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The ordinary Chow Chern map on isomorphism classes

Statement

Assume AC. For a smooth equidimensional finite-type scheme $X$ over a field and $r \geq 1$ , write ${C H}^{i} (X) = A^{i} (X)$ for the group of codimension- $i$ cycles modulo rational equivalence. The tuple map $C_{r} : {V e c t}_{r} (X) \to \prod_{i = 1}^{r} {C H}^{i} (X)$ sends $[E]_{i s o}$ to $(c_{i} (E))_{i}$ . It is well-defined and uses the quotient-projective-bundle convention: if $ξ = c_{1} (O (1))$ , its coefficients satisfy $ξ^{r} + \sum_{i = 1}^{r} (- 1)^{i} π^{*} c_{i} (E) ξ^{r - i} = 0$ . Equality of total Chern polynomials $1 + \sum c_{i} (E) t^{i}$ is exactly equality of these tuples.

Facts & Assumptions

Given: The data and hypotheses in the statement.

Goal: Establish the stated conclusion.

[F1]

Bundle isomorphism classes are a set. Isomorphism classes of vector bundles

[F2]

The Chow Chern classes are the unique coefficients of the quotient-projective-bundle relation and lie in $A^{i} (X)$ . Chern classes of a vector bundle on a smooth scheme

Proof

1.1F1F2

An isomorphism $E ≅ F$ identifies their quotient projective bundles, projections and tautological quotient lines. The projective relations therefore identify, and coefficient uniqueness gives $c_{i} (E) = c_{i} (F)$ . The tuple consequently descends to the set in [F1].

2.1F2step 1.1∎

For a formal variable $t$ , a polynomial is its finite list of coefficients. Thus equality of the total polynomials is equivalent to equality in each degree $1, \dots, r$ , precisely equality of $C_{r}$ . This fixes the actual operation and its sign normalization.

Depends on

Used by · 5

Dependency tree · 3 levels

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