Homological Algebra for Commutative Monoids? - MathOverflow

Homological Algebra for Commutative Monoids?

2009-10-13T16:23:02Z

Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of short exact sequence. Can we use this to define ext groups which classify extensions? What works and what doesn't work and why?

Answer by Mikael Vejdemo-Johansson for Homological Algebra for Commutative Monoids?

2009-10-13T18:02:57Z

Note that Homological algebra for abelian groups is really about homological algebra in the category of Z-modules.

The immediate neighbouring area I know of, where we take something not a module over some sort of algeba, is for group cohomology - where the standard construction is to grab the group ring, and then consider the abelian category of modules over the group ring.

So - the issue at hand, essentially, is to find yourself an abelian category that reflects properties of the monoids you want to study. Having short exact sequences is a good start, but you really want an abelian category to make the standard constructions from homological algebra work.

And if the monoids themselves don't provide all you want, you can always start looking at monoid rings instead: given M, form kM as the k-vector space spanned by a formal basis element for each element of M, and introduce an algebra structure by the monoid multiplication. After this, you can proceed in analogy to group cohomology.

Answer by Scott Carnahan for Homological Algebra for Commutative Monoids?

2009-10-13T19:21:28Z

Commutative monoids (aka N-modules) play a key role in logarithmic geometry (Ogus should have a book coming out soon), but I think the focus there is on a restricted class known as integral cones (aka fine saturated, or fs monoids). These have decent homological properties. I think the general setting has problems, since group completion can kill things.

Answer by Eric Wofsey for Homological Algebra for Commutative Monoids?

2009-10-13T20:37:37Z

One interpretation of the "right" category for doing homological algebra of commutative monoids would be the category of simplicial commutative monoids. Now up to homotopy, a space is a commutative monoid iff it is a commutative group iff it is a product of Eilenberg-MacLane spaces. There are some details to check, but this should imply that the forgetful functor from simplicial abelian groups to simplicial commutative monoids is a Quillen equivalence for the standard model structures, with the inverse being group-completion. This correspondence would indicate that the correct notion of "homology" of a simplicial monoid would be the homology of its group-completion as a chain complex.

However, this is presumably not what you're looking for, since you care about the monoids themselves, not just their group-completions.

Answer by Josh Shadlen for Homological Algebra for Commutative Monoids?

2009-10-16T15:31:13Z

I can't give a substantive answer to this question, but differential graded abelian monoids have been sighted in the wild here:

A Chain Functor For Bordism Author(s): Stanley O. Kochman Source: Transactions of the American Mathematical Society, Vol. 239 (May, 1978), pp. 167-196 Published by: American Mathematical Society Stable URL: http://www.jstor.org/stable/1997852

Answer by Yemon Choi for Homological Algebra for Commutative Monoids?

2009-10-20T19:44:59Z

I might be misunderstanding your question - are you asking about a (co)homology theory for commutative monoids, or about trying to do homological constructions in the category of commutative monoids?

In the former case, I think Grillet has some work on this, see e.g.

Grillet, Pierre-Antoine(1-TULN) Commutative semigroup cohomology. (English summary) Comm. Algebra 23 (1995), no. 10, 3573--3587.

The idea is basically what Mikael describes, although I only know the more primitive version which is cotriple/comonad cohomology as set up by Barr and Beck. Presumably the problem with trying to apply that set up is ensuring there are enough abelian group objects to use as coefficient modules.

Answer by Charles Rezk for Homological Algebra for Commutative Monoids?

2009-10-21T01:02:06Z

A comment on Eric's answer (which is mostly the answer I would give): what is true is that any connected commutative monoid space is (weakly) equivalent to a topological abelian group. A non-connected topological commutative monoid (or equivalently, a non-connected simplicial commutative monoid) contains some more information than its group completion.

So there will be a nice theory of "homological algebra", via simplicial objects, which is closely related to, but not equivalent to, homological algebra of abelian groups. I'm not aware that anyone has examined this closely, before.

Answer by Peter Arndt for Homological Algebra for Commutative Monoids?

2009-10-21T22:37:06Z

Your question can be understood as how to do Homological Algebra over the Field with one Element.

Deitmar, in http://arxiv.org/abs/math/0608179 , section 6, gives an example of what can go wrong if you try to do sheaf cohomology directly via resolutions...

You might also want to look at his http://arxiv.org/abs/math/0605429 ; in order to construct K-theory of monoids he sets up an analogue of the Q-construction. The Hom-sets in the resulting category are sort of Exts, maybe something to start with...

Durov, in http://arxiv.org/abs/0704.2030 , follows the simplicial approach for commutative monads, of which commutative monoids are a special case

Answer by Shizhuo Zhang for Homological Algebra for Commutative Monoids?

2009-11-17T08:52:09Z

category of commutative monoids plays central roles in algebraic geometry over F_un.

In fact, for such kind of category , Durov use homotopy theory in this paper:

http://arxiv.org/abs/0704.2030

But algebraic geometry over F_un is essentially study the geometry of right exact category(which is a category with subcanonical grothendieck topology) over category with only one object, in general, the correct things we should study is right exact category over category with initial object.

So, what we need is homological algebra framework on right exact category(in particular, topos, quasi abelian category, abelian category and so on).

In fact, Alexander Rosenberg built Homological algebra on noncommutative "space" (i.e.grothendieck topos)as a noncommutative version of Grothendieck tohoku lecture. He also introduced the Higher K theory for right exact category which led the universal K theory for abelian category and exact category and long exact sequence of Higher K functor of exact categories.

Here is the link of the paper: http://publications.ictp.it/lns/vol23.html

http://users.ictp.it/~pub_off/lectures/lns023/Rosenberg/Rosenberg.pdf

Answer by marc for Homological Algebra for Commutative Monoids?

2009-11-17T18:21:59Z

Presummably you could use the forgetful functor to sets and its adjoint to construct a homology theory (using the walking adjunction, simplicial sets, etc) analogously to how group homology arises from the forgetful-free adjunction to set.

The details are probably rather dense and tedious though.

Answer by SixWingedSeraph for Homological Algebra for Commutative Monoids?

2009-12-09T02:54:45Z

This is an answer to one part of your question. The paper “Extension Theories for Monoids” by Charles Wells, Semigroup Forum 16 (1978), 13-35, gives a precise answer to the specific question: How does the Beck cohomology theory for monoids classify extensions of monoids? (It classifies Leech extensions.) The paper with corrections and a list of subsequent papers related to it may be found here. Beck's thesis is now online here.

Answer by aleph0 for Homological Algebra for Commutative Monoids?

2009-12-12T01:38:51Z

I used to think about this problem in relation to a chain theory for bordism ( as mentioned by Josh Shadlen above).
The problems you have with monoids is first and foremost that the category is not balanced. That means that you can have an epimorphism that is also a monomorphism but NOT an isomorphism. eg. the inclusion of N --> Z.
Subsequently most constructions that you would like to make - notably short exact sequences and the snake lemma - fail at some level.
I made a few notes on this as part of my investigation into bordism theory / homework assignment here.
In this case, we have complexes of free abelian monoids whose homology takes it's values in abelian groups and yet, the long exact sequence does not come from a short exact sequence of monoid complexes.

My references include:

[Bau89] Friedrich W. Bauer. Generalised homology theories and chain complexes. Annali di Mathematica pura ed applicata, CLV:143–191, 1989.

[Bau95] Friedrich W. Bauer. Bordism theories and chain complexes. Journal of Pura and Applied Algebra, 102:251–272, 1995.

[BCF63] R.O. Burdick, P.E. Conner, and E.E. Floyd. Chain theories and their derived homology. Proceedings of the AMS, 19(5):1115–1118, Oct. 1963.

[Koc78] S. O. Kochman. A chain functor for bordism. Transactions of the American Mathematical Society, 239:167–196, 1978.

Answer by Reid Barton for Homological Algebra for Commutative Monoids?

2010-02-06T16:22:36Z

I agree with Charles and Eric that the natural setting for your question is the model category of simplicial commutative monoids. However, my earlier guess that the resulting homotopy theory would admit a simple algebraic description (as in the case of simplicial abelian groups) was wrong. For instance, consider the free commutative monoid functor from simplicial sets to simplicial commutative monoids and apply it to a space $X$ which I will assume for convenience to be connected and reduced ($X$ has only a single 0-simplex). The resulting simplicial commutative monoid $Y$ has $\pi_0(Y) = \mathbb{N}$. The element $1 \in \mathbb{N}$ is distinguished as the unique generator, and the connected component of $Y$ corresponding to $1 \in \pi_0(Y)$ is just $X$. So, we can recover $X$ from $Y$.

In other words, when Charles says "A non-connected topological commutative monoid (or equivalently, a non-connected simplicial commutative monoid) contains some more information than its group completion", it is actually quite a lot more information, enough to encode an entire homotopy type!

If we apply the reduced chain complex functor to one of these free simplicial commutative monoids, we'll just get $\mathbb{N}$ in degree 0 and zero in higher degree. The problem is that knowing the kernel of a map is 0 tells us very little (every map in the image of the free commutative monoid functor has kernel 0).