27
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ Of course I'd be happy if the "ext groups" weren't groups but monoids. $\endgroup$
    – Chris Schommer-Pries
    Commented Oct 13, 2009 at 16:38
  • $\begingroup$ Added the "f-1" tag since I believe this question and its answers are interesting for people working over the field with one element. $\endgroup$
    – javier
    Commented Dec 8, 2009 at 21:31
  • 1
    $\begingroup$ Monoids have an identity by definition. Also, it's not immediately straighforward that we have a good notion of a kernel either without categorifying to some extent, as indicated in the answer. $\endgroup$
    – Harry Gindi
    Commented Dec 9, 2009 at 11:08

13 Answers 13

Reset to default
8
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • 4
    $\begingroup$ I'm not sure about your first sentence. The monoids in these papers are playing the roles of the commutative rings in ordinary algebraic geometry, not the modules. Modules become sets. As far as I can tell, "homological algebra over the field with one element" is a fancy phrase for homotopy theory. $\endgroup$
    – Reid Barton
    Commented Nov 17, 2009 at 19:22
  • 1
    $\begingroup$ True, one would have to consider "Andre-Quillen-homology" over the field with one element, but I definitely agree that this is now just fancy language. $\endgroup$
    – Peter Arndt
    Commented Nov 19, 2009 at 1:16
9
$\begingroup$

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.

Improve this answer
$\endgroup$
9
$\begingroup$

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.

Improve this answer
$\endgroup$
0
7
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This seems very promising. Thanks for the reference! $\endgroup$
    – Chris Schommer-Pries
    Commented Dec 9, 2009 at 16:26
  • $\begingroup$ The entire paper “Extension Theories for Monoids” is now available here: cwru.edu/artsci/math/wells/pub/pdf/ExtThMon.pdf Look at Grillet's paper, too. I have not read the Novikov paper. $\endgroup$
    – SixWingedSeraph
    Commented Dec 10, 2009 at 16:06
  • $\begingroup$ The reference to Grillet's and Novikov's papers is in the preamble to the Extension Theories paper on line as in the previous comment. $\endgroup$
    – SixWingedSeraph
    Commented Dec 10, 2009 at 16:12
6
$\begingroup$

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.

Improve this answer
$\endgroup$
0
6
$\begingroup$

Jaret Flores's PhD thesis,

Jaret Flores, Homological Algebra for Commutative Monoids, arXiv:1503.02309.

treats precisely this topic.

Improve this answer
$\endgroup$
5
$\begingroup$

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).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I just pointed out this same example on my simplicial comm. monoid question. Should have checked here first... $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 6, 2010 at 20:48
4
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ As far as I can tell, just looking the monoid ring associated to a monoid doesn't cut it. For example let N be the natural numbers and consider the set E= N x N with multiplication (a,b) (x,y) = (a + bx, by) This is a monoid and is an extension of (N,x) by (N, +). The monoid ring of (N, x) is a polynomial ring on infinitely many generators (one for each prime). I don't see any way to extract this extension from the homological algebra of this ring. $\endgroup$
    – Chris Schommer-Pries
    Commented Oct 13, 2009 at 19:32
4
$\begingroup$

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.

Improve this answer
$\endgroup$
3
$\begingroup$

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.

Improve this answer
$\endgroup$
3
$\begingroup$

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

Improve this answer
$\endgroup$
2
$\begingroup$

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

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.