(P, ≤ P ) P ≤ P P x ∈ P x ≤ P x x ≤ P y y ≤ P x x = y x ≤ P y y ≤ P z x ≤ P z ≤ ≤ P x ≥ y y ≤ P x x<y x ≤ p y x 6= y x, y ∈ P x ≤ P y y ≤ P x n ∈ N P = {i ∈ N :1 ≤ i ≤ n} x, y ∈ P x ≤ P y x ≤ y 1 < 2 < 3 < ··· < (n - 1) <n n ∈ N P = {A : A ⊆{1,...,n}} x, y ∈ P x ≤ P y x ⊆ y (P, ≤ P ) {1,...,n} B n n =2: P = {∅, {1}, {2}, {1, 2}} ∅≤ P {1}≤ P {1, 2} ∅≤ P {2}≤ P {1, 2} n ∈ N P = {i ∈ N : i n} x, y ∈ P x ≤ P y x y n = 18 P = {1, 2, 3, 6, 9, 18} 1 ≤ P 1, 1 ≤ P 2, 1 ≤ P 3,..., 1 ≤ P 18 6 ≤ P 6, 6 ≤ P 18 2 ≤ P 2, 2 ≤ P 6, 2 ≤ P 18 9 ≤ P 9, 9 ≤ P 18 3 ≤ P 6, 3 ≤ P 9, 3 ≤ P 18 18 ≤ P 18
26
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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.
(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.
(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
(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.
(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
68
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
(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.
70
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.
(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
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
(3.85) 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.�