7
$\begingroup$

In algebraic geometry there is a correspondence between Weil divisors, Cartier divisors and line bundles. Over an integral separated locally factorial noetherian scheme, the group of isomorphism classes of Weil divisors and the group of isomorphism classes of Cartier divisors are isomorphic. Moreover, in a general scheme, there is a bijection between Cartier divisors and invertible subsheaves of the sheaf of total quotient rings.

In my (admittedly little) experience, it seems that in "nice" situations (when dealing with smooth varieties, for example) we prefer to use Weil divisors as it is more geometric and that we resort to Cartier divisors otherwise.

I wonder why don't we always use line bundles (= invertible sheaves). Let me explain!

  • Contrarily to Cartier divisors, line bundles are indeed very geometric objects. Perhaps as geometric as Weil divisors.
  • As with Cartier divisors, we don't have to deal with the trouble of using orders to define principal divisors.
  • The pullback works at the level of line bundles, where for Cartier divisors we have to use the "moving lemma" to define the pullback on the Picard group.

What do we lose, or becomes harder, when using line bundles?

$\endgroup$
2
  • 4
    $\begingroup$ I also have a limited experience with the subject, but let me comment on the fact that for instance in the context of Riemann surfaces (which I am more familiar with) the divisor associated to a line bundle provides a quite concrete way of thinking about the line bundle (sheaf of sections = meromorphic functions with prescribed poles and zeros, and all line bundles arise in this way). So I believe in general the two pictures complement each others in nontrivial ways, but I am eager to see some more qualified and more appropriate answer :) $\endgroup$
    – Giulio R
    Mar 19, 2021 at 17:07
  • 2
    $\begingroup$ In practice it is often much easier to describe a Weil divisor than to give a concrete, computationally tractable description of a line bundle (even on a singular variety; normal is frequently enough). Tremendous amounts of effort can then go into things like taking the limit (but as Cartier divisors) of a family of Weil divisors and computing what that actually looks like, for instance in moduli theory. $\endgroup$ Mar 20, 2021 at 0:23

1 Answer 1

5
+150
$\begingroup$

Question: "I wonder why don't we always use line bundles (= invertible sheaves)?"

Answer: If $k$ is a field and $X$ is a regular scheme of finite type over $k$ there is the Chern character isomorphism

$$ch_i: \operatorname{K}_0(X)_{\mathbb{Q}}^{(i)} \cong \operatorname{CH}^i(X)_{\mathbb{Q}}$$

where $\operatorname{K}_0(X)_{\mathbb{Q}}^{(i)}$ is the $i$'th eigenspace of the Grothendieck group of finite rank vector bundles on $X$ with respect to the Adams operations, and $\operatorname{CH}^i(X)_{\mathbb{Q}}$ is the $i$'th Chow group (with rational coefficients). This isomorphism sends the class of a linebundle $[L]$ to its Weil divisor $D(L)$: It induce an isomorphism

$$Pic(X) \cong Cl(X)$$

where $Pic(X)$ is the Picard group of $X$ and $Cl(X)$ is the group of Weil divisors (or Cartier divisors) (HH.Prop.II.6.16). Hence if you work with rational coefficients you may identify the Grothendieck group with the Chow group. The Grothendieck group is defined in terms of the "category of finite rank vector bundles on $X$" and the Chow group is defined in terms of the "set of closed integral subschemes of $X$". Hence when working with the Grothendieck (and higher algebraic K-groups) group your proofs and constructions become "more category theoretical". When working with the Chow group (and higher Chow groups) you work with the "set of closed integral subschemes" of $X$, hence there are "fewer" set theoretical complications. The Chern character exists for higher algebraic K-groups and higher Chow groups

$$ ch^m_i: \operatorname{K}_m(X)_{\mathbb{Q}}^{(i)} \cong \operatorname{CH}^i(X,m)_{\mathbb{Q}}$$

and higher algebraic K-theory is defined in terms of the category $Vect(X)$ of finite rank vector bundles on $X$, and the higher Chow groups are defined in terms of an explicit complex defined in terms of subschemes of $X$. Again algebraic K-theory is "more category theoretical" and Chow theory is "less category theoretical". When $X$ is regular there is an intersection product on the Chow group and this differs from the multiplication on the Grothendieck group. The intersection product is "more geometric", the tensor product is "more general".

Question: "I wonder why don't we always use line bundles (= invertible sheaves)?"

Answer: Algebraic "K-theory is about vector bundles" and "Chow theory is about cycles and subschemes" and in some cases these two theories are equivalent - the equivalence is given by the chern character. But the theories do not exclude each other in general.

Note: There is an exercise in Hartshorne (Ex II.6.11) where they calculate the Grothendieck group of a non-singular curve $C$. They prove using Proposition II.6.16 that there is an isomorphism

$$f:\operatorname{K}_0(C) \cong \mathbb{Z} \oplus Cl(C)$$

where $Cl(C)$ is the group of Weil divisors on $C$. By definition (Exercise HH:II.6.10) the Grothendieck group $\operatorname{K}_0(C)$ is the free abelian group on isomorphism classes $[E]$ where $E$ is a coherent sheaf on $C$, modulo exact sequences. If $E$ is a finite rank locally free sheaf on $C$ it follows

$$f([E]):=(rk(E), D(\wedge^e E))$$

where $e:=rk(E)$ and $D(\wedge^e E)$ is the Weil divisor of $\wedge^e E$. Since $rk(E)=e$ it follows $\wedge^e E$ is a linebundle and you may take its corresponding Weil divisor. This correspondence between algebraic cycles/divisors and vector bundles is a much studied correspondence, and in the case of a curve $C$ it follows

$$\operatorname{CH}^*(C) \cong \mathbb{Z} \oplus Cl(C)$$

is an isomorphism. In this decomposition you should view the $\mathbb{Z}$-summand as $\mathbb{Z}[C]$ where $[C]:=1$ is the multiplicative unit in the Chow group. Similarly for the Grothendieck group: In the direct sum decomposition $\operatorname{K}_0(C)\cong \mathbb{Z}\oplus Pic(C)$ you should view the $\mathbb{Z}$-summand as $\mathbb{Z}[\mathcal{O}_C]$ where $[\mathcal{O}_C]:=1$ is the multiplicative unit. The structure sheaf $\mathcal{O}_C$ has the property that for any class $[E]\in \operatorname{K}_0(C)$ we get

$$ [\mathcal{O}_C][E]:=[\mathcal{O}_C \otimes E]=[E].$$

Hence via the exercise in HH you may view Proposition HH.II.6.16 as a "special case" of the Chern character construction above.

$\endgroup$
1
  • 2
    $\begingroup$ Dear @hm2020, I confess that I perhaps don't understand the main idea behind your answer. You say that algebraic K-theory is about line sheaves (vector bundles in general), that Chow theory is about Weil divisors (cycles and subschemes in general) and that they are equivalent sometimes. Well... this is exactly because line bundles and Weil divisors are equivalent sometimes. Surely this is because I don't understand algebraic K-theory nor Chow theory, but it's not clear to me that this adds something to the discussion. $\endgroup$
    – Gabriel
    Mar 27, 2021 at 13:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .