , are two left exact functors. Suppose has a class of -acyclic objects and has a class of -acyclic objects such that is -acyclic.

Theorem 10 (Grothendieck spectral sequence) For any , there exists a spectral sequence .

Let be a smooth fiber bundle with smooth compact fibers . Then the Grothendieck spectral sequence associated to applied to the constant sheaf becomes the classical Leray spectral sequence where is the local system on with fiber .

Example 5 For the Hopf fibration , and are constant sheaves as .
Definition 23 A family of projective manifolds is a proper, holomorphic submersion of smooth varieties factoring through with fibers smooth projective varieties.
Remark 8 Ehresmann's theorem implies this is a smooth fibration.

Deligne proved the degeneracy of the Leray spectral sequence for a family of projective manifolds.

Theorem 11 (Deligne) Let be a family of projective manifolds.
1. Version I: the Leray spectral sequence degenerates at the -page. In particular, ; Version II: the derived pushforward and .
2. The monodromy representation for any is semisimple.
Remark 9 is NOT true for the Hopf fibration as one can check directly. It is a real algebraic proper submersion, but not holomorphic.
Remark 10 We need smooth projective condition on the fibers, since the usage of the Hard Lefschetz Theorem is crucial.
Remark 11 It is important that the coefficients of the cohomologies are (at least characteristic 0). For example, in characteristic 2, the matrix is not semisimple (consider , is a rank two local system and does not split).
Proof (Version II implies Version I) Let . The Leray spectral sequence is the spectral sequence applied to and it is degenerate if is concentrated in a single degree. If , then the Leray sequence is the direct sum of the spectral sequence of . Hence the Leray spectral sequence degenerates. ¡õ
Remark 12 Since , we know that . Moreover, for , , the cup product is given by .
Proof (Version II) Suppose . Let be the first Chern class of the hyperplane section. By the previous remark, we have and induces . On each fiber, is an isomorphism by the Hard Lefschetz. Therefore is an isomorphism on each stalk, thus is an isomorphism. Now applying the following Key Lemma to , and the Lefschetz operator, we know that . ¡õ
Lemma 5 (Key Lemma) Let be an abelian category. Let and such that the induced maps are isomorphisms. Then .
Proof (van den Bergh) By induction downward on , where is an integer such that for any . Then is the isomorphism induced by (by taking ). Using We get a map and . And . Using properties of triangles, we get . To finish the proof, we check that whenever . ¡õ

Next part of this course is to define more operations on sheaves and establish the Poincare duality of sheaves for singular varieties.

## Operations on sheaves

Let be a commutative ring and . We define be the sheaf . Then . Because is a left exact functor, we obtain a right derived functor .

Similarly, we define by . Then . Because is right exact, we obtain a left derived functor . is called flat if is exact. Note that is flat if and only if is a flat -module for any . Flat sheaves form an acyclic class for .

Proposition 4 .
Corollary 7 .
Proof If is locally free and injective, then is injective. So The general case follows by taking a locally free resolution of and an injective resolution of . ¡õ
Definition 24 Let be a map of topological spaces. For any open subset , let . Then forms a subsheaf of , called the direct image with compact support. Note that is a left exact functor.
Definition 25 Define . Then for the morphism .
Proposition 5 Let , where locally compact. Then for any , is an isomorphism.
Proof For any , there exists an open neighborhood of and such that is proper. It follows that has compact support . We define . One can check that is injective. For surjectivity, we use the following lemma. Hence is an isomorphism. ¡õ
Lemma 6 If is Hausdorff (resp. paracompact), is compact (resp. closed). Then is an isomorphism (i.e. we do not need to sheafify).
Example 6 Consider . Then as the embedding is never proper.
Definition 26 A sheaf is soft if for any compact, the restriction map is surjective.
Remark 13 Flabby sheaves are soft.
Lemma 7 is soft if and only if for any closed subset , the restriction map is surjective.
Corollary 8 If is soft, then for any locally closed embedding , is also soft.
Proof For any closed, we have the surjection by the softness of . ¡õ
Proposition 6 If is exact with soft. Then is also exact. In particular, is exact.
Proof By Corollary 8, is soft for any . Since it is enough to check the exactness on stalks, we reduce to the exactness of by Proposition 5. By the left exactness of , we only need to check the surjectivity of . Let . Choose a compact open subset . Replace by and by , we may assume that is compact. Giving is the same as giving a finite compact cover of and such that . One can check for some . By the softness of , we get a global section maps to . Replace by , then . Now an induction shows that can be glued to be a section of . ¡õ
Proposition 7 If is exact and are soft, then is soft too.
Proof For any closed, we have the following diagram Hence is soft. ¡õ

The above two propositions together imply the following theorem.

Theorem 12 Soft sheaves form an acyclic class for .

In fact more is true:

Proposition 8 If is locally compact and countable at (its one-point compactification is Hausdorff). Then soft sheaves form an acyclic class for .
Example 7 The de Rham resolution is a soft resolution, hence .

By Remark 13, we know there exists enough soft sheaves. So we have the right derived functor . In particular, we have the derived functor . Define the cohomology with compact support . Note that .

Theorem 13 (Proper Base change) If we have a Cartesian diagram Then there exists a canonical isomorphism .
Proof There exists a canonical map . By adjunction of , have , which corresponds to . This induces an isomorphism. ¡õ
Proposition 9 (Projection formula) There exists a natural map . It is an isomorphism if is flat.
Corollary 9
1. .
2. .
Example 8 (Stalks of ) is naturally isomorphic to the sheafification of the presheaf . Denote . As is exact, we know that . For example, let and , then .
Example 9 (Stalks of ) Using the base change we know that .

We have seen in the exercise that if is a closed embedding, then has a right adjoint , where is functor of taking the sheaf of sections with support inside . On the contrary, suppose , then does not admit a right adjoint. Does admit a right adjoint in the bounded below derived category? Or more generally, does admit a right adjoint functor for a continuous map between locally compact spaces? The answer is YES.

Definition 27 There exists a right adjoint to . We call the exceptional inverse image functor. (See the next section for a brief discussion of .)
Definition 28 We define the dualizing sheaf , where .

For a oriented manifold, . In particular, by adjunction we have On the other hand, So in this way we recover the Poincare duality. (In general, for unoriented manifolds, the dualizing sheaf is the orientation sheaf shifted by the dimension.)

More generally, for any . We have

Definition 29 We define the dualizing functor
Question Given a "nice" singular space , can we associate to some canonical object in that is self-dual, i.e., ? If it is the case, then we will obtain a desired analog of the Poincare duality for singular spaces.

## Verdier duality

The original proof existence of is due to Verdier, using that we already know that exists for a locally closed embedding and then gluing them together. It is difficult since the derived category does not have good gluing property. Instead of Verdier's approach, we will give a proof due to A. Neeman.

Definition 30 A triangulated category is an additive category with an additive automorphism and a set of distinguished triangles satisfying the following condition:
1. If a triangle is isomorphic to a distinguished triangle, then it is distinguished.
2. is distinguished.
3. For any map , there exists an distinguished triangle .
4. The following diagram commutes:
5. Rotations of distinguished triangles are distinguished.
6. The octahedral axiom:
Theorem 14 For any abelian category , the categories and are triangulated categories.
Definition 31 A triangulated category (with arbitrary coproducts) is called well generated if there exists a set of objects of such that
1. For any , if and only if for any , .
2. For any set of maps in , if is surjective for any and , then is also surjective.
3. There is a cardinal such that all objects in are -small, namely any map factors through , where and .
Theorem 15 (Alonso-Jeremias-Souto, Neeman) The unbounded derived theory of a Grothendieck abelian category is well generated. (A Grothendieck abelian category is an abelian category with generators such that small colimits and filtered colimits are exact.)
Remark 14 is a Grothendieck abelian category, hence the unbounded derived category is well generated.
Theorem 16 (Brown representability) Let be two triangulated categories and be well generated and with arbitrary coproducts. Then a functor admits a right adjoint if and only if commutes with coproducts.
Theorem 17 (Spaltenstein) For any locally compact spaces, is defined on all of .
Lemma 8 commutes with arbitrary direct sums.
Remark 15 This is not true for . For example, consider and be the skyscraper sheaf concentrated on . Then but .

By the lemma and the Brown representability, we conclude that has a right adjoint functor . To get a bounded functor , we need further boundedness of .

Definition 32 The dimension with compact support for locally compact is the smallest such that for any and any .
Proposition 10
1. .
2. If is closed, then .
3. is local, namely if for any , there exists a neighborhood of such that , then . In particular, for any -dimensional manifold.
4. For and , then for any and . Moreover, .

Now assume that . Let , then by adjunction Suppose , then we know that . So for , we have , hence . It follows that , hence the adjunction can be defined on . This adjoint pair is called the (global) Verdier duality.

Let . Then there is a canonical map Deriving this, for any , we get Replacing by for some , we obtain

Proposition 11 (Local Verdier duality) The map is an isomorphism.
Proof We check that is an isomorphism on each open : by global Verdier duality, the right-hand-side is isomorphic to The proposition follows. ¡õ

We also have the following similar useful identity and base change.

Proposition 12
Proposition 13 If and assume that has fibers of finite dimensions (hence so does ), then .
Proof The idea is to use the adjunction . ¡õ

## Contraction of curves on complex surfaces

Throughout this section, we assume that the coefficient ring is a field. Let be a quotient map such that and , namely a contraction of the union of curves on a complex surface to a single point .

Theorem 18 (Grauert) is holomorphic if and only if the intersection matrix is negative definite.
Example 10 The contraction has , hence is not holomorphic.
Example 11 Let be a line bundle over such that , then the contraction of its zero-section on the total space of has negative definite intersection pairing , hence is holomorphic.
Example 12 Consider the blowup , the contraction of ( is the strict transform of . is the exceptional divisor) has , hence is not holomorphic.
Example 13 Consider the blowup has , hence is holomorphic.

Now restrict to one of our four examples. Let and , . The following is a fact from basic algebraic topology.

Theorem 19 (Lefschetz duality) .

Since retracts onto , we have . Let be the cycle class map given by and and . There is a long exact sequence associated to , by the Lefschetz duality we have the following identification:

Note that , so is an isomorphism if and only if it is surjective if and only it is injective, which is also equivalent to say that is an isomorphism, or , or , or is an isomorphism, if and only if is non-degenerate (e.g., when and is holomorphic.)

Question What does this mean in ?

Consider . Then

Consider the truncation triangle

Question Does this split? Namely, does there exist such that is an isomorphism?

By the exact sequence of sheaves We know that for any , we have a distinguished triangle Taking its cohomology, we have a long exact sequence Hence is actually the relative cohomology.

has support on a single point . By applying to the above distinguished triangle,we get Hence must factor through . Taking , we get . By base change, . We conclude that the sequence splits if and only if is an isomorphism.

Question When does the complex split?
Proposition 14 If the intersection pairing is invertible, then . (Note and .)
Proof When is invertible, we want a splitting . Since is supported on . Then exists if and only if there exists a map , which live in . Thus it is equivalent to giving a map such that is the identity map, which is equivalent to saying that is an isomorphism, is invertible. ¡õ
Remark 16 When is holomorphic and , then by Grauert, is negative definite, hence invertible.

Truncating the adjunction map , we get , which is an isomorphism in . We can check this on locally, as and , which an isomorphism if and only if is an isomorphism.

Definition 33 We denote , , . More generally, for any smooth , we denote . Then if is holomorphic and , then .
Example 14 In the above non-holomorphic examples, does not split. However, one can always split off a skyscraper sheaf of rank .

Here is another approach. Consider the adjunction map .

Question When does it split?

Again, truncating the adjunction map , we get as for . Note that and for . We obtain that We get given by . Applying to the triangle, we get which sends to . Then lifts if and only if , if and only if is an isomorphism.

Thus exists if and only is invertible. Similarly form the triangle , we know that is an isomorphism. Therefore exists (and is unique) if and only if is an isomorphism and (in this case ).

## Borel-Moore homology and dualizing functor

Last time we studied the pushforward of constant sheaves and how they decompose. Now let us step back to duality.

We have seen that cohomology can be naturally expressed in terms of sheaves.

Let be the chain complex of possibly infinite singular (simplicial) chains on together with the usual differential such that for any compact , there exists at most finitely many such that with . Its homology is called the Borel-Moore homology.

Example 15
1. In , a ray is a 1-chain. Its boundary is a single point and itself is a boundary. Hence and . Also, .
2. Consider a three rays in plane branched at one point. Then its Borel-Moore and .
Theorem 20 Let . is in fact a complex of flabby sheaves, so .
Remark 17 If is compact, then .

The Poincare duality for smooth oriented manifolds can be also stated as .

The universal coefficients theorem says that By the Poincare duality, we obtain where can be also identified as .

We want a general notion of a dual complex for any such that

Definition 34 Given a complex , we let be the complex of presheaves and define to be its sheafification. In fact, .

Let , then the Borel-Moore homology is the dual of its cohomology with compact support. Hence . It follows from the calculation of basic Borel-Moore homology calculation that

Corollary 10 Let be a smooth manifold of dimension . Then , where is the orientation sheaf.

We would like to say that gives us a dualizing functor , with . Unfortunately, is a bit too wild for this to be true. The first problem is that the image of the functor may not lie in . This is not a major problem (e.g. for , which works out). The second problem is that there may exist bad sheaves on nice spaces. For example, let and be locally constant sheaf on and consider the sheaf on . We would like to eliminate those problems.

Definition 35 An analytic space is a subset of an analytic manifold of cut out by analytic functions. A subanalytic space is one cut out by analytic equalities and inequalities.
Definition 36 A sheaf on a subanalytic space is called constructible if there exists a subanalytic locally finite stratification (i.e., is a subanalytic subspace which is also a manifold) such that is a local system, i.e., a locally constant sheaf.
Definition 37 Let . We say is (cohomologically) constructible if is constructible.
Definition 38 The constructible derived category is defined to be the full subcategory of consisting of cohomologically constructible complexes.
Theorem 21
1. Let be an analytic map, then all preserve .
2. In , is a dualizing functor, i.e., .

Given . We have as , hence . Similarly, we have

## Stratification

Suppose we have a locally finite covering of smooth subanalytic subsets , we say that it is a (Whitney) stratification if it satisfies and the Whitney conditions A and B. The condition A says that is contained the limit of the tangent spaces for any and in the strata. It follows under these conditions that there exists a neighborhood of such that is strata-preserving homeomorphic to , where is the link of . Let be a transverse slice to the strata containing , then . The following are the basic facts about stratification.

Theorem 22
1. Any locally finite covering by subanalytic subsets can be refined to a stratification.
2. Any algebraic variety admits a stratification by locally closed subvarieties.
3. Any map of varieties can be stratified (namely, there exists stratifications of and such that the preiamge of a strata is a union of strata such that is a submersion and is locally constant over .)

We have seen that , When is smooth and oriented, we have , in fact for any smooth morphism . Let , we obtain the Poincare duality This can be interpreted as the statement that is self dual.

An analog of Poincare duality on stratified space of even real dimension should then be such that and an open smooth (strata) such that .

We have seen that for a smooth projective map , and by the Hard Lefschetz. Also, we have seen that for a contraction of curves , . As is self dual, is self dual (as is proper) and is self dual, we know that is self dual.

In general for proper, we cannot hope , but instead we would hope that , where is the form of .

## Poincare duality for singular spaces

For singular, usually . In order to obtain the Poincare duality for singular spaces, we need to find some such that and for some open such that . This is the goal for today.

We will go by induction. Let , where is smooth closed (but is not necessarily smooth). Let is open and . Assume there exists a stratification of such that is smooth closed stratum and is constructible with respect to the stratification.

Definition 39 An extension of is a pair , where and an isomorphism.
Remark 18 Extensions of form a category.
Lemma 9 Fix , then there exists a natual bijection given by sending to .
Proof Associated to we have an adjunction triangle Applying , we get Rotating the triangle we obtain the desired map for one direction.

For the other direction, the adjunction triangle for is Starting from , we get where . Let . ¡õ

Example 16 The sheaf correpsonds to . Namley, .

The sheaf correpsonds to .

From , we get corresponding to , hence we get . So for to be self-dual, we need and . So we want to find such that . The (only) way to find a splitting is by trucation: As is locally constant on , we know that Decompose as and , where is contractible and open. Then , where 's are all constant sheaves.

Question What is the "costalk", namely of a constant sheaf on a smooth of dimension ?
Answer As the dualizing sheaf of is , which is also equal to . Also . Hence .

Therefore we . Applying to , we obtain that Hence So So we should take , namely .

To summarize, the above procudure works for even and . Starting with a self-dual sheaf , we get sheaf given by the extension corresponding to the distinguished triangle

In fact, we will see that defines a functor .

Proposition 15 Suppose we have the following diagram of distinguished triangles Then the following are equivalent:
1. .
2. There exists such that the first square commutes.
3. There exists such that the second square commutes.
4. There exists a morphism of triangle .

Moreover, if these condition holds and , then and are unique.

Proof Consider the long exact sequence of , we know (a) is equivalent to (b) and the uniqueness. From the triangulated category axiom, we know that (b) is equivalent to (d). A dual argument implies that (c) is equivalent to (d). ¡õ
Corollary 11 Let be a distinguished triangulated and , then
1. The cone of it is unique up to unique isomorphism.
2. is the unique map such that is distinguished.
Proposition 16 is a functor.
Proof Since and as and . By Corollary 11, we know is functorial. ¡õ
Remark 19 .
Lemma 10 has a nontrivial summand with support in ifa nd only if can be expresseed as , where , and .
Proof If , then is the same as and .

For the other direction, we know is a direct summand of as the map is the identity. ¡õ

Corollary 12 If for any , then has no summands with support in .
Proof If not, then . But is nonzero for some . ¡õ
Remark 20 for any does not imply that . For example, has a nonzero element corresponding to . The corresponding map has trival cohomology but itself is not zero.
Corollary 13 has no summands with support in .
Proof Since for any and for any . Hence for any . ¡õ
Definition 40 Let be stratified by decreasing dimensions (so is open in ). Let is locally constant on . Define the intersection cohomology complex , where . Write .

We saw that . Also, is indecomposable if and only if is indecomposable.

Definition 41 Define the intersection cohomology and .

As is self-dual, we get the Poincare-Verdier-Goresky-MacPherson duality

Now many results can be extended to singular varieties using intesection cohomology.

Theorem 23 The Lefschetz hyperplane thoerem is true for any projective variety with replaced by .
Theorem 24 The Hard Lefschetz theorem is true for any projective variety with replaced by .

Recall that for a projective smooth morphism , we have , where is a semisimple local system by Deligne's theorem. Moreover, .

Theorem 25 (Decomposition theorem, BBD) If is proper and is smooth, then

Our next goal is to "filter" in such a way that the -th "associated graded piece" is . Namely, the following relative Hard Lefschetz holds: and .

## -structures

Let be a triangulated category. Let and be full categories of . Let and .

Definition 42 is a -structure on if
1. and .
2. if and .
3. For , there exists a distinguished triangle , where and . (Uniqueness also follows.)
Definition 43 is called the heart or core of the -structure.
Example 17 Let be an abelian categroy, then and gives a -structure on with heart equivalent to .
Proposition 17
1. There exists a functor which is right adjoint to the inclusion . Similarly for a functor .
2. There exists a unique such that is a distinguished triangle.
Proof Define and . We want to show that given , there exists a canonical map . Applying to , we get . Since and , we know , hence is a functor.

The same argument shows that . Since , we know that there exists a unique . ¡õ

Corollary 14
1. If , then . Similarly for .
2. Let . Then if and only if . Similarly for .
3. If is a distinguished triangle in and , then .
Proof For (a), use the adjunction from the last proposition. For (b), use (a). For (c), and the long exact sequence implies that . ¡õ
Proposition 18
1. If , then . Similarly for .
2. If , then .
3. More generally, .
Definition 44 Let be the heart. We define , where and . So if and only if for every .
Theorem 26 The heart is an abelian categroy.
Proof Note that for any , the distinguished triangle shows that . Thus is additive. For in , the distinguished triangle shows that . We claim that and . This claim can be checked using and . It remains to check that . Define such that is a distinguished triangle. Then and . By completing into a tetrahedron, we get an triangle Hence and . Hence and . ¡õ
Proposition 19 The functor is cohomological, i.e., for every triangle in , we obtain a long exact sequence of
Proof By rotation, it suffices to show that is exact.
1. Assume , we shall show that is exact. In this case, we have and for any . Applying , we know the short exact sequence.
2. Assume that only . We shall show the short exact sequence . By applying for any , we know . By completing into a tetrahedron, we have the triangle . By the first step, we know is exact.
3. Run the same argument, we know that if , then is exact.
4. Let be arbitrary. Let such that is a distinguished triangle. Then is exact by the third step. We also have the triangle , hence is exact by the second step. It follows that is exact. ¡õ

## Perverse -structures

Definition 45 Let . We say satisfies the support condition if for any . We define to be the full subcategory of objects satisfying the support condition.
Definition 46 We say satisfies the cosupport condition if . We define to be the full subcategory of objects satisfying the cosupport condition.
Theorem 27 is a -structure on (called the perverse -structure).
Remark 21 is in the heart of this perverse -structure.
Definition 47 We define to be the categroy of perverse sheaves in the heart of the perverse -structure.
Question Why do we define the perverse -structure in this way?

Here is another approach. Let be a fixed stratification and be the complexes constructible with respect with this stratification. We would like to construct a self-dual -structure on individually using descending induction on the strata.

On the top strata, we define to be the complex of sheaves on with local systems on . Let , for a local system on , we have , where . So is not self-dual, but instead is self-dual.

Now let and be the open and closed embeddings obtained from the strata. Suppose we already have a self-dual -structure on and also a self-dual -structure on given by , where . We want a self-dual -structure on such that and This is self-dual by definition, so we only need to check it is actually a -structure.

Theorem 28 is a -structure on .
Proof
1. Let and . Applying to the adjunction distinguished triangle of , we obtain Since and by construction, we know that .
2. Since shifts commute with restrtion, we know that .
3. For , we construct . We then construct such that and such that By completing the tetrahedron, we check that and . In fact, we have and , hence . Also, and , hence . ¡õ

It follows from the inductive construction that and This explains why we defined perverse -structures in such a way.

Remark 22 Actually the above proof gives us a general procedure to glue any -structures.

Notice that when is smooth, consists of complexes where are local systems. Then is th self-dual -structure on .

Remark 23 We are making an implicit assumption that is a field.

Let be the degenerate -structure on . By gluing with , we obtain a -structure on . Let be the corresponding truncating functor. Then is the right adjoint of the inclusion of objects such that and Dually, define by using the -structure glued from . We have

Lemma 11 Let and . Then there exists a unique up to a unique isomorphism extension of such that and . Namely
Proof Use Lemma 9 and notice that . ¡õ
Remark 24
1. Suppose . When , we have , hence we do not expect to be perverse. Similarly, if , we do not expect is perverse. If , then .
2. . In fact, by construction and . This corresponds to the case .
3. For . On the open part , so . Since (as , we know it is equal to .
4. For , we have .
5. We obtain functors and and are adjoint pairs.

From and , we have a morphism of functors .

Proposition 20 The image functor is .
Proof Using the triangle for , we have a short exact sequence Similarly we have It follows that by identifying with . ¡õ

## Simple objects

Theorem 29 The simple objects in are the -sheaves.

From the strata and , we have and . Let be the essential image of in , in other words, the full subcategory with objects such that . Consider and the triangle , applying , we have a long exact sequence We know that is the maximal quotient of with support in . Similarly, is the maximal subobject of with support in .

Proposition 21 The functor factors through the Serre quotient categroy . Moreover, is an equivalence of categories.
Proof is faithful: Let and . Let be a lift of . Since , we know that , so , therefore .

is essentially surjective and full: As . ¡õ

Proposition 22
1. For , is the unique extension of which has no nontrivial subquotients with support in .
2. The simple objects of are
1. for simple,
2. for simple.
Proof
1. Recall that is the maximal quotient ojbect of with support in and is the maximal suboject of with support in . So if has no subquotients with support in , then . Hence and , which is equivalent to . Now use the triangle .
2. Any simple object in is either
1. the image of a simple object in ,
2. an extension of a simple object in which has no nontrivial subobjects with support in , i.e. for simple. ¡õ

## Operations on perverse sheaves

Suppose and for any . Then Hence . By duality, we know that for , .

Lemma 12 Let and be an adjoint pair of triangulated functors. Then (right -exact) if and only if (left -exact).
Corollary 15 and .

If is proper, then , hence . Recall that if is smooth of relative dimension , then . Hence and they take perverse sheaves to perverse sheaves (-exact). In particular, when is etale, we know that is -exact.

Recall that if is smooth and affine, then for any . The following is a generalization of this fact. The proof can be given by generalizing the original proof using Morse theory.

Theorem 30 If is affine, then and .
Definition 48 Let be proper. is locally closed subvarieties and . We say is semismall (resp. small) if for any (resp. holds for any ).
Remark 25 If is semismall, then it is generically finite ().

Now suppose is semismall.

Proposition 23 .
Proposition 24 Let be smooth. Then .
Proof has stalks . So it is zero if . So as . Namely . One can similarly check that . ¡õ

We showed that . Let . If, then , hence . In other words, the fibers over have dimension . is a local system on , hence if and only if for all . This explaines why we defined "semismall" in such a way.

Example 18 Let and . Then , where consists of and consists of . Let be the projection. Then , is the glueing of two spheres along a point. is a single point. We have the following table

We know that is semismall and . By decomposition theorem, we know that . Comparing the stalk complexes of , and

with the above table, we know that , , . We have , . The Weyl group acts on .

## Kazhdan-Lusztig conjecture

Fix a connected reductive algebraic group and . Let be the associated Weyl group.

Definition 49 The Hecke algebra is defined to be where
Remark 26 is a deformation of the group algebra of .
Theorem 31 (Iwahori) Let , . Then . ??
Remark 27 , hence is invertible.
Definition 50 The endomorphism of given by and is an involution, denoted by . We call is self-dual if .
Theorem 32 (Kazhdan-Lusztig) There exists a unique baissi of such that
1. Each is self-dual.
2. , where .
Conjecture 1 If
1. .
2. , where self-dual.

Then .

The idea of the solution relates the with the geometry of . By the Bruhat decomposition, , where each . Write . Let be the constructible derived category with respect to the Bruhat stratification. For , by the decomposition theorem, we know that .

Definition 51 We define by .
Example 19 , then .
Theorem 33 .

Then the Kazhdan-Lusztig conjecture follows from this geometric characterization.

Example 20 , . Then .
Example 21 For , we have . Then , .
<definition> For is called -even (resp. odd) if for all odd (resp. even) (Equivalently, for all and odd (resp. even). is called -parity if .
Lemma 13 Let be a distinguished triangle in . If are -even, then so is . Moreover, .
Proof Applying and using the long exact sequence in cohomology, we know that and . ¡õ
Remark 28 In general, it is not true that is additive on distinguished triangle.

Let be the parabolic subgroup (e.g., block upper triangular matrices for ). Let .

Lemma 14 (Push-Pull (Springer, Brylinski, MacPherson)) If is -parity, then .
Proof Let . We will induct on .

When . We have . So it suffices to check on . reduces to the case of . For :

1. , The corresponding .
2. , . The corresponding .

Now suppose . Pick maximal. Using the adjunction triangle where . is -even. And is -even since is. Applying , we know that by induction. ¡õ

Now we will use the Push-Pull Lemma to prove the main theorem, namely .

Proof First we will inductively show that is self-dual. Choose a reduced expression for . By the by Push-Pull Lemma, The net effect of this push-pull sequence is pulling back the constant sheaf all the way up and pushing forward through the Bott-Samelson resolution, Its image is and is in fact a resolution of singularities. By base change, we know which is isomorphic to by the decomposition theorem, where is a self-dual vector space. Now the self-duality of follows from the induction hypothesis and the self-duality of .

Let . If , then . If , then . If , then for any . Then the second part follows from the definition of . ¡õ