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A pointwise one-dimensional FI-module that is not finitely generated

Statement refuted

Refuted claim: an FI-module over a field whose values $V_{S}$ are finite-dimensional for every finite set $S$ is finitely generated.

The witness is built below: over a field $k$ , take $V_{S} = k e_{S}$ one-dimensional for every finite set $S$ , with bijections acting as the identity identification of the lines and non-bijections acting as $0$ . Then $\dim_{k} V_{S} = 1$ for all $S$ , but no finite set of elements generates $V$ (Span, finite generation and generation degree of an FI-module).

Facts & Assumptions

Given: A field $k$ and, for every finite set $S$ , the one-dimensional $k$ -vector space $V_{S} : = k e_{S}$ with distinguished nonzero vector $e_{S}$ ; for an injection $f ⁣ : S \to S^{'}$ let $V (f) ⁣ : V_{S} \to V_{S^{'}}$ send $e_{S} \mapsto e_{S^{'}}$ when $f$ is a bijection and $e_{S} \mapsto 0$ otherwise.

[F1]

An FI-module over $k$ is a covariant functor from the category of finite sets and injections to $k$ -modules, so $V ({i d}_{S}) = {i d}_{V_{S}}$ and $V (g \circ f) = V (g) \circ V (f)$ for composable injections (The category FI and FI-modules over a commutative ring).

[F2]

For a set $Σ$ of supported elements $(S, s)$ , ${span}_{V} (Σ)_{T}$ is the $k$ -span of the elements $f_{*} (s)$ with $(S, s) \in Σ$ and $f ⁣ : S \to T$ an injection; $V$ is finitely generated when ${span}_{V} (Σ) = V$ for some finite $Σ$ (Span, finite generation and generation degree of an FI-module).

[F3]

A composite of bijections is a bijection; if $g \circ f$ and $f$ are bijections then $g$ is a bijection; and if $f$ is not a bijection then $g \circ f$ is not a bijection (Injection, surjection, bijection).

[F4]

For every $n \geq 0$ the set $[n] = {0, 1, \dots, n - 1}$ has exactly $n$ elements (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

Counterexample

technique · direct

1.1F1F3givenalgebra

$V$ is a functor: $V ({i d}_{S})$ sends $e_{S}$ to $e_{S}$ because ${i d}_{S}$ is a bijection; and for composable injections $f ⁣ : S \to T$ , $g ⁣ : T \to U$ the two sides of $V (g \circ f) = V (g) V (f)$ agree on $e_{S}$ in each case — if $f$ and $g$ are bijections both sides send $e_{S} \mapsto e_{U}$ ; if $g$ fails to be a bijection then $g \circ f$ fails to be a bijection, so both sides are $0$ ; and if $f$ fails to be a bijection then $g \circ f$ fails to be a bijection and both sides are $0$ . Hence $V$ is an FI-module and $\dim_{k} V_{S} = 1$ for every finite set $S$ .

1.2F3given

Suppose $Σ = {(S_{1}, s_{1}), \dots, (S_{r}, s_{r})}$ is a finite set of supported elements that generates $V$ , with $s_{i} \in V_{S_{i}}$ , and put $M : = \max_{i} ∣ S_{i} ∣$ (if $r = 0$ , then ${span}_{V} (\emptyset) = 0 \neq V$ , so $r \geq 1$ ). For every finite set $T$ with $∣ T ∣ > M$ , every injection $f ⁣ : S_{i} \to T$ satisfies $∣ S_{i} ∣ \leq M < ∣ T ∣$ , so $f$ is not a bijection and $f_{*} (s_{i}) = V (f) (s_{i}) = 0$ .

2.1F2F4step 1.2

By the explicit description of the span, step 1.2 shows ${span}_{V} (Σ)_{T} = 0$ for every finite set $T$ with $∣ T ∣ > M$ . Taking $T = [M + 1]$ , which has $M + 1 > M$ elements, gives ${span}_{V} (Σ)_{T} = 0$ while $V_{T} = k e_{T} \neq 0$ ; hence ${span}_{V} (Σ) \neq V$ and no finite set of elements generates $V$ .

3.1F2step 1.1step 2.1∎

Thus $V$ has $\dim_{k} V_{S} = 1$ for every finite set $S$ by step 1.1, yet is not finitely generated by step 2.1: pointwise finite-dimensionality does not imply finite generation, and the refuted claim fails.

Depends on

Used by

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Dependency tree · two levels

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Sources

Thomas Church, Jordan S. Ellenberg and Benson Farb, “FI-modules and stability for representations of symmetric groups”, arXiv:1204.4533v4 (2014); published in Duke Math. J. 164 (2015) 1833–1910 (standard reference, not scraped)
Thomas Church, Jordan S. Ellenberg, Benson Farb and Rohit Nagpal, “FI-modules over Noetherian rings”, Geom. Topol. 18 (2014) 2951–2984; author-hosted full text (standard reference, not scraped)