Re: Tensor of monads

[Ronnie Brown <ronnie.profbrown@btinternet.com>]

There is another possibly relevant area. It used to be standard (e.g. 
Huppert, Endliche Gruppen) that the only tensor product of groups was 
the usual tensor product of their abelianisations. This is because if 
b:G \times G \to H is a bimorphism then by expanding
b(gg',hh') in two ways, and applying cancellation, you get a 
commutativity condition. However with Jean-Louis Loday we realised, as 
others had before us,  that another interesting condition is for b to be 
a biderivation, since this is one of the rules satisfied by the 
commutator map [ , ] : G \times G \to G. The universal object for 
biderivations is then the nonabelian tensor square G \otimes G. This 
idea applies to other areas such as Lie algebras. A bibliography of 100 
items is on
www.bangor.ac.uk/r.brown/nonabtens.html
Any possibility for monads?? This may be wild, but on the other hand........

On another tack, my memory is, and this puzzled me at first,  that the 
paper
Loday, Jean-Louis 
<http://0-ams.mpim-bonn.mpg.de.unicat.bangor.ac.uk/mathscinet/search/publications.html?pg1=IID&s1=115225>
$K$-théorie algébrique et représentations de groupes. *(French)*
/Ann. Sci. École Norm. Sup. (4)/ 
<http://0-ams.mpim-bonn.mpg.de.unicat.bangor.ac.uk/mathscinet/search/journaldoc.html?cn=Ann_Sci_Ecole_Norm_Sup_4> 
* 9 * (1976), no. 3, 309--377.

uses a multiplication induced essentially by a structure of a monoid 
with a compatible structure of semigroup; so the Eckmann-Hilton argument 
does not apply!

Ronnie Brown


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