geometric lattice

geometric lattice

A

lattice

is said to be

geometric

if it is

1.

algebraic

(

http://planetmath.org/AlgebraicLattice

),

2.

semimodular (

http://planetmath.org/SemimodularLattice

), and

3.

each

compact element

is a join of atoms.

By the definition of

compactness

, the last condition is

equivalent

to “each compact element is a finite join of atoms”.

Three examples that come to mind are

•

the

power set

of a set;

•

an

incidence geometry

with the

empty set

adjoined to form the bottom

element

; and

•

a

projective geometry

(the lattice of

subspaces

of a vector space).

From the last two examples, one sees how the name “geometric” lattice is derived.

To generate geometric lattices from existing ones, one has the following

Theorem

.

Any

lattice interval

of a

geometric lattice

is also geometric.

Proof.

Let

L

be a geometric lattice and

I

=

[

x

,

y

]

a lattice interval of

L

. We first prove that

I

is algebraic, that is,

I

is both

complete

and that every element is a join of compact elements. Since

L

is complete, both

⋁

S

and

⋀

S

exist in

L

for any subset

S

⊆

L

. Since

x

≤

s

≤

y

for each

s

∈

S

,

⋁

S

and

⋀

S

are in fact in

I

. So

I

is a

complete lattice

.

Now, suppose that

a

∈

I

. Since

L

is algebraic,

a

is a join of compact elements in

L

:

a

=

⋁

i

a

i

, where each

a

i

is

compact

in

L

. Since

a

i

≤

y

, the elements

b

i

:=

a

i

∨

x

are in

I

for each

i

. So

a

=

a

∨

x

=

(

⋁

i

a

i

)

∨

x

=

⋁

i

(

a

i

∨

x

)

=

⋁

i

b

i

. We want to show that each

b

i

is compact in

I

. Since

a

i

is compact in

L

,

a

i

=

⋁

k

=

1

m

α

k

, where

α

k

are atoms in

L

. Then

b

i

=

(

⋁

k

=

1

m

α

k

)

∨

x

=

⋁

k

=

1

m

(

α

k

∨

x

)

. Let

S

be a subset of

I

such that

α

k

∨

x

≤

⋁

S

. Since

α

k

≤

⋁

S

and

α

k

is an atom in

L

and hence compact, there is a finite subset

F

⊆

S

such that

α

k

≤

⋁

F

. Because

F

⊆

I

,

x

≤

⋁

F

, and so

α

k

∨

x

≤

⋁

F

, meaning that

α

k

∨

x

is compact in

I

. This shows that

b

i

, as a finite join of compact elements in

I

, is compact in

I

as well. In turn, this shows that

a

is a join of compact elements in

I

.

Since

I

is both complete and each of its elements is a join of compact elements,

I

is algebraic.

Next, we show that

I

is semimodular. If

c

,

d

∈

I

with

c

∧

d

≺

c

(

c

∧

d

is

covered (

http://planetmath.org/CoveringRelation

)

by

c

). Since

L

is semimodular,

d

≺

c

∨

d

. As

c

∨

d

is the least

upper bound

of

{

c

,

d

}

,

c

∨

d

≤

y

, and thus

c

∨

d

∈

I

. So

I

is semimodular.

Finally, we show that every compact element of

I

is a finite join of atoms in

I

. Suppose

a

∈

I

is compact. Then certainly

a

≤

⋁

I

. Consequently,

a

≤

⋁

J

for some finite subset

J

of

I

. But since

L

is

atomistic

, each element in

J

is a join of atoms in

L

. Take the join of each of the atoms with

x

, we get either

x

or an atom in

I

. Thus, each element in

J

is a join of atoms in

I

and hence

a

is a join of atoms in

I

.
∎

Note that in the above proof,

b

i

is in fact a finite join of atoms in

I

, for if

α

k

≤

x

, then

α

k

∨

x

=

x

. Otherwise,

α

k

∨

x

covers

x

(since

L

is semimodular), which means that

α

k

∨

x

is an atom in

I

.

Remark

. In

matroid

theory, where geometric lattices play an important role, lattices considered are generally assumed to be finite. Therefore, any lattice in this

context

is automatically complete and every element is compact. As a result, any finite lattice is geometric if it is semimodular and atomistic.

Title

geometric lattice

Canonical name

GeometricLattice

Date of creation

2013-03-22 15:57:32

Last modified on

2013-03-22 15:57:32

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

12

Author

CWoo (3771)

Entry type

Definition

Classification

msc 05B35

Classification

msc 06C10

Classification

msc 51D25

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