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Cech cohomology of line bundles on the projective line

Statement

Assume the Axiom of Choice, inherited from sheaf cohomology.

For d∈Z, let OP1(d) have standard frames e0,e1 on U0={x≠0} and U1={y≠0} with e1=zde0 on the overlap, where z=y/x and w=1/z. Then dim⁡H0(P1,O(d))=max⁡(d+1,0),dim⁡H1(P1,O(d))=max⁡(−d−1,0), and Hq=0 for q≥2. For d≥0, restrictions to U0 of global sections on P1, written in the frame e0, are precisely the polynomials of degree at most d; for d≤−2, H1 has Laurent basis zd+1,zd+2,…,z−1.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), P1 with the two standard affine charts, an integer d, the line-bundle transition e1=zde0, and sheaf cohomology with the supplied injective resolution datum.

[F1]

Under the assumed Axiom of Choice inherited from the definition of sheaf cohomology, the two-chart affine cover computes cohomology of a quasi-coherent sheaf on the separated scheme P1 (Cech cohomology computes quasi-coherent cohomology on a separated scheme).

[F2]

The sheaf O(d) is invertible, hence quasi-coherent, with the standard transition rule (Invertible sheaves).

Proof

1.1F1F2givenalgebra

The Cech complex for U0,U1 has C0=C[z]e0⊕C[w]e1,C1=C[z,z−1]e0, with differential (f0,f1)↦zdf1(1/z)−f0(z). Its kernel consists of polynomials f0(z) that also lie in zdC[z−1]. If d≥0, these are exactly the span of 1,z,…,zd; if d<0, the exponent ranges {0,1,…} and {d,d−1,…} are disjoint, so the kernel is zero. This proves the assertion for H0.

2.1F1step 1.1algebra∎

The degree-one cohomology is the Laurent-polynomial quotient C[z,z−1]/(C[z]+zdC[z−1]). The first summand spans exponents n≥0 and the second spans n≤d. If d≥−1, these ranges cover every integer, so the quotient is zero. If d≤−2, the missing exponents are exactly d+1,d+2,…,−1, which gives the stated basis and dimension −d−1. The Cech complex has no degree above one, so Hq=0 for q≥2.

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