ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02

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One cocycle carried through the residue realization of Serre duality

Example

Assume the Axiom of Choice as inherited from the cited duality, residue and cohomology suppliers (The Axiom of Choice). Let $k$ be a perfect field, let $C = P_{k}^{1}$ have coordinate $t = x_{1} / x_{0}$ on $U_{0}$ , fix $d \geq 0$ , and put $L = O (- d - 2)$ . Its dual canonical twist is $ω_{C} \otimes L^{- 1} ≅ O (d)$ .

For $1 \leq j \leq d + 1$ , the local tail in the $U_{0}$ frame of $L$ , $c_{j} = [t^{- j} \otimes x_{0}^{- d - 2}] \in H^{1} (C, L),$ is a finite-support principal-part class. For $0 \leq m \leq d$ , the section of the dual twist is $σ_{m} = t^{m} d t \otimes x_{0}^{d + 2} on U_{0}, σ_{m} = - s^{d - m} d s \otimes x_{1}^{d + 2} on U_{1},$ where $s = 1 / t$ . The second expression shows it is regular at infinity and corresponds to $x_{0}^{d - m} x_{1}^{m}$ under $ω_{C} ≅ O (- 2)$ .

The positive residue pairing evaluates at the supported rational point: $⟨ c_{j}, σ_{m} ⟩ = {res}_{0} (t^{m - j} d t) = δ_{j, m + 1} .$ With rows $j = 1, \dots, d + 1$ and columns $m = 0, \dots, d$ , this is the diagonal identity matrix. Reversing the section order to $m = d, \dots, 0$ makes it anti-diagonal. Since the sections form a basis, this invertible matrix also proves that the classes $c_{j}$ form a basis. The fixed normalized Serre pairing is its negative, so its matrix has entries $- δ_{j, m + 1}$ ; it is perfect. At $d = 0$ its value on $c_{1}$ and $σ_{0}$ is $- 1$ , which equals $1$ in characteristic two.

The Axiom of Choice enters through the cited suppliers; the displayed computations make no additional choices.

Facts & Assumptions

Given: the Axiom of Choice, a perfect field $k$ , $C = P_{k}^{1}$ with coordinate $t = x_{1} / x_{0}$ , an integer $d \geq 0$ , and $L = O (- d - 2)$ .

[F1]

The Axiom of Choice is inherited through the duality, residue and projective-cohomology suppliers; this computation makes no further selection. (The Axiom of Choice)

[F2]

The canonical bundle is $ω_{C} ≅ O (- 2)$ . Under the standard charts, $d t = - s^{- 2} d s$ and the canonical-bundle identification sends $d t$ to $x_{0}^{- 2}$ and $d s$ to $- x_{1}^{- 2}$ . The sheaves $O (r)$ are invertible, satisfy $O (r)^{- 1} ≅ O (- r)$ , and their frames obey $x_{0}^{r} = s^{r} x_{1}^{r}$ on the overlap. (Canonical bundle and canonical divisors, Twisting sheaf on Proj)

[F3]

The cohomology of the twists gives $H^{1} (O (- d - 2))$ the Laurent basis $x_{0}^{e_{0}} x_{1}^{e_{1}}$ with $e_{0}, e_{1} < 0$ and $e_{0} + e_{1} = - d - 2$ , and gives $H^{0} (O (d))$ the basis $x_{0}^{d - m} x_{1}^{m}$ , $0 \leq m \leq d$ . (Cohomology of O(d) on projective space)

[F4]

Principal parts compute $H^{1}$ as finite-support local tails modulo principal parts of global meromorphic sections (including zero); in particular each tail $t^{- j}$ in the frame $x_{0}^{- d - 2}$ represents the class $c_{j}$ . (H^1 of a line bundle on a curve as principal parts modulo meromorphic and regular sections)

[F5]

At a rational point with parameter $t$ , residue is the coefficient of $t^{- 1} d t$ . Over a perfect field, the positive residue pairing is the sum of these local residues. (Residue of a rational differential at a separable closed point, The residue pairing of a line bundle with the dual canonical twist, Perfect fields: every irreducible polynomial is separable)

[F6]

Over a perfect field, the fixed normalized Serre pairing is the negative of the positive residue pairing. (Serre duality for line bundles on a smooth proper curve, and the residue realization, Normalization of the trace for Serre duality on a curve)

[F7]

For a smooth proper geometrically integral curve, $h^{1} (C, ω_{C}) = h^{0} (C, O_{C})$ . (h^1 of a line bundle equals the dimension of the space of dual sections)

Verification

Proof technique: represent the tails and global sections in their actual line-bundle frames, evaluate the positive local coefficient, and apply the fixed-trace sign comparison.

1.1F1F2F3F4

By [F2], $ω_{C} \otimes L^{- 1} ≅ O (d)$ , and [F3] gives $\dim_{k} H^{1} (C, L) = \dim_{k} H^{0} (C, O (d)) = d + 1$ . For each $1 \leq j \leq d + 1$ , the tail $t^{- j}$ in the $U_{0}$ frame has finite support at the rational origin and represents $c_{j}$ by [F4].

2.1F1F2F3step 1.1

The standard section $x_{0}^{d - m} x_{1}^{m}$ is $t^{m} x_{0}^{d}$ on $U_{0}$ , so under $d t \mapsto x_{0}^{- 2}$ it is represented by $σ_{m} = t^{m} d t \otimes x_{0}^{d + 2}$ . With $s = 1 / t$ , $d t = - s^{- 2} d s$ and $x_{0}^{d + 2} = s^{d + 2} x_{1}^{d + 2}$ , this becomes $- s^{d - m} d s \otimes x_{1}^{d + 2}$ , regular for $0 \leq m \leq d$ . Thus the $σ_{m}$ form the displayed section basis.

3.1F1F5F6step 1.1step 2.1

Multiplying the local representative of $c_{j}$ by $σ_{m}$ cancels the line-bundle frames and gives $t^{m - j} d t$ at the origin. By [F5], its positive residue is $1$ when $m - j = - 1$ and $0$ otherwise, namely $δ_{j, m + 1}$ . By [F6], the fixed normalized Serre value is $- δ_{j, m + 1}$ .

4.1F1F3step 1.1step 2.1step 3.1algebra

For ascending section order $m = 0, \dots, d$ , the matrix $δ_{j, m + 1}$ is diagonal; for reversed order $m = d, \dots, 0$ it is anti-diagonal. Since the sections form a basis by step 2.1, invertibility of this positive residue matrix proves that the $c_{j}$ are independent; their number is $d + 1 = \dim_{k} H^{1} (C, L)$ by step 1.1, so they form a basis. The normalized matrix is its negative and is invertible, so both pairings are perfect.

5.1F1F2F6F7step 3.1∎

For $d = 0$ , [F7] gives $h^{1} (C, ω_{C}) = 1$ , and step 3.1 gives normalized trace $- 1$ on $[t^{- 1} \otimes x_{0}^{- 2}]$ paired with $d t \otimes x_{0}^{2}$ . In characteristic two this value is $1$ .

Depends on

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Sources

John Tate, Residues of differentials on curves, Ann. Sci. E.N.S. (4) 1 (1968) 149-159 (standard reference, not scraped)
MIT 18.725 Algebraic Geometry (Fall 2015) course notes, Lectures 24-25 (standard reference, not scraped)