I found the algebraic equivalence relation is closely related to deformation, the most intuitive description that I found is in Griffiths' Topics in Transcendental Algebraic Geometric Chapter 1 which is defined as follows:

(Definition 1) Let $X$ be a projective (algebraic) variety. Two cycles $Z_1$ and $Z_2$ on $\mathrm{X}$ are algebraically equivalent if, roughly speaking, one can be deformed into the other via an algebraic family of cycles on $X$.

To be more precise, there is an algebraic variety $S$ and an algebraic cycle $T$ in $S \times X$ such that $Z_1$ and $Z_2$ are the restrictions of $T$ to two fibers of the projection $\pi: S \times X \rightarrow S$.

The picture shows as follows

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In the case of divisor, there is another definition as follows

(Definition 2) Let $X$ be projective (algebraic) variety, assume $D,D'$ be two Cartier divisor, we say $D\sim D'$ being algebraic equivalence if the linear bundle $D-D'\in \operatorname{Pic}^0(X)$. If we mod out it we get $$\text{NS}(X) = \operatorname{Pic}(X)/\operatorname{Pic}^0(X)$$

The question is: Are these two definitions the same?

Easy to see from the second one that algebraic equivalence relation is also numerical equivalence. If we can prove definition 1 = definition 2, then it will imply that the intersection number is deformation invariance, correct?

(see related question: https://math.stackexchange.com/a/2679928/360262)

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    $\begingroup$ I believe they are, but an explanation might hinge on your definition of $\operatorname{Pic}^0$. $\endgroup$ Jan 16 at 15:41

1 Answer 1


$\DeclareMathOperator{\Pic}{Pic} \newcommand{\an}{\textrm{an}}$ This is still a WIP, but maybe you have an idea how to continue. I remembered that for divisors, algebraic equivalence is the same as homological equivalence. So I looked at Fulton's Intersection Theory, especially part 19.3, and at Voisin's Hodge Theory and Complex Algebraic Geometry, I + II. Both write something, but neither gives a complete argument.

For now assume that $X$ is a smooth projective variety over $\mathbb C$.

Let's start with the Picard group. By definition, the Picard group $\Pic(X)$ parametrizes algebraic line bundles, or divisors modulo rational equivalence. It is a basic fact when learning sheaf cohomology, that $$\Pic(X) \cong H^1(X, \mathcal O_X^*).$$ As we are working over $\mathbb C$, we can also think of $X$ as a complex manifold, and consider the exponential sequence of analytic sheaves $$0 \to \underline{\mathbb Z} \to \mathcal O_X^{\an} \xrightarrow{e} \mathcal O_X^{\an *} \to 0,$$ where $e(f) = e^{2\pi i f}$, for a holomorphic function $f:U \to \mathbb C$ on $U \subset \mathbb C$. The exponential sequence gives a long exact sequence in cohomology $$0 = H^0(X, \mathcal O_X^{\an *}) \to H^1(X, \mathbb Z) \to H^1(\mathcal O_X^\an) \to H^1(\mathcal O_X^{\an*}) \xrightarrow{c_1} H^2(X, \mathbb Z) \to H^2(\mathcal O_X^\an).\tag{1}\label{eq1}$$ Note that by a direct application of the GAGA theorems, $H^i(\mathcal O_X^\an) = H^i(\mathcal O_X)$. This does not apply directly to $H^1(\mathcal O_X^{\an*})$, as the sheaf $\mathcal O_X^{\an*}$ is not coherent. But the cohomology group $H^1(\mathcal O_X^{\an*})$ classifies analytic line bundles, and by GAGA, analytic line bundles are algebraic, hence parametrised by $H^1(\mathcal O_X^*)$. So in total $$H^1(\mathcal O_X^{\an*}) = H^1(\mathcal O_X^*) = \operatorname{Pic}(X).$$ The map $c_1: H^1(\mathcal O_X^*) \to H^2(X, \mathbb Z)$ is the first Chern class of a line bundle, and by definition, its kernel is the group $$\Pic^0(X) \subset \Pic(X).$$ We say that two divisors are homologically equivalent, if they have the same Chern class in $H^2(X, \mathbb Z)$, or equivalently if $$D - D' \in \Pic^0(X).$$ By the exactness of \eqref{eq1}, $\Pic^0(X)$ is also isomorphic to $H^1(\mathcal O_X) / H^1(X, \mathbb Z)$. Hodge theory tells us that this is an abelian variety, so it is connected. Voisin writes in volume II, Chapter 8.2:

we know by the theory of the Picard group that the divisors homologous to $0$ are parametrised up to rational equivalence by a connected algebraic variety $\Pic^0(X)$, so they are algebraically equivalent to $0$.

Is that enough for you?

  • $\begingroup$ Thank you red_trumpet, excellent solution. That's exactly what I want, I believe there are some book about deformation theory of line bundle will discuss Voisin's claim. $\endgroup$
    – yi li
    Jan 17 at 8:42
  • $\begingroup$ I come up with this question when reading the paper BDPP13. Sorry I don't have sufficient knowledge to prove Voisin's claim right now. $\endgroup$
    – yi li
    Jan 17 at 8:53
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    $\begingroup$ @yili We would need a line bundle $\mathcal L$ on $X \times \operatorname{Pic}^0(X)$ such that $\mathcal L|_{X \times \{L\}} \cong L$ for each line bundle $L \in \operatorname{Pic}^0(X)$. I'm not sure how to construct such a thing, and I asked an additional question regarding this. $\endgroup$ Jan 17 at 12:37
  • $\begingroup$ Ok thank you red_trumpet $\endgroup$
    – yi li
    Jan 17 at 14:18

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