Chapter 3
Partially Ordered Sets (Posets)
3.1 Posets
(3.1) Denition (3.1.1.)
where
P
≤P
is a set and
A
partially ordered set (poset)
P satisfying
is a pair
(P, ≤P )
is a binary relation on
x ∈ P , x ≤P x (reflexivity);
x ≤P y and y ≤P x, then x = y (anti-symmetry);
x ≤P y and y ≤P z , then x ≤P z (transitivity).
(1) for all
(2) if
(3) if
Where there is no possibility of confusion, we will write
x≥y
means
y ≤P x
x<y
means
x ≤p y
x, y ∈ P are comparable
incomparable.
Two elements
called
if
and
x ≤P y
≤
instead of
≤P .
Also
x 6= y
or
y ≤P x.
Otherwise, they are
(3.2) Example (3.1.2.)
(1) For
n∈N
let
P = {i ∈ N : 1 ≤ i ≤ n}.
Given
x, y ∈ P ,
dene
x ≤P y
if
x ≤ y.
1 < 2 < 3 < · · · < (n − 1) < n
Each pair of elements is comparable. (Totally
(2)
ordered set.)
Given n ∈ N, let P = {A : A ⊆ {1, . . . , n}}. For x, y ∈ P , dene 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}
n ∈ N let P = {i ∈ N : i divides n}. For x, y ∈ P
divides y . E.g. if n = 18, then P = {1, 2, 3, 6, 9, 18} and
(3) Given
dene
x ≤P y
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
53
if
x
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(P, ≤P ) is denoted by Dn .
n ∈ N let P = { partitions of {1, . . . , n}}. For x, y ∈ P dene x ≤P y if
every 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 , we
have 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 renement and is denoted by Πn .
Here
(4) Given
(3.3) Denition
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 then
x ≤P y .
When we say
x, y ∈ P
For
subposet,
where
we mean an
(P, ≤)
induced subposet.
is a poset, and
x ≤ y,
the
interval
[x, y] = {z ∈ P : x ≤ z ≤ y}
is a subposet. The smallest interval is
(3.4) Denition
If every interval
[x, x] = {x}.
[x, y] of (P, ≤P ) is nite, then P
is called a
finite poset.
If
Q
and
is a subposet of
x, z ∈ Q.
P,
then
Q
is called
convex
if
y∈Q
whenever
locally
x<y<z
in
P
Interval posets are always convex.
Open interval:
(x, y) = {z ∈ P : x < z < y}
(3.5) Denition
not exist
z∈P
(3.6) Remark
(3.7) Denition
x∈P
Given
such that
x, y ∈ P , we
x < z < y.
say that
y covers x
if
and there does
A locally nite poset is completely determined by its cover relations.
The
Hasse diagram of a nite poset P
is the graph with vertices
and
if
if
x < y , then y is drawn above x in the diagram;
y covers x, then there is an edge between x and y
(3.8) Example
If
P = {a, b, c, d, e, f }
in the diagram.
and
a < b, a < c, a < d, b < e, e < f, c < f, d < f
then the Hasse diagram of
P
is
~•00
~~ 00
~
e•
00
0
•c •d
•@@ b @@ •
f
a
54
x<y
CHAPTER 3.
(3.9) Example
All dierent (non-isomorphic) posets on 4-elements
•
• •
• • • •
•
•
•@
~~ @@@
~~
•@@ • ~•
@@ ~~
•~
•
•
•
•
•
•
(3.10) Question
•
•00
000 •
•
•
@@
~~
•~ •
@@
•
•
•
•@
~~ @@
•~
•
•@@ ~•
•99
99
•
•999
9
•
•@@@ ~•
~
•~
•
•@@ •
•
PARTIALLY ORDERED SETS (POSETS)
•
•
•0
•@000
~~ @@0
•~
•
Why are
and
•@
~~ @@
~
•@@@ ~•
~~
•00
•
00 •
0 •
• •
•
•
~•@@
~~ @@
~
•@@
•
@@ ~~~
~
•999
9
•
•999
•
9 •
not in the above diagrams?
•
(3.11) Example (From example 3.1.2.)
(1) If
(2)
n=4
and
B4 = (P, ≤)
P = {1, 2, 3, 4},
the Hasse diagram of the total order on
P = {A : A ⊆ {1, 2, 3, 4}}
where
P
is
•4
•3
•2
•1
and elements are ordered by
inclusion.
P = {∅, 1, 2, 3, 4, 12, . . . , 134, 1234}
The Hasse diagram:
o•@O1O234
oo~o~~ @@O@OOOO
o
o
@@ OOO
oo ~~
@@
OOO
ooo ~~~
o
@
OO
o
o
~
o
io•O1O34O ~•@2@34
UUUUoo•U124
UUUU
•U123
i
i
o
~
i
O
o
o
O~O~ @@
oo UUUUUU iUiUiUioUioUioUo
~~
iiUUoUoUoU UUUUU ~~~ OOOOO@@@
~~ooooo
i
i
~
i
i
OOO@@
UUUU
U~
o
~o~ooo
iiii ooo
UU ~~UUUUU
O'
i
•@~Oo1@O2OO •Ui13iUUUU ~•oU14UUUU
•
•
24
23
o~•34
o
i
i
o
o
i
o
o
i
U
U
@@ OO
~
~
i
U
U
o
oo~~
o
@@ OOOO ~~~ UUUUUUU oUoUoiUoiUiUiUiUii
o
UU
@@ ~O~OO
ooo ~~
ooUiUi
@ ~~ OO iooioioioiii UUUUUU oUoUoUoUoUU ~~~
•3
•O1OOO •@2@
o•4
~~ oooo
OOO @@
~
o
~ o
OOO @@
OOO@@ ~~o~oooo
OO@ ~o~oo
•
D12 = (P, ≤)
x divides y .
(3) Consider
P
if
where
∅
P = {i : i
divides
12} = {1, 2, 3, 4, 6, 12}. x ≤ y
1 < 2, 3, 4, 6, 12
4 < 12
2 < 4, 6, 12
6 < 12
in
3 < 6, 12
Hasse diagram for
D12 :
12
~•@@
~~ @@@
~
~
4•
oo•6
ooo
o
o
•@o
•
2 @@@ ~~~ 3
@ ~~
•
1
55
CHAPTER 3.
(4)
PARTIALLY ORDERED SETS (POSETS)
Π3 = (P, ≤)
where
P
is the set of all partitions of
{1, 2, 3},
P = {{123}, {1, 23}, {2, 13}, {3, 12}, {1, 2, 3}}
=x1
=x2
=x3
=x4
=x5
Then
x5 < x1 , x2 , x3 , x4
x2 , x3 , x4 < x1
Hasse diagram:
ss
sss
s
s
s
{1, 23}
{123}
KKK
KKK
KK
{2, 13}
KKK
KKK
KK
{3, 12}
s
sss
s
s
ss
{1, 2, 3}
(3.12) Denition
Given poset
(3.13) Denition
A
(P, ≤), if there exists an element a ∈ P such that
a ≤ x for all x ∈ P , then a is a smallest element of the poset P and is denoted 0̂.
Similarly, if there is a largest element a such that x ≤ a for all x ∈ P , then a is
written 1̂.
A subset
of
C
chain
of a poset
P
is a poset in which every 2 elements are comparable.
is a
chain
if
C
is a chain when regarded as a subposet
P.
C in a poset is saturated if it is not amissing any elements within it,
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 xi
for all 0 ≤ i < n.
A chain
i.e.
(3.14) Denition
The
The
length (rank)
length
of a nite chain is
of a nite poset
P
`(P ) = max{`(C) : C
Length of the
A chain
C
`(C) = |C| − 1.
is
interval [x, y] in P is `(x, y).
P is maximal if no more
in a poset
chain in
P}
elements can be added to it.
(3.15) Example
P :
f
•@
~~ @@ g
~
•
•@@@
d ~~~
•
•h
~
~
•~c
e
`(P ) = 5
•b
•a
C1 = {a, b, c, d, e, f }
C2 = {h, g, f }
56
is a maximal chain,
is a maximal chain,
`(C1 ) = 5
`(C2 ) = 2
CHAPTER 3.
(3.16) Denition
PARTIALLY ORDERED SETS (POSETS)
If every maximal chain of
P
has the same length
n,
graded of length n.
Notice that if
P
is graded, then there is a unique rank function
then
P is
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
If
x ≤ y,
for
then
x ∈ P,
then we say element
x has rank i.
`(x, y) = ρ(y) − ρ(x).
(3.17) Denition
If
P
is a graded poset of rank
F (P, q) =
n
X
n,
then the polynomial
pi q i ,
i=0
pi = #
where
elements of rank
i
in
P,
is called the
of P .
(3.18) Remark
rank generating function
All posets in Example 3.1.2. were graded.
(3.19) Example
Let
P
be the poset with Hasse diagram
rank:
4
3
2
1
0
p0 = 1,
p1 = 2,
(3.20) Example
chain
⇒
graded.
p2 = 3,
p3 = 2,
•<<
<<
•
•<<
<<
•<S<SSSS•<< •
<< SSS<< •<< S•
<< •
p4 = 1;
P on {1, . . . , n} with 1 < 2 < . . . < n.
`(C) = n − 1, ρ(x) = x − 1, ρ(P ) = n − 1.
Consider poset
(3.21) Example
The poset
Bn = (P, ≤),
P = {A : A ⊆ {1, . . . , n}},
A chain in
Bn
F (P, q) = 1 + 2q + 3q 2 + 2q 3 + q 4
One maximal
♦
where
x≤y ⇔ x⊆y
is a sequence
{A1 , . . . , Ak }
Maximal chain e.g.
such that
A1 ⊂ A2 ⊂ · · · ⊂ Ak ⊆ {1, . . . , n}
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
Note that if
x ∈ P,
ρ(x) = |x|
(3.22) Example
For
Dn = (Pn , ≤)
x ≤ y ⇔ x divides y .
n = 12
is graded of length
n.
x), e.g. ρ({1, . . . , n}) = n.
n
n
X
X
n i
F (Bn , q) =
pi q i =
q = (1 + q)n
i
then
(no. of elements in
i=0
then
Bn
Thus
i=0
where
Pn = {1 ≤ i ≤ n : i
divides
n}.
If
x, y ∈ Pn ,
12 = 22 31 := pα1 1 pα2 2 ,
we have
12=22 31
•@
~~ @@@@
~~
~
4=22 30 •
o•6=21 31
ooo
o
o
o
•@o@
•
2=21 30 @@@ ~~~ 3=20 31
~
~
•
D12 :
1=20 30
and
|D12 | = (2 + 1)(1 + 1) = 6.
Claim
Proof.
Poset
If
Dn
is graded.
n = pα1 1 pα2 2 · · · pαmm
where
p1 , . . . , p n
are primes, then
Pn = {pβ1 1 · · · pβmm : 0 ≤ β1 ≤ α1 , . . . , 0 ≤ βm ≤ αm }
|Pn | = (1 + α1 )(1 + α2 ) · · · (1 + αm ). If x ∈ Pn and x = pβ1 1 · · · pβmm ,
β1 + · · · + βm . In particular ρ(Pn ) = α1 + α2 + · · · + αm .
then
and
ρ(x) =
We have
F (Dn , q) =
α1 +···+α
X m
pj q j =
j=0
=
β1 =0
q α1 +1
−1
q−1
(3.23) Denition
α1
X
A
···
αm
X
···

1q β1 +···+βm = 
βm =0
q αm +1
−1
q−1
multichain
α1
X
β1 =0


αm
X
q β1  · · · 

q βm 
βm =0
of a poset
P
is a chain in which elements are
allowed to be repeated.
~•00
~~ 00
~
c•
00
0
•d
•@
b @@@ •a
e
(3.24) Example
Poset
C = {a, b, b, b, c, c, e}
`(C) = 7 − 1 = 6.
58
P
with Hasse diagram:
is a multichain. Length of
C 0 = {x0 ≤ x1 ≤ · · · ≤ xn }
is
n.
Hence,
♦
CHAPTER 3.
(3.25) Denition
pair of elements of
An
A
(3.26) Example
antichain
PARTIALLY ORDERED SETS (POSETS)
of a poset
P
is a subset
A = {c, d}
A = {b, d}
A=∅
A = {a}
A = {b}
A = {c}
A = {d}
A = {e}
P.
♦
(3.27) Denition
x∈I
and
An order ideal (downset)
y ≤P x, then y ∈ I .
dual order ideal (upset) is a subset I ⊆ P
y ∈ I.
A
(3.28) Example (cf.
I2 = {b, c, d, e}
I1 = {a, b, c, d}
(3.24))
is an upset of
For a nite poset
of
such that every
Continuing last example,
are the antichains of
that if
A⊆P
are incomparable.
P
of poset
P
such that is
is a downset of
P
is a subset
x∈I
and
I⊆P
such
x ≤P y , then
and not an upset.
♦
and not a downset.
P , there is a 1-1 correspondence between antichains of P
and downsets
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
∅
∅
{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}
(I)
59
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(3.31) Remark
The set of all order ideals (downsets) of a poset
inclusion, forms a poset
(3.32) Example
P,
when ordered by
J(P ).
Continuing the last example,
{a, b, c, d, e}
•1̂
i.e.
•@
~~ @@@
~
@
~
•~
oo•
o
o
o
ooo
•@o@
•
@@ ~~~
@ ~~
•
{a,
P
n b, c, Pd}
PPP
PPP
{a, b, c}
fff{a, b, d}
fffff
f
f
f
f
f
fffff
{a, d}
{a, b}PP
PPP
nn
n
n
PPP
nn
P
nnn
n
nnn
nnn
•0̂
{a}
∅
(3.33) Notation
order ideal
A = {x1 , . . . , xk }
corresponding to A.
If
(3.34) Denition
is an antichain, then we write
hx1 , . . . , xk i
for the
(P. ≤P ) and (Q, ≤Q ) are isomorphic, P ∼
= Q,
bijection ϕ : P → Q such that
Two posets
there exists order preserving
if
x ≤P y ⇔ ϕ(x) ≤Q ϕ(y)
(3.35) Denition
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) Denition
a
z
y,
then
(a poset), an element
x and y such that z ≤ w
least upper bound.
is an upper bound of
x
z∈P
is called
upper bound of x and y if x ≤ z and y ≤ z ;
lower bound of x and y if z ≤ x and z ≤ y .
z
is called a
Similarly, if
and
y,
z
then
z
is called a
If the least upper bound of
If the greatest lower bound of
y ,
x inf
for all other upper bounds
x and y such that w ≤ z for
greatest lower bound.
is a lower bound of
sup
60
x, y ∈ P
an
If
of
Given
x
and
y
w
of
x and
all other lower bounds
exists, then it is denoted
x∨y
( x join
w
y , x
join together).
y ,
where they meet).
x
and
y
exists, then it is denoted
x∧y
( x meet
y ,
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(3.38) Example
•h
g
~•@@
~~ @@@
~
~
e•
o•f
ooo
o
o
oo
•@o
~•
c @@
@@ ~~~ d
~
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
•b
•a
(3.39) Denition
called a
If
P
is a poset and for all
x, y ∈ P , x ∨ y
and
x∧y
exist, then
P
is
lattice.
(3.40) Remark
In a lattice
tativity are true for the
∨
and
(L, ≤) many properties
∧ operations.
such as associativity and commu-
(3.41) Example
•
•a
~•@@
~~ @@
~
•
•
•b
is
not a lattice since a ∨ x and a ∧ x does not exist for any x ∈ P \ {a}
is a lattice, since
•a
a≤b
and
b ≤ b,
giving
a∨b=b
and
a∧b=a
c
•@
~~ @@@
~~
•a
b
•
is
not a lattice, since a ∧ b is undened.
(3.42) Example
Out of the 16 dierent posets on 4 element (see Example 3.9), only
2 are lattices:
(3.43) Remark
If
|L|
~•@@
~~ @@
~
•@@
•
@@ ~~~
~
•
All nite lattices (|L|
and
< ∞)
have a
•
•
•
•
0̂
and a
1̂.
is innite, then this need not be true.
(3.44) Example
If
L = (Z2 , ≤)
(x, y) ≤ (x0 , y 0 ) ⇔ x ≤ x0
which has no 0̂ and no 1̂.
where
have:
@
@ @
@
@ @ @
@ @@ @ @
@
@
@ @@ @ @@ @
@ @
@
@
@ @@ @ @@@@ @ @
@ @ @ @ @ @ @@
@ @@ @ @ @ @ @
@ @@ @ @@@@@@
@ @@ @ @ @
@ @@ @ @
@ @@@@
@@
@
and
y ≤ y0,
we
♦
61
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(3.45) Denition
that
P
is a
If P is a poset
meet-semilattice.
P is a poset
join-semilattice.
Simlarly, if
x∨y
and
(3.46) Useful Theorem
If
and every
P
is a
x, y ∈ P
exists for all
have a meet,
x, y ∈ P ,
x ∧ y,
then we say
then we say that
P
is a
nite meet (join) semilattice with a 1̂ (0̂), then P
is a lattice.
All dierent lattices on 5 points:
•
•
•
•
•
~•@@
~~ @@
~
•@@
•
@@ ~~~
~
•
•
~•@@
~~ @@
~
•@@
•
@@ ~~~
~
•
•0
~~ 00
~
00
•
•
•@@@ ~•@@
@@
~~
•@~@ • ~•
@@ ~~
•~
•
•
They are all graded, except the last one.
(3.47) Denition
(1)
(2)
(3)
(4)
A lattice is an
algebra L = (L, ∨, ∧)
satisfying, for
x, y, z ∈ L,
x ∧ x = x and x ∨ x = x;
x ∧ y = y ∧ x and x ∨ y = y ∨ x;
x ∧ (y ∧ z) = (x ∧ y) ∧ z and x ∨ (y ∨ z) = (x ∨ y) ∨ z ;
x ∧ (x ∨ y) = x and x ∨ (x ∧ y) = x.
3.3 Modular Lattices
(3.48) Denition
function
ρ
A nite lattice
L
is called
modular
ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y),
In the case where one is considering
to
if it is graded and the rank
satises
(L, ∨, ∧)
(L, ∨, ∧)
∀x, y ∈ L
(3.1)
as an algebra, this condition is equivalent
also satisfying
(5) for all
x, y, z ∈ L
with
x ≤ z , x ∨ (y ∧ z) = (x ∨ y) ∧ z .
g
(3.49) Example
Consider lattice
L
with Hasse diagram
•F
xx FFF
FF
xx
x
•4x •4e4 •f
d 44
4 44 •4
b 44
•c
4 •
.
a
It is graded with ρ(a) = 0, ρ(b) = ρ(c) = 1, . . ..
ρ(g) = 3, ρ(d ∧ f ) = ρ(a) = 0. Thus
Notice that
ρ(d) = ρ(f ) = 2, ρ(d ∨ f ) =
ρ(d) + ρ(f ) = 4 6= 3 = ρ(d ∨ f ) + ρ(d ∧ f )
62
CHAPTER 3.
So
L
PARTIALLY ORDERED SETS (POSETS)
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
Proof. (P, ∨, ∧)
The total order
is a lattice with
P = ({1, . . . , n}, ≤)
x ∨ y = max(x, y)
and
is a modular lattice.
x ∧ y = min(x, y). It is easy to
ρ(x) = x − 1.
check that conditions (1) through (4) hold. The lattice is graded with rank
For
x, y ∈ P
condition (3.1) holds since
∀x, y ∈ P
x + y = min(x, y) + max(x, y),
(3.51) Proposition
if
x ⊆ y . (Bn
Proof.
The poset
Bn = ({1, . . . , n}, ≤) is a modular lattice,
{1, . . . , n} ordered by inclusion.)
x∧y = x∩y
If we set
and
x∨y = x∪y
x, y ∈ Bn
for
conditions (1) through (4), and thus is a lattice. The lattice
ρ(x) = |x|.
where
x ≤P
is the poset of all subsets of
Since
x, y ⊆ {1, . . . , n}
then
(Bn , ∨, ∧)
satises
Bn
is graded with rank
and
x ≤ z,
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
{1, . . . , n}
and
x ⊆ z.
x, y, z ∈ Bn
then
x, y, 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
to be the poset of subspaces of
(3.52) Proposition
Proof.
Ln (q)
It is easy to see that
vector space over the eld
Vn (q)
where
x≤y
if
x
Fq
(or
GF (q)).
y.
Dene
Ln (q)
is a subspace of
is a modular lattice.
∅
is the
0̂
Ln (q) and that Vn (q)
that x ∧ y = x ∩ y , which
of
x, y ∈ Ln (q), then one can easily argue
Vn (q). So Ln (q) is a meet-semilattice. Since
is the
1̂
of
Ln (q).
If
is also a subspace of
it is nite, by the useful theorem (3.46) it is
a lattice.
The lattice
If
Ln (q)
x, y ∈ Ln (q),
is graded with rank
then
ρ(x) = dim(x),
the dimension of the subspace
x ∨ y = x + y = {u + v : u ∈ x, v ∈ y}.
x.
From linear algebra, one
knows
dim x + dim y = dim(x ∩ y) + dim(x + y)
Hence
ρ(x) + ρ(y) = ρ(x ∧ y) + ρ(x ∨ y)
63
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
3.4 Distributive Lattices
(3.53) Denition
(6')
(L, ∨, ∧) is called distributive if for all x, y, z ∈ L,

x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z),


A nite lattice
imply one another
or
(6)
x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).
(3.54) Proposition
Proof.


If
(L, ∨, ∧)
If
L
is a nite distributive lattice, then
is a distributive lattice, then for all
L
is modular.
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
Proof.
The lattice
Bn
From Prop. (3.51) we know
is distributive.
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
(3.56) Proposition
Proof.
Bn
is distributive.
The lattice
Ln (q)
is
not distributive.
Later...
(3.57) Theorem (Fundamental theorem of nite distributive lattices, FTFDL)
a nite distributive lattice, then there exists a poset
P
such that
L∼
= J(P ).
If
L is
3.5 Incidence Algebra of a Poset
(3.58) Denition
Let
P
be a locally nite poset. The
set of intervals of P
Int(P ) = {[x, y] : x, y ∈ P, x ≤ y}
64
is
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(3.59) Denition
The incidence algebra
f : Int(P ) → C, with multiplication
X
f (x, z)g(z, y),
(f g)(x, y) =
of
P,
denoted
I(P ),
is the algebra of
all functions
where
f (x, y) := f ([x, y])
x≤z≤y
(3.60) Remark
It is easy to see that the identity of this algebra is
(
1(x, y) = δ(x, y) =
(3.61) Proposition
(1')
(1)
(1 ')
(2)
Let
f ∈ I(P ).
1
if
x=y
0
if
x<y
The following statements are equivalent:
f has left inverse;
f has right inverse;
f has a 2-sided inverse;
f (x, x) 6= 0 for all x ∈ P .
Proof. f g = δ = 1 is equivalent to
f g(x, x) = 1 for all x ∈ P ;
f g(x, y) = 0 for all x < y ∈ P .
Notice that
(f g)(x, y) =
X
f (x, z)g(z, y)
x≤z≤y
= f (x, x)g(x, y) +
X
f (x, z)g(z, y) = 0
∀x < y
x<z≤y
which is equivalent to
g(x, y) = −f (x, x)−1
X
f (x, z)g(z, y)
x<z≤y
So this value is OK so long as
f
has a right inverse
Applying the same argument to
f
f (x, x) 6= 0.
hf = δ
has a left inverse
Thus
⇔ f (x, x) 6= 0 ∀x ∈ P
shows
⇔ f (x, x) 6= 0 ∀x ∈ P
So
f
has a right inverse
x, one sees
g(x, y) = h(x, y) = f (x, y)−1 .
By looking at the covers of
quently
⇔ f (x, x) 6= 0 ∀x ∈ P ⇔ f
that
f (x, x)−1
has a left inverse
depends only on
[x, y]
and conse-
65
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
3.6 The Zeta Function
(3.62) Denition
The
zeta function
ζ : I(P ) → C,
i.e.
ζ
is
ζ(x, y) = 1
of every interval is equal to 1.
(3.63) Remark
Notice that
ζ 2 (x, y) = ζζ(x, y) =
X
ζ(x, z)ζ(z, y)
x≤z≤y
X
=
1 = |[x, y]|
x≤z≤y
= |{z ∈ P : x ≤ z
Consider
and
z ≤ y}|
k ∈ N:
X
ζ k (x, y) =
X
ζ(x0 , x1 )ζ(x1 , x2 ) · · · ζ(xk−1 , xk ) =
x=x0 ≤x1 ≤···≤xk =y
1
x=x0 ≤x1 ≤···≤xk =y
= |{(x0 , . . . , xk ) : x = x0 , y = xk , x0 ≤ x1 ≤ · · · ≤ xk }|
= the
number of multichains of length
(3.64) Proposition
from
x
Proof.
to
If
k ∈ N,
then
k
(ζ − 1)k (x, y)
from
x
to
y
is the number of chains of length
k
y.
Notice rst that
(
(ζ − 1)(x, y) = ζ(x, y) − 1(x, y) =
0
if
x=y
1
if
x<y
Using this,
X
(ζ − 1)k (x, y) =
(ζ − 1)(x0 , x1 ) · (ζ − 1)(x1 , x2 ) · · · (ζ − 1)(xk−1 , xk )
x=x0 ≤x1 ≤···≤xk =y
X
=
1
x=x0 <x1 <···<xk =y
= number
(3.65) Proposition
of chains in
The function
P
of length
k
(2 − ζ)−1 ∈ I(P )
from
x
to
y
counts the number of chains in an
interval.
Proof.
First
(
(2 − ζ)(x, y) = 2δ(x, y) − ζ(x, y) =
66
1
if
x=y
−1
if
x<y
CHAPTER 3.
Since
Fix
(2 − ζ)(x, x) = 1
x, y ∈ P
with
for all
x≤y
x ∈ P,
and let
l
PARTIALLY ORDERED SETS (POSETS)
the inverse
(2 − ζ)−1
exists.
be the length of the longest chain in
[x, y].
From the
previous proposition we know that
(ζ − 1)l+k (u, v) = 0
for all
x≤u≤v≤y
k∈N
and
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
(3.66) Denition
ζ
satises
ζ(x, x) = 1
The inverse
µ=
for all
ζ −1 of
P
x∈P
and so,
is called the
ζ −1
exists.
Möbius function
of
P
and is given by
µ(x, x) = 1
µ(x, y) = −
X
µ(x, z)
for all
x∈P
for all
x<y∈P
x≤z<y
f
(3.67) Example
Intervals of
Let
P
be the poset with Hasse diagram
•@@
@@ e
•@@
~~~~ @@
•@ •c nnn•
b @@@
d
nnn
•na
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
X
µ(a, b) = −
µ(a, z) = −µ(a, a) = −1
a≤z<b
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) = 1
µ(a, f ) = −µ(a, a) − µ(a, b) − µ(a, c) − µ(a, d) − µ(a, e) = −1 − 3(−1) − 1 = 1
µ(b, f ) = −µ(b, b) = −1
µ(c, e) = −µ(c, c) = −1
µ(c, f ) = −µ(c, c) − µ(c, e) = −1 − (−1) = 0
µ(d, e) = −1
µ(d, f ) = 0
µ(e, f ) = −µ(e, e) = −1
The point of all this:
(3.68) Proposition (Möbius Inversion Formula, M.I.F.)
every order ideal
hxi
g(x) =
is nite. Let
X
f, g : P → C.
f (y), ∀x ∈ P ⇔ f (x) =
y≤x
(3.69) Lemma
Let
µ
Let
P
be a poset in which
Then
X
g(y)µ(y, x), ∀x ∈ P
y≤x
be the Möbius function of a poset
µ(a, b) = −
X
P.
If
a < b,
then
µ(z, b)
a<z≤b
Proof.
From the denition of
µ
we have
µ(a, a) = 1
X
µ(a, b) = −
µ(a, z),
and
if
a<b
a≤z<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],
I(P ) remain
= µ(x, y) in P .
Alternatively: all functions in
poset). So
µP ∗ (y, x)
∗
in P
Proof (of M.I.F.). ⇒:
Let
f :P →C
g(x) =
X
y≤x
68
the result then holds.
the same when we move from
to
P∗
(dual
and dene
f (y)
P
∀x ∈ P
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
Then the sum
X
µ(y, x)g(y) =
ẏ≤x
X
µ(y, x)
ẏ≤x
X
f (z) =
ż≤y
= f (x)µ(x, x) +
µ(y, x)
z≤ẏ≤x
X
f (z)
X
f (z)
ż≤x
X
ż<x
µ(y, x) = f (x) +
z≤ẏ≤x
X
f (z) · (0)
ż<x
∀x ∈ P
= f (x)
⇐:
X
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)
every dual order ideal is nite. Let
g(x) =
X
f, g : P → C.
f (y) ∀x ∈ P
⇔
Let
P
be a poset in which
Then
f (x) =
x≤ẏ
X
µ(x, y)g(y) ∀x ∈ P
x≤ẏ
3.8 Applications of the Möbius Inversion Formula
(3.71) Example
(1) Consider the total order on
P = (N, ≤).
P:
For each
i ∈ N, hii = {1 ≤ j ≤ i}
is
nite. The Möbius function of
If
µ(i, i) = 1
i≥1
µ(i, i + 1) = −µ(i, i) = −1
i≥1
µ(i, i + k) = 0
k > 1, i ≥ 1
f, g : P → C,
then M.I.F. gives
g(x) =
X
f (y) =
y≤x
x
X
f (y)
y=1
i
f (x) =
X
µ(y, x)g(y) =
x
X
µ(y, x)g(y) = µ(x, x)g(x) + µ(x − 1, x)g(x − 1)
y=1
y≤x
= g(x) − g(x − 1)
X1 , . . . , Xn ⊆ S
P be the poset of all intersections of
1̂ = X1 ∪ · · · ∪ Xn ∈ P . We want to give
an expression for |X1 ∪ · · · ∪ Xn | in terms og the sizes |Xi | and their intersections.
Dene two functions on P :
(2) Let
be nite sets and let
these sets, ordered by inclusion. Also let
g(x) = #
elements in set
x
f (x) = #
elements in set
x
that do not belong to any set
x0 < x
69
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
g(x) =
X
f (x)
y≤x
Notice that
f (1̂) = 0
so
0 = f (1̂) =
X
g(y)µ(y, 1̂)
by M.I.F.
y≤1̂
Thus
0=
X
|y|µ(y, 1̂) = |1̂|µ(1̂, 1̂) +
y≤1̂
X
|y|µ(y, 1̂)
y<1̂
i.e.
|X1 ∪ · · · ∪ Xn | = −
X
|y|µ(y, 1̂)
y<1̂
If we let
n = 3:
X1 ∪X2 ∪X3 =1̂
•@
~~ @@@
~
@
~ β =X
2 @2
~~
β1 =X1 •@@ ~•@@ ~• β3 =X3
@~@~ @~@~
~ @@ ~~ @@
~
•@~@ •~α2 =X
•1 ∩X3
α1 =X1 ∩X2
@@
~~ α3 =X2 ∩X3
@@ ~~~
•~
X1 ∩X2 ∩X3 =0̂
then
µ(βi , 1̂) = −1,
µ(αi , 1̂) = 1,
µ(0̂, 1̂) = −1
Thus
|X1 ∪ X2 ∪ X3 | = −|X1 |µ(X1 , 1̂) − |X2 |µ(X2 , 1̂) − |X3 |µ(X3 , 1̂)
− |X1 ∩ X2 |µ(X1 ∩ X2 , 1̂) − · · · − |X2 ∩ X3 |µ(X2 ∩ X3 , 1̂)
− |X1 ∩ X2 ∩ X3 |µ(X1 ∩ X2 ∩ X3 , 1̂)
= |X1 | + |X2 | + |X3 | − |X1 ∩ X2 | − |X1 ∩ X3 | − |X2 ∩ X3 |
+ |X1 ∩ X2 ∩ X3 |
(3) The poset
Lemma
(Bn ≤)
and the Principle of Inclusion-Exclusion:
P = (Bn , ≤)
x ≤ y ∈ Bn
Let
inclusion. For
be the poset of all subsets of
{1, . . . , n}
ordered by
µ(x, y) = (−1)|y|−|x|
Proof.
x ≤ y and r = ρ(y) − ρ(x) = |y| − |x|. It is an easy
exercise 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.)
•{1}=1̂
For n = 1:
⇒ µ(0, 1̂) = (−1)1−0 .
•∅
For n = 2:
Same.
70
Let
x, y ∈ Bn
with
CHAPTER 3.
By induction on
[0, 1̂].
For
n,
the above map
PARTIALLY ORDERED SETS (POSETS)
ψx
shows the result to be true for all
[0, 1̂],
µ(0, 1̂) = −
X
µ(0̂, x) = −
x<1̂
X
(−1)|x| = −
n−1
X
i=0
x<1̂
n
n
[x, y] 6=
n
(−1)i
i
n
= − [(1 − 1) − (−1) ] = (−1)
Bn
So, Möbius inversion for
g(x) =
X
f, g : Bn → C
X
(−1)|x|−|y| g(y) ∀x ∈ Bn
f (x) =
tells us that for
f (y) ∀x ∈ Bn
⇔
y≤x
y≤x
g(x) =
X
f (y) ∀x ∈ Bn
⇔
f (x) =
An example of a
D(n) = #
(−1)|y|−|x| g(y) ∀x ∈ Bn
x≤y
x≤y
Let
X
(f, g)
Bn :
{1, . . . , n} = #{π ∈ Sn : π1 6= 1, . . . , πn 6= n}
D(0) = 1, D(1) = 0, D(2) = 1, D(3) = 2. For π ∈ Sn let
pair on
derangements of
(no xed points). E.g.
fix(π) = {i : πi = i}.
Dene
f (T ) = |{π ∈ Sn : fix(π) = T }|
g(T ) = |{π ∈ Sn : fix(π) ⊇ T }|
T ⊆ {1, . . . , n}, T ∈ Bn .
f (∅) = f (0̂).
for all
(Note:
g(T ) = (n − |T |)! ).
We want to nd
Since
g(T ) =
X
f (T 0 )
∀T ∈ Bn
T ⊆T 0
the D.M.I.F. gives
f (T ) =
X
(−1)|T
0 |−|T |
g(T 0 )
∀T ∈ Bn
T ⊆T 0
Thus
f (∅) =
X
0
(−1)|T | (n − |T 0 |)! =
∅⊆T 0
n
X
n X
n
(−1)i (n − i)!
i
i=0
n
X
n!
(−1)i
(−1)i (n − i)! = n!
(n − i)!i!
i!
i=0
i=0
1
1
1
(−1)n
= n! 1 − + − + · · · +
1 2! 3!
n!
n
1
1
(−1)
= n!
− + ··· +
2! 3!
n!
=
E.g. for
n = 3:
3!( 2!1 −
1
3! )
= 3 − 1 = 2.
As another example, recall problem sheet 5, q. 3:
Given
X ⊆ {1, . . . , n − 1},
let
α(X) = |{π ∈ Sn : des π ⊆ X}|
71
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
X = {x1 , . . . , xk }, then
n
α(X) =
x1 , x2 − x1 , . . . , n − xk
It was shown that if
What is
β(X) = |{π ∈ Sn : des π = X}| ?
α, β : Bn−1 → C
From the Möbius Inversion Formula:
α(X) =
X
β(Y )
for all
X ⊆ {1, . . . , n − 1}
Y ⊆X
and thus
β(X) =
X
α(Y )µ(Y, X) =
Y ⊆X
X
Y ⊆X
X
=
k−j
(−1)
1≤i1 <i2 <···<ij ≤k
(4)
α(Y )(−1)|X|−|Y |
n
,
xi1 , xi2 − xi1 , . . . , n − xik
Classical Möbius function in Number theory:
of
(Dn , ≤)
k = |X|
is the lattice of divisors
n.
µ(1, 1) = 1 = µ(2, 2) = µ(3, 3) = . . .
2 31
D12 :
12=2
~•@@@
~
~~
@@
o•6=21 31
o
oo
o
o
o
•@o@
•
2=21 30 @@@ ~~~ 3=20 31
~
•~
4=22 30
•~
µ(1, 2) = −1 = µ(1, 3)
µ(1, 2 × 2) = −µ(1, 1) − µ(1, 2) = 0
µ(1, 2 × 3) = −µ(1, 1) − µ(1, 2) − µ(1, 3) = 1
µ(1, 2 × 2 × 3) = −µ(1, 1) − µ(1, 2) − µ(1, 3)
1=20 30
−µ(1, 2 × 2) − µ(1, 2 × 3)
=1−1=0
Lemma
x, y ∈ Dn , x ≤ y , then

t
y

 (−1) if /x is a product
1
if x = y
µ(y/x) = µ(x, y) =


0
otherwise.
Set
D=
Given
of
t
distinct primes
innite 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) =
X
f (d)
for all integers
m,
d|m
then
f (m) =
X
d|m
72
g(d)µ(m/d)
∀m ∈ N
CHAPTER 3.
As an example, let
f (x) = x.
g(m) =
PARTIALLY ORDERED SETS (POSETS)
Then
X
d=
sum of divisors of
m
d|m
and
m = f (m) =
X
g(d)µ(m/d) .
d|m
f (x) = 1,
Or if
then
g(m) =
X
1=#
of divisors of
m
d|m
and
1 = f (m) =
X
g(d)µ(m/d)
d|m
Ln (q)
(5) The Möbius function of
partitions of
for
(lattice of subspaces of
Vn (q))
and
Πn
(lattice of
{1, . . . , n}):
Ln (q),
n
µ(0̂, 1̂) = (−1)n q ( 2 )
for
Πn ,
µ(0̂, 1̂) = (−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.)
covered by
Let
L
be a nite lattice with
1̂
covering
{x1 , . . . , xk }
and
0̂
{y1 , . . . , ym }.
(3.73) Remark
{x1 , . . . , xk }
If
x1 ∧ x2 ∧ . . . ∧ xk 6= 0̂
then
µ(0̂, 1̂) = 0
If
y1 ∨ y2 ∨ . . . ∨ ym 6= 1̂
then
µ(0̂, 1̂) = 0
The elements
covered by
1̂
(3.74) Lemma (3.9.2.)
µ(x, y) =
{y1 , . . . , ym } covering 0̂ are called atoms. The elements
coatoms.
are called
Let
L
be a nite distributive lattice. For
x, y ∈ L,

`(x,y)

 (−1)
[x, y] is a Boolean algebra
(i.e. ∼
= Bk for some k)


otherwise
0
if
73
CHAPTER 3.
Proof.
PARTIALLY ORDERED SETS (POSETS)
L∼
= J(P ) for some poset P . If x, y ∈ L with
x ≤ y , then there are antichains X, Y in P such that y corresponds to the order ideal
hY i and X corresponds to the order ideal hXi. So, since x ≤ y , hXi ⊆P hY i and Y \ X
is an antichain of P . Thus in J(P ) the sublattice [hXi, hY i] ∼
= [0̂, hY \ Xi]. Using this
L
Recall that if
is distributive, then
argument it is clear that
hY i
is the join of covers of
From the previous lemma (3.72), if
µ(x, y) = 0.
hXi ⇔ [x, y]
hY i
is a boolean algebra
is not the join of the covers of
hXi,
Hence the result.
(3.75) Lemma (3.9.3.)
Let
0̂ = x0 < x1 < · · · < xi = 1̂
then
P
be a nite poset with a
i.
of length
0̂
and
1̂.
Let
ci = #
chains
Then
µ(0̂, 1̂) = c0 − c1 + c2 − c3 + c4 − · · ·
(Related to the Euler characteristic).
Proof.
The function
µ = (1 + (ζ − 1))−1 ,
so
µ(0̂, 1̂) = (1 + (ζ − 1))−1 (0̂, 1̂) = (1 − (ζ − 1) + (ζ − 1)2 − (ζ − 1)3 + · · · )(0̂, 1̂)
= 1(0̂, 1̂) − (ζ − 1)(0̂, 1̂) + (ζ − 1)2 (0̂, 1̂) − · · ·
= c0 − c1 + c2 − c3 + · · ·
since
ci = (ζ − 1)i (0̂, 1̂)
by Prop. (3.64).
3.10 Young's lattice
From chapter 1, we have
Par(n) =
set of partitions of the integer
n
Dene
Par = Par(0) ∪ Par(1) ∪ · · ·
Then
The sets
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}
Par(n) are easily visualised as Young diagrams. Given λ, µ ∈ Par,
λ is contained in the shape of µ. For example
if the shape of
32 =
74
⊆
= 4211
dene
λ⊆µ
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
but
6⊆
21 =
Let
Y = (Par, ⊆)
Par
be the poset on
= 111
with this inclusion/containment order
•NNN
p•NN
p•
NNN
ppp 21 NNNNN
ppp 3
p
p
p
p
NNN
p
NNN
NN ppppp
N pppp
•NNN
•
p
pp 2
NNN
11
ppp
NNN
p
p
NN ppp
111
•1
•∅
(3.76) Theorem
Proof.
Given
contains both
Y
is a lattice.
λ = (λ1 , λ2 , . . .), µ = (µ1 , µ2 , . . .) ∈ Y ,
λ and µ is
the smallest shape (diagram) that
λ ∨ µ = (max(λ1 , µ1 ), max(λ2 , µ2 ), . . .)
and similarly
λ ∧ µ = (min(λ1 , µ1 ), (λ2 , µ2 ), . . .)
It is straightforward to check that conditions (1)(4) hold for
(3.77) Remark
The rank function
ρ:Y→N
(Y, ∨, ∧).
is given by
ρ(λ) = λ1 + λ2 + · · ·
By noticing that
|{λ ∈ Par : ρ(λ) = n}| = p(n) = #
the rank generating function of
Y
partitions of the integer
n
is
F (Y, q) =
+∞
X
n
p(n)q =
n=0
+∞
Y
n=0
1
1 − qn
(see problem sheet 4).
(3.78) Proposition
Maximal chains
dence with all standard Young tableaux of
(3.79) Remark
i,
put
i
in
are in 1-1 correspon-
Draw the corresponding Young diagram at each element in Young's
lattice. Choose your element
step
Y , where ρ(λ) = n,
shape λ (SYTλ ).
[0̂, λ]
λ
in the lattice. As you proceed from
0̂
up to
λ,
in each
in the box that was added to the Young diagram from last step, e.g.
1 3
∅→ 1 → 1 → 1 3 → 2
=λ
2
2
4
if we choose
λ = 211
and our route is
0̂ → 1 → 11 → 21 → 211.
(Draw this!)
75
CHAPTER 3.
Proof.
Let
PARTIALLY ORDERED SETS (POSETS)
0̂ = y0 <· y1 <· · · · <· yn = λ
yi and yi ⊂ yi+1 , label
one more cell than
0 ≤ i < n. The
of shape λ. The
yi+1 has exactly
i + 1. Do this for all
be a maximal chain. Since
the cell
yi+1 − yi
with
resulting labeling of the Young diagram is a standard Young tableaux
inverse map is easy to describe. A consequence of this is the number of
[0̂, λ] is f λ (calculated by the hook-length formula). Also the number
of maximal chains [0̂, λ] for all λ where ρ(λ) = n, is equal to In , the number of involutions
maximal chains in
{π ∈ Sn : π 2 = id}
(3.80) Denition
with
dominance order (Par(n), ≤) is such that if λ, µ ∈ Par(n)
λ = (λ1 , λ2 , . . .) and µ = (µ1 , µ2 , . . .) then we dene
The
λ≤µ
⇔
λ 1 + · · · + λ i ≤ µ 1 + · · · + µi
∀i
3.11 Linear extensions
(3.81) Denition
P
Let
(P, ≤P )
be a poset with
|P | = n.
A
linear extension
of
is a bijective order preserving map
σ : P → {1, . . . , n}
such that
σ −1 (i) <P σ −1 (j)
if
The number of linear extensions of
(3.82) Example
If
Consider P:
is denoted
c •@@
•d
a•
•b
However if
which is
σ
e(P ).
then
σ(P )
is
3 •@@
•4
1•
•2
and
@@
σ is
a linear
P.
is chosen to be
(σ(a), σ(b), σ(c), σ(d)) = (3, 2, 1, 4)
then
σ(P )
not a linear extension.
All linear extensions of
•4
3 •@@
•4
4 •@@
•3
4 •@@
•3
4 •@@
•2
1•
•2
2•
•1
1•
•2
2•
•1
3•
•1
@@
e(P ) = 5.
is
1 •@@
•4
3•
•2
@@
P:
3 •@@
σ=(1,2,3,4)
Thus
i < j.
@@
σ = (σ(a), σ(b), σ(c), σ(d)) = (1, 2, 3, 4)
extension of
76
P
then
@@
σ=(2,1,3,4)
@@
σ=(1,2,4,3)
@@
σ=(2,1,4,3)
@@
σ=(3,1,4,2)
♦
CHAPTER 3.
PARTIALLY ORDERED SETS (POSETS)
(3.83) Remark
P = •|
• {z· · ·
•}
n
P =

• 



• 



.
.
.
e(P ) = n!
e(P ) = 1
n



• 




•
(3.84) Proposition
e(P ) = #
Proof.
maximal chains in
J(P )
Homework.
(3.85) Denition
then
P
P
If a poset P on elements {1, . . . , n} is such that i <P j ⇒ i < j ,
natural poset on {1, . . . , n}. In this case, each linear extension of
−1 (1), σ −1 (2), . . . , σ −1 (n)) ∈ S .
identied with a permutation (σ
n
is called a
may be
(3.86) Denition
If
P
Hölder set of P .
(3.87) Example
For
{1, . . . , n}, then the set of permutations
P is denoted L(P ) and is called the Jordan-
is a natural poset on
σ −1 corresponding to linear extensions of
P = 3•@@@ •4 , L(P ) = {1234, 2134, 1243, 2143, 2413}.
@
•2
1•
♦
3.12 Rank-selection
Let
P
be a nite graded poset of rank
n with a 0̂ and 1̂. Let ρ be the rank function of P .
(3.88) Denition
Let S ⊆ {0, . . . , n} and PS = {x ∈ P : ρ(x) ∈ S}. PS is called the
S -rank-selected subposet of P . Let α(P, S) = # maximal chains in PS . Since
P has a 0̂ and 1̂, the quantity α(P, S) can be restricted to S ⊆ {1, . . . , n − 1}.
Dene for
S ⊆ {0, . . . , n},
β(P, S) =
X
(−1)|S|−|T | α(P, T ) .
T ⊆S
The Möbius inversion formula gives
α(P, S) =
X
β(P, T ) .
T ⊆S
77
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 that
α(L, S) is the number of chains I1 , . . . , Ik of order ideals of P such that |Ii | = ai for all
i = 1, . . . , k . For such a chain
arrange the elements of
arrange the elements of
I1 in increasing order;
I2 − I1 in increasing order;
.
.
.
arrange the elements of
This maps chains
P − Ik
in increasing order.
I ∈ LS → π ∈ L(P )
and this map is bijective. Also
Des(π) ⊆ S .
If we dene
γ(L, S) = {π ∈ L(P ) : Des(π) = S}
then
α(L, S) =
X
γ(L, S)
T ⊆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 :
•
|
• {z· · ·
n
•}
⇒
J(P ) = Bn
From the previous proposition, and using the fact that
78
L(P ) = Sn ,
the result follows.