# A "boundary map" for the algebraic equivalence relation of cycles

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 $$s,t\in C$$?

• Have you tried the case $X=C$, $p=0$ and $W$, the diagonal in $C\times C$? Nov 18, 2018 at 18:59
• I'm trying but it seems quite difficult. Nov 18, 2018 at 20:18
• 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 $\mathbb{P}^1$. Nov 19, 2018 at 9:13

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 $$s,t\in C$$.
• Why should $\partial_C$ be compatible with the hyperelliptic involution? Nov 19, 2018 at 9:20