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The projective-space twist pairing in Serre duality

Statement

Assume the Axiom of Choice. Let $k$ be a field and consider $P_{k}^{2}$ with the residue trace of Residue pairing between H^0 and top cohomology of projective space. Then the class $x_{0}^{2} \in H^{0} (P_{k}^{2}, O (2))$ pairs to $1$ with the class $x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1} \in H^{2} (P_{k}^{2}, O (- 5))$ and to $0$ with every other monomial basis class of $H^{2} (P_{k}^{2}, O (- 5))$ ; in particular the pairing $H^{0} (O (2)) \times H^{2} (O (- 5)) \to k$ of Serre duality for twisting sheaves on projective space is nondegenerate on the basis monomial $x_{0}^{2}$ , and the six monomial basis classes of $H^{2} (O (- 5))$ are each detected by a degree-two monomial.

Facts & Assumptions

Given: the field $k$ , the projective plane $P_{k}^{2}$ with homogeneous coordinates $x_{0}, x_{1}, x_{2}$ , and the two monomial classes displayed in the statement.

[F1]

For $n \geq 1$ and $d \geq 0$ the pairing $H^{0} (P_{k}^{n}, O (d)) \times H^{n} (P_{k}^{n}, O (- n - 1 - d)) \to k$ is the coefficient of $(x_{0} \dots x_{n})^{- 1}$ in the product of the representing monomials; $H^{0} (O (d))$ has as a $k$ -basis the degree- $d$ monomials and $H^{n} (O (- n - 1 - d))$ the all-negative monomials of total degree $- n - 1 - d$ . (Residue pairing between H^0 and top cohomology of projective space)

[F2]

For every $d$ and $q$ the evaluation pairing $H^{q} (O (d)) \times H^{n - q} (O (- d - n - 1)) \to k$ using the residue trace is perfect. (Serre duality for twisting sheaves on projective space)

[F3]

The Axiom of Choice is The Axiom of Choice.

Proof

1.1F1

The two bases. For $n = 2$ and $d = 2$ the first space $H^{0} (P_{k}^{2}, O (2))$ has the six degree-two monomials $x_{0}^{2}, x_{0} x_{1}, x_{0} x_{2}, x_{1}^{2}, x_{1} x_{2}, x_{2}^{2}$ as a basis, and the second space $H^{2} (P_{k}^{2}, O (- 5))$ has the six all-negative monomials of total degree $- 5$ as a basis, namely $x_{0}^{e_{0}} x_{1}^{e_{1}} x_{2}^{e_{2}}$ with $e_{0}, e_{1}, e_{2} < 0$ and $e_{0} + e_{1} + e_{2} = - 5$ : writing $f_{i} = - e_{i} \geq 1$ gives $f_{0} + f_{1} + f_{2} = 5$ , so there are $(\binom{5 - 1}{2}) = 6$ such classes, among them $x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1}$ .

1.2F1F2

The nonzero pairing. Multiplying the two displayed classes gives $x_{0}^{2} \cdot x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1} = x_{0}^{- 1} x_{1}^{- 1} x_{2}^{- 1},$ whose coefficient in $x_{0}^{- 1} x_{1}^{- 1} x_{2}^{- 1}$ is $1$ ; by the coefficient description of the pairing in [F1] one has $⟨ x_{0}^{2}, x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1} ⟩ = 1$ , and $t (x_{0}^{2} \cup x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1}) = 1$ for the residue trace $t$ of [F2].

2.1F1step 1.1step 1.2

All cross pairings vanish. Let $x_{0}^{e_{0}} x_{1}^{e_{1}} x_{2}^{e_{2}}$ be a basis class of $H^{2} (O (- 5))$ different from $x_{0}^{- 3} x_{1}^{- 1} x_{2}^{- 1}$ . Its product with $x_{0}^{2}$ has exponent vector $(2 + e_{0}, e_{1}, e_{2})$ , and the coefficient of $x_{0}^{- 1} x_{1}^{- 1} x_{2}^{- 1}$ in that monomial is $1$ exactly when $(2 + e_{0}, e_{1}, e_{2}) = (- 1, - 1, - 1)$ , that is $e_{0} = - 3$ , $e_{1} = e_{2} = - 1$ ; if $e_{0} \neq - 3$ the first coordinate is wrong and the pairing is $0$ , while if $e_{0} = - 3$ the total-degree condition $e_{0} + e_{1} + e_{2} = - 5$ forces $e_{1} + e_{2} = - 2$ with $e_{1}, e_{2} < 0$ , so $e_{1} = e_{2} = - 1$ and the class is the one already treated. Hence the pairing of $x_{0}^{2}$ with every other basis class is $0$ by the coefficient description of [F1].

3.1F1F2F3step 1.2step 2.1∎

Nondegeneracy on the displayed class. Step 1.2 exhibits a class pairing to $1$ with $x_{0}^{2}$ , so $x_{0}^{2}$ is detected by the pairing; conversely, for each of the six basis classes $x^{\underline{e}}$ of $H^{2} (O (- 5))$ the monomial $x^{\underline{a}}$ of degree $2$ with $a_{i} = - 1 - e_{i} \geq 0$ satisfies $⟨ x^{\underline{a}}, x^{\underline{e}} ⟩ = 1$ by the bijection of [F1], so every basis class is detected, in accordance with the perfectness of the pairing recorded in [F2]. The Axiom of Choice is assumed in the statement and declared as the dependency The Axiom of Choice ([F3]); it is inherited through the two named suppliers and no further choice is made.

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Sources

R. Hartshorne, Algebraic Geometry (standard reference, not scraped)