Chapter 3
Partially Ordered Sets (Posets)
3.1 Posets
(3.1) De�nition (3.1.1.) A partially ordered set (poset) is a pair (P,≤P )where P is a set and ≤P is a binary relation on P satisfying
(1) for all x ∈ P , x ≤P x (reflexivity);
(2) if x ≤P y and y ≤P x, then x = y (anti-symmetry);
(3) if x ≤P y and y ≤P z, then x ≤P z (transitivity).
Where there is no possibility of confusion, we will write ≤ instead of ≤P . Also
x ≥ y means y ≤P xx < y means x ≤p y and x 6= y
Two elements x, y ∈ P are comparable if x ≤P y or y ≤P x. Otherwise, they arecalled incomparable.
(3.2) Example (3.1.2.)
(1) For n ∈ N let P = {i ∈ N : 1 ≤ i ≤ n}. Given x, y ∈ P , de�ne x ≤P y if x ≤ y.
1 < 2 < 3 < · · · < (n− 1) < n
Each pair of elements is comparable. (Totally ordered set.)
(2) Given n ∈ N, let P = {A : A ⊆ {1, . . . , n}}. For x, y ∈ P , de�ne x ≤P y if x ⊆ y.In this case (P,≤P ) is the poset of all subsets of {1, . . . , n} ordered by inclusion.
It is denoted by Bn.
n = 2 : P = {∅, {1}, {2}, {1, 2}} ∅ ≤P {1} ≤P {1, 2}∅ ≤P {2} ≤P {1, 2}
(3) Given n ∈ N let P = {i ∈ N : i divides n}. For x, y ∈ P de�ne x ≤P y if xdivides y. E.g. if n = 18, then P = {1, 2, 3, 6, 9, 18} and
1 ≤P 1, 1 ≤P 2, 1 ≤P 3, . . . , 1 ≤P 18 6 ≤P 6, 6 ≤P 182 ≤P 2, 2 ≤P 6, 2 ≤P 18 9 ≤P 9, 9 ≤P 183 ≤P 6, 3 ≤P 9, 3 ≤P 18 18 ≤P 18
53
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
Here (P,≤P ) is denoted by Dn.
(4) Given n ∈ N let P = { partitions of {1, . . . , n}}. For x, y ∈ P de�ne x ≤P y ifevery block of x is contained in a block of y. If n = 4, we have x = {13, 2, 4}and y = {13, 24} ∈ P . Since every block of x is contained in a block of y, wehave x ≤P y. We say that x is a refinement of y. Here (P,≤P ) is the poset of
partitions of {1, . . . , n} ordered by re�nement and is denoted by Πn.
(3.3) De�nition There are two types of subposets of (P,≤P ).
� An induced subposet (Q,≤Q) of (P,≤P ) is such that Q ⊆ P and x ≤Q y ⇔x ≤P y. E.g. choose Q = {1, 2, 3} ⊆ {1, 2, 3, . . .} in example (3) above.
� A weak subposet of (P,≤P ) is (Q,≤Q) where Q ⊆ P and if x ≤Q y thenx ≤P y.
When we say subposet, we mean an induced subposet.
For x, y ∈ P where (P,≤) is a poset, and x ≤ y, the interval
[x, y] = {z ∈ P : x ≤ z ≤ y}
is a subposet. The smallest interval is [x, x] = {x}.
(3.4) De�nition If every interval [x, y] of (P,≤P ) is �nite, then P is called a locallyfinite poset.
If Q is a subposet of P , then Q is called convex if y ∈ Q whenever x < y < z in Pand x, z ∈ Q. Interval posets are always convex.
Open interval:
(x, y) = {z ∈ P : x < z < y}
(3.5) De�nition Given x, y ∈ P , we say that y covers x if x < y and there doesnot exist z ∈ P such that x < z < y.
(3.6) Remark A locally �nite poset is completely determined by its cover relations.
(3.7) De�nition TheHasse diagram of a �nite poset P is the graph with verticesx ∈ P and
� if x < y, then y is drawn above x in the diagram;
� if y covers x, then there is an edge between x and y in the diagram.
(3.8) Example If P = {a, b, c, d, e, f} and
a < b, a < c, a < d, b < e, e < f, c < f, d < f
then the Hasse diagram of P is
•f
0000000~~~~
•e•c •d
�������•b @@@@
•a
54
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.9) Example All di�erent (non-isomorphic) posets on 4-elements
• • • ••• •
•
• •• •
•
�����00000•
• •
•00000 •
�����•
•
•• ••
•@@@@ •
• ••
@@@@ •~~~~
• •
• @@@ •~~~••
•• @@@
~~~• •
•~~~~ @@@@
•@@@@ •
~~~~
•
•9999
•9999 •
����•
•���� 9999
• •9999
•
•~~~~ @@@@
• • •• • ••
~~~~@@@@
••••
(3.10) Question Why are•
�����00000
• @@@~~~• •
and• @@@
~~~• @@@ •
~~~•
not in the above diagrams?
(3.11) Example (From example 3.1.2.)
(1) If n = 4 and P = {1, 2, 3, 4}, the Hasse diagram of the total order on P is
•4•3•2•1
(2) B4 = (P,≤) where P = {A : A ⊆ {1, 2, 3, 4}} and elements are ordered byinclusion.
P = {∅, 1, 2, 3, 4, 12, . . . , 134, 1234}The Hasse diagram:
•
ooooooooooooooo
~~~~~~~~~
@@@@@@@@@
OOOOOOOOOOOOOOO1234
•
~~~~~~~~~
UUUUUUUUUUUUUUUUUUUUUU123 •
ooooooooooooooo
UUUUUUUUUUUUUUUUUUUUUU124 •
iiiiiiiiiiiiiiiiiiiiii
ooooooooooooooo
''OOOOOOOOOOOOOOO134 •
~~~~~~~~~
@@@@@@@@@234
•
@@@@@@@@@
OOOOOOOOOOOOOOO12 •13
UUUUUUUUUUUUUUUUUUUUUU •
~~~~~~~~~
UUUUUUUUUUUUUUUUUUUUUU14 •
ooooooooooooooo 23 •
iiiiiiiiiiiiiiiiiiiiii 24 •
ooooooooooooooo
~~~~~~~~~ 34
•
OOOOOOOOOOOOOOO1 •
@@@@@@@@@2 •
~~~~~~~~~ 3 •
ooooooooooooooo 4
•∅
~~~~~~~~~
(3) Consider D12 = (P,≤) where P = {i : i divides 12} = {1, 2, 3, 4, 6, 12}. x ≤ y inP if x divides y.
1 < 2, 3, 4, 6, 12 4 < 122 < 4, 6, 12 6 < 12
3 < 6, 12
Hasse diagram for D12:
•~~~~~
12
@@@@@
•4 •oooooooo 6
•2 @@@@@ •
3~~~~~
•1
~~~~~
55
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(4) Π3 = (P,≤) where P is the set of all partitions of {1, 2, 3},
P = {{123}=x1
, {1, 23}=x2
, {2, 13}=x3
, {3, 12}=x4
, {1, 2, 3}=x5
}
Thenx5 < x1, x2, x3, x4 x2, x3, x4 < x1
Hasse diagram:
{123}
ssssssss
KKKKKKKK
{1, 23}
KKKKKKKK{2, 13} {3, 12}
ssssssss
{1, 2, 3}
(3.12) De�nition Given poset (P,≤), if there exists an element a ∈ P such thata ≤ x for all x ∈ P , then a is a smallest element of the poset P and is denoted 0.Similarly, if there is a largest element a such that x ≤ a for all x ∈ P , then a iswritten 1.
(3.13) De�nition A chain is a poset in which every 2 elements are comparable.
� A subset C of a poset P is a chain if C is a chain when regarded as a subposetof P .
� A chain C in a poset is saturated if it is not amissing any elements within it,i.e. 6 ∃z ∈ P − C such that x < z < y for some x, y ∈ C� In a locally �nite poset, a chain x0 < x1 < · · · < xn is saturated ⇔ xi+1 covers xifor all 0 ≤ i < n.
(3.14) De�nition The length of a �nite chain is `(C) = |C| − 1.The length (rank) of a �nite poset P is
`(P ) = max{`(C) : C chain in P}
Length of the interval [x, y] in P is `(x, y).A chain C in a poset P is maximal if no more elements can be added to it.
(3.15) Example
P : •f@@@~~~•e
~~~ •g@@@•d
~~~ •h@@@
•c•b•a
`(P ) = 5
C1 = {a, b, c, d, e, f} is a maximal chain, `(C1) = 5C2 = {h, g, f} is a maximal chain, `(C2) = 2
56
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.16) De�nition If every maximal chain of P has the same length n, then P is
graded of length n.
Notice that if P is graded, then there is a unique rank function ρ : P → {0, 1, . . . , n}such that
� ρ(x) = 0 if x is a minimal element of P ,
� ρ(y) = ρ(x) + 1 if y covers x in P .
If ρ(x) = i for x ∈ P , then we say element x has rank i.
If x ≤ y, then `(x, y) = ρ(y)− ρ(x).
(3.17) De�nition If P is a graded poset of rank n, then the polynomial
F (P, q) =n∑i=0
piqi,
where pi = # elements of rank i in P , is called the rank generating function
of P .
(3.18) Remark All posets in Example 3.1.2. were graded.
(3.19) Example Let P be the poset with Hasse diagram
rank:
4 •���� <<<<
3 •���� •
<<<<����
2 •<<<<
SSSSSSSSS •���� <<<< •
����
1 •<<<< •
����
0 •
p0 = 1, p1 = 2, p2 = 3, p3 = 2, p4 = 1; F (P, q) = 1 + 2q + 3q2 + 2q3 + q4
(3.20) Example Consider poset P on {1, . . . , n} with 1 < 2 < . . . < n. One maximalchain ⇒ graded. `(C) = n− 1, ρ(x) = x− 1, ρ(P ) = n− 1. ♦
(3.21) Example The poset Bn = (P,≤), where
P = {A : A ⊆ {1, . . . , n}}, x ≤ y ⇔ x ⊆ y
A chain in Bn is a sequence
{A1, . . . , Ak} such that A1 ⊂ A2 ⊂ · · · ⊂ Ak ⊆ {1, . . . , n}
Maximal chain e.g. C:
∅ < {1} < {1, 2} < {1, 2, 3} < . . . < {1, . . . , n}
We see that`(C) = n+ 1− 1 = n
57
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
which applies to all maximal chains. Thus the poset Bn is graded of length n.
Note that if x ∈ P , then ρ(x) = |x| (no. of elements in x), e.g. ρ({1, . . . , n}) = n. Thus
F (Bn, q) =n∑i=0
piqi =
n∑i=0
(n
i
)qi = (1 + q)n
(3.22) Example Dn = (Pn,≤) where Pn = {1 ≤ i ≤ n : i divides n}. If x, y ∈ Pn,then x ≤ y ⇔ x divides y.
For n = 12 we have 12 = 2231 := pα11 pα2
2 ,
D12 : •~~~~~
12=2231
@@@@@
•4=2230 •oooooooo 6=2131
•2=2130
@@@@@ •3=2031
~~~~~
•1=2030
~~~~~
and |D12| = (2 + 1)(1 + 1) = 6.
Claim Poset Dn is graded.
Proof. If n = pα11 pα2
2 · · · pαmm where p1, . . . , pn are primes, then
Pn = {pβ11 · · · p
βmm : 0 ≤ β1 ≤ α1, . . . , 0 ≤ βm ≤ αm}
and |Pn| = (1 + α1)(1 + α2) · · · (1 + αm). If x ∈ Pn and x = pβ11 · · · p
βmm , then ρ(x) =
β1 + · · ·+ βm. In particular ρ(Pn) = α1 + α2 + · · ·+ αm. �
We have
F (Dn, q) =α1+···+αm∑
j=0
pjqj =
α1∑β1=0
· · ·αm∑βm=0
1qβ1+···+βm =
α1∑β1=0
qβ1
· · · αm∑βm=0
qβm
=(qα1+1 − 1q − 1
)· · ·(qαm+1 − 1q − 1
)
(3.23) De�nition A multichain of a poset P is a chain in which elements areallowed to be repeated.
(3.24) Example Poset P with Hasse diagram:
•e
0000000~~~~
•c•d
�������•b @@@@
•a
C = {a, b, b, b, c, c, e} is a multichain. Length of C ′ = {x0 ≤ x1 ≤ · · · ≤ xn} is n. Hence,`(C) = 7− 1 = 6. ♦
58
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.25) De�nition An antichain of a poset P is a subset A ⊆ P such that everypair of elements of A are incomparable.
(3.26) Example Continuing last example,
A = {c, d} A = {b, d} A = ∅ A = {a}A = {b} A = {c} A = {d} A = {e}
are the antichains of P . ♦
(3.27) De�nition An order ideal (downset) of poset P is a subset I ⊆ P suchthat if x ∈ I and y ≤P x, then y ∈ I.
A dual order ideal (upset) is a subset I ⊆ P such that is x ∈ I and x ≤P y, theny ∈ I.
(3.28) Example (cf. (3.24)) I1 = {a, b, c, d} is a downset of P and not an upset.
I2 = {b, c, d, e} is an upset of P and not a downset. ♦
For a �nite poset P , there is a 1-1 correspondence between antichains of P and downsetsof P :
(3.29) Proposition
� If A is antichain of P (�nite) then
I(A) = {x : x ≤ y and y ∈ A}
is an order ideal.
� If I is an order ideal, then
A(I) = { maximal x ∈ I}
is an antichain.
(3.30) Example (cf. (3.24))
Antichain (A) Order ideal (I)
∅ ∅{a} {a}{b} {a, b}{c} {a, b, c}{d} {a, d}{e} {a, b, c, d, e}{b, d} {a, b, d}{c, d} {a, b, c, d}
59
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.31) Remark The set of all order ideals (downsets) of a poset P , when ordered byinclusion, forms a poset J(P ).
(3.32) Example Continuing the last example,
{a, b, c, d, e}
{a, b, c, d}
nnnnnnnnPPPPPPPP
{a, b, c} {a, b, d}
ffffffffffffffffff
{a, b}PPPPPPPPP {a, d}
nnnnnnnnn
{a}
∅
i.e. •1
•~~~~~
@@@@@
• •ooooooooo
•@@@@@ •
~~~~~
•
•0
(3.33) Notation If A = {x1, . . . , xk} is an antichain, then we write 〈x1, . . . , xk〉 for theorder ideal corresponding to A.
(3.34) De�nition Two posets (P. ≤P ) and (Q,≤Q) are isomorphic, P ∼= Q, ifthere exists order preserving bijection ϕ : P → Q such that
x ≤P y ⇔ ϕ(x) ≤Q ϕ(y)
(3.35) De�nition The dual of a poset P is P ∗ such that
x ≤P y ⇔ y ≤P ∗ x
(3.36) Remark The Hasse diagram of P ∗ = the Hasse diagram of P upside down.
3.2 Lattices
(3.37) De�nition Given x, y ∈ P (a poset), an element z ∈ P is called
� an upper bound of x and y if x ≤ z and y ≤ z;� a lower bound of x and y if z ≤ x and z ≤ y.
If z is an upper bound of x and y such that z ≤ w for all other upper bounds w of x andy, then z is called a least upper bound.
Similarly, if z is a lower bound of x and y such that w ≤ z for all other lower bounds wof x and y, then z is called a greatest lower bound.
� If the least upper bound of x and y exists, then it is denoted x ∨ y (�x join y�, �xsup y�, �join together�).
� If the greatest lower bound of x and y exists, then it is denoted x∧ y (�x meet y�,�x inf y�, �where they meet�).
60
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.38) Example
•h
•~~~~~
g
@@@@@
•e •ooooooooo f
•c
@@@@@ •d
~~~~~
•b
•a
f is the least upper bound of c and d.b is the greatest lower bound of c and d.
c ∨ d = f, c ∧ d = b
(3.39) De�nition If P is a poset and for all x, y ∈ P , x∨ y and x∧ y exist, then P iscalled a lattice.
(3.40) Remark In a lattice (L,≤) many properties such as associativity and commu-tativity are true for the ∨ and ∧ operations.
(3.41) Example
••a •
~~~~ @@@@
• •is not a lattice since a ∨ x and a ∧ x does not exist for any x ∈ P \ {a}
•b•a
is a lattice, since a ≤ b and b ≤ b, giving a ∨ b = b and a ∧ b = a
•~~~~
c
@@@@
•a
~~~~ •b
@@@@ is not a lattice, since a ∧ b is unde�ned.
(3.42) Example Out of the 16 di�erent posets on 4 element (see Example 3.9), only2 are lattices:
•~~~~ @@@@
•@@@@ •
~~~~
•and
••••
(3.43) Remark All �nite lattices (|L| <∞) have a 0 and a 1.
If |L| is in�nite, then this need not be true.
(3.44) Example If L = (Z2,≤) where (x, y) ≤ (x′, y′) ⇔ x ≤ x′ and y ≤ y′, wehave:
��������
��������
��������
��������
��������
��������
��������
��������
��������
��������@
@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
@@@@@@@@
which has no 0 and no 1. ♦
61
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.45) De�nition If P is a poset and every x, y ∈ P have a meet, x ∧ y, then we saythat P is a meet-semilattice.
Simlarly, if P is a poset and x ∨ y exists for all x, y ∈ P , then we say that P is ajoin-semilattice.
(3.46) Useful Theorem If P is a �nite meet (join) semilattice with a 1 (0), then Pis a lattice.
All di�erent lattices on 5 points:
•••••
••
~~~~ @@@@
•@@@@ •
~~~~
•
•~~~~ @@@@
•@@@@ •
~~~~
••
•~~~~ @@@@
•@@@@ • •
~~~~
•
•~~~
00000•
•
�����• @@@•
They are all graded, except the last one.
(3.47) De�nition A lattice is an algebra L = (L,∨,∧) satisfying, for x, y, z ∈ L,
(1) x ∧ x = x and x ∨ x = x;
(2) x ∧ y = y ∧ x and x ∨ y = y ∨ x;(3) x ∧ (y ∧ z) = (x ∧ y) ∧ z and x ∨ (y ∨ z) = (x ∨ y) ∨ z;(4) x ∧ (x ∨ y) = x and x ∨ (x ∧ y) = x.
3.3 Modular Lattices
(3.48) De�nition A �nite lattice L is called modular if it is graded and the rankfunction ρ satis�es
ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y), ∀x, y ∈ L (3.1)
In the case where one is considering (L,∨,∧) as an algebra, this condition is equivalentto (L,∨,∧) also satisfying
(5) for all x, y, z ∈ L with x ≤ z, x ∨ (y ∧ z) = (x ∨ y) ∧ z.
(3.49) Example Consider lattice L with Hasse diagram
•xxxxxx
g
FFFFFF
•d
4444 •
e
4444 •f
•b
4444 •c
•a
.
It is graded with ρ(a) = 0, ρ(b) = ρ(c) = 1, . . .. Notice that ρ(d) = ρ(f) = 2, ρ(d ∨ f) =ρ(g) = 3, ρ(d ∧ f) = ρ(a) = 0. Thus
ρ(d) + ρ(f) = 4 6= 3 = ρ(d ∨ f) + ρ(d ∧ f)
62
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
So L is not modular.
Alternatively, in terms of condition (5), choose x = b, y = f, z = d. Then x ≤ z and
x ∨ (y ∧ z) = b ∨ (f ∧ d) = b ∨ a = b 6= d = g ∧ d = (b ∨ f) ∧ d = (x ∨ y) ∧ z
(3.50) Proposition The total order P = ({1, . . . , n},≤) is a modular lattice.
Proof. (P,∨,∧) is a lattice with x ∨ y = max(x, y) and x ∧ y = min(x, y). It is easy tocheck that conditions (1) through (4) hold. The lattice is graded with rank ρ(x) = x−1.For x, y ∈ P condition (3.1) holds since
x+ y = min(x, y) + max(x, y), ∀x, y ∈ P
(3.51) Proposition The poset Bn = ({1, . . . , n},≤) is a modular lattice, where x ≤Pif x ⊆ y. (Bn is the poset of all subsets of {1, . . . , n} ordered by inclusion.)
Proof. If we set x ∧ y = x ∩ y and x ∨ y = x ∪ y for x, y ∈ Bn then (Bn,∨,∧) satis�esconditions (1) through (4), and thus is a lattice. The lattice Bn is graded with rankρ(x) = |x|. Since x, y ⊆ {1, . . . , n} we know
|x|+ |y| = |x ∩ y|+ |x ∪ y|
i.e.ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y)
thereby satisfying condition (3.1). Alternatively, if x, y, z ∈ Bn and x ≤ z, then x, y, z ⊆{1, . . . , n} and x ⊆ z. So
x ∨ (y ∧ z) = x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z) = (x ∪ y) ∩ (z)= (x ∨ y) ∧ z
i.e. condition (5) is true. �
Let Vn(q) be the n-dimensional vector space over the �eld Fq (or GF (q)). De�ne Ln(q)to be the poset of subspaces of Vn(q) where x ≤ y if x is a subspace of y.
(3.52) Proposition Ln(q) is a modular lattice.
Proof. It is easy to see that ∅ is the 0 of Ln(q) and that Vn(q) is the 1 of Ln(q). Ifx, y ∈ Ln(q), then one can easily argue that x ∧ y = x ∩ y, which is also a subspace ofVn(q). So Ln(q) is a meet-semilattice. Since it is �nite, by the useful theorem (3.46) it isa lattice.
The lattice Ln(q) is graded with rank ρ(x) = dim(x), the dimension of the subspace x.
If x, y ∈ Ln(q), then x ∨ y = x+ y = {u + v : u ∈ x,v ∈ y}. From linear algebra, oneknows
dimx+ dim y = dim(x ∩ y) + dim(x+ y)
Henceρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y)
63
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
3.4 Distributive Lattices
(3.53) De�nition A �nite lattice (L,∨,∧) is called distributive if for all x, y, z ∈ L,
(6') x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z),
or
(6�) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).
imply one another
(3.54) Proposition If L is a �nite distributive lattice, then L is modular.
Proof. If (L,∨,∧) is a distributive lattice, then for all x, y, z ∈ L
x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)
If x ≤ z, then x ∨ z = z. So
x ∨ (y ∧ z) = (x ∨ y) ∧ z
and condition (5) holds. �
(3.55) Proposition The lattice Bn is distributive.
Proof. From Prop. (3.51) we know Bn is a modular lattice. In order to show it is dis-tributive, we must show either (6') or (6�) holds.
If x, y, z ∈ Bn, then
x ∨ (y ∧ z) = x ∪ (y ∩ z)= (x ∪ y) ∩ (x ∪ z) (since ∩& ∪ distributive)= (x ∨ y) ∧ (x ∨ z)
Therefore (6') is true and Bn is distributive. �
(3.56) Proposition The lattice Ln(q) is not distributive.
Proof. Later... �
(3.57) Theorem (Fundamental theorem of �nite distributive lattices, FTFDL) If L is
a �nite distributive lattice, then there exists a poset P such that L ∼= J(P ).
3.5 Incidence Algebra of a Poset
(3.58) De�nition Let P be a locally �nite poset. The set of intervals of P is
Int(P ) = {[x, y] : x, y ∈ P, x ≤ y}
64
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.59) De�nition The incidence algebra of P , denoted I(P ), is the algebra ofall functions f : Int(P )→ C, with multiplication
(fg)(x, y) =∑x≤z≤y
f(x, z)g(z, y), where f(x, y) := f([x, y])
(3.60) Remark It is easy to see that the identity of this algebra is
1(x, y) = δ(x, y) =
{1 if x = y
0 if x < y
(3.61) Proposition Let f ∈ I(P ). The following statements are equivalent:
(1') f has left inverse;
(1�) f has right inverse;
(1� ') f has a 2-sided inverse;
(2) f(x, x) 6= 0 for all x ∈ P .
Proof. fg = δ = 1 is equivalent to
� fg(x, x) = 1 for all x ∈ P ;� fg(x, y) = 0 for all x < y ∈ P .
Notice that
(fg)(x, y) =∑x≤z≤y
f(x, z)g(z, y)
= f(x, x)g(x, y) +∑x<z≤y
f(x, z)g(z, y) = 0 ∀x < y
which is equivalent to
g(x, y) = −f(x, x)−1∑x<z≤y
f(x, z)g(z, y)
So this �value� is OK so long as f(x, x) 6= 0. Thus
f has a right inverse ⇔ f(x, x) 6= 0 ∀x ∈ P
Applying the same argument to hf = δ shows
f has a left inverse ⇔ f(x, x) 6= 0 ∀x ∈ P
So
f has a right inverse ⇔ f(x, x) 6= 0 ∀x ∈ P ⇔ f has a left inverse
By looking at the covers of x, one sees that f(x, x)−1 depends only on [x, y] and conse-quently g(x, y) = h(x, y) = f(x, y)−1. �
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
3.6 The Zeta Function
(3.62) De�nition The zeta function is
ζ : I(P )→ C, ζ(x, y) = 1
i.e. ζ of every interval is equal to 1.
(3.63) Remark Notice that
ζ2(x, y) = ζζ(x, y) =∑x≤z≤y
ζ(x, z)ζ(z, y)
=∑x≤z≤y
1 = |[x, y]|
= |{z ∈ P : x ≤ z and z ≤ y}|
Consider k ∈ N:
ζk(x, y) =∑
x=x0≤x1≤···≤xk=yζ(x0, x1)ζ(x1, x2) · · · ζ(xk−1, xk) =
∑x=x0≤x1≤···≤xk=y
1
= |{(x0, . . . , xk) : x = x0, y = xk, x0 ≤ x1 ≤ · · · ≤ xk}|
= the number of multichains of length k from x to y
(3.64) Proposition If k ∈ N, then (ζ − 1)k(x, y) is the number of chains of length kfrom x to y.
Proof. Notice �rst that
(ζ − 1)(x, y) = ζ(x, y)− 1(x, y) =
{0 if x = y
1 if x < y
Using this,
(ζ − 1)k(x, y) =∑
x=x0≤x1≤···≤xk=y(ζ − 1)(x0, x1) · (ζ − 1)(x1, x2) · · · (ζ − 1)(xk−1, xk)
=∑
x=x0<x1<···<xk=y1
= number of chains in P of length k from x to y
(3.65) Proposition The function (2− ζ)−1 ∈ I(P ) counts the number of chains in an
interval.
Proof. First
(2− ζ)(x, y) = 2δ(x, y)− ζ(x, y) =
{1 if x = y
−1 if x < y
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
Since (2− ζ)(x, x) = 1 for all x ∈ P , the inverse (2− ζ)−1 exists.
Fix x, y ∈ P with x ≤ y and let l be the length of the longest chain in [x, y]. From theprevious proposition we know that
(ζ − 1)l+k(u, v) = 0 for all x ≤ u ≤ v ≤ y and k ∈ N
Using this we have
(2− ζ)[1 + (ζ − 1) + · · ·+ (ζ − 1)l
](u, v)
=(1− (ζ − 1)
)[1 + (ζ − 1) + · · ·+ (ζ − 1)l
](u, v)
=(1 + (ζ − 1) + · · ·+ (ζ − 1)l − (ζ − 1)− · · · − (ζ − 1)l − (ζ − 1)l+1
)(u, v)
=(1− (ζ − 1)l+1)(u, v)
= 1(u, v)= δ(u, v)
Thus(2− ζ)−1 = 1 + (ζ − 1) + · · ·+ (ζ − 1)l
on all intervals of [x, y]. This gives
(2− ζ)−1(x, y) = total number of chains in [x, y]
3.7 The Möbius Inversion Formula
The zeta function ζ satis�es ζ(x, x) = 1 for all x ∈ P and so, ζ−1 exists.
(3.66) De�nition The inverse µ = ζ−1 of P is called the Möbius function of Pand is given by
µ(x, x) = 1 for all x ∈ P
µ(x, y) = −∑x≤z<y
µ(x, z) for all x < y ∈ P
(3.67) Example Let P be the poset with Hasse diagram
•
�������f
@@@@
•~~~~
e
@@@@
•b @@@@ •c •
dnnnnnnn
•a
Intervals of P :
[a, a], [b, b], . . . , [f, f ][a, b], [a, c], [a, d], [a, e], [a, f ][b, f ], [c, e], [c, f ][d, e], [d, f ][e, f ]
67
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
So
µ(a, a) = µ(b, b) = · · · = µ(f, f) = 1
µ(a, b) = −∑a≤z<b
µ(a, z) = −µ(a, a) = −1
µ(a, c) = −µ(a, a) = −1 = µ(a, d)µ(a, e) = −µ(a, a)− µ(a, c)− µ(a, d) = −1− (−1)− (−1) = 1µ(a, f) = −µ(a, a)− µ(a, b)− µ(a, c)− µ(a, d)− µ(a, e) = −1− 3(−1)− 1 = 1µ(b, f) = −µ(b, b) = −1µ(c, e) = −µ(c, c) = −1µ(c, f) = −µ(c, c)− µ(c, e) = −1− (−1) = 0µ(d, e) = −1µ(d, f) = 0µ(e, f) = −µ(e, e) = −1
The point of all this:
(3.68) Proposition (Möbius Inversion Formula, M.I.F.) Let P be a poset in which
every order ideal 〈x〉 is �nite. Let f, g : P → C. Then
g(x) =∑y≤x
f(y), ∀x ∈ P ⇔ f(x) =∑y≤x
g(y)µ(y, x), ∀x ∈ P
(3.69) Lemma Let µ be the Möbius function of a poset P . If a < b, then
µ(a, b) = −∑a<z≤b
µ(z, b)
Proof. From the de�nition of µ we have µ(a, a) = 1 and
µ(a, b) = −∑a≤z<b
µ(a, z), if a < b
Notice that if b covers a, a < b, then
µ(a, b) = −µ(a, a) = −1 = −µ(b, b)
By induction of the longest chain in [a, b], the result then holds.
Alternatively: all functions in I(P ) remain the same when we move from P to P ∗ (dualposet). So µP ∗(y, x) in P ∗ = µ(x, y) in P . �
Proof (of M.I.F.). �⇒�: Let f : P → C and de�ne
g(x) =∑y≤x
f(y) ∀x ∈ P
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
Then the sum∑y≤x
µ(y, x)g(y) =∑y≤x
µ(y, x)∑z≤y
f(z) =∑z≤x
f(z)∑z≤y≤x
µ(y, x)
= f(x)µ(x, x) +∑z<x
f(z)∑z≤y≤x
µ(y, x) = f(x) +∑z<x
f(z) · (0)
= f(x) ∀x ∈ P
�⇐�: Handled in exactly the same way. �
The dual form of the M.I.F. is also useful:
(3.70) Proposition (Möbius inversion formula, dual form) Let P be a poset in which
every dual order ideal is �nite. Let f, g : P → C. Then
g(x) =∑x≤y
f(y) ∀x ∈ P ⇔ f(x) =∑x≤y
µ(x, y)g(y) ∀x ∈ P
3.8 Applications of the Möbius Inversion Formula
(3.71) Example
(1) Consider the total order on P = (N,≤). For each i ∈ N, 〈i〉 = {1 ≤ j ≤ i} is�nite. The Möbius function of P :
µ(i, i) = 1 i ≥ 1µ(i, i+ 1) = −µ(i, i) = −1 i ≥ 1µ(i, i+ k) = 0 k > 1, i ≥ 1
If f, g : P → C, then M.I.F. gives
g(x) =∑y≤x
f(y) =x∑y=1
f(y)
i�
f(x) =∑y≤x
µ(y, x)g(y) =x∑y=1
µ(y, x)g(y) = µ(x, x)g(x) + µ(x− 1, x)g(x− 1)
= g(x)− g(x− 1)
(2) Let X1, . . . , Xn ⊆ S be �nite sets and let P be the poset of all intersections ofthese sets, ordered by inclusion. Also let 1 = X1 ∪ · · · ∪Xn ∈ P . We want to givean expression for |X1∪ · · · ∪Xn| in terms og the sizes |Xi| and their intersections.De�ne two functions on P :
g(x) = # elements in set x
f(x) = # elements in set x that do not belong to any set x′ < x
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
g(x) =∑y≤x
f(x)
Notice that
f(1) = 0
so
0 = f(1) =∑y≤1
g(y)µ(y, 1) by M.I.F.
Thus
0 =∑y≤1
|y|µ(y, 1) = |1|µ(1, 1) +∑y<1
|y|µ(y, 1)
i.e.
|X1 ∪ · · · ∪Xn| = −∑y<1
|y|µ(y, 1)
If we let n = 3:
•
~~~~~~X1∪X2∪X3=1
@@@@@@
•β1=X1
@@@@@@ •
~~~~~~β2=X2
@@@@@@ •
~~~~~~ β3=X3
•α1=X1∩X2
@@@@@@ •α2=X1∩X3•α3=X2∩X3
~~~~~~
•X1∩X2∩X3=0
~~~~~~
then
µ(βi, 1) = −1, µ(αi, 1) = 1, µ(0, 1) = −1
Thus
|X1 ∪X2 ∪X3| = −|X1|µ(X1, 1)− |X2|µ(X2, 1)− |X3|µ(X3, 1)
− |X1 ∩X2|µ(X1 ∩X2, 1)− · · · − |X2 ∩X3|µ(X2 ∩X3, 1)
− |X1 ∩X2 ∩X3|µ(X1 ∩X2 ∩X3, 1)= |X1|+ |X2|+ |X3| − |X1 ∩X2| − |X1 ∩X3| − |X2 ∩X3|
+ |X1 ∩X2 ∩X3|
(3) The poset (Bn ≤) and the Principle of Inclusion-Exclusion:
Lemma Let P = (Bn,≤) be the poset of all subsets of {1, . . . , n} ordered by
inclusion. For x ≤ y ∈ Bnµ(x, y) = (−1)|y|−|x|
Proof. Let x, y ∈ Bn with x ≤ y and r = ρ(y) − ρ(x) = |y| − |x|. It is an easyexercise to see that the interval [x, y] of Bn is isomorphic to the poset (Br,≤).(Consider the map ψx : [x, y]→ Br, ψx(z) = z − x.)
For n = 1:•{1}=1
•∅⇒ µ(0, 1) = (−1)1−0.
For n = 2: Same.
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
By induction on n, the above map ψx shows the result to be true for all [x, y] 6=[0, 1]. For [0, 1],
µ(0, 1) = −∑x<1
µ(0, x) = −∑x<1
(−1)|x| = −n−1∑i=0
(n
i
)(−1)i
= − [(1− 1)n − (−1)n] = (−1)n
So, Möbius inversion for Bn tells us that for f, g : Bn → C
g(x) =∑y≤x
f(y) ∀x ∈ Bn ⇔ f(x) =∑y≤x
(−1)|x|−|y|g(y) ∀x ∈ Bn
g(x) =∑x≤y
f(y) ∀x ∈ Bn ⇔ f(x) =∑x≤y
(−1)|y|−|x|g(y) ∀x ∈ Bn
An example of a (f, g) pair on Bn:Let D(n) = # derangements of {1, . . . , n} = #{π ∈ Sn : π1 6= 1, . . . , πn 6= n}(no �xed points). E.g. D(0) = 1, D(1) = 0, D(2) = 1, D(3) = 2. For π ∈ Sn letfix(π) = {i : πi = i}. De�ne
f(T ) = |{π ∈ Sn : fix(π) = T}|
g(T ) = |{π ∈ Sn : fix(π) ⊇ T}|
for all T ⊆ {1, . . . , n}, T ∈ Bn. (Note: g(T ) = (n − |T |)! ). We want to �ndf(∅) = f(0).Since
g(T ) =∑T⊆T ′
f(T ′) ∀T ∈ Bn
the D.M.I.F. gives
f(T ) =∑T⊆T ′
(−1)|T′|−|T |g(T ′) ∀T ∈ Bn
Thus
f(∅) =∑∅⊆T ′
(−1)|T′|(n− |T ′|)! =
n∑i=0
(n
i
)(−1)i(n− i)!
=n∑i=0
n!(n− i)!i!
(−1)i(n− i)! = n!n∑i=0
(−1)i
i!
= n!(
1− 11
+12!− 1
3!+ · · ·+ (−1)n
n!
)= n!
(12!− 1
3!+ · · ·+ (−1)n
n!
)E.g. for n = 3: 3!( 1
2! −13!) = 3− 1 = 2.
As another example, recall problem sheet 5, q. 3: Given X ⊆ {1, . . . , n− 1}, let
α(X) = |{π ∈ Sn : desπ ⊆ X}|
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
It was shown that if X = {x1, . . . , xk}, then
α(X) =(
n
x1, x2 − x1, . . . , n− xk
)What is
β(X) = |{π ∈ Sn : desπ = X}| ?
From the Möbius Inversion Formula: α, β : Bn−1 → C
α(X) =∑Y⊆X
β(Y ) for all X ⊆ {1, . . . , n− 1}
and thus
β(X) =∑Y⊆X
α(Y )µ(Y,X) =∑Y⊆X
α(Y )(−1)|X|−|Y |
=∑
1≤i1<i2<···<ij≤k(−1)k−j
(n
xi1 , xi2 − xi1 , . . . , n− xik
), k = |X|
(4) Classical Möbius function in Number theory: (Dn,≤) is the lattice of divisorsof n.
D12 : •~~~~~
12=2231
@@@@@
•4=2230 •oooooooo 6=2131
•2=2130
@@@@@ •3=2031
~~~~~
•1=2030
~~~~~
µ(1, 1) = 1 = µ(2, 2) = µ(3, 3) = . . .
µ(1, 2) = −1 = µ(1, 3)µ(1, 2× 2) = −µ(1, 1)− µ(1, 2) = 0µ(1, 2× 3) = −µ(1, 1)− µ(1, 2)− µ(1, 3) = 1µ(1, 2× 2× 3) = −µ(1, 1)− µ(1, 2)− µ(1, 3)
−µ(1, 2× 2)− µ(1, 2× 3)= 1− 1 = 0
Lemma Given x, y ∈ Dn, x ≤ y, then
µ(y/x) = µ(x, y) =
(−1)t if y/x is a product of t distinct primes
1 if x = y
0 otherwise.
Set D = in�nite lattice of divisors of the integers; �divisor� lattice. Then
{µ(1), µ(2), µ(3), µ(4), µ(5), µ(6), µ(7), . . .} = {1,−1,−1, 0,−1, 1,−1, . . .}
From this, if
g(m) =∑d|m
f(d) for all integers m,
then
f(m) =∑d|m
g(d)µ(m/d) ∀m ∈ N
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
As an example, let f(x) = x. Then
g(m) =∑d|m
d = sum of divisors of m
and
m = f(m) =∑d|m
g(d)µ(m/d) .
Or if f(x) = 1, then
g(m) =∑d|m
1 = # of divisors of m
and
1 = f(m) =∑d|m
g(d)µ(m/d)
(5) The Möbius function of Ln(q) (lattice of subspaces of Vn(q)) and Πn (lattice ofpartitions of {1, . . . , n}):
� for Ln(q),
µ(0, 1) = (−1)nq(n2)
� for Πn,
µ(0, 1) = (−1)n−1(n− 1)!
3.9 The Möbius Function of Lattices
In particular cases the Möbius function of a lattice may be easy to calculate.
(3.72) Lemma (3.9.1.) Let L be a �nite lattice with 1 covering {x1, . . . , xk} and 0covered by {y1, . . . , ym}.
If x1 ∧ x2 ∧ . . . ∧ xk 6= 0 then µ(0, 1) = 0
If y1 ∨ y2 ∨ . . . ∨ ym 6= 1 then µ(0, 1) = 0
(3.73) Remark The elements {y1, . . . , ym} covering 0 are called atoms. The elements{x1, . . . , xk} covered by 1 are called coatoms.
(3.74) Lemma (3.9.2.) Let L be a �nite distributive lattice. For x, y ∈ L,
µ(x, y) =
(−1)`(x,y) if [x, y] is a Boolean algebra
(i.e. ∼= Bk for some k)0 otherwise
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
Proof. Recall that if L is distributive, then L ∼= J(P ) for some poset P . If x, y ∈ L withx ≤ y, then there are antichains X,Y in P such that y corresponds to the order ideal〈Y 〉 and X corresponds to the order ideal 〈X〉. So, since x ≤ y, 〈X〉 ⊆P 〈Y 〉 and Y \Xis an antichain of P . Thus in J(P ) the sublattice [〈X〉, 〈Y 〉] ∼= [0, 〈Y \ X〉]. Using thisargument it is clear that
〈Y 〉 is the join of covers of 〈X〉 ⇔ [x, y] is a boolean algebra
From the previous lemma (3.72), if 〈Y 〉 is not the join of the covers of 〈X〉, thenµ(x, y) = 0. Hence the result. �
(3.75) Lemma (3.9.3.) Let P be a �nite poset with a 0 and 1. Let ci = # chains
0 = x0 < x1 < · · · < xi = 1 of length i. Then
µ(0, 1) = c0 − c1 + c2 − c3 + c4 − · · ·
(Related to the Euler characteristic).
Proof. The function µ = (1 + (ζ − 1))−1, so
µ(0, 1) = (1 + (ζ − 1))−1(0, 1) = (1− (ζ − 1) + (ζ − 1)2 − (ζ − 1)3 + · · · )(0, 1)
= 1(0, 1)− (ζ − 1)(0, 1) + (ζ − 1)2(0, 1)− · · ·= c0 − c1 + c2 − c3 + · · ·
since ci = (ζ − 1)i(0, 1) by Prop. (3.64). �
3.10 Young's lattice
From chapter 1, we have
Par(n) = set of partitions of the integer n
De�nePar = Par(0) ∪ Par(1) ∪ · · ·
Then
Par(0) = ∅ Par(3) = {3, 21, 111}Par(1) = {1} Par(4) = {4, 31, 22, 211, 1111}Par(2) = {2, 11} Par(5) = {5, 41, 32, 311, 221, 2111, 11111}
The sets Par(n) are easily visualised as Young diagrams. Given λ, µ ∈ Par, de�ne λ ⊆ µif the shape of λ is contained in the shape of µ. For example
32 = ⊆ = 4211
74
CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
but
21 = 6⊆ = 111
Let Y = (Par,⊆) be the poset on Par with this inclusion/containment order
•111
•21
ppppppppppp •3
•11
NNNNNNNNNNN
ppppppppppp •
NNNNNNNNNNN2
ppppppppppp
•
NNNNNNNNNNN1
ppppppppppp
•∅
(3.76) Theorem Y is a lattice.
Proof. Given λ = (λ1, λ2, . . .), µ = (µ1, µ2, . . .) ∈ Y, the smallest shape (diagram) thatcontains both λ and µ is
λ ∨ µ = (max(λ1, µ1),max(λ2, µ2), . . .)
and similarlyλ ∧ µ = (min(λ1, µ1), (λ2, µ2), . . .)
It is straightforward to check that conditions (1)�(4) hold for (Y,∨,∧). �
(3.77) Remark The rank function ρ : Y → N is given by
ρ(λ) = λ1 + λ2 + · · ·
By noticing that
|{λ ∈ Par : ρ(λ) = n}| = p(n) = # partitions of the integer n
the rank generating function of Y is
F (Y, q) =+∞∑n=0
p(n)qn =+∞∏n=0
11− qn
(see problem sheet 4).
(3.78) Proposition Maximal chains [0, λ] in Y, where ρ(λ) = n, are in 1-1 correspon-
dence with all standard Young tableaux of shape λ (SYTλ).
(3.79) Remark Draw the corresponding Young diagram at each element in Young'slattice. Choose your element λ in the lattice. As you proceed from 0 up to λ, in eachstep i, put i in the box that was added to the Young diagram from last step, e.g.
∅ → 1 → 12→ 1 3
2→
1 324
= λ
if we choose λ = 211 and our route is 0→ 1→ 11→ 21→ 211. (Draw this!)
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
Proof. Let 0 = y0 <· y1 <· · · · <· yn = λ be a maximal chain. Since yi+1 has exactlyone more cell than yi and yi ⊂ yi+1, label the cell yi+1 − yi with i + 1. Do this for all0 ≤ i < n. The resulting labeling of the Young diagram is a standard Young tableauxof shape λ. The inverse map is easy to describe. A consequence of this is the number ofmaximal chains in [0, λ] is fλ (calculated by the hook-length formula). Also the numberof maximal chains [0, λ] for all λ where ρ(λ) = n, is equal to In, the number of involutions
{π ∈ Sn : π2 = id}
(3.80) De�nition The dominance order (Par(n),≤) is such that if λ, µ ∈ Par(n)with λ = (λ1, λ2, . . .) and µ = (µ1, µ2, . . .) then we de�ne
λ ≤ µ ⇔ λ1 + · · ·+ λi ≤ µ1 + · · ·+ µi ∀i
3.11 Linear extensions
(3.81) De�nition Let (P,≤P ) be a poset with |P | = n. A linear extension ofP is a bijective order preserving map
σ : P → {1, . . . , n}
such that
if σ−1(i) <P σ−1(j) then i < j.
The number of linear extensions of P is denoted e(P ).
(3.82) Example Consider P: •c@@@@ •d
•a •b
If σ = (σ(a), σ(b), σ(c), σ(d)) = (1, 2, 3, 4) then σ(P ) is •3@@@@ •4
•1 •2and σ is a linear
extension of P .
However if σ is chosen to be (σ(a), σ(b), σ(c), σ(d)) = (3, 2, 1, 4) then σ(P ) is •1@@@@ •4
•3 •2which is not a linear extension.
All linear extensions of P :
•3@@@@ •4
•1 •2•3
@@@@ •4•2 •1
•4@@@@ •3
•1 •2•4
@@@@ •3•2 •1
•4@@@@ •2
•3 •1
σ=(1,2,3,4) σ=(2,1,3,4) σ=(1,2,4,3) σ=(2,1,4,3) σ=(3,1,4,2)
Thus e(P ) = 5. ♦
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.83) Remark
P = • • · · · •︸ ︷︷ ︸n
e(P ) = n!
P =
••...
••
n e(P ) = 1
(3.84) Proposition
e(P ) = # maximal chains in J(P )
Proof. Homework. �
(3.85) De�nition If a poset P on elements {1, . . . , n} is such that i <P j ⇒ i < j,then P is called a natural poset on {1, . . . , n}. In this case, each linear extension ofP may be identi�ed with a permutation (σ−1(1), σ−1(2), . . . , σ−1(n)) ∈ Sn.
(3.86) De�nition If P is a natural poset on {1, . . . , n}, then the set of permutationsσ−1 corresponding to linear extensions of P is denoted L(P ) and is called the Jordan-Hölder set of P .
(3.87) Example For P = •3@@@@ •4
•1 •2, L(P ) = {1234, 2134, 1243, 2143, 2413}. ♦
3.12 Rank-selection
Let P be a �nite graded poset of rank n with a 0 and 1. Let ρ be the rank function of P .
(3.88) De�nition Let S ⊆ {0, . . . , n} and PS = {x ∈ P : ρ(x) ∈ S}. PS is called theS-rank-selected subposet of P . Let α(P, S) = # maximal chains in PS . SinceP has a 0 and 1, the quantity α(P, S) can be restricted to S ⊆ {1, . . . , n− 1}.
De�ne for S ⊆ {0, . . . , n},
β(P, S) =∑T⊆S
(−1)|S|−|T |α(P, T ) .
The Möbius inversion formula gives
α(P, S) =∑T⊆S
β(P, T ) .
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CHAPTER 3. PARTIALLY ORDERED SETS (POSETS)
(3.89) Proposition Let P be a �nite poset and L = J(P ) with ρ(L) = n. For S ⊆{1, . . . , n− 1},
β(L, S) = #{π ∈ L(P ) : Des(π) = S}
Proof. Suppose S = {a1, . . . , ak} ⊆ {1, . . . , n − 1}. The structure of J(P ) tells us thatα(L, S) is the number of chains I1, . . . , Ik of order ideals of P such that |Ii| = ai for alli = 1, . . . , k. For such a chain
� arrange the elements of I1 in increasing order;
� arrange the elements of I2 − I1 in increasing order;
�...
� arrange the elements of P − Ik in increasing order.
This maps chains I ∈ LS → π ∈ L(P ) and this map is bijective. Also Des(π) ⊆ S.
If we de�neγ(L, S) = {π ∈ L(P ) : Des(π) = S}
thenα(L, S) =
∑T⊆S
γ(L, S)
By comparison, we have that γ(L, S) = β(L, S). �
(3.90) Proposition
β(Bn, S) = #{π ∈ Sn : Des(π) ∈ Sn}
Proof. Let P be the n element antichain
P : • • · · · •︸ ︷︷ ︸n
⇒ J(P ) = Bn
From the previous proposition, and using the fact that L(P ) = Sn, the result follows.�
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