modules over a group

modules over a group

[Michael Barr]

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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Re: modules over a group

[Paul Blain Levy]

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
>



[For admin and other information see: http://www.mta.ca/~cat-dist/ ]

Re: modules over a group

[Michael Barr]

Okay, your argument seems sound.  But consider the following.  If Z is a 
free group, say Z = F(X), then you can show that for any Z-module M (if M 
is a pi-module a homomorphism Z --> pi makes M into a Z-module of course), 
then for either a right or a 2-sided module, it is trivial to see that 
Der(Z,M) = M^X.  That is, any function X --> M extends in a unique way to 
a derivation Z --> M.  Now resolve an arbitrary Z with G^2Z ===> GZ ---> Z 
and then we get an equalizer Der(Z,M) ---> Der(GZ,M) ===> Der(G^2Z,M) or 
Der(Z,M) ---> Der(GC,M) ===> Der(G^2Z,M) which is Der(GZ,M) ---> M^Z ===> 
M^{GZ} for whichever kind of module you have.  But, the ismorphisms are 
not natural (except with respect to maps of the basis, which G\epsilon is, 
but \epsilon G isn't) and so the two versions of Der(Z,M) may well be 
different and hence the two versions of Y.

I guess Jon didn't correctly describe the category of pi-modules.

Michael

----- Original Message -----
From: "George Janelidze" <George.Janelidze@uct.ac.za>
To: "Michael Barr" <barr@math.mcgill.ca>, "Categories list\" <Categories list>" <categories@mta.ca>
Sent: Saturday, August 26, 2017 6:47:42 AM
Subject: Re: categories: modules over a group

Dear Michael,

Although the answer to your question is "Of course not!", it is a very
interesting question because it inspires very interesting questions about
internal actions in semi-abelian categories. In this message I shall,
however, only explain why is it "Of course not!" (with "!").

1. The category of pi-modules can be identified with the category of
Z[pi]-modules, where Z[pi] is the group ring over pi. Now, suppose, for
simplicity, that pi is abelian. Then you question can be stated as:

Are the rings Z[pi] and Z[pi x pi] Morita equivalent?

Of course not, because two commutative rings are Morita equivalent if and
only if they are isomorphic. Well, for some funny groups they are, but I
know you are not asking asking about funny groups.

2. Your functor from 2-sided modules to split epis is fine, but it is not an
equivalence, e.g. because your structure on Y will not allow you to recover
the pi-module structure on M from the Beck module structure on Y-->pi. In
fact one can prove the following:

Two 2-sided pi-module structures * and # on M give the same Beck modules
structures on a chosen pi-->Y-->pi if and only if

x*m*(inverse of x) = x#m#(inverse of x) for every x in pi and every m in M.

For example if we assume that both left actions are trivial, then the right
actions will have to coincide, as in Beck's theory.

Remark: but you never mention what happened after Beck - I mean Bourn
protomodularity and internal actions...

Best regards,
George


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]

Re: modules over a group

[George Janelidze]

Could you please make more precise what you are saying?

Meanwhile, let me go back to your:

> ....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.

and put my "of course not" also here:

Let pi = be the 2-element cyclic group, written as pi={1,x}, and let A and B
be 2-sided pi-modules defined as follows:

- A and B have the same underlying group, which is the additive group of
integers;

- pi acts trivially on A, that is, xa=a=ax;

- pi acts on B by xb=-b=bx.

Then your functor will send A and B to the same pi-module, while A and B are
obviously not isomorphic. Therefore your functor is not an equivalence.

George


From: Michael Barr
Sent: Saturday, August 26, 2017 7:00 PM
To: George Janelidze
Cc: Categories list" <Categories list>
Subject: Re: categories: modules over a group


Okay, your argument seems sound. But consider the following. If Z is a free
group, say Z = F(X), then you can show that for any Z-module M (if M is a
pi-module a homomorphism Z --> pi makes M into a Z-module of course), then
for either a right or a 2-sided module, it is trivial to see that Der(Z,M) =
M^X. That is, any function X --> M extends in a unique way to a derivation
Z --> M. Now resolve an arbitrary Z with G^2Z ===> GZ ---> Z and then  we get
an equalizer Der(Z,M) ---> Der(GZ,M) ===> Der(G^2Z,M) or Der(Z,M) --->
Der(GC,M) ===> Der(G^2Z,M) which is Der(GZ,M) ---> M^Z ===> M^{GZ} for
whichever kind of module you have. But, the ismorphisms are not natural
(except with respect to maps of the basis, which G\epsilon is, but \epsilon
G isn't) and so the two versions of Der(Z,M) may well be different and hence
the two versions of Y.

I guess Jon didn't correctly describe the category of pi-modules.

Michael

----- Original Message -----
From: "George Janelidze" <George.Janelidze@uct.ac.za>
To: "Michael Barr" <barr@math.mcgill.ca>, "Categories list\" <Categories
list>" <categories@mta.ca>
Sent: Saturday, August 26, 2017 6:47:42 AM
Subject: Re: categories: modules over a group

Dear Michael,

Although the answer to your question is "Of course not!", it is a very
interesting question because it inspires very interesting questions about
internal actions in semi-abelian categories. In this message I shall,
however, only explain why is it "Of course not!" (with "!").

1. The category of pi-modules can be identified with the category of
Z[pi]-modules, where Z[pi] is the group ring over pi. Now, suppose, for
simplicity, that pi is abelian. Then you question can be stated as:

Are the rings Z[pi] and Z[pi x pi] Morita equivalent?

Of course not, because two commutative rings are Morita equivalent if and
only if they are isomorphic. Well, for some funny groups they are, but I
know you are not asking asking about funny groups.

2. Your functor from 2-sided modules to split epis is fine, but it is not an
equivalence, e.g. because your structure on Y will not allow you to recover
the pi-module structure on M from the Beck module structure on Y-->pi. In
fact one can prove the following:

Two 2-sided pi-module structures * and # on M give the same Beck modules
structures on a chosen pi-->Y-->pi if and only if

x*m*(inverse of x) = x#m#(inverse of x) for every x in pi and every m in M.

For example if we assume that both left actions are trivial, then the right
actions will have to coincide, as in Beck's theory.

Remark: but you never mention what happened after Beck - I mean Bourn
protomodularity and internal actions...

Best regards,
George

--------------------------------------------------
From: "Michael Barr" <barr@math.mcgill.ca>
Sent: Saturday, August 26, 2017 12:18 AM
To: "Categories list" <Categories list>" <categories@mta.ca>
Subject: categories: modules over a group

> 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
>
>
> [For admin and other information see: http://www.mta.ca/~cat-dist/ ]

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Re: modules over a group

[Patrik Eklund]

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
>>
>
>
>
> [For admin and other information see: http://www.mta.ca/~cat-dist/ ]


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