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To explain my problem, I must give a lemma:

Let $X$, $Y$, $Z$ be curves over $k$ (of characteristic 0) such that the genus of $Z$ is greater than 2, and $\pi : X \to Y$, $\phi : X \to Z$ two non-constant morphisms. If $\phi^\star(H^0(Z,\Omega))\subseteq\pi^\star(H^0(Y,\Omega))$, where $\Omega$ denotes the sheaf of regular 1-forms in each case, then there exists a non-constant morphism $u: Y \to Z$ such that $\phi = u \circ \pi$.

Now, in a proof, I saw the use of this lemma, except that the hypothesis was the inclusion $\mathrm{Image}(\mathrm{Jac}(Z) \to \mathrm{Jac}(X)) \subseteq \mathrm{Image}(\mathrm{Jac}(Y) \to \mathrm{Jac}(X))$, instead of $\phi^\star(H^0(Z,\Omega))\subseteq\pi^\star(H^0(Y,\Omega))$. I can guess it is equivalent, but why? Is it related to Grothendieck's duality? Did I miss something obvious?

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2 Answers 2

up vote 3 down vote accepted

Since you are in char zero, you can assume the ground field is the complex numbers. The inclusion of jacobians follows from the inclusion of spaces of differentials via the description in terms of periods and calculus.

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I think the questioner wants to understand the deduction the other way, i.e. to go from the statement about Jacobians to the statement about differentials. –  Emerton Aug 30 '10 at 15:34
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Perhaps writing the formulas $$Jac(X) = H^0(X,\Omega_X^1)^*/H_1(X,\mathbb{Z})$$ $$T^*_0Jac(X)= H^0(X,\Omega^1),$$ will give the questioner enough to work with. –  Donu Arapura Aug 30 '10 at 16:19
    
Then you take the derivative of the maps between the jacobians, again using the complex analytic description. $(\int_0^P \omega)' = \omega$ –  Felipe Voloch Aug 30 '10 at 16:21
    
What is the meaning of $T_0$? –  Bernikov Aug 31 '10 at 2:49
    
T_0 is the tangent space at zero. –  Felipe Voloch Aug 31 '10 at 2:54

you do not give your definition of jac, and you mention duality, so I presume your question concerns the variance of these functors, in which case some duality is involved. I.e. if you consider jac as the albanese, i.e. the quotient of the dual of the differentials by the integer homology, then it is covariant, and the induced map from jax(x) to jax(y) pulls back differentials from y to x as the coderivative map of cotangent spaces at the origin. This is the realization of jac via periods of differentials.

If you consider jac as the picard variety, i.e. the quotient of the first sheaf cohomology with coefficients in the structure sheaf O by the first integer cohomology, it is contravariant and induces a map jax(Y) to jac(X) which is probably the one in the alternate hypothesis you mention. Fortunately the Jacobian is "self dual", i.e. it is principally polarized, so both points of view are equivalent. This self duality is also called the abstract "Abel's theorem" and is a consequence of Poincare duality on appropriate subspaces of the topological first complex homology and cohomology spaces.

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my suggestion is that to answer the original question, it is crucial to define the maps from Jac(Z) and Jac(Y) to Jac(X), since the descriptions of Jac given in answer 1 above in terms of differential forms, i.e. as the (covariant) albanese variety, do not obviously induce maps in the other (contravariant) direction. –  roy smith Sep 2 '10 at 13:27

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