SEMISMALL MORPHISMS YEHAO ZHOU 1. Motivation Let's recall some results concerning t-exactness of pushforward from last time: Proposition 1.1. Let f : X ! Y be a morphism between varieties, then (1) If f is quasi-finite, then f∗ is left t-exact and f! is right t-exact; (2) If f is affine, then f∗ is right t-exact and f! is left t-exact; (3) As a result, if f is quasi-finite affine, then f∗ and f! are t-exact. Unfortunately, proper morphisms, which commonly show up in applications (e.g. reso- lutions of singularities), are not in this list (except for finite morphisms). Heuristically, this comes from the fact that for a proper variety Z of dimension d, the top degree co- homology H2d(Z; C) 6= 0 (Lemma 2.3), so if there is a proper morphism f : X ! Y such that there is a subvariety W ⊂ Y with dimensions of fibers over W being ≥ d, 2d−dim X p ≤0 then the H (fjf −1W )∗C[dim X] 6= 0, i.e. (fjf −1W )∗C[dim X] 2 Dc (Y ) only if dim W ≥ 2d − dim X. So we restrict to proper morphisms with a nice property on the dimension of fibers: Definition 1.2. Let X be an irreducible variety, f : X ! Y is called semismall if there exists a stratification fYtgt2T of Y , such that for each stratum Yt and each point y 2 Yt\f(X) −1 2 dim f (y) + dim Yt ≤ dim X f is called small with respect to a dense open subset W ⊂ Y if 8y 2 W , f −1(y) is a finite set, and there exists a stratification fYtgt2T , such that W is a union of strata and such that 8y 2 Yt \ f(X) ⊂ Y − W −1 2 dim f (y) + dim Yt < dim X Proposition 1.3 (Equivalent definition of semismallness). Let X be an irreducible variety. Then f : X ! Y is semismall if and only if dim X ×Y X ≤ dim X Proof. If f is semismall, then there is a stratification fYtgt2T , such that for each stratum Yt and each point y 2 Yt \ f(X) −1 −1 dim(f × f) (y) + dim Yt = 2 dim f (y) + dim Yt ≤ dim X −1 −1 which implies that dim(f × f) Yt ≤ dim X. Since X ×Y X = qt(f × f) Yt, we conclude that dim X ×Y X ≤ dim X. Let's prove the "if" part by constructing a stratification. Let's prove a lemma first. 1 2 YEHAO ZHOU Lemma 1.4. If f : X ! Y is a morphism, then there exists a stratification fYtgt2T such −1 that 8t 2 T , the fibers of fjf Yt are either empty or of dimensions d(t) which only depend on t, and −1 −1 dim f Yt = d(t) + dim Yt ; dim(f × f) Yt = 2d(t) + dim Yt Proof. We prove it by Noetherian induction. First of all, we replace Y with the closure of f(X), endowed with reduced structure sheaf. Suppose that the lemma is true for any proper closed subvariety of Y . According to the generic flatness theorem [GD65] (EGA IV2, Th´eor`eme6.9.1) there is an irreducible open subvariety U ⊂ Y such that fjf −1U is flat. Shrink U if necessary, we assume that each irreducible component of f −1U is mapped surjectively to U. Then −1 we apply Theorem 15.1 of [Mat89] to fjf −1U and we see that 8y 2 U and 8x 2 f (y), −1 −1 dim OX;x ≤ dim f (y) + dim OY;y, and there is at least one x 2 f (y) such that the equality is obtained. In other words −1 −1 sup dim OX;x = dim f (y) + dim OY;y = dim f (y) + dim U x2f −1(y) Since each irreducible components of f −1U is mapped surjectively to U, there exists x 2 f −1(y) such that x lies in the component with maximal dimension, i.e. dim f −1U, so −1 sup dim OX;x = dim f U x2f −1(y) As a consequence, dim f −1(y) = dim f −1U − dim U is constant on U. −1 −1 Apply Theorem 15.1 of [Mat89] to the flat morphism (f × f)jf U×Y f U , we see that −1 −1 −1 sup dim OX×Y X;x = dim f (y) × f (y) + dim OY;y = 2 dim f (y) + dim U x2(f×f)−1(y) The RHS is constant on U, so −1 −1 dim(f × f) U = sup sup dim OX×Y X;x = 2 dim f (y) + dim U y2U x2(f×f)−1(y) According to the lemma, there is a stratification fYtgt2T such that 8t 2 T , fibers of −1 fjf Yt are either empty or of dimensions d(t) which only depend on t and −1 dim(f × f) Yt = 2d(t) + dim Yt −1 If dim X ×Y X ≤ dim X, then (f × f) Yt ≤ dim X ×Y X ≤ dim X, which implies that 2d(t) + dim Yt ≤ dim X, so f is semismall. Exercise 1.5. For any f : X ! Y , inequality dim X ×Y X ≥ dim X always hold. Exercise 1.6. Consider morphisms f : X ! Y and g : Y ! Z, if g ◦ f is semismall, then f is semismall. Exercise 1.7 (Composition of semismall maps can fail to be semismall). Blowing up subvariety fx = w = 0g inside the affine cone fxy = zwg ⊂ C4, followed by blowing up the preimage of f0g, each blow up is semismall, but the composition is not. Show that the composition is directly blowing up f0g. SEMISMALL MORPHISMS 3 2. Semismall and small pushforward Theorem 2.1. Assume that X is smooth, f : X ! Y is a semismall morphism, then p ≤0 p ≤0 (1) If F 2 Dloc(X), then f!F 2 Dc (Y ); p ≥0 p ≥0 (2) If F 2 Dloc(X), then f∗F 2 Dc (Y ). Note that (2) follows from (1) by Verdier duality. p ≤0 Proof. It suffices to prove that if F 2 Loc(X)[dim X] then f!F 2 Dc (Y ), i.e. i dim Supp(H f!F ) ≤ −i From the definition of semismallness, we know that there is a stratification fYtgt2T , such that −1 for each stratum it : Yt ,! Y and each point y 2 Yt \ f(X), 2 dim f (y) + dim Yt ≤ dim X, −1 −1 or equivalently − dim X + 2 dim f (y) ≤ − dim Yt. We see that F jf Yt is of cohomological i degree − dim X, hence the homological bound of f! [Bai18], i.e. H f!K = 0 for i > 2d and K 2 Shvc(X), implies that ∗ ≤− dim Yt ∼ −1 −1 it f!F = (fjf Yt )!F jf Yt 2 Dc (Yt) ∗ i i ∗ Since it H f!F = H it f!F , we have following inequality on strata ∗ i dim Supp(it H f!F ) ≤ −i i The LHS is dim(Supp(H f!F ) \ Yt). Let t runs through T , then we have the desired inequality i dim Supp(H f!F ) ≤ −i Corollary 2.2. Assume that X is smooth, f : X ! Y is a proper morphism, TFAE: (a) f is semismall; (b) f∗ : Loc(X)[dim X] ! Perv(Y ); (c) f∗C[dim X] 2 Perv(Y ). Proof. By the previous theorem, (a) implies (b). (b) implies (c) tautologically. It remains to prove that (c) implies (a). Assume that (a) is false. −1 By Lemma 1.4, there is a stratification fYtgt2T such that 8t 2 T , fibers of fjf Yt are either empty or of dimensions d(t) which only depends on t and −1 dim(f × f) Yt = 2d(t) + dim Yt Since f is not semismall, there must be at least one stratum Yt such that 2d(t) + dim Yt > −1 dim X, or equivalently − dim X + 2 dim f (y) > − dim Yt. We need the following lemma Lemma 2.3. If X is a proper variety of dimension d, then H2d(X; C) 6= 0. 2d −2d Proof. By Verdier duality, the statement H (X; C) 6= 0 is equivalent to H (X; !X ) 6= 0, where !X is the dualizing complex on X. Take a smooth open subvariety j : U,! X with complement i : Z,! X such that dim Z < d, consider the distinguished triangle i∗!Z ! !X ! j∗!U ! 4 YEHAO ZHOU Apply the functor RΓ(X; −) and we get exact sequence −2d −2d −2d −2d+1 H (Z; !Z ) ! H (X; !X ) ! H (U; !U ) ! H (Z; !Z ) _ _ Note that RΓ(Z; !Z ) = RΓc(Z; C) RΓ(Z; C) because Z is proper, and since dim Z < i −2d+1 −2d d,H (Z; C) = 0 for i > 2d − 2, hence H (Z; !Z ) = H (Z; !Z ) = 0. We see that −2d ∼ −2d H (X; !X ) = H (U; !U ). Since there is at least one connected component of U with −2d 0 dimension d, denote it by U0, then !U0 = C[2d], so H (U0;!U0 ) = H (U0; C) = C 6= 0, −2d hence we conclude that H (X; !X ) 6= 0. − dim X+2 dim f −1(y) ∗ −1 Apply the lemma and we get H (fjf Yt )∗C 6= 0, so that it f∗C[dim X] 2= ≤− dim Yt Dc (Yt), hence f∗C[dim X] 2= Perv(Y ), i.e. (c) is false. It turns out that we can say more for proper small morphisms. Theorem 2.4.