For a smooth (complex) projective surface $S$ with non trivial Albanese variety, are there simple conditions for the existence a universal codimension $2$ cycle i.e. a $Z\in \mathrm{CH}^2(Alb(S)\times S)$ such that the induced morphism $Alb(S)\rightarrow Alb(S)$ is the identity?
3
-
$\begingroup$ Do you want sufficient conditions? It is true if $S$ is a product of curves whose Jacobians share no nontrivial factors. $\endgroup$– Jason StarrCommented Jan 30, 2017 at 15:46
-
$\begingroup$ Thank you for your answer and for the example. Yes, I would like sufficient conditions or some references to the study of instances (other than your example) of this property. $\endgroup$– pi_1Commented Jan 30, 2017 at 15:50
-
$\begingroup$ I guess the hypothesis about common factors is superfluous. Sorry about that. $\endgroup$– Jason StarrCommented Jan 30, 2017 at 16:09
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
For rationally connected varieties, which have trivial Albanese, there exists a universal $0$-cycle, but that is not very interesting.
For Abelian varieties, which are their own Albanese, the universal cycle is given by the diagonal cycle.
We know that universal codimension $1$ cycles always exist, which means that for curves there is always a universal $0$-cycle.
The existence of a universal $0$-cycle is stable under products.
Now for a more non-trivial result. Let $S$ be a variety and $i: S \to Alb(S)$ its Albanese embedding, with $d = dim(Alb(S))$. Suppose that there are closed subvarieties $Z_1 \dotsc Z_n$ of $S$ so that the induced map $e: \prod_i Z_i \to Alb(S)$ is birational. For each $Z_i$, there is an obvious cycle $W_i \in Ch^d(S \times Z_i)$ so that the induced map $f_{W_i}: Z_i \to Alb(S)$ is given by the inclusion into $S$ followed by the Albanese embedding. This gives rise to a cycle $W = \oplus_i W_i \in Ch^d(S \times \prod_i Z_i)$ such that $f_W = e$. Let $e^{-1}$ be a birational inverse to $e$, and consider the cycle $(e^{-1})^*(W) \in Ch^d(S \times Alb(S))$. It must give rise to the identity map as required.
In more general situations unfortunately the best you can get is a morphism $\prod_i Z_i \to Alb(s)$ which is generically finite of degree $m$. This can give you cycles which induce an isogeny $Alb(S) \to Alb(S)$ of degree $m$. If there is a group acting on $\prod_i Z_i$ preserving the cycle $\oplus_i W_i$ then you may sometimes be able to pass to the quotient.
This result can be directly applied to the case of curves: the Jacobian of a curve $C$ of genus $g$ is birational to its $g^{th}$ symmetric power, and the obvious cycle in $Ch^d(C \times C^g)$ descends to a cycle in $Ch^d(C \times J(C))$, being invariant under the action of the symmetric group.
Claire Voisin has also shown that there are surfaces with no universal $0$-cycle, namely a surface $S$ related to quartic double solids with at most $7$ nodes.
