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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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Incidence convolution is associative and distributes over pointwise addition

Statement

For a locally finite poset PP, a commutative ring RR, and f,g,h∈I(P,R)f,g,h\in I(P,R), incidence convolution satisfies

(f∗g)∗h=f∗(g∗h),(f*g)*h=f*(g*h),

and both distributive laws over pointwise addition.

Facts & Assumptions

Given: A locally finite poset PP, a commutative ring RR, incidence functions f,g,hf,g,h, and a comparable pair x≤yx\le y.

[F1]

(a∗b)(x,y)=∑x≤z≤ya(x,z)b(z,y)(a*b)(x,y)=\sum_{x\le z\le y}a(x,z)b(z,y), and [x,y][x,y] is finite (The incidence functions I(P,R)I(P,R) of a locally finite poset and their convolution).

[L1]

Finite sums in a commutative monoid may be reindexed, split, and interchanged by the finite Fubini rule (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F2]

In a ring, multiplication is associative and distributes over addition on both sides; in a commutative ring the order of factors may also be exchanged (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring).

Proof

technique · direct
1.1

Expanding the left bracketing and distributing the factor h(v,y)h(v,y) through the inner sum gives ((f∗g)∗h)(x,y)=∑x≤v≤y∑x≤u≤vf(x,u)g(u,v)h(v,y)((f*g)*h)(x,y)=\sum_{x\le v\le y}\sum_{x\le u\le v}f(x,u)g(u,v)h(v,y).

F1F2
1.2

Put D:={(u,v)∈[x,y]2:u≤v}D:=\{(u,v)\in[x,y]^2:u\le v\}. Expanding the right bracketing gives (f∗(g∗h))(x,y)=∑x≤u≤y∑u≤v≤yf(x,u)g(u,v)h(v,y)(f*(g*h))(x,y)=\sum_{x\le u\le y}\sum_{u\le v\le y}f(x,u)g(u,v)h(v,y).

F1F2
1.3

For every x≤yx\le y, ((f+g)∗h)(x,y)=∑x≤z≤y(f(x,z)+g(x,z))h(z,y)=∑x≤z≤yf(x,z)h(z,y)+∑x≤z≤yg(x,z)h(z,y)((f+g)*h)(x,y)=\sum_{x\le z\le y}(f(x,z)+g(x,z))h(z,y)=\sum_{x\le z\le y}f(x,z)h(z,y)+\sum_{x\le z\le y}g(x,z)h(z,y) by distributivity in RR and additivity of a finite sum; hence (f+g)∗h=f∗h+g∗h(f+g)*h=f*h+g*h.

F1F2L1
1.4

The same calculation with the sum in the right factor gives f∗(g+h)=f∗g+f∗hf*(g+h)=f*g+f*h.

F1F2L1
2.1

Extend the displayed summand by 0R0_R from DD to [x,y]2[x,y]^2. Splitting each finite inner sum into the admissible indices and the zero terms identifies steps 1.1 and 1.2 with its two iterated sums over [x,y]2[x,y]^2. Finite Fubini makes those iterated sums equal.

step 1.1step 1.2L1
3.1

Since steps 2.1 and 1.2 agree for every comparable (x,y)(x,y), (f∗g)∗h=f∗(g∗h)(f*g)*h=f*(g*h).

step 2.1step 1.2
4.1

Steps 3.1, 1.3 and 1.4 prove associativity and both distributive laws.

step 3.1step 1.3step 1.4∎

Depends on

Used by

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