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## Re: [univalg] Epimorphisms of finite modular lattices

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• 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
Message 1 of 5 , Jul 28, 2012
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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
Message 2 of 5 , Jul 28, 2012
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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
>
>
>
>
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