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The Hodge conjecture for the projective line

Example

For $X = P_{C}^{1}$ , every rational Hodge class in degree $0$ or $2$ is the class of a rational algebraic cycle: the degree- $0$ class is a rational multiple of $[X]$ , and every degree- $2$ rational class is a rational multiple of the class of a point. Thus the Hodge conjecture holds for the projective line in both possible codimensions.

Facts & Assumptions

Given: $X = P_{C}^{1}$ with its standard complex orientation. For this example, codimension- $p$ cycles are finite integer linear combinations of codimension- $p$ irreducible closed subvarieties; rational equivalence is generated by divisors of rational functions on codimension- $(p - 1)$ subvarieties; ${C H}^{p} (X)_{Q}$ is the resulting quotient tensored with $Q$ ; and the cycle-class map sends an irreducible subvariety to its Poincaré-dual cohomology class.

[F1]

A Hodge class of codimension $p$ is a rational class in $H^{2 p} (X; Q)$ whose image in complex cohomology belongs to $H^{p, p} (X)$ ; the Hodge-type convention is the usual decomposition of complex differential forms by holomorphic and antiholomorphic degree.

Proof

1.1givenF1algebra

The projective line is the Riemann sphere, hence is topologically $S^{2}$ with one cell in dimensions $0$ and $2$ and no other cells. Therefore $H^{0} (X; Q) = Q$ and $H^{2} (X; Q) = Q$ , with $H^{1} (X; Q) = 0$ . Every smooth $2$ -form on a complex curve has type $(1, 1)$ because there are no nonzero $(2, 0)$ or $(0, 2)$ forms; the constant function represents the type- $(0, 0)$ generator of $H^{0}$ . Thus the Hodge classes in codimensions $0$ and $1$ are exactly all rational classes in those degrees.

1.2givenalgebra

Codimension- $1$ cycles are zero-cycles $D = \sum_{a \in C} m_{a} [a] + m_{\infty} [\infty]$ with finitely many nonzero integer coefficients. The degree $\deg D = \sum_{a} m_{a} + m_{\infty}$ is unchanged by rational equivalence because the divisor of a rational function on $P^{1}$ has degree zero. Conversely, if $\deg D = 0$ , then $m_{\infty} = - \sum_{a} m_{a}$ and the rational function $f (z) = \prod_{a} (z - a)^{m_{a}}$ has divisor exactly $D$ (negative exponents are allowed in a rational function); hence $D$ is rationally equivalent to zero. It follows that ${C H}^{1} (X) ≅ Z$ by degree, generated by $[0]$ .

2.1givenstep 1.1algebra

The only codimension- $0$ irreducible closed subvariety is $X$ itself, and there are no codimension- $(- 1)$ subvarieties that could generate a rational-equivalence relation; hence ${C H}^{0} (X)_{Q} = Q [X]$ . The cycle class of $[X]$ is the generator of $H^{0} (X; Q)$ , so the codimension- $0$ cycle map is surjective.

3.1step 1.1step 2.1step 1.2givenalgebra∎

The cycle class of a point is the generator of $H^{2} (S^{2}; Z)$ by the definition as Poincaré dual of the positively oriented point, so after tensoring with $Q$ the codimension- $1$ cycle map is an isomorphism ${C H}^{1} (X)_{Q} \to H^{2} (X; Q)$ . Together with step 2.1, both possible Hodge-class degrees are algebraic. No claim in dimensions greater than one follows from this calculation.

Remarks

This is a checked special case only. The general question is recorded separately as Hodge conjecture for smooth projective complex varieties ‡; that link is context and is not a logical premise.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.