2
$\begingroup$

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 $s,t\in C$?

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

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 $s,t\in C$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Why should $\partial_C$ be compatible with the hyperelliptic involution? $\endgroup$ – Francesco Polizzi Nov 19 '18 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.