# Relation between the cohomology group of a curve and the cohomology group of its jacobian

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.

• 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 $H^1(C,\mathbb{C})$.
– abx
Nov 5, 2021 at 16:15
• 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? Nov 5, 2021 at 16:57
• And welcome to MO! Nov 5, 2021 at 16:57
• 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. Nov 5, 2021 at 20:20
• Dear @abx thank you for your comment. Nov 5, 2021 at 20:31

$$\def\Alb{\text{Alb}}\def\Pic{\text{Pic}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cO{\mathcal{O}}$$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 $$[x]-[x_0]$$.

• 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? Nov 12, 2021 at 1:22
• @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. Mar 6, 2023 at 15:28