0
$\begingroup$

The following question can be thought as a sequel of this one.

Here I'm looking for a big list of example of weak algebraic structures: here weak means that the structure (i.e. operations) need not to satisfy equations but rather they must satisfy equations up to higher equivalence, the sort of conditions that can be expressed via commutative diagrams up to higher equivalence.

In particular I'm interested in example of structures of this sort in which we can do explicit calculations that in theory can be implemented in a computer.

It would be nice if every example come equipped with some reference where said example is presented.

Improve this question
$\endgroup$
5
  • 6
    $\begingroup$ Almost everything you can think of is weak, and you have to do work to make something not weak. Tensor product of vector spaces? Weak! Composition of homotopies? Weak! It's just that there's often a contractible space of choices, so you can stricitify with no harm. $\endgroup$
    – Noah Snyder
    Dec 20, 2011 at 16:23
  • $\begingroup$ @NoahSnyder This arise a problem: what does it mean strictify? We know of course that every bicategory is biequivalent to a strict one, but we also know that this does not hold for higher categories. So the point is what does strictify means and how do you strictify in higher dimension? $\endgroup$
    – Giorgio Mossa
    Dec 20, 2011 at 18:40
  • 2
    $\begingroup$ I don't see the point to this kind of question. Actually, I don't think this is a question at all. $\endgroup$
    – Fernando Muro
    Dec 20, 2011 at 20:59
  • $\begingroup$ I agree with Fernando, this is way too vague to be considered a real question. $\endgroup$
    – Vladimir Dotsenko
    Dec 21, 2011 at 10:53
  • 1
    $\begingroup$ Noah, why not post your answer as an answer? The site works best that way. $\endgroup$
    – Joel David Hamkins
    Dec 21, 2011 at 13:03

1 Answer 1

Reset to default
2
$\begingroup$

I am not so certain this question is unreasonable, but may need to be expressed differently.

We have found that some questions of say computing resolutions are well expressed in terms of computing contracting homotopies. Thus instead of killing kernels, as is traditional, we find a home for a contracting homotopy. See the paper

Brown, R. and Razak~Salleh, A. Free crossed resolutions of groups and presentations of modules of identities among relations. LMS J. Comput. Math. 2 (1999) 28--61 (electronic),

whose techniques have been developed by Graham Ellis in his packages HAP.

The ideas for the above paper were suggested by the field of Homological Perturbation Theory, in which homotopies are crucial, and which has been expressed in computational terms, in work of Huebschmann/Kadeishvili, Gugenheim/Lambe/Stasheff and for example

Larry Lambe, Leif Johansson and Emil Skoldberg, On Constructing Resolutions Over the Polynomial Algebra, Special Issue of Homology, Homotopy and Applications in honor of Jan-Erik's Roos' 65th birthday, Homology, Homotopy and Applications, vol. 4, no. 2, pp. 315-336, (2002).

and Lambe's home page at

http://pages.bangor.ac.uk/~mas019/pubs.html

All this suggests that if you want to compute with weak structures you will be involved with computing the fillers of the diagrams. That raises the problem of the algebra and computation of geometric diagrams. An answer was attempted in David Jones thesis

Jones, D.W., A general theory of polyhedral sets and the corresponding $T$-complexes. Dissertationes Math. (Rozprawy Mat.) 266 (1988) 110.

see the ncat-lab page on ''T-complex'' for a download.

There are aspects of dealing with weak structures in one formulation of Schreier theory:

Brown, R. and Porter, T., ``On the Schreier theory of non-abelian extensions: generalisations and computations''. Proceedings Royal Irish Academy 96A (1996) 213-227.

The standard free crossed resolution of a group gives another method of handling the `weak associativity' which you get in a group extension, and is described by a factor set.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.