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PTFA finite projective base change and Euler characteristic

Statement

Let $k$ be a commutative field, let $G$ be PTFA, put $A = k [G]$ , and supply its Ore fraction division ring $K$ . Subscripts $k$ and $K$ mean tensor base change along augmentation $A \to k$ and inclusion $A \to K$ , respectively.

For every finitely generated projective right- $A$ module $P$ , $\dim_{K} P_{K} = \dim_{k} P_{k} < \infty .$ For a homomorphism $f : P \to Q$ between such modules, $\dim_{K} im f_{K} \geq \dim_{k} im f_{k} .$ Consequently, for any bounded complex $P_{*}$ of finitely generated projective right- $A$ modules, $\sum_{i} (- 1)^{i} \dim_{K} H_{i} (P_{K}) = \sum_{i} (- 1)^{i} \dim_{k} H_{i} (P_{k}) .$ If additionally the complex is nonnegative, $\dim_{k} H_{0} (P_{k}) = 1$ and $H_{0} (P_{K}) = 0$ , then $\dim_{K} H_{1} (P_{K}) \leq \dim_{k} H_{1} (P_{k}) - 1.$ No choice principle or free basis for a projective module is assumed.

Facts & Assumptions

Given: The field, group, fraction division ring and the finite projective modules or bounded complex in the relevant assertion.

[F1]

PTFA augmentation detects projective injectivity, and finite matrices over $A$ have $K$ -rank at least their augmented $k$ -rank, without a choice principle. PTFA augmentation detects projective injectivity without choice

[F2]

Projectivity is the individual lifting property for surjections, with right modules read over the opposite ring. Tensor base change preserves compositions, sums and split identities. Projective modules and the lifting property, The opposite ring $R^{o p}$ , The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums

Proof

1.1F1F2givenconstructalgebra

Choose a finite generating list for $P$ , giving a surjection $π : A^{n} \to P$ . Projectivity gives one section $s : P \to A^{n}$ . The matrix $E = s π$ is idempotent, and after either base change $P$ identifies with the image of that idempotent. Put $r = {rank}_{k} E_{k}$ . By [F1], ${rank}_{K} E_{K} \geq r$ and ${rank}_{K} (1 - E)_{K} \geq n - r$ . Idempotency gives the actual complementary decomposition $K^{n} = im E_{K} \oplus im (1 - E)_{K}$ : every vector is the sum of its two displayed images, and their intersection is zero after applying $E_{K}$ . Their dimensions therefore add to $n$ , forcing equality in both inequalities. Hence $\dim_{K} P_{K} = r = \dim_{k} P_{k}$ . Only one finite generating list and one projectivity section were used.

2.1F1F2step 1.1constructalgebra

Take finite split surjections $π_{P} : A^{m} \to P$ and $π_{Q} : A^{n} \to Q$ , with sections $s_{P}, s_{Q}$ . The finite matrix $T = s_{Q} f π_{P}$ has, after either base change, the same image dimension as $f$ : the map $π_{P}$ remains surjective and $s_{Q}$ remains injective by their split identities. Applying [F1] to $T$ yields $\dim_{K} im f_{K} \geq \dim_{k} im f_{k}$ . This argument does not treat $P$ or $Q$ as free.

3.1step 1.1step 2.1constructalgebra

For a bounded complex after either base change all chain groups are finite-dimensional. If $d_{i} : P_{i} \to P_{i - 1}$ is its differential, finite-dimensional rank-nullity and cycles modulo boundaries give $\dim (P_{i})_{*} = rank (d_{i})_{*} + \dim H_{i} (P_{*}) + rank (d_{i + 1})_{*},$ where $*$ is $k$ or $K$ . Summing with signs over the finite range cancels every differential rank, so the alternating homology dimension equals the alternating chain dimension. The latter sums are equal under the two base changes by step 1.1. This proves the asserted Euler equality, with no boundedness or generation conclusion inferred for an unrelated singular complex.

4.1F2step 1.1step 2.1step 3.1algebra∎

In the nonnegative case, $d_{0} = 0$ . Write $n_{i} = \dim_{k} (P_{i})_{k} = \dim_{K} (P_{i})_{K}$ . The degree-zero hypotheses give $rank (d_{1})_{k} = n_{0} - 1$ and $rank (d_{1})_{K} = n_{0}$ . Step 2.1 gives $rank (d_{2})_{K} \geq rank (d_{2})_{k}$ . Therefore $\dim_{K} H_{1} (P_{K}) = n_{1} - n_{0} - rank (d_{2})_{K} \leq n_{1} - n_{0} - rank (d_{2})_{k} = \dim_{k} H_{1} (P_{k}) - 1.$ Every section and generating set needed for the bounded complex belongs to one finite list; finite elimination over $k$ and $K$ requires no choice axiom. This proves all assertions in their stated projective scope.

Depends on

Used by

Dependency tree · two levels

15 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

Cochran, Orr and Teichner, Knot concordance, Whitney towers and L2-signatures (standard reference, not scraped)