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Prime-power cyclic covers of a slice-disk exterior over the defining field

Statement

Assume AC. Let $D \subset B^{4}$ be a smooth proper disk, let $p$ be prime, $q = p^{k}$ with $k \geq 1$ , and let $F_{p}$ be the field with $p$ elements. For the connected $q$ -fold cyclic cover $X_{D, q}$ of the disk exterior, $H_{0} (X_{D, q}; F_{p}) ≅ F_{p}$ , $H_{1} (X_{D, q}; F_{p}) ≅ F_{p}$ , and $H_{i} (X_{D, q}; F_{p}) = 0$ for every $i \geq 2$ .

Statement references

The deck action and finite cyclic quotients

The disk exterior has the homology of a circle

Charles Livingston and Swatee Naik, Introduction to Knot Concordance, §7.3, pp. 133–134 and §7.8, pp. 142–145

Axiom of Choice

Facts & Assumptions

Given: Full AC, a smooth proper disk $D$ , its exterior $X_{D}$ , the infinite cyclic cover $X_{D, \infty}$ with deck generator $t$ , and $q = p^{k}$ .

[A1]

Full AC is assumed wherever the cover and orientation conventions use the choice axiom. The Axiom of Choice

[F1]

$H_{0} (X_{D}; F_{p}) = F_{p}$ , $H_{1} (X_{D}; F_{p}) = F_{p}$ generated by the meridian, and $H_{i} (X_{D}; F_{p}) = 0$ for $i \geq 2$ . This follows from the integral homology of the exterior and the universal coefficient theorem for homology over a PID. The exterior of a smooth proper disk has the homology of a circle The universal coefficient theorem for homology over a PID

[F2]

The infinite cyclic cover and its finite quotient are those defined by the positive-meridian deck action in Infinite and finite cyclic covers of a slice-disk exterior. Their chain groups give short exact sequences $0 \to C_{*} (X_{D, \infty}; F_{p}) \overset{t - 1}{\to} C_{*} (X_{D, \infty}; F_{p}) \to C_{*} (X_{D}; F_{p}) \to 0$ and $0 \to C_{*} (X_{D, \infty}; F_{p}) \overset{t^{q} - 1}{\to} C_{*} (X_{D, \infty}; F_{p}) \to C_{*} (X_{D, q}; F_{p}) \to 0.$ Each singular simplex has a free orbit of lifts; quotienting its orbit module by $t - 1$ or $t^{q} - 1$ gives the singular chains of the corresponding quotient cover. The long exact homology sequences are supplied by The long exact sequence in homology.

Proof

1.1F1F2givenalgebraA1

The deck transformation acts trivially on $H_{0} (X_{D, \infty}; F_{p}) ≅ F_{p}$ because the cover is connected. From the first sequence in [F2], its homology exact sequence and [F1] show that $t - 1$ is an isomorphism on $H_{i} (X_{D, \infty}; F_{p})$ for every $i \geq 1$ : for $i \geq 2$ both neighboring homology groups of $X_{D}$ vanish, while in degree one exactness uses $H_{2} (X_{D}) = 0$ , $H_{1} (X_{D}) = F_{p}$ , and the zero map $t - 1$ on $H_{0} (X_{D, \infty})$ . Thus the degree-one quotient and kernel both vanish.

2.1F1F2step 1.1algebra

In characteristic $p$ , repeated Frobenius gives $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ . Therefore $t^{q} - 1$ is an isomorphism on every positive-degree homology group of $X_{D, \infty}$ and is zero on its degree-zero homology. The long exact sequence of the second sequence in [F2] now gives $H_{1} (X_{D, q}; F_{p}) ≅ H_{0} (X_{D, \infty}; F_{p}) ≅ F_{p}$ , while the quotient cover is connected and hence $H_{0} (X_{D, q}; F_{p}) ≅ F_{p}$ .

3.1F2step 1.1step 2.1algebra∎

For each $i \geq 2$ , both the cokernel on $H_{i} (X_{D, \infty})$ and the kernel on $H_{i - 1} (X_{D, \infty})$ of $t^{q} - 1$ vanish by step 2.1 and the isomorphisms of step 1.1. Exactness therefore gives $H_{i} (X_{D, q}; F_{p}) = 0$ . This includes all higher degrees and completes the calculation.

Depends on

Used by

A prime-power branched cover of a smooth proper disk is a rational homology ball Lemma

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Charles Livingston and Swatee Naik, Introduction to Knot Concordance (standard reference, not scraped)