Simple objects on Z Fix any i : Z X closed and complementary j : U - PowerPoint PPT Presentation

Simple objects on Z ◮ Fix any i : Z ֒ → X closed and complementary j : U → X open for today. At the end of last time we say we would like j ∗ = j ! i ∗ = i ! → Perv( U ) → 0 ′′ “0 → Perv( Z ) − − − → Perv( X ) − − − ◮ In fact, I think there are good reason that the above won’t be true with appropriate interpretation (See next problem set) ◮ The above is true on the level of pointed set (or groupoid?). Indeed, if j ∗ F = 0 for F ∈ Perv( X ), then F = i ∗ i ∗ F is in the essential image i ∗ (Perv( Z )). ◮ We furthermore note that i ∗ = i ! : Perv( Z ) → Perv( X ) is fully faithful, for p H 0 ◦ i ∗ is the inverse functor from the essential image back to Perv( Z ). ◮ Moreover, suppose G is a simple object in Perv( Z ), we claim that i ∗ G is simple.

Simple objects on U Suppose G is a simple object in Perv( Z ), we claim that i ∗ G is simple. ◮ One important thing I failed to stress last time: a t-exact (resp. left, right exact) functor of two triangulated category equipped with their own t-structures induces exact (resp. left, right exact) functor on the core/heart. ◮ Now for G ∈ Perv( Z ) simple, if 0 → F ′ → i ∗ G → F ′′ → 0 in Perv( X ) we would have 0 → j ∗ F ′ → 0 → j ∗ F ′′ → 0 so j ∗ F ′ = j ∗ F ′′ = 0, i.e. F ′ , F ′′ are in the essential image of Perv( Z ). But then fully faithfulness shows F ′ = 0 or F ′′ = 0. This proves that i ∗ G is simple. ◮ On the other hand, let us assume that F ∈ Perv( X ) is simple, and j ∗ F � = 0. We claim that: ◮ Theorem. The set of (isomorphism classes of) such simple F is in bijection via j ∗ with simple objects Perv( U ). ◮ What do we do? So the most interesting question is probably: given simple G ∈ Perv( U ), how do we find a simple F with F| U = G ? ◮ Naive approach will be either j ! G or j ∗ G . We have j ! G ∈ p D ≤ 0 ( X ) and j ∗ G ∈ p D ≥ 0 ( X ) and not perverse in general.

Middle extension Naive approach will be either j ! G or j ∗ G . We have j ! G ∈ p D ≤ 0 ( X ) and j ∗ G ∈ p D ≥ 0 ( X ) and not perverse in general. ◮ A simple fix: one may take either p τ ≥ 0 j ! G = p H 0 ( j ! G ) and/or p τ ≤ 0 j ∗ G = p H 0 ( j ∗ G ). However as we will see in the problem session, they are usually not simple when G is. ◮ What do we do? Let’s try to fix any of them. Say we look at any, e.g. p τ ≤ 0 j ∗ G . ◮ How did we construct it last time? Since j ∗ G has the correct perverse degree on U (i.e. correct degree with respect to the perverse t-structure), the trouble happens on Z , and we define p τ ≤ 0 j ∗ G as the objects that fits into +1 p τ ≤ 0 j ∗ G → j ∗ G → i ∗ p τ ≥ 1 i ∗ j ∗ G − − → ◮ Now, if we want something whose p H 0 “only comes from G ,” it seems like we should get rid of the perverse degree 0 part from Z , suggesting to look at the object K that fits into +1 p τ ≥ 0 i ∗ j ∗ G K → j ∗ G → i ∗ − − →

Middle extension, II +1 K → j ∗ G → i ∗ p τ ≥ 0 i ∗ j ∗ G → . − − ◮ Proposition. Fix G ∈ Perv( U ). Suppose K is an object in D ( X ) with given isomorphism j ∗ K ∼ = G . TFAE: (i) K ∈ Perv( X ) is the object fitting the distinguished triangle above. (ii) i ∗ K ∈ p τ ≤− 1 ( Z ) and i ! K ∈ p τ ≥ 1 ( Z ). (iii) K ∈ Perv( X ) does not have any non-trivial sub-object nor quotient object from Perv( Z ). (iv) K is the image in Perv( X ) of the natural map p H 0 ( j ! G ) → p H 0 ( j ∗ G ). ◮ Let us prove (ii) implies (i). From j ∗ K ∼ = G we have by adjunction K → j ∗ G = j ∗ j ∗ K which fits into a distinguished triangle +1 i ∗ i ! K → K → j ∗ j ∗ K − − → . We would like to prove that j ∗ G = j ∗ j ∗ K → i ∗ i ! K [1] is equal to j ∗ G → i ∗ p τ ≥ 0 i ∗ j ∗ G . ◮ By adjunction, we have to prove that the second map in the triangle +1 i ∗ K → i ∗ j ∗ G → i ! K [1] − − → is equal to i ∗ j ∗ G → p τ ≥ 0 i ∗ j ∗ G . ◮ Such a statement is equivalent to i ∗ K ∈ p D ≤− 1 ( Z ) and i ! K [1] ∈ p D ≥ 0 ( Z ). This is exactly (ii).

Middle extension, III +1 (i) K ∈ Perv( X ) is the object fitting K → j ∗ G → i ∗ p τ ≥ 0 i ∗ j ∗ G → . − − (ii) i ∗ K ∈ p τ ≤− 1 ( Z ) and i ! K ∈ p τ ≥ 1 ( Z ). (iii) K ∈ Perv( X ) has no non-trivial sub/quotient object from Perv( Z ). (iv) K is the image in Perv( X ) of the natural map p H 0 ( j ! G ) → p H 0 ( j ∗ G ). ◮ Next prove (iii) ⇒ (ii). It is known that i ∗ K ∈ p D ≤ 0 ( Z ). Suppose (iii) holds but on the contrary i ∗ K �∈ p D ≤− 1 ( Z ). Then adjunction gives a non-trivial morphism K → i ∗ p H 0 ( i ∗ K ) in Perv( X ) whose image is a quotient object of K supported on Z . This contradiction ⇒ (ii). and its dual version proves (iii) = ◮ Now we prove (iv) ⇒ (iii). Suppose (iv) but we have an epimorphism K → i ∗ A with non-zero A ∈ Perv( Z ). We have j ! G → p H 0 ( j ! G ) → K → i ∗ A . The whole composition is zero as it’s adjoint to G → j ! i ∗ A = 0. Since i ∗ A ∈ p D ≥ 0 ( X ), this implies that the composition p H 0 ( j ! G ) → K → i ∗ A is also zero. But this contradicts with that p H 0 ( j ! G ) → K is surjective. ◮ So we have (iv) ⇒ (iii) ⇒ (ii) ⇒ (i). But both (i) and (iv) uniquely characterize K and thus (i) ⇔ (iv) and we are done.

Middle extension, IV +1 (i) K ∈ Perv( X ) is the object fitting K → j ∗ G → i ∗ p τ ≥ 0 i ∗ j ∗ G → . − − (ii) i ∗ K ∈ p τ ≤− 1 ( Z ) and i ! K ∈ p τ ≥ 1 ( Z ). (iii) K ∈ Perv( X ) has no non-trivial sub/quotient object from Perv( Z ). (iv) K is the image in Perv( X ) of the natural map p H 0 ( j ! G ) → p H 0 ( j ∗ G ). ◮ We have shown that for G ∈ Perv( U ), there exists a unique extension K ∈ Perv( X ) satisfying the nice properties. Since properties (i) and (iv) are functorial in G , this defines a functor j ! ∗ : Perv( U ) → Perv( X ) for an Zariski open j : U ֒ → X . The object K = j ! ∗ G is typically called the IC/intermediate/middle extension of G . ◮ We have that G is simple iff j ! ∗ G . The only if statement follows from applying the t-exact functor j ∗ = j ! . For the if statement, suppose j ! ∗ G was not simple, by (iii) it has 0 → F ′ → j ! ∗ G → F ′′ → 0 in Perv( X ) with F ′ , F ′′ not supported on Z , i.e. j ∗ F ′ , j ∗ F ′′ non-zero. ◮ Applying the t-exact functor j ∗ then gives 0 → j ∗ F ′ → G → j ∗ F ′′ → 0, a contradiction.

Middle extension, V ◮ We have shown that for G ∈ Perv( U ), there exists a unique extension K ∈ Perv( X ) satisfying properties (i)-(iv). Since (i) and (iv) are functorial in G , this defines a functor j ! ∗ : Perv( U ) → Perv( X ) for an Zariski open j : U ֒ → X . The object K = j ! ∗ G is typically called the IC/intermediate/middle extension of G . ◮ We have by property (ii), (iii), (iv) above that j ! ∗ ◦ D U = D X ◦ j ! ∗ . ◮ Lemma. j ! ∗ is left and right exact (but not middle exact, see problem set 3). ◮ Proof. We prove that j ! ∗ is left exact. The right exact statement follows either applying D X to the argument, or to the exact sequence. Now suppose we have G ′ → G a monomorphism in Perv( U ) and we would like to show that j ! ∗ G ′ → j ! ∗ G is a monomorphism. Suppose we have F → j ! ∗ G ′ → j ! ∗ G so that the composition is zero. We have to show F → j ! ∗ G ′ is zero. ◮ Applying j ∗ gives j ∗ F → G zero. Since G ′ → G is a monomorphism, we have j ∗ F → G ′ zero too. By adjunction this means the composition F → j ! ∗ G ′ → j ∗ G is zero. But j ! ∗ G ′ → j ∗ G is a monomorphism then implies F → j ! ∗ G ′ is zero, as desired.

Simple objects ◮ Lemma. G ∈ Perv( U ) is simple iff j ! ∗ G ∈ Perv( X ) is. ◮ For the only if statement, if j ! ∗ G was not simple, by (iii) it has sub and quotient object not supported on Z . But then j ∗ again shows G is not simple. ◮ For the if statement, suppose otherwise G has a proper sub-object → G . Then we have j ! ∗ A → j ! ∗ G . This map is both non-trivial A ֒ and not surjective as both can be detected by j ∗ = j ! . Hence j ! ∗ G is not simple.

Simple objects, II ◮ Suppose now X is irreducible smooth and L is a local system. Then L [dim X ] ∈ Perv( X ). ◮ Lemma. L is simple as a local system, i.e. an object of Rep( π 1 ( X )) iff L [dim U ] is simple in Perv( X ). ◮ A short exact sequence 0 → L ′ → L → L ′′ → 0 in Rep( π 1 ( X )) induces a distinguished triangle in D ( X ) and thus another short exact sequence 0 → L ′ [dim X ] → L [dim X ] → L ′′ [dim X ] → 0 in Perv( X ). This proves the if part. ◮ For the only if part, we note that L [dim X ] satisfies the property (ii) in the criterion of middle extension; i ∗ L [dim X ] ∈ p D ≤− 1 ( Z ) for any i : Z → X with dim Z < dim X . Hence if L [dim X ] has a proper sub-object E in Perv( X ), it’s restriction to any open set must be non-trivial. ◮ We may then find a small enough Zariski open U so that j ∗ E = L ′ [dim X ] for some L ′ ∈ Rep( π 1 ( U )) and we have L ′ ֒ → L| U . But this contradicts with the simplicity of L as π 1 ( U ) → π 1 ( X ) is surjective; any loop can avoid real codimension two strata.