Timeline for Cycle classes that are killed by pushing forward from normalization
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2014 at 8:08 | comment | added | Dan Petersen | @MikhailBondarko Thanks, that's nice! The application I had in mind for this result was a bit half-baked, so I'm sorry to say that I don't personally need this written down. I can't really judge how interesting it is on its own. | |
| Jul 15, 2014 at 8:12 | comment | added | Mikhail Bondarko | I can probably prove that any element of this kernel is the image of an element of $CH_k(X'\times_X X')$ via $p_{1*}-p_{2*}$ where $p_i$ are the projections. Should I write this down? | |
| Oct 19, 2013 at 18:47 | comment | added | Jason Starr | I think that you are correct. Perhaps the example can be salvaged if $C$ is a hyperelliptic curve admitting correspondences "without valence", i.e., if there are "extra" numerical equivalence classes. I will see if I can fix this. | |
| Oct 19, 2013 at 6:12 | comment | added | Dan Petersen | @Jason: I tried computing $\mathrm{CH}_1^{\mathrm{num}}(Y')$ in your example and it seems to me that $[\Delta]+[\Delta']$ is numerically equivalent to $2[F_1] + 2[F_2]$. So I don't think it has rank four. | |
| Oct 18, 2013 at 15:30 | comment | added | Dan Petersen | @Jason: Thanks a lot! I will try to understand your example. | |
| Oct 18, 2013 at 15:10 | comment | added | Jason Starr | (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., $[\Delta]-[F_1]-[F_2]$. | |
| Oct 18, 2013 at 15:07 | comment | added | Jason Starr | (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 $i$, $[\Delta']$. | |
| Oct 18, 2013 at 14:58 | comment | added | Jason Starr | 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 $Y$. | |
| Oct 18, 2013 at 14:43 | comment | added | Jason Starr | 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. | |
| Oct 18, 2013 at 13:51 | history | asked | Dan Petersen | CC BY-SA 3.0 |