2
$\begingroup$

Let $X$ be an algebraic variety over an algebraically closed field $k$ of characteristic $0$. A locally trivial $\mathbf{A}^n$-fibration is a morphism $\pi \colon Y \to X$ such that $\pi^{-1}(U)\cong U\times \mathbf{A}^n$, and that $\pi\colon U\times \mathbf{A}^n \to U$ is the projection onto the second factor, for every open set $U\subseteq X$ in some Zariski open covering of $X$.

It is proved in Fulton's Intersection Theory that any such $\pi$ induces surjections on the Chow groups via pull-back $\pi^*\colon A^*(X) \to A^*(Y)$, and that this map is an isomorphism if $\pi$ admits a structure of vector bundle.

I would like to know whether $\pi^*$ is an isomorphism in general or not.

I suspect it to be so: Recall that $M^c(X)$ is the motive with compact support of $X$ in the sense of Voevodsky. He proved that there are

  1. Localisation sequences for any closed subscheme $Z\subset X$ and its complement $U$ $$M^c(Z) \to M^c(X) \to M^c(U) \to, $$
  2. Pull-back morphisms for any flat morphism $f\colon Y\to X$ of equidimension $n$ $$M^c(X)(n)[2n]\to M^c(Y),$$
  3. Isomorphisms $M^c(X)(n)[2n]\cong M^c(X\times \mathbf{A}^n)$, and
  4. Comparisons $\mathrm{Hom}_{\mathbf{DM}^-}(\mathbf{Z}(k)[2k], M^c(X))\cong A^{\dim X-k}(X)$.

It appears to me that one can apply 1., 2., and 3. to obtain an isomorphism $M^c(X)(n)[2n]\to M^c(Y)$ for the locally trivial $\mathbf{A}^n$-fibration $\pi$, and then use 4. to conclude the isomorphisms $$A^{k}(X)\cong\mathrm{Hom}_{\mathbf{DM}^-}(\mathbf{Z}(\dim X-k)[2(\dim X-k)], M^c(X))\cong\mathrm{Hom}_{\mathbf{DM}^-}(\mathbf{Z}(\dim Y-k)[2(\dim Y-k)], M^c(X)(n)[2n])\cong\mathrm{Hom}_{\mathbf{DM}^-}(\mathbf{Z}(\dim Y-k)[2(\dim Y-k)], M^c(Y))\cong A^{k}(Y).$$ Is anything wrong with the above arguments?

$\endgroup$
  • $\begingroup$ Singularities?? $\endgroup$ – Konrad Voelkel May 11 '15 at 12:16
  • 1
    $\begingroup$ I didn't find any assumption on the singularities imposed in each of the statements 1., 2., 3., 4. in the literature, either in Mozza-Voevodsky-Weibel, or in Voevodsky-Suslin-Friedlander. Only the resolution of singularity is assumed for the ground field. $\endgroup$ – Wille Liou May 11 '15 at 15:02
  • $\begingroup$ I would prefer to write the right hand side of (4) as $A_k(X)$ (since $X$ does not have to be equi-dimensional); everything else seems to be fine. $\endgroup$ – Mikhail Bondarko May 12 '15 at 18:04
  • $\begingroup$ Oh yeah, I should have written in that way. Thanks. $\endgroup$ – Wille Liou May 12 '15 at 23:07
1
$\begingroup$

Not sure if your argument is correct, but the statement is Lemma 2.2 in Totaro's Group cohomology and algebraic cycles.

$\endgroup$
  • $\begingroup$ Thanks. I just want to make sure whether I use the results correctly, as I haven't gone through all the materials of motivic cohomology. $\endgroup$ – Wille Liou May 11 '15 at 14:55

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.