Tensor For F , F Sh( X ) we have F F Sh( X ). Since tensor - - PowerPoint PPT Presentation

Tensor

◮ For F, F′ ∈ Sh(X) we have F ⊗ F′ ∈ Sh(X). Since tensor commutes with taking stalk, the non-exactness comes from that Fx might not torsion-free, and this can be taken care by a simple resolution 0 → P′ → P → F → 0 where P is e.g. direct sum of ZU indexed by (U, s) with U ֒ → X open, s ∈ F(U). ◮ Hence we can derive ⊗: we will continue to denote it by · ⊗ · : D∗(Sh(X)) × D∗(Sh(X)) → D∗(Sh(X)) where ∗ = ∅, +, − or b. ◮ Lemma. For F ∈ D+(Sh(X)), G ∈ D+(Sh(Y )), and f : X → Y a map of locally compact Hausdorff spaces with Rkf! = 0 for k ≫ 0, we have Rf!(F ⊗ f ∗G) = Rf!F ⊗ G. ◮ This is because · ⊗ f ∗G preserves f -softness. ◮ Lemma. For G, G′ ∈ D+(Sh(Y )), there is a morphism f !G ⊗ f ∗G′ → f !(G ⊗ G′) ◮ This follows from Rf!(f !G ⊗ f ∗G′) = Rf!f !G ⊗ G′ → G ⊗ G′ and adjunction again.

f ! for smooth morphisms

◮ Last time, we proved the Poincar´ e-Verdier duality that f !Z Y = ZX[n] for a topological submersion of (real) dimension n assuming Y is locally compact Hausdorff and locally contractible. ◮ Corollary. We also have f !G = OY /X ⊗ f ∗G[n] for general G ∈ D+(Sh(Y )). ◮ For general f : X → Y the functor f ! is local on X: Indeed, for j : U ֒ → X open we have HomSh(X)(j!F′, F) = HomSh(U)(F′, j∗F). Since j! is exact this gives j! = j∗. The same can be checked for topological submersion of dimension 0, giving n = 0 case of above. ◮ We have (f !G)|U = j!f !G = (f ◦ j)!G, which is what we want. ◮ Another preparation: for any Cartesian diagram X ′ X Y ′ Y

p f ′ f q

we have a natural transformation p∗f ! → (f ′)!q∗.

f ! for smooth morphisms, II

f !G = OY /X ⊗ f ∗G[n] for a topological submersion of (real) dimension n assuming ...

◮ We have the natural map f !ZY ⊗ f ∗G → f !G. Since we know f !ZY = OY /X[n], it remains to prove this is an (quasi-)isomorphism. ◮ Since f ! is local on X, we may assume X = Y × Rn. Consider the Cartesian diagram X Rn Y pt

f p

◮ Take U ⊂ Rn and V ⊂ Y open we have RΓ(U × V ; f !G) = RHomSh(X)(ZU×V , f !G) = RHomSh(Y )(Rf!ZU×V , G) = RHomSh(Y )(H∗

c (U; Z) ⊗ ZV , G)

= RHomAb(H∗

c (U; Z), Z) ⊗ RHomSh(Y )(ZV , G) =

RΓ(U; p!Z) ⊗ RΓ(V ; G) Since p!Z is just a shift of ZRn, the last item sheafifies to RΓ(U × V ; f !ZY ⊗ f ∗G) and we have proved the corollary.

Computation of Rf!

◮ It relies on the computation of Rf! locally on X. Thanks to proper base change it is reduced to Hk

c (Rn; Z) =

0, if k = n Z, if k = n ◮ At the same time, even earlier we assumed that f! has finite cohomological dimension. Again by proper base change this reduces to that Hk

c (f −1(y); F) = Hk c (X; i!F) = 0 ∀k ≫ 0 where

i : f −1(y) → X and i! = i∗. ◮ That is, we can reduce to the fiber and then again reduce to H∗

c of

some sheaf on the whole space.

Distinguished triangles

◮ Recall that for j : U ֒ → X open and F ∈ Sh(X) we put FU := j!j∗F = j!j!F. ◮ Now for U, V ⊂ X open we have Mayer-Vietoris 0 → FU∩V → FU ⊕ FV → FU∪V → 0 where the natural maps can be realized by adjunction j!j! → id. ◮ This extends from Sh(X) to C(Sh(X)) and short exact sequence in C(Sh(X)) gives distinguished triangles in K(Sh(X)) and D(Sh(X)). Since derived functor preserves distinguished triangles, we have H∗

c (U ∩ V ; F) → H∗ c (U; F) ⊕ H∗ c (U; F) → H∗ c (U ∪ V ; F) +1

− − → ◮ Likewise, for closed subset E1, E2 ⊂ X using id → i∗i∗ = i!i∗, we have 0 → FE1∪E2 → FE1 ⊕ FE2 → FE1∩E2 → 0 H∗

c (E1 ∪ E2; F) → H∗ c (E1; F) ⊕ H∗ c (E2; F) → H∗ c (E1 ∩ E2; F) +1

− − → ◮ One more: Suppose j : U ֒ → X is open and i :֒ → Z ֒ → X is the

• complement. We have

0 → FU → F → FZ → 0 and H∗

c (U; F) → H∗ c (X; F) → H∗ c (Z; F) +1

− − →

Compute H∗([0, 1]; −)

◮ Prop. For any F ∈ Sh([0, 1]), we have Hk

c ([0, 1]; F) = Hk([0, 1]; F) = 0 for k ≥ 2. Also

H1([0, 1]; Z(0,1)) = Z, i.e. H∗

c (R1; Z) = Z[1].

◮ Suppose s ∈ Hk([0, 1]; F) is non-zero for some k > 1. For any t ∈ (0, 1) we have the Mayer-Vietoris 0 = Hk−1({a}; F) → Hk([0, 1]; F) → Hk([0, a]; F) ⊕ Hk([a, 1]; F) Hence the restriction of s to either [0, t] or [t, 1] is non-zero. We can decompose closed interval like this forever, and they have to “converge” to some point x. ◮ But for some injective resolution 0 → F → I0 → ... → Ik−1 → Ik → Ik+1 → .., this s is represented by some ker(Ik([a, b]) → Ik+1([a, b]) which has to come from Ik−1([a, b]) when a, b → x. This shows s|[a,b] = 0 for [a, b] → x and thus we have a contradiction.

Compute H∗

c

Hk−1({a}; F) δ − → Hk([0, 1]; F) → Hk([0, a]; F) ⊕ Hk([a, 1]; F)

◮ When k = 1 and F = Z(0,1), we have 0 → H0({a}; Z)

δ

− → Hk([0, 1]; Z(0,1)) → Hk([0, t]; Z(0,1)) ⊕ Hk([t, 1]; Z(0,1)). By the same trick we can show the last morphism is zero and the middle morphism is an isomorphism, hence the assertion. ◮ With careful bookkeeping, one verifies that when [0, 1] is reversed (i.e. t → 1 − t), there is a − sign on δ. ◮ Given the proposition, suppose we have any X that we find “n-dimensional” so that we can map X → [0, 1] with “(n − 1)-dimensional” fiber (or, decompose X using Mayer-Vietoris and have each part with this propety). Proper base change then allows us to do induction and prove Hk

c (X; F) = 0 for all

F ∈ Sh(X), k > n. ◮ We have H∗

c (Rn; Z) = Z[n] by induction. And again careful

bookkeeping shows that any linear transformation T on Rn induces sgn(T) on H∗

c (Rn; Z).