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In an exercise of Voisin book, says:

Let $j:C\rightarrow S$ the inclusion of a smooth curve on a smooth connected projective surface. Set $H=ker(j_*:H^1(C,\mathbb{Z})\rightarrow H^3(S,\mathbb{Z}))$.

We also write $A\subset J(C)$ for the Abelian subvariety corresponding to the Hodge substructure $H$.

I do not understand how the correspondence between Abelian subvarieties and Hodge substructures goes, so I will be grateful if someone can suggest to me a reference to learn it. Thank you

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2 Answers 2

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Hodge substructures of $H^1(C,\mathbb Z)$ and Abelian subvarieties of $J(C)$ ar essentially the same thing:

In general, let $V,\omega$ be a free $\mathbb Z$-module of even rank with an integral symplectic form. A polarized Hodge structure of weight 1 on $V$ is the data of a decomposition $V_{\mathbb C} := V\otimes_{\mathbb Z} \mathbb C = V^{1,0} \oplus V^{0,1}$ with $V^{0,1} = \overline{V^{1,0}}$ and a positivity condition with respect to $\omega$. This is canonically equivalent to the data of a linear complex structure $I$ on $V_{\mathbb R} := V\otimes_{\mathbb Z} \mathbb R$ for which $\omega$ is of positive. The associated abelian variety is $V_{\mathbb R} / V_{\mathbb Z}$, with this linear complex structure.

A Hodge substructure is a $\mathbb Z$-submodule $W\subset V$ such that $W\otimes \mathbb C$ is compatible with the Hodge decomposition, i.e. $$W_\mathbb C:= W \otimes \mathbb C = W_\mathbb C \cap V^{1,0} \oplus W _\mathbb C \cap V^{0,1}~.$$ Equivalently, $W_\mathbb R := W\otimes \mathbb R$ is $I$-invariant. Then $W_\mathbb R/W$ is an abelian subvariety of $V_\mathbb R / V$. Conversely, every abelian subvariety is a the quotient of a $I$-invariant subspace.

In your case, $V= H^1(C,\mathbb Z)$, $V^{1,0} = H^{1,0}(C)$, the associated abelian variety is $J(C)$ and $H= W$. What remains to understand is that $H$ is a Hodge substructure, and this is because $j_*$ is compatible with the Hodge decomposition (it maps $H^{1,0}$ to $H^{2,1}$ and $H^{0,1}$ to $H^{1,2}$).

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  • $\begingroup$ Dear @Nicolas, for $V^{0,1}=\overline{V^{0,1}}$, Do you mean $V^{0,1}=\overline{V^{1,0}}$? $\endgroup$
    – Roxana
    Commented Sep 7, 2021 at 15:59
  • $\begingroup$ Yes indeed. Let me edit my answer. $\endgroup$
    – Nicolast
    Commented Sep 9, 2021 at 7:16
  • $\begingroup$ Dear @Nicolas, Thank you a lot! $\endgroup$
    – Roxana
    Commented Sep 9, 2021 at 12:03
  • $\begingroup$ Dear @Nicolas, if in your explanation above I change $\mathbb{Z}$ for $\mathbb{Q}$ all you said is still true? $\endgroup$
    – Roxana
    Commented Feb 15, 2022 at 7:12
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See ncatlab and the reference to the Peters-Steenbrink for more details.

More concretely, consider the complex torus $$J(H):=\dfrac{H_{\mathbb{C}}}{H_{\mathbb{Z}}+F^1}$$

View the torus in the Jacobian of the curve $C$, $J(C):=J(H^1(C, \mathbb{Z}))$.

Now by the assumptions, one shows that the hodge structure is polarizable, which shows that the corresponding complex torus $J(H)$ is algebraic, i.e. an abelian variety. You can find more discussion on this in Peters-Steenbrink Chapter 7, 7.1.2.

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  • $\begingroup$ Dear @Evans thank you! $\endgroup$
    – Roxana
    Commented Sep 7, 2021 at 17:30

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