Re: modules over a group

[Paul Blain Levy <P.B.Levy@cs.bham.ac.uk>]

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/ ]