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    The Jacobian, Abel's theorem, and the Abel--Jacobi map

    T-U5QSY

    The Jacobian geometrizes degree-zero divisors: Abel kills exactly the principal ones, and Jacobi inversion reaches every divisor class.Suggest a slogan

    For \(X\) a smooth projective curve of genus \(g\) over \(\CC\), \[ \Jac(X) = \dualof{H^0(X,\Omega^1)} / H_1(X,\ZZ) , \] a complex torus of dimension \(g\), and an abelian variety. The Abel–Jacobi map sends a degree-zero divisor \(\sum (p_i - q_i)\) to \(\sum \int_{q_i}^{p_i}\), a linear functional on \(H^0(\Omega^1)\) taken modulo periods.

    (Abel)

    A divisor of degree zero is principal exactly when its image under the Abel–Jacobi map is zero. Hence \(\Pic^0(X) \cong \Jac(X)\).

    (Jacobi inversion)

    The Abel–Jacobi map is surjective, so \(\Jac(X)\) is exactly the group of degree-zero divisor classes.

    The significance is the sentence the two theorems combine to: the Jacobian is a projective variety whose points are the degree-zero line bundles on \(X\), so a discrete-looking classification problem becomes a geometric object of dimension \(g\).

    Fixing a base point \(p_0\) gives \(X \to \Jac(X)\), \(p \mapsto [p - p_0]\), which is an embedding for \(g \geq 1\). In genus \(1\) it is an isomorphism: an elliptic curve is its own Jacobian, and the group law on \(E\) is the statement that \([p]+[q]=[r]+[0]\) exactly when \(p+q=r\). Everything special about elliptic curves is this coincidence of \(X\) with \(\Jac X\).

    State Abel's theorem.
    What is the significance of the Jacobian?
    What kind of map is the Abel--Jacobi map, and what is it in genus $1$?

    Dependencies

    • T-MWDVLRiemann--Roch and Serre duality on a curveuses
    • PR-Y5S7VThe degree of a divisor on a projective curveuses

    Backlinks

    • T-CRVUNIFEvery elliptic curve over $\CC$ is a torus $\CC/\Lambda$related-to
    • T-CRVJACFUNThe Jacobian represents $\Pic^0(X/-)$, and is a smooth proper group scheme of dimension $g$related-to
    • PR-CRVGRPThe group law on an elliptic curve, and multiplication by $n$uses

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