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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