False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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False: convolution defines an incidence algebra for every poset

Statement

For every poset $P$ and every commutative ring $R$ , the formula

$(f * g) (x, y) = \sum_{x \leq z \leq y} f (x, z) g (z, y)$

defines a convolution operation on all functions on comparable pairs.

Facts & Assumptions

Given: A countably infinite set $M$ , two further elements $⊥, ⊤$ , a nonzero commutative ring $R$ , and the poset $P : = {⊥, ⊤} \cup M$ in which $⊥ < m < ⊤$ for every $m \in M$ and distinct elements of $M$ are incomparable.

[F1]

Incidence convolution is defined by a finite commutative-monoid sum over $[x, y]$ , under the hypothesis that the poset is locally finite (The incidence functions $I (P, R)$ of a locally finite poset and their convolution).

[F2]

A partial order is reflexive, antisymmetric and transitive, and local finiteness means that every interval $[x, y]$ is finite (Partial order and partially ordered set, Intervals in a poset; locally finite, lower-finite and upper-finite posets).

Refutation

technique · direct

1.1

The displayed relation on $P$ is reflexive, antisymmetric and transitive: the only strict comparisons are from $⊥$ to a middle element, from a middle element to $⊤$ , and from $⊥$ to $⊤$ . Thus $P$ is a poset.

givenF2

1.2

Its interval $[⊥, ⊤]$ is all of $P$ and contains the countably infinite set $M$ , so it is infinite and $P$ is not locally finite.

givenF2

2.1

Let $f$ and $g$ be the constant-one functions on comparable pairs. At the endpoint, the proposed convolution asks for $\sum_{z \in P} 1_{R}$ , an infinite sum that is not supplied by the additive group or ring axioms and is not the finite sum in [F1].

step 1.2F1

3.1

Therefore the formula does not define convolution for every poset; local finiteness is a genuine well-definedness hypothesis.

step 1.2step 2.1∎

Depends on

Used by

A poset with a bottom, a top and countably many incomparable middle elements has an infinite interval, so convolution of constant-one functions is not defined Counterexample

Dependency tree · two levels

10 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

F. Gotti, Incidence Algebras, MIT 18.211 notes (standard reference, not scraped)
Y. Guan and Y. Zhang, Additive Biderivations of Incidence Algebras, §2.1 (standard reference, not scraped)