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The low-dimensional quotient follows from the stage sequence

PDF p. 11: Proposition 5.5; pp. 11–12

Statement

Assume the Axiom of Choice and the fixed site and universe bounds of the supplied constructions. Let $k$ be algebraically closed and $X$ a normal finite-type $k$ -scheme. If $d = \dim X \leq 3$ , the actual projection induces an isomorphism $F_{m o t}^{2} K_{0} (X) / F_{m o t}^{3} K_{0} (X) \overset{{\overline{p}}_{2}}{\to} H_{m o t}^{4, 2} (X, Z), {\overline{p}}_{2} = - {\overline{c}}_{2} .$ If $d \leq 2$ , the third image subgroup itself is zero.

Facts & Assumptions

[A1]

Assume the Axiom of Choice as formulated in The Axiom of Choice and the stated size bounds. Choice and the existing size bounds are inherited from the stage-vanishing and stage-to-image premises in [F2] at steps 1.1–2.1 and the perfect-image comparison in [F3] at step 2.1.

[F1]

Stage projections, kernels and image groups gives the exact sequence ending in $H_{m o t}^{4, 2} (X, Z) \to π_{- 1} {Fil}^{3} K (X)$ .

[F2]

High filtration vanishing says $π_{j} {Fil}^{i} K = 0$ for $j < 0$ and $i \geq d$ , and $π_{0} {Fil}^{i} K = 0$ for $i > d$ . Graded filtration projections identifies the second stage with its image under the present normal hypotheses.

[F3]

The perfect image filtration has the stated second projection and kernel identifies the actual transported map with $- c_{2}$ and its kernel with the third image subgroup under the stated Choice and site/universe size hypotheses.

Proof

Given: $X$ in the stated normal finite-type algebraically closed scope, of dimension at most three, with Choice and the fixed site and universe bounds as in [A1].

Goal: Obtain the quotient without a later surjectivity consumer.

1.1A1F1F2

Since $3 \geq d$ , [F2] makes the final obstruction group $π_{- 1} {Fil}^{3} K (X)$ zero. Exactness of [F1] makes the second-stage projection surjective. Transfer this assertion to $F^{2}$ using its stage-image isomorphism in [F2].

2.1A1F2F3step 1.1∎

Its kernel is $F^{3}$ and its value is $- c_{2}$ by [F3]. The first isomorphism theorem proves the displayed quotient identification with that exact sign. If $d \leq 2$ , then $3 > d$ and [F2] makes the third stage's $π_{0}$ , hence its image subgroup, zero. No later second-Chern-surjectivity proposition is used.

Depends on

Used by · 1

Second projection is minus second Chern class

Dependency tree · 3 levels

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