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Let $X$ be a smooth scheme (maybe some other condition needed) then the first Chern class map gives an isomorphism between the equivalent classes of line bundle and the linear equivalent classes of divisors: $$ c_1:{\rm Pic}(X)\to {\rm CH^1}(X) $$

Is there an analog of this for higher dimension? Is it true that the Chern class map gives isomorphism between the equivalent classes of vector bundle and the rational equivalent classes of algebraic cycles?

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  • $\begingroup$ If I am right, $CH^1(X)$ is the class group of Weil divisors, so it is not isomorphic to the Picard group unless the scheme is a smooth variety (probably integral and normal suffices). This is contained, e.g., in Hartshorne II.6. I do not think there is any analogue in higher dimension, you should replace the Picard group with a moduli space of sheaves which is not even of finite type in general. But there may be something I am not aware of... $\endgroup$ – Cla Mar 15 '18 at 11:52
  • $\begingroup$ By the way, what do you mean by equivalence relation of vector bundles? $\endgroup$ – Cla Mar 15 '18 at 11:53
  • $\begingroup$ @Cla Yeah I mean smooth scheme... Sorry about missing this important condition. For the equivalence I mean isomorphic as sheaves. $\endgroup$ – Akatsuki Mar 15 '18 at 16:23
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As mentioned in the comments, works well only for smooth varieties. If $X$ is smooth, one has a filtration $\{F^rK_0(X)\}$ on $K^0(X)$ given by sheaves of support of codimension at least $r$. Then, you have the Chern class map $c_r:F^r/F^{r+1}\to CH^r(X)$ and the natural surjective map $\phi_r$ in the other direction. The composite (in both directions) is multiplication by $(r-1)!$ (why you get the isomorphism above for Picard group). This is called Riemann-Roch without denominators. So, if you tensor with $\mathbb{Q}$, $c_r$ is an isomorphism.

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  • $\begingroup$ Very interesting Mohan, thank you. Which is a reference for this? $\endgroup$ – Cla Mar 15 '18 at 14:35
  • $\begingroup$ @Cla Many places including Grothendieck as well as Futon's Intersection Theory. $\endgroup$ – Mohan Mar 15 '18 at 16:05
  • $\begingroup$ @Mohan Could you explain what is the definition of $K_0(X)$ (I guess this is the isomorphic class of vector bundle?) and $K^0(X)$? $\endgroup$ – Akatsuki Mar 16 '18 at 16:29
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    $\begingroup$ @Akatsuki No, it is the Grothendieck group. Many non-isomorphic vector bundles can give the same element in $K^0$. Isomorphism classes are described by `moduli spaces', which are in general quite ill-behaved. $\endgroup$ – Mohan Mar 16 '18 at 17:08
  • $\begingroup$ @Mohan Okay... And what is the difference between $K_0$ and $K^0$? $\endgroup$ – Akatsuki Mar 16 '18 at 18:36

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