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The Möbius function of a product poset is the product of the Möbius functions

Statement

Let $(P,\le_P)$ and $(Q,\le_Q)$ be locally finite posets. Define the product order on $P\times Q$ by

$(p,q)\le(p',q')\quad\Longleftrightarrow\quad p\le_Pp'\text{ and }q\le_Qq'.$

This relation is a partial order, the product poset is locally finite, and for comparable pairs

$\mu_{P\times Q}\big((p,q),(p',q')\big)=\mu_P(p,p')\mu_Q(q,q').$

Facts & Assumptions

Given: Locally finite posets $P,Q$ and elements $p\le_Pp'$ , $q\le_Qq'$ .

[F1]

A partial order is reflexive, antisymmetric and transitive (Partial order and partially ordered set).

[F2]

Local finiteness means every closed interval is finite (Intervals in a poset; locally finite, lower-finite and upper-finite posets).

[L1]

A Cartesian product of finite sets is finite (The product rule: $\lvert A \times B\rvert = \lvert A\rvert\,\lvert B\rvert$ , and $\big\lvert\prod_{i<m} A_i\big\rvert = \prod_{i<m}\lvert A_i\rvert$ ).

[L2]

Finite Fubini interchanges a sum over a finite Cartesian product with its two iterated sums (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[L3]

The Möbius function is the unique integer-valued function with diagonal value $1$ and vanishing interval sums off the diagonal (The Möbius recurrence: $\mu_P(x,x)=1$ and both interval sums of $\mu_P$ vanish when $x<y$ , The integers form a commutative ring).

Proof

technique · direct

1.1

The product relation is reflexive because both coordinate orders are reflexive; it is antisymmetric because two opposite product inequalities give equality in each coordinate; and it is transitive because coordinatewise inequalities compose. Hence it is a partial order.

1.2

Its intervals are exactly Cartesian products: $[(p,q),(p',q')]=[p,p']_P\times[q,q']_Q$ . Both factors are finite by local finiteness, so the interval is finite by [L1]; thus $P\times Q$ is locally finite.

F2L1

1.3

Define $\nu((p,q),(p',q')):=\mu_P(p,p')\mu_Q(q,q')$ . On the diagonal, $\nu((p,q),(p,q))=1\cdot1=1$ .

2.1

For a nontrivial product interval, finite Fubini gives $\sum_{(p,q)\le(u,v)\le(p',q')}\nu((p,q),(u,v))=\left(\sum_{p\le_Pu\le_Pp'}\mu_P(p,u)\right)\left(\sum_{q\le_Qv\le_Qq'}\mu_Q(q,v)\right)$ .

step 1.2L2L3

3.1

Each factor in step 2.1 is $1$ when its endpoints agree and $0$ otherwise. Since the product interval is nontrivial, at least one coordinate pair has distinct endpoints, so the product is $0$ .

step 2.1L3

4.1

Thus $\nu$ has the diagonal and recurrence properties of the Möbius function on $P\times Q$ , and uniqueness in [L3] gives $\nu=\mu_{P\times Q}$ .

step 1.3step 3.1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

F. Gotti, Incidence Algebras, MIT 18.211 notes (standard reference, not scraped)
R. Stanley, Enumerative Combinatorics, Volume 1, §§3.6–3.8 (standard reference, not scraped)