# Maps of algebraic curves (and their Jacobian)

When people consider a map $\varphi: C \rightarrow C$ between algebraic curves and they mention the "associated map" on the Jacobian of $C$. Which map do they mean? Do they mean $\varphi^{\*}$ or $\varphi_*$ (or some other map)?

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Does it matter? The principal polarization implies that either one of $\phi^\ast$ and $\phi_\ast$ determines the other... –  Dan Petersen Jun 14 '12 at 18:40
Typically, one extends the map linearly to divisors, i.e. $\sum n_iP_i \mapsto \sum n_i\phi(P_i)$, so this is $\phi_*$. But, as Dan pointed out, usually it doesn't matter. –  Felipe Voloch Jun 14 '12 at 19:22

We have to remind ourselves who the Jacobian $J(C)$ is. One quick and dirty answer is that the Jacobian of a curve $C$--seen as a riemann surface--is just the quotient $H_1(C, \mathbb{R}) /H_1(C, \mathbb{Z})$ -- this is an insufficient definition since I have not specified either a metric or a Kahler form. But it's sufficient to convince us that the induced map $J(\phi): J(C) \to J(C')$ ought to be $H_1(\phi)$ factored through the quotient. Again, the above description of $J(C)$ is incomplete but sufficient our purpose.
Any riemannian manifold has a meaningful associated Jacobi variety. In the case of complex curves, there's that alternative description as the quotient $H^0(C, \Omega^1)^* / jH_0(C, \mathbb{Z})$ where $j: C \to H_1(C, \mathbb{Z})$ is the Abel-Jacobi period mapping (defined up to a choice of base point on $C$--but all homotopy equivalent). Here we find that the induced map $J'(C)$ between jacobians is $\phi^*$ where one deserves to convince themselves that map actually factors through the quotient.
As Bill Clinton once said, it depends on the meaning of "is". The jacobian is what you said only over $\mathbb{C}$. –  Felipe Voloch Jun 14 '12 at 20:28
Ah, quite right. What i've described is almost exclusively over $\mathbb{C}$. Over arbitrary fields I have no idea. –  J. Martel Jun 15 '12 at 0:11