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Let $J_C$ be the Jacobian of a smooth projective curve $C$ over $\mathbb{C}$. I would like understand the isomorphism between $H^1(J_C,\mathbb{C})$ and $H^1(C,\mathbb{C})$. I read in a paper that this isomorphism can be easily achieved by the Hodge-theoretical methods, but they do not give any reference.

Maybe someone can give any reference or explanation about it. Sorry if it is a very basic question, I do not a lot about Hodge theory. So a detailed explanation will be very useful for me.

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    $\begingroup$ This is indeed very basic. $J_C$ is by definition $H^0(C,K_C)^*/H_1(C,\mathbb{Z})$, so $H_1(J_C,\mathbb{Z})$ is canonically isomorphic to $H_1(C,\mathbb{Z})$, hence $H^1(J_C,\mathbb{C})$ to $H^1(C,\mathbb{C})$. $\endgroup$
    – abx
    Nov 5, 2021 at 16:15
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    $\begingroup$ I was going to leave a much friendlier version of this comment and then a student came in. So, trying now: What background are you coming from here? Are you defining $J_C$ complex analytically as something like $H^0(C, \Omega^1)^{\vee}/H_1(C, \mathbb{Z})$, or are you defining it as something like the group of degree $0$ line bundles? Are you defining cohomology as something topological, or as something like de Rham cohomology? $\endgroup$ Nov 5, 2021 at 16:57
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    $\begingroup$ And welcome to MO! $\endgroup$ Nov 5, 2021 at 16:57
  • $\begingroup$ Dear @DavidESpeyer In the paper I am reading they define $J_C$ as an abelian variety isomorphic to $A_0(C)$ the group of $0$-cycles of degree zero on $C$ modulo rational equivalence, and the cohomology as something like de Rham cohomology. $\endgroup$
    – Roxana
    Nov 5, 2021 at 20:20
  • $\begingroup$ Dear @abx thank you for your comment. $\endgroup$
    – Roxana
    Nov 5, 2021 at 20:31

1 Answer 1

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$\def\Alb{\text{Alb}}\def\Pic{\text{Pic}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cO{\mathcal{O}}$There are two abelian varieties associated to a smooth projective connected $n$-fold $X$: The Albanese variety $\Alb(X)$ and the Picard variety $\Pic^0(X)$. The Albanese has a natural map $X \to \Alb(X)$ which induces an isomorphism $H^1(\Alb(X)) \to H^1(X)$; the Picard variety parametrizes $(n-1)$-cycles of degree $0$ up to rational equivalence. For a curve, we have $\Alb(X) \cong \Pic^0(X)$, and we call both of these the Jacobian. So we have to understand three things: The Albanese variety, the Picard variety, and the isomorphism.

This causes the ambiguity about whether this question is straightforward or not: It is pretty straightforward that $H^1(X) \cong H^1(\Alb(X))$. But the description in terms of $0$-cycles is a description of $\Pic^0(X)$. What I'm describing in this answer is a modern (or at least, 20-th century version) of the Abel-Jacobi theorem which gives a criterion for determining when two $0$-cycles on a curve are rationally equivalent in terms of integrating holomorphic $1$-forms.

The Albanese variety: Let's start by just thinking about a smooth connected manifold $M$. We get a map $H_1(M, \ZZ) \longrightarrow H^1_{DR}(M, \RR)^{\vee}$ by sending a $1$-cycle $\gamma$ to the linear functional $\omega \mapsto \int_{\gamma} \omega$. The kernel is the torsion part of $H_1(M, \ZZ)$, so the image is the torsion free quotient $H_1(M, \ZZ)_{tf}$. Define $\Alb(M)$ to be $H^1_{DR}(M, \RR)^{\vee}/H_1(M, \ZZ)_{tf}$. Topologically, this is just a torus whose dimension is the first betti number of $M$. We clearly have a natural isomorphism $H_1(\Alb(M), \ZZ) \cong H_1(M, \ZZ)_{tf}$ and so $H^1(\Alb(M), \ZZ) = \text{Hom}(H_1(M, \ZZ)_{tf}, \ZZ) \cong H^1(M, \ZZ)$.

We can get a map $M \to \Alb(M)$ as follows: Choose a vector space $V$ of closed $1$-forms on $M$ which maps isomorphically to $H^1_{DR}(M)$, and choose a base point $x_0 \in M$. For any $x \in M$, choose a path $\beta$ from $x_0$ to $x$. Then $\eta \mapsto \int_{\beta} \eta$ is a linear functional on $V$. If $\beta'$ is a different path from $x_0$ to $x$, then $\beta' - \beta = \gamma$ for a $1$-cycle $\gamma$, so $\int_{\beta'} \eta = \int_{\beta} \eta + \int_{\gamma} \eta$. In other words, replacing $\beta$ by $\beta'$ changes the linear functional $\int_{\beta} (\cdot)$ by an element of $H_1(X, \ZZ)$. So the class of $\int_{\beta} (\cdot)$ in $V^{\vee}/H_1(X, \ZZ)$ depends only on $x$, and we get a map $M \to V^{\vee}/H_1(X, \ZZ)$ sending $x$ to $\int_{\beta} (\cdot)$. Since we choose $V$ to be isomrophic to $H^1_{DR}(M)$, we get a map $M \to \Alb(M)$, and it is easy to check that this map induces the isomorphism $H^1(\Alb(M), \ZZ) \cong H^1(M, \ZZ)$.

Everything becomes more canonical if $M$, which I'll now call $X$, is a connected compact Kahler manifold, for example, a smooth connected complex variety. In that case, Hodge theory tells us that $H^1(X, \CC) = H^{10}(X) \oplus H^{01}(X)$, where $H^{10}(X)$ is the global holomorphic $1$-forms, $H^{10}(X) = H^0(X, \Omega^1)$. Concretely, this isomorphism says that we can take our $V$ to be the real parts and the imaginary parts of holomorphic $1$-forms (or, equivalently, we can take $V$ to be the real harmonic $1$-forms). Thus, $\Alb(X) = H^0(X, \Omega^1)^{\vee}/H_1(X, \ZZ)_{tf}$. Now $H^0(X, \Omega^1)^{\vee}$ becomes a complex vector space, so $\Alb(X)$ is not merely a real manifold but a complex manifold.

The Picard variety Let $X$ be a smooth complex manifold. Divisors on $X$, modulo rational equivalence, are the same as line bundles on $X$, and are the same as classes in $H^1(X, \cO^{\ast})$. (This should be in most algebraic geometry textbooks.) I am going to work in the analytic world here, so $\cO$ is the sheaf of holomorphic functions; $\cO^{\ast}$ is the sheaf of non-vanishing holomorphic functions and my topology is the analytic topology.

We have the exponential sequence of sheaves $0 \to \underline{\ZZ} \overset{2 \pi i}{\longrightarrow} \cO \overset{\exp}{\longrightarrow} \cO^{\ast} \to 1$, where $\underline{\ZZ}$ is locally constant $\ZZ$-valued functions. So we get a long exact sequence of cohomology which includes $$H^1(X, \underline{\ZZ}) \longrightarrow H^1(X, \cO) \longrightarrow H^1(X, \cO^{\ast}) \longrightarrow H^2(X, \ZZ).$$ The kernel of the map to $H^2(X, \ZZ)$ are called the cycles of degree $0$ and denoted $\Pic^0(X)$, so we have $\Pic^0(X) \cong H^1(X, \cO) / H^1(X, \underline{\ZZ})$.

Again, things are nicer if $X$ is connected compact Kahler. Then $H^1(X, \cO) = H^{01}(X)$ and the map $H^1(X, \ZZ) \to H^1(X, \cO)$ is the composition of $H^1(X, 2 \pi i \ZZ) \subset H^1(X, \CC) \to H^{01}(X)$ where the second map is the projection onto the second summand of the Hodge decomposition. (I'm going to start getting sloppy about dropping the $2 \pi i$.) In particular, it follows from Hodge theory that the image of $H^1(X, \ZZ)$ is a discrete, cocompact lattice in $H^{1}(X, \cO)$, so the quotient $H^1(X, \cO) / H^1(X, \underline{\ZZ})$ is a compact complex manifold.

The case of curves So far, we have two abelian varieties: $$\Alb(X) = H^0(X, \Omega^1)^{\vee}/H_1(X, \ZZ)_{tf} = H^{10}(X)^{\vee}/H_1(X, \ZZ)_{tf}$$ and $$\Pic(X) = H^1(X, \cO)/H^1(X, \ZZ) = H^{01}(X)/H^1(X, \ZZ).$$ But, if $X$ is a curve, then Poincare duality gives an isomorphism $H_1(X, \ZZ) \cong H^1(X, \ZZ)$ and Serre duality gives an isomorphism $H^{10}(X)^{\vee} \cong H^{01}(X)$. (In fact, Serre duality is just the Poincare duality pairing restricted to the two Hodge summands of $H^1(X, \CC)$.)

After checking enough compatibility of diagrams, this gives an isomorphism $\Alb(X) \cong \Pic^0(X)$, and gives that the map $X \to \Alb(X)$ we defined by integration (using the base point $x_0$) matches the map $X \to \Pic^0(X)$ sending $x$ to the divisor $[x]-[x_0]$.

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  • $\begingroup$ Dear @DavidESpeyer thank you for this very clear and amazing explanation! It is very helpful for me. I have another question :). I am reading a Hartshorne's paper "equivalence relations on algebraic cycles" and in page 133 he says: the 0-cycles of degree $0$ on a non-singular projective variety $X$ over $\mathbb{C}$, modulo rational equivalence (sometimes denoted by $CH_0(X)_{deg=0}$) coincides with the $0$-cycles algebraically equivalent to $0$, modulo rational equivalence. Do you know about it? or any reference? $\endgroup$
    – Roxana
    Nov 12, 2021 at 1:22
  • $\begingroup$ @Roxana This is probably a little late, but the point is that any zero-cycle is contained in a curve, which lets us reduce to the case of a curve, and the result then follows from the Picard variety theory of curves - i.e. the fact that the space parameterizing degree 0 zero-cycles on a curve modulo rational equivalence is connected, which is one of the things proved in David's answer. $\endgroup$
    – Will Sawin
    Mar 6, 2023 at 15:28

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