In what follows by projective variety I will mean the zero locus of homegeneous polynomials in some projective space. Anyway, feel free to deal with a sufficiently good scheme if you want.

Let $X$ be a projective variety. Denote by $\mathcal{Z}_{p}( X)$ the group of the $p$-algebraic cycles of $X$. We say that a $p$-cycle $\gamma\in\mathcal{Z}_p(X) $ is algebraically equivalent to zero if there exist a non-singular irreducible projective curve $C$, two points $s,t\in C$ and a finite number of $(p+1)$-prime cycles (irriducible subavrieties of $C\times X$) $V_i\in \mathcal{Z}_{p+1}(C\times X)$ such that $$\gamma=\sum_i[V_i(s)]-[V_i(t)],$$ where by $[V_i(c)]$ we mean the cycle associated to the fiber of the restriction to $V_i$ of the projection $C\times X\longrightarrow C$. I am trying to understand if this definition can be given by mean of a "boundary map". Just to clarify what I mean let me consider the case $C=\mathbb P^1$ (rational equivalence). In this case we can introduce the boundary map $$\partial\colon \mathcal{Z}_{p+1}(\mathbb P^1\times X)\longrightarrow \mathcal{Z}_p(X)$$ defined as follows. If $W\subseteq \mathbb P^1\times X$ is a irreducible projective variety whose image via the projection $\mathbb P^1\times X\longrightarrow \mathbb P^1$ is the whole $\mathbb P^1$ then $$\partial W=[W(s)]-[W(t)],$$ otherwise $\partial W=0$. Now the map $\partial$ extends by linearity on all of $\mathcal{Z}_{p+1}(\mathbb P^1\times X)$. A cycle $\gamma \in\mathcal{Z}_p(X)$ is said to be rationally equivalent to the zero if it lies in the image of $\partial$.

In this case the image of $\partial$ does not depend on the choice of the points $s,t\in\mathbb P^1$. Indeed, for any pair of points $s',t'\in\mathbb P^1$ we can find an automorphism $\phi$ of $\mathbb P^1$ taking $s'$ to $s$ and $t'$ to $t$ in such a way that $[W(s')]-[W(t')]=[f(W(s))]-[f(W(t))]$, where $f= \phi\times id\colon \mathbb P^1\times X\longrightarrow \mathbb P^1\times X$.

Question: Is it possible to introcuce a boundary map $\partial_C$ for any non-singular irreducible projective curve $C$ in such a way that the image of $\partial_C $ does not depend on the choice $s,t\in C$?

  • $\begingroup$ Have you tried the case $X=C$, $p=0$ and $W$, the diagonal in $C\times C$? $\endgroup$
    – Mohan
    Nov 18, 2018 at 18:59
  • $\begingroup$ I'm trying but it seems quite difficult. $\endgroup$ Nov 18, 2018 at 20:18
  • 1
    $\begingroup$ Well, a general curve of genus at least $3$ has trivial automorphism group, so it seems quite difficult to replicate in any way the argument you used for $\mathbb{P}^1$. $\endgroup$ Nov 19, 2018 at 9:13

1 Answer 1


If I understand your question correctly the answer is no. For example, let $C$ be a hyperelliptic curve and let $s,t\in C$ be two of the ramification points of the hyperelliptic map $\pi\colon C \rightarrow \mathbb{P}^1$. Then if $\alpha\in Z_p(X)$ is in the image of $\partial_C$ it follows that $2\alpha$ is in the image of $\partial_{\mathbb{P}^1}$.

Consider the case $X=C$ and $p=0$. If a divisor $D=a_1p_1 +\cdots a_n p_n$ is in the image of $\partial_C$ then $2D$ is linearly equivalent to 0.

If instead you take general points $s,t\in C$ then the divisor $D=s-t$ is obviously in the image of $\partial_C$ for this new equivalence relation. Choosing $s,t$ whose difference is not a 2-torsion divisor (always possible over an algebraicly closed field when the genus of $C$ is greater than 1), we see that the definition of $\partial_C$ depends on the choice of $s,t\in C$.

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    $\begingroup$ Why should $\partial_C$ be compatible with the hyperelliptic involution? $\endgroup$ Nov 19, 2018 at 9:20

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