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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