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
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