- Dear all,

In this question, I use the category-theoretic definition

of an epimorphism.

http://en.wikipedia.org/wiki/Epimorphism

In the category FDL of finite distributive latices with

0,1-morphisms a morphism f:L -> K is an epimorphism

if and only the if height of the range of f is equal to the

height of K.

This follows from the facts that

* FDL is dually isomorphic to the category FinPos

of finite posets and

* that a morphism in FinPos

is monic iff it is an injective mapping.

So, there are non-surjective epimorphisms in FDL;

for example

L = { {} , {1} ,{1,2} }

K = 2^{1,2}

f:L->K f(x)=x

Question: is there a similar characterization of

epimorphisms in the category FML of finite modular

lattices with 0,1-morphisms? The characterization

I mention above clearly fails. To see this,

let M=M_3 and embed K->M by g_1,g_2 so that

g_1\neq g_2 but they coincide on the range of f.

An easier question: is there a non-surjective epimorphism

in FML?

Have a nice day,

--

Gejza Jenca - There is a countable, simple modular lattice L of dimension 4 such that if a is covered by b in L and f : L -> M is an embedding into a modular lattice M, then the interval [f(a), f(b)] is distributive. Using this one can show that epimorphisms are not onto for modular lattices (and for finite dimensional modular lattices). But I'm not sure about finite modular lattices.

Ralph Freese

On Jun 13, 2012, at 8:31 AM, Gejza Jenca <gejza.jenca@...> wrote:

> Dear all,

>

> In this question, I use the category-theoretic definition

> of an epimorphism.

>

> http://en.wikipedia.org/wiki/Epimorphism

>

> In the category FDL of finite distributive latices with

> 0,1-morphisms a morphism f:L -> K is an epimorphism

> if and only the if height of the range of f is equal to the

> height of K.

>

> This follows from the facts that

>

> * FDL is dually isomorphic to the category FinPos

> of finite posets and

> * that a morphism in FinPos

> is monic iff it is an injective mapping.

>

> So, there are non-surjective epimorphisms in FDL;

> for example

>

> L = { {} , {1} ,{1,2} }

> K = 2^{1,2}

>

> f:L->K f(x)=x

>

> Question: is there a similar characterization of

> epimorphisms in the category FML of finite modular

> lattices with 0,1-morphisms? The characterization

> I mention above clearly fails. To see this,

> let M=M_3 and embed K->M by g_1,g_2 so that

> g_1\neq g_2 but they coincide on the range of f.

>

> An easier question: is there a non-surjective epimorphism

> in FML?

>

>

> Have a nice day,

> --

> Gejza Jenca

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

> - 2012/6/17 Ralph Freese <ralph@...>
>

Thank you.

>

>

>

> There is a countable, simple modular lattice L of dimension 4 such that if a

> is covered by b in L and f : L -> M is an embedding into a modular lattice

> M, then the interval [f(a), f(b)] is distributive. Using this one can show

> that epimorphisms are not onto for modular lattices (and for finite

> dimensional modular lattices). But I'm not sure about finite modular

> lattices.

>

> Ralph Freese

>

Could you please provide a reference for the fact you mentioned?

--

Gejza Jenca - First let me apologize for being so slow in responding---I've been

traveling.

The paper is "The variety of modular lattices is not generated by its

finite members," Trans AMS 255, 1979. The lattice is easy to describe: let

F and F' be countable fields of characteristics p and q, and let L and L'

be the lattice of subspaces of a 4 dimensional vectors space over F (resp.

F'). The two dimensional intervals of both lattices are M_\omega and so we

can glue (via Hall-Dilworth) an upper interval of L to a lower interval

of L'. The result is a 5-generated, simple modular lattice M of length 6.

It has the property that prime (covering) intervals "stay distributive" in

extensions. "Stays distributive" is the property defined in the message of

2012/6/17 below. The proof is somewhat technical.

When a lattice L has a covering with this "stays distributive" property,

then the embedding of L(\leq) = {(a,b) in L^2 : a \leq b} into L^2 is an

epimorphism that is not onto.

Talking with Keith Kearnes we realized that a nondesarguean

projective plane (as a modular lattice) also has this "stays distributive"

property, showing epimorphisms need not be onto in the category of finite

modular lattices as well.

The advantage of the lattice M above is that, while it is not a projective

modular lattice, much of its structure can be pulled back through

homomorphisms. Using this it is possible to show that there are elements s

< t in FM(5) such that the interval [s,t] is distributive (and "stays

distributive"). If you now take a larger generating set X and for each x

not in the original 5 generators, let x' = (x \join s) \meet t, then these

generate a sublattice that is the free distributive lattice. Thus every

sublattice of a free distributive lattice can be embedded into a free

modular lattice. (On he other hand distributive sublattices of free

lattices are quite restricted; in particular at most countable.)

The lattice M is finite dimensional but not in the variety generated by

all finite modular lattices. Christian Herrmann has shown that the variety

of modular lattices is not generated by its finite dimension members. So

the varieties of modular lattices generated by the finite members, the

finite dimension members, and all members are distinct.

Ralph

On Mon, 18 Jun 2012, Gejza Jenca wrote:

> 2012/6/17 Ralph Freese <ralph@...>

>>

>>

>>

>>

>> There is a countable, simple modular lattice L of dimension 4 such that if a

>> is covered by b in L and f : L -> M is an embedding into a modular lattice

>> M, then the interval [f(a), f(b)] is distributive. Using this one can show

>> that epimorphisms are not onto for modular lattices (and for finite

>> dimensional modular lattices). But I'm not sure about finite modular

>> lattices.

>>

>> Ralph Freese

>>

>

> Thank you.

>

> Could you please provide a reference for the fact you mentioned?

>

> --

> Gejza Jenca

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

> - PS. I should have mentioned that this ability to control the equations

within an entire interval in the embedding's target began with Bjarni

Jonsson, who showed that if a and b are elements of M_3, a covered by b,

and f: M_3 to L (L modular), the [f(a),f(b)] is arguesian.

Bjarni asked the question which distributive lattices can be embedded into

a free modular lattice and this is the genesis of the results of the paper

described below. And despite the results below, that question is still

open.

Ralph

On Sat, 28 Jul 2012, Ralph Freese wrote:

>

> First let me apologize for being so slow in responding---I've been

> traveling.

>

> The paper is "The variety of modular lattices is not generated by its

> finite members," Trans AMS 255, 1979. The lattice is easy to describe: let

> F and F' be countable fields of characteristics p and q, and let L and L'

> be the lattice of subspaces of a 4 dimensional vectors space over F (resp.

> F'). The two dimensional intervals of both lattices are M_\omega and so we

> can glue (via Hall-Dilworth) an upper interval of L to a lower interval

> of L'. The result is a 5-generated, simple modular lattice M of length 6.

> It has the property that prime (covering) intervals "stay distributive" in

> extensions. "Stays distributive" is the property defined in the message of

> 2012/6/17 below. The proof is somewhat technical.

>

> When a lattice L has a covering with this "stays distributive" property,

> then the embedding of L(\leq) = {(a,b) in L^2 : a \leq b} into L^2 is an

> epimorphism that is not onto.

>

> Talking with Keith Kearnes we realized that a nondesarguean

> projective plane (as a modular lattice) also has this "stays distributive"

> property, showing epimorphisms need not be onto in the category of finite

> modular lattices as well.

>

> The advantage of the lattice M above is that, while it is not a projective

> modular lattice, much of its structure can be pulled back through

> homomorphisms. Using this it is possible to show that there are elements s

> < t in FM(5) such that the interval [s,t] is distributive (and "stays

> distributive"). If you now take a larger generating set X and for each x

> not in the original 5 generators, let x' = (x \join s) \meet t, then these

> generate a sublattice that is the free distributive lattice. Thus every

> sublattice of a free distributive lattice can be embedded into a free

> modular lattice. (On he other hand distributive sublattices of free

> lattices are quite restricted; in particular at most countable.)

>

> The lattice M is finite dimensional but not in the variety generated by

> all finite modular lattices. Christian Herrmann has shown that the variety

> of modular lattices is not generated by its finite dimension members. So

> the varieties of modular lattices generated by the finite members, the

> finite dimension members, and all members are distinct.

>

> Ralph

>

>

> On Mon, 18 Jun 2012, Gejza Jenca wrote:

>

>> 2012/6/17 Ralph Freese <ralph@...>

>>>

>>>

>>>

>>>

>>> There is a countable, simple modular lattice L of dimension 4 such that if a

>>> is covered by b in L and f : L -> M is an embedding into a modular lattice

>>> M, then the interval [f(a), f(b)] is distributive. Using this one can show

>>> that epimorphisms are not onto for modular lattices (and for finite

>>> dimensional modular lattices). But I'm not sure about finite modular

>>> lattices.

>>>

>>> Ralph Freese

>>>

>>

>> Thank you.

>>

>> Could you please provide a reference for the fact you mentioned?

>>

>> --

>> Gejza Jenca

>>

>>

>> ------------------------------------

>>

>> Yahoo! Groups Links

>>

>>

>>

>>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

>