Let $X$ be a non-normal algebraic variety and $f \colon X' \to X$ its normalization. Is there a general description $\mathrm{ker}\left(\mathrm{CH}_k(X') \to \mathrm{CH}_k(X)\right)$? Are there hypotheses that ensure that it is generated by cycles $[Z]$ such that $\dim f(Z) < \dim Z$, and differences $[Z]-[W]$ where $f(Z)=f(W)$?
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asked Oct 18, 2013 at 13:51
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3$\begingroup$ There are no cycles $Z$ such that $\text{dim} f(Z)$ is strictly less than $\text{dim}(Z)$, since $f$ is finite. I believe there are examples where the kernel is not generated only by $[Z]-[W]$ with $f(Z)=f(W)$. I will try to come up with a specific example. $\endgroup$– Jason StarrCommented Oct 18, 2013 at 14:43
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$\begingroup$ I think this gives a counterexample. I am posting as a comment, because it does not address the actual question: "hypotheses that ensure it is generated by ...". Let $C$ be a very general hyperelliptic curve of genus $g>1$. Let $i:C\to C$ be the hyperelliptic involution. Let $f:C\to \mathbb{P}^1$ be the quotient. Let $Y'$ be $C\times C$ with $\mathbb{Z}/2\times \mathbb{Z}/2$ acting via $(i,\text{Id})$ and $(\text{Id},i)$. Let $g:Y'\to Y$ be the quotient, $\mathbb{P}^1\times \mathbb{P}^1$. Let $X'$ be $\mathbb{P}^1\times Y'$, and let $f:X'\to X$ contract $\{0\}\times Y'$ to $Y$. $\endgroup$– Jason StarrCommented Oct 18, 2013 at 14:58
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$\begingroup$ (cont.) Now consider $1$-cycles in $X'$. The kernel will basically be the kernel of the pushforward map of $1$-cycles $g_*:\text{CH}_1(Y')\to \text{CH}_1(Y)$. If $g(Z)$ equals $g(W)$, then $W$ is a translate of $Z$ by an element $g$ in the group. Thus $[W]-[Z]$ is $g^*[Z]-[Z]$. If the kernel was generated by such elements, the same would be true after passing to numerical equivalence classes. But I believe this already fails there. The group of numerical equivalence classes is generated by the two fibers, $[F_1]$ and $[F_2]$, the diagonal $[\Delta]$, and the graph of $i$, $[\Delta']$. $\endgroup$– Jason StarrCommented Oct 18, 2013 at 15:07
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$\begingroup$ (cont. 2) Now $[F_1]$ and $[F_2]$ are invariant under the group action, and $[\Delta]$ is permuted with $[\Delta']$. So every class $[W]-[Z]$ is numerically equivalent to a multiple of $[\Delta]-[\Delta']$. Unfortunately, the quotient of the numerical group by this cyclic subgroup has rank $3$, whereas the rankof the numerical group of $Y=\mathbb{P}^1\times \mathbb{P}^1$ is only $2$. So the kernel does contain more elements, e.g., $[\Delta]-[F_1]-[F_2]$. $\endgroup$– Jason StarrCommented Oct 18, 2013 at 15:10
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$\begingroup$ @Jason: Thanks a lot! I will try to understand your example. $\endgroup$– Dan PetersenCommented Oct 18, 2013 at 15:30
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