# Motivation for the Jacobian Variety

I've been to many talks in Number Theory and for some reason I've yet to fully grasp, we all seem to like Jacobian Varieties a lot. I know that they are Abelian varieties, which give information about their respective curve, but I'm not sure what information exactly. I know of the analytic description of the Jacobian, but I'm still not exactly sure why the Jacobian is so studied.

In his AMS article, What is a motive Barry Mazur seems to suggest that Jacobians encapsulate all cohomology theories. Is this true? How can I see this?

• Kleiman's chapter in FGA explained (available separately on arXiv) contains a great and very detailed historical account (albeit possibly a bit tangential to your main question) of the development of the Picard scheme (as well as details for the construction in a very general setting). – R. van Dobben de Bruyn Jul 6 at 20:12

If you are a number theorist, you presumably like class groups? Let $$X$$ be a curve defined over $$\mathbb{F}_p$$, let $$J$$ be its Jacobian and let $$x$$ be an $$\mathbb{F}_p$$ point of $$X$$. Let $$A$$ be the coordinate ring of the affine curve $$X \setminus \{ x \}$$. Then the class group of $$A$$ is $$J(\mathbb{F}_p)$$. (And similar statements can be made for deleting more than one point, or deleting points defined over extensions of $$\mathbb{F}_p$$.)

• Why do you need to delete a point? – Rdrr Jul 8 at 1:18
• Answer 1 to get an affine variety. If someone is truly coming from classical number theory, they may only know class groups of rings, not Pic. Answer 2 if you take $Pic(X)$, you get $\mathbb{Z} \times J(\mathbb{F}_p)$, not $J(\mathbb{F}_p)$. – David E Speyer Jul 8 at 10:51
• Could you give a reference for the fact that the class group of $A$ is $J(\mathbb{F}_p)$? Thanks. – Ang Jul 15 at 17:18
• @Ang I don't have a reference off the top of my head. We have $\mathrm{Pic}(X \setminus \{ x \}) \cong \mathrm{Pic}^0(X)$, since every divisor on $X \setminus \{ x \}$ can be extended to a degree $0$ divisor on $X$ in a unique way. (Here it matters that $X$ is an $\mathbb{F}_p$ point.) The fact that $J(\mathbb{F}_p) \cong \mathrm{Pic}^0(X)$ is the defining property of the Picard functor. – David E Speyer Jul 15 at 17:30

Suppose $$X/\mathbb{Q}$$ is a (smooth, projective, geometrically integral) curve of genus $$g\geq 2$$ and $$J/\mathbb{Q}$$ its Jacobian variety. If one is interested in determining the (finite, by Faltings) set of rational points $$X(\mathbb{Q})$$, then it can be useful to compute $$J(\mathbb{Q})$$ first. The latter is easier because $$J(\mathbb{Q})$$ is a finitely generated abelian group, and descent theory analogous to elliptic curves allows us to often do this in practice. If we pick a point $$P\in X(\mathbb{Q})$$ then we have an associated embedding $$i_P: X \hookrightarrow J$$. In favorable situations studying this embedding allows us to determine $$X(\mathbb{Q})$$ from $$J(\mathbb{Q})$$. For example, the method of Chabauty-Coleman gives a very concrete instance of this when the rank of $$J(\mathbb{Q})$$ is less than $$g$$ (for a friendly introduction to this method see the nice survey of McCallum-Poonen).

The moral is: by replacing $$X$$ by $$J$$, we somehow have made the geometry harder but the arithmetic easier.

The relation with motives can be explained in relatively concrete terms. The $$\ell$$-adic cohomology groups $$H^i(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$$ are zero if $$i\neq 0,1$$, $$2$$ and isomorphic to $$\mathbb{Q}_l, \mathbb{Q}_l(-1)$$ if $$i=0, 2$$ respectively. (The minus $$-1$$ denotes the Tate twist.) So the only interesting degree is $$i=1$$, and pulling back via $$i_P$$ will induce an isomorphism $$H^1(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l) \simeq H^1(J_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$$. This last group (with its Galois action) is isomorphic to the dual of the $$\ell$$-adic Tate module of $$J$$. So $$J$$ and its torsion points encapsulate all the cohomological information of $$X$$. Similar statements will hold for other Weil cohomology theories: the only interesting degree is $$1$$ and $$i_P$$ will induce an isomorphism on $$H^1$$.

Edit: as pointed out in the comments, the geometry of $$J$$ is arguably easier than that of $$X$$. A better moral is thus maybe that we have made the space we're considering larger but richer in structure.

• I must respectfully disagree that replacing $C$ with $J$ "makes the geometry harder." It does increase the dimension, which one could argues makes the geometry harder, but it introduces a group structure, and I'd sugestt that the geometry of a high dimensional group variety (especially one that's compact) is much less difficult than the geometry of lower dimensional varieties having less structure. Or even ignoring the group structure, $J$ has Kodaira dimension 0, while $C$ has Kodaira dimension 1, again suggesting that $J$'s geometry is simpler than $C$'s. – Joe Silverman Jul 6 at 20:53
• Thanks for the comment, I'll edit my vague moral to make it more accurate. – Jef Jul 6 at 20:57
• What are the favourable situations that allow us to determine $J(\mathbb{Q})$ from $X(\mathbb{Q})$? Also, why does $i_P$ become an isomorphism on $H^1$? – Rdrr Jul 8 at 1:18

As outlined by the other answers, the Jacobian $$J_X$$ of a curve $$X$$ defined over $$\mathbb{F}_q$$ indeed encapsulates all cohomology information of $$X$$. In particular one can read the zeta function $$\zeta_X$$ directly on $$J_X$$: the numerator of $$\zeta_X$$ is simply the (reciprocal) polynomial of the Frobenius $$\pi_q$$ acting on $$J_X$$.

In particular André Weil's original proof of the Hasse-Weil bound for curves used Jacobians (implicitely). That was a big motivation in his Foundations of algebraic geometry: the algebraic construction of Jacobians over any field.

By the way over $$\mathbb{C}$$ the Abel-Jacobi map shows that the Jacobian of $$X$$ is intimately related to the study of abelian integrals. I think historically that was the prime motivation to study Jacobians. A fun fact is that modular functions coming from hyperelliptic integrals can be used to solve algebraic equations. Cf the appendix of Mumford's TATA2.