Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

Assume you have a cosimplicial group $G$, so that for each $n \ge 0$ there is a group $G_n$, and you have the usual cofaces and codegeneracies.

Is there a known way to associate to this a collection of homology/homotopy groups in a sensible way?

"Sensible" means at least that it should provide a generalisation of the following two particular cases:

(1) When each $G_n$ is abelian, one can get a cochain complex (I believe this is called the Dold-Kan construction), and one can consider its cohomology groups.

(2) (I apologize for the vagueness of this one.) In low degrees, one can sometimes use a couple of tricks. I have, for example, come across the following situation: the map $x\mapsto d^0(x) d^2(x)$ was a homomorphism $G_2 \to G_3$, as was the map $x\mapsto d^3(x)d^1(x)$; so I could consider their equalizer. Moreover the map $x\mapsto d^0(x)d^1(x)^{-1}d^2(x)$ was a homomorphism $G_1 \to G_2$, whose image was normal in the preceding equalizer. Taking the quotient gave a generalisation of $H^2$ in this lucky situation. The "general" definition which I'm asking for, should it exist, would hopefully coincide with this equalizer trick whenever it makes sense.

Let me also point out that, in the example above, I had originally started with some bigger cosimplicial group $(\Gamma_n)$ and then decided to restrict to the smaller $(G_n)$ precisely so that I could use this little trick. I do believe that the details of this example are completely irrelevant to the general discussion; I mention it because someone might know how to go from a cosimplicial group to a "nicer" one somehow.

Thank you for reading this.

Pierre

share|improve this question
    
Do you want homology or cohomology? Your title is asking for one, and the body of the question seems to indicate the other. –  David Roberts Jan 6 '11 at 3:23
    
@David: I meant, quite informally, any invariants which would look familiar; the rest of the question was meant to make this a little more precise. As you are pointing out, in the abelian case one gets a co-chain complex so perhaps the word cohomology should be used, sorry for the confusion (mind you, a cochain complex defines a chain complex so immediately (by negating the degrees) that I've never felt the need to make a strong distinction). –  Pierre Jan 6 '11 at 4:09
    
I should point out to anyone reading, however, that the distinction between cosimplicial and simplicial is absolutely essential: for simplicial groups there is a nice construction explained eg at ncatlab.org/nlab/show/Moore+complex –  Pierre Jan 6 '11 at 4:09
    
Let me also mention that the groups associated with a simplicial group (which may be called homology groups, especially if the simplicial group is abelian) are in fact isomorphic to the homotopy groups of (the realization of) the underlying simplicial set. –  Tom Goodwillie Jan 6 '11 at 4:20

2 Answers 2

up vote 9 down vote accepted

If we put some (to me) pretty reasonable-looking axioms, then the answer is no.

For a cosimplicial group $G$ let $h_0G$ be the equalizer of the maps $d_0,d_1:G_0\to G_1$.

Suppose there is another functor $h_1$ from cosimplicial groups to groups such that for a short exact sequence $$1\to G\to H\to K\to 1$$ of cosimplicial groups the exact sequence $$1\to h_0G\to h_0H\to h_0K$$ extends to an exact sequence $$ 1\to h_0G\to h_0H\to h_0K\to h_1G\to h_1H\to h_1K $$ with the middle map natural in the obvious sense and the other new maps given by the functoriality of $h_1$. Suppose that $h_1$ coincides with the usual thing in the case of cosimplicial abelian groups, and suppose also that $h_1G$ is trivial when both $G_0$ and $G_1$ are trivial.

Make a cosimplicial group $H$ where $H_0$ has order $2$ and $H_1$ is nonabelian of order $6$ and the two face maps are not equal. Let $G$ be the kernel of abelianization $H\to H^{ab}$. Then $h_0H$ is trivial, $h_0H^{ab}$ has order $2$, so $h_1G$ must have a subgroup of order $2$. On the other hand $G_0$ is trivial and $G_1$ has order $3$, which implies that $h_1G$ injects into $h_1G^{ab}$, which has order $3$. Contradiction.

share|improve this answer

You will probably find interesting the paper by Mariam Pirashvili called "Second cohomotopy and nonabelian cohomology". Cohomotopy groups $\pi^nG^\bullet$ of cosimplicial groups $G^\bullet$ have been considered in very low dimensions $n=0,1$ since long time ago, e.g. in the context of non-abelian cohomology. Actually, the only group is $\pi^0G^\bullet$, which is the coequalizer of the cofaces

$$G^0\mathop{\rightrightarrows}\limits^{d^0}_{d^1}G^1.$$ Then $\pi^1G^\bullet$ is just a pointed set and there seems to be no way of going higher in general (unless of course $G^\bullet$ is abelian, which is a very strong restriction). Nevertheless, M. Pirashvili has found a nice condition, satisfied by many examples of interest, under which the previous $\pi^0G^\bullet$ and $\pi^1G^\bullet$ are an abelian group and a group, respectively, and one can define a pointed set $\pi^2G^\bullet$. There are also nice exact sequences associated to cosimplicial subgroups, etc.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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