Re: modules over a group

[Patrik Eklund <peklund@cs.umu.se>]

Hi Michael and Paul,

These are indeed interesting observations and potential developments.

Let me try to connect this also to previous postings on automata and
"Turing categories". In that context, A and B are alphabets and states,
and maybe in rather abstract forms, but A and B are fundamentally
different. Whereas A is intuitively closer to terms so that TA can be a
construction by a term monad, B are statements and the "statement
functor" is usually not seen to be extendable to a monad (otherwise
there are only terms and no statements).

  From application point of of view, and indeed referring to possible
categorization of automata, T is more of a generalized powerfunctor, so
the term monad is hidden somewhere inside X, Y or Z. If so, and if that
term monad is made explicit, distributive laws will come into play.

Generalized powers may involve many-valuedness and as related to truth
value algebras. Such things invites to believing that applications
related to many-valuedness in ontology may become feasible. In some
previous postings I have tried to advertise that health care is a
typical area where this kind of algebraically explained many-valuedness
seems to come into play.

Funny, maybe, but the underpinnings may be quite serious, by the looks
of it. Maybe even dramatic e.g. for an extension of the notion of
evidence in evidence-based medicine.

Best,

Patrik



On 2017-08-26 13:35, Paul Blain Levy wrote:
> Funny you should ask that; I was wondering about a similar question.
>
> Suppose T is a monad on a category V.
>
> Suppose that V has a monoidal structure *.
>
> A right strength for T is a natural transformation TX * Y --> T(X * Y)
> satisfying 4 equations.
>
> A left strength for T is a natural transformation X * TY --> T(X * Y)
> satisfying 4 equations.
>
> A bistrength for T consists of a right strength and left strength such
> that the two maps X * TY * Z --> T(X * Y * Z) are equal.
>
> A bistrength is commutative if the two maps TX * TY --> T(X * Y) are
> equal.
>
> [Kock proved that Monad with commutative bistrength = Monoidal monad.
> I
> recently learnt this from Paul Taylor.]
>
> If * has a symmetry sigma, then any right strength  t gives rise to a
> left strength t-sigma, and (t, t-sigma) forms a bistrength.
>
> But is it possible that T has another left strength s such that (t,s)
> is
> a bistrength?
>
> What if we assume the symmetric monoidal structure to be cartesian?
>
> What if we assume that both (t,t-sigma) and (t,s) are commutative?
>
> Paul
>
>
>
>
> On 25/08/17 23:18, barr@math.mcgill.ca wrote:
>> If pi is a group, are the categories of right pi-modules and of
>> 2-sided
>> pi-modules equivalent?
>>
>> The reason I raise this question is that I was looking at Example 5,
>> p.
>> 43 in the TAC reprint of Jon Beck's thesis.  He identifies the abelian
>> group objects of Gp/pi as right pi-modules via the following
>> construction.  Let M be a right pi-module and let Y = pi x M with
>> multiplication (x,m)(x',m') = (xx',m+m'x).  This is an abelian group
>> object in Gp/pi and for any Z --> pi, Hom(Z,Y) = Der(Z,M) the group of
>> derivations d: Z --> M meaning that d(zz') = (dz)z' + d(z').  Here M
>> becomes a right Z-module using the map Z --> Y.  For the converse, if
>> Y
>> --> pi is an abelian group object in Gp/pi, then the kernel of the map
>> is an abelian normal subgroup M of Y and the conjugation action of Y
>> on
>> M descends to pi since M is abelian and this conjugation action is a
>> right pi-module structure (see Beck's thesis for all details).
>>
>> But suppose M is a 2-sided pi-module.  Now let Y = pi x M and define
>> (x,m)(x',m') = (xx',mx'+xm').  This is just as above an abelian group
>> structure in Gp/pi.  For any Z --> pi, Hom(Z,Y) = Der(Z,M) as above
>> but
>> now Der means 2-sided derivations d(zz') = (dz)z' + z(dz'), still an
>> abelian group.  But the kernel of Y --> pi is M but now with a new
>> right
>> module structure m*x = x^{-1}mx.  So there is a functor from 2-sided
>> to
>> right modules by replacing the 2-sided operation by conjugation which
>> is
>> now a right operation, but could this possibly be an equivalence?  I
>> don't see how.
>>
>> Michael
>>
>
>
>
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