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Serre duality for twisting sheaves on projective space

Statement

Assume the Axiom of Choice. Let $k$ be a field, $n \geq 0$ and let $ω_{P^{n}} = O (- n - 1)$ be the dualizing line bundle of $P_{k}^{n}$ with the Laurent-coefficient residue trace $t_{P^{n}} : H^{n} (P_{k}^{n}, O (- n - 1)) ⟶ k$ of Dualizing line bundle and trace datum of a smooth projective variety. Then for every integer $d$ and every $q \in {0, \dots, n}$ the evaluation pairing $H^{q} (P_{k}^{n}, O (d)) \times H^{n - q} (P_{k}^{n}, O (- d - n - 1)) ⟶ k, (α, β) ⟼ t_{P^{n}} (α \cup β),$ formed with the cup product for the multiplication pairing $O (d) \otimes_{Z} O (- d - n - 1) \to O (- n - 1)$ , is a perfect pairing of $k$ -vector spaces.

Facts & Assumptions

Given: the field $k$ , the integer $n \geq 0$ , the projective space $P_{k}^{n}$ with twisting sheaves $O (d)$ , the dualizing bundle $ω_{P^{n}} = O (- n - 1)$ , its residue trace $t_{P^{n}}$ , and the in-run cohomology computation Cohomology of O(d) on projective space.

[F1]

The dualizing line bundle of $P_{k}^{n}$ is $ω_{P^{n}} = O (- n - 1)$ , and the residue trace $t_{P^{n}}$ is the $k$ -linear map $H^{n} (P_{k}^{n}, O (- n - 1)) \to k$ which on the monomial basis sends the class with Laurent tail $(x_{0} \dots x_{n})^{- 1}$ to $1$ . (Dualizing line bundle and trace datum of a smooth projective variety)

[F2]

For abelian sheaves with a tensor pairing $μ : F \otimes_{Z} G \to H$ on a space $X$ there is a cup product $H^{p} (X, F) \times H^{q} (X, G) \to H^{p + q} (X, H)$ , bilinear and natural in the pairing and sheaf maps. The constant class $1 \in H^{0} (X, Z_{X})$ acts by both the left and right tensor-unit isomorphisms. (Cup product in sheaf cohomology, Cup-product laws)

[F3]

For $n \geq 1$ and $d \geq 0$ the pairing $H^{0} (P_{k}^{n}, O (d)) \times H^{n} (P_{k}^{n}, O (- n - 1 - d)) \to k$ given by the cup product for $O (d) \otimes O (- n - 1 - d) \to O (- n - 1)$ followed by $t_{P^{n}}$ , equivalently by the coefficient of $(x_{0} \dots x_{n})^{- 1}$ in the product of monomials, is a perfect $k$ -bilinear pairing, compatible with multiplication by homogeneous polynomials; for $n = 0$ it is ordinary multiplication $k \times k \to k$ . (Residue pairing between H^0 and top cohomology of projective space)

[F4]

The Axiom of Choice is The Axiom of Choice.

[F5]

For $n \geq 1$ and any integer $m$ , $H^{q} (P_{k}^{n}, O (m)) = 0$ for $0 < q < n$ ; $H^{0} (O (m)) = 0$ when $m < 0$ ; and $H^{n} (O (m)) = 0$ when $m > - n - 1$ . For $n = 0$ , $P_{k}^{0} = Spec k$ , every twist is trivial and only $H^{0} ≅ k$ is nonzero. (Cohomology of O(d) on projective space)

Proof

1.1F1F2

The pairing is well defined. Sheaf multiplication $O (d) \otimes_{Z} O (- d - n - 1) \to O (- n - 1)$ gives by [F2] a bilinear cup product into $H^{n} (O (- n - 1))$ ; composing with the $k$ -linear residue trace [F1] gives the displayed pairing. It is $k$ -bilinear: multiplication by $λ \in k$ on either twist sheaf commutes with the tensor pairing, so naturality of the cup product [F2] carries the scalar action on either argument to multiplication by $λ$ on the target.

1.2F5

Vanishing in the middle degrees. By [F5], for $n \geq 1$ both $H^{q} (O (d))$ and $H^{n - q} (O (- d - n - 1))$ vanish whenever $0 < q < n$ , because both cohomological degrees lie strictly between $0$ and $n$ . Their zero-space pairing is perfect. The remaining degrees are $q = 0$ and $q = n$ .

1.3F3F5

The case $q = 0$ . If $d \geq 0$ , this is precisely the perfect residue pairing of [F3]. If $d < 0$ , then $H^{0} (O (d)) = 0$ by [F5], while $- d - n - 1 > - n - 1$ , so $H^{n} (O (- d - n - 1)) = 0$ by [F5]; the pairing of two zero spaces is perfect.

1.4F2F3F5

The case $q = n$ . If $d \leq - n - 1$ , put $d^{'} = - d - n - 1 \geq 0$ . For $s \in H^{0} (O (d^{'}))$ and $η \in H^{n} (O (- n - 1 - d^{'}))$ , [F3] identifies $t_{P^{n}} (s \cup η)$ with the perfect residue pairing. The exchanged cup product $η \cup s$ has the same image: the section $s$ defines a sheaf map $σ_{s} : Z_{P^{n}} \to O (d^{'})$ and multiplication $μ_{s} : O (- n - 1 - d^{'}) \to O (- n - 1)$ ; naturality in [F2] applied to $id \otimes σ_{s}$ and the right-unit class gives $η \cup s = H^{n} (μ_{s}) (η)$ , while the left-unit argument of [F3] gives $s \cup η = H^{n} (μ_{s}) (η)$ because sheaf multiplication is commutative. Thus the exchanged pairing is perfect. If $d > - n - 1$ , then $H^{n} (O (d)) = 0$ and $- d - n - 1 < 0$ gives $H^{0} (O (- d - n - 1)) = 0$ by [F5], so the pairing is perfect vacuously.

1.5F1F2F3F5

The case $n = 0$ . By [F5], $P_{k}^{0} = Spec k$ and every twist has $H^{0} ≅ k$ with no higher cohomology. The trace [F1] and degree-zero cup product [F2] make the pairing ordinary multiplication $k \times k \to k$ , perfect with dual basis $1$ , as also recorded in [F3].

2.1F1F2F3F4F5step 1.1step 1.2step 1.3step 1.4step 1.5∎

Conclusion. Steps 1.1–1.5 cover bilinearity, the middle degrees, both extremes for $n \geq 1$ , and $n = 0$ . AC [F4] is inherited through the cup product [F2] and the projective-space cohomology and residue suppliers [F3, F5]; no additional selection is made.

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Sources

R. Hartshorne, Algebraic Geometry (standard reference, not scraped)
The Stacks Project, Cohomology of Schemes (standard reference, not scraped)