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Let $X$ be a commutative H-space. A group completion is an H-map $X\to Y$, where $Y$ is another H-space, such that

  • $\pi_0(Y)$ is a group
  • The Pontrjagin ring $H(Y; R)$ is the localization of the Pontrjagin ring $H_*(X; R)$ at the multiplicative submonoid $\pi_0(X)$ for every coefficient ring $R$.

Perhaps most interesting is the case where $X$ is a commutative monoid or more generally an $E_\infty$-space. In this case, May gives a functorial group completion $B_0$ via the two-sided bar construction which is an infinite-loop space. If I understand it right, it is defined as $colim_j \Omega^j|\Sigma^j(C_j\times C')^\bullet X|$, where $C_j\times C'$ denotes the monad associated to the product of the little $j$-cube operad and an $E_\infty$-operad and $\bullet$ the simplicial variable. I have some question concerning group completions:

1) Are all group completions equivalent? That is, does there always exist a homotopy equivalence of H-spaces between them?

2) Does the group completion preserve homotopy limits? For example, does the May functor preserve homotopy limits of $E_\infty$-spaces?

3) Suppose one knows that all loop spaces of $X$ are infinite loop spaces. Is there a simple relationship between the infinite-loop space $B_0 X$ and the loop spaces of $X$? Especially, I am interested in the homotopy groups of $B_0 X$.

An answer to any of these question would be helpful to me.

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Question (1) leads me to a many more questions. Like: suppose we have a cofibrant simplicial object in unital monoids and we levelwise adjoin inverses; is that a "group completion" in this sense of localizing homology? This is much harder when you can't play games with the classifying space. Re question (3): I'm not sure what you mean by "all loop spaces of X are infinite loop spaces" - do you mean e.g. that the loop space based at the identity is an infinite loop space? –  Tyler Lawson Aug 28 '10 at 17:21
My feeling is that in order for the answer to (1) to be "Yes" you should allow local coefficient systems in homology. Does anyone have a more precise thought about this point? –  Dan Ramras Aug 28 '10 at 22:16
@Tyler: Eg the loop space based at the identity. More precisely: for every $x \in X$, the loop space $\Omega(X,x)$ is an infinite loop space. –  Lennart Meier Aug 29 '10 at 7:50
I should add that I want the $E_\infty$-structure on $\Omega(X,x)$ coming from the infinite-loop space structure to be compatible with that coming from X. –  Lennart Meier Aug 29 '10 at 10:06
How do you obtain an $E_\infty$-structure on the loop space based at a nonidentity point? –  Tyler Lawson Aug 29 '10 at 12:35
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1 Answer

up vote 8 down vote accepted

A well-written discussion of the group completion can be found on pp. 89--95 of J.F. Adam: Infinite loop spaces, Ann. of Math. studies 90 (even though he only discusses a particular group completion of a monoid). In particular you assumption of commutativity comes in under the assumption that $\pi_0(M)$ is commutative which makes localisation with respect to it well-behaved (commutativity is not the most general condition what is needed is some kind of Øre condition).

In any case if you really want conclusions on the homotopy equivalence level I think you need to put yourself in some nice situation for instance requiring that all spaces be homotopy equivalent to CW-spaces. If you don't want that you should replace homotopy equivalences by weak equivalences, if not you will probably find yourself in a lot of trouble. In any case I will assume that we are dealing with spaces homotopy equivalent to CW-complexes.

Starting with 1) a first note is that your conditions does not have to involve an arbitrary ring $R$. It is enough to have $R=\mathbb Z$ and one should interpret the localisation in the way (for instance) Adams does: $H_\ast(X,\mathbb Z)=\bigoplus_\alpha H_\ast(X_\alpha,\mathbb Z)$, where $\alpha$ runs over $\pi_0(X)$, and a $\beta$ maps $H_\ast(X_\alpha,\mathbb Z)$ to $H_\ast(X_{\alpha\beta},\mathbb Z)$. Then your group completion condition is that the natural map $\mathbb Z[\pi_0(Y)]\bigotimes_{\mathbb Z[\pi_0(X)]} H_\ast(X,\mathbb Z)\rightarrow H_\ast(Y,\mathbb Z)$ should be an isomorphism. This then implies the same for any coefficient group (and when the coefficient group is a ring $R$ you get your condition). (Note that for this formula to even make sense we need at least associativity for the action of $\pi_0(X)$ on the homology. This is implied by the associativity of the Pontryagin product of $H_*(X,\mathbb Z)$ which in turn is implied by the homotopy associativity of the H-space structure.)

Turning now to 1) it follows from standard obstruction theory. In fact maps into simple (hope I got this terminology right!) homotopy types, i.e., spaces for which the action of the fundamental groups on the homotopy groups is trivial (in particular the fundamental group itself is commutative). The reason is that the Postnikov tower of such a space consists of principal fibrations and the lifting problem for maps into principal fibrations is controlled by cohomology groups with ordinary coefficients. Hence no local systems are needed (they would be if non-simple spaces were involved). The point now is that H-spaces are simple so we get a homotopy equivalence between any two group completions and as everything behaves well with respect to products these equivalences are H-maps.

Addendum: As for 2) it seems to me that this question for homotopy limits can only be solved under supplementary conditions. The reason is that under some conditions we have the Bousfield-Kan spectral sequence (see Bousfield, Kan: Homotopy limits, completions and localizations, SLN 304) which shows that $\varprojlim^s(\pi_s X_i)$ for all $s$ will in general contribute to $\pi_0$ of the homotopy limit. As the higher homotopy groups can change rather drastically on group completion it seems difficult to say anything in general (the restriction to cosimplicial spaces which the OP makes in comments doesn't help as all homotopy limits can be given as homotopy limits over $\Delta$. Incidentally, for homotopy colimits you should be in better shape. There is however an initial problem (which also exists in the homotopy limit case): If you do not assume that the particular group completions you choose have any functorial properties it is not clear that a diagram over a category will give you a diagram when you group complete. This can be solved by either assuming that in your particular situation you have enough functoriality to get that (which seems to be the case for for instance May's setup) or accepting "homotopy everything" commutative diagrams which you should get by the obstruction theory above. If this problem is somehow solved you should be able to conclude by the Bousfield-Kan spectral sequence $\injlim^\ast H_*(X_i,\mathbb Z)\implies H_*(\mathrm{hocolim}X_i,\mathbb Z)$. We have that localisation is exact and commutes with the higher derived colimits so that we get upon localisation a spectral sequence that maps to the Bousfield-Kan spectral sequence for $\{Y_i\}$ and is an isomorphism on the $E_2$-term and hence is so also at the convergent.

As for 3) I don't altogether understand it. Possibly the following gives some kind of answer. For the H-space $\coprod_n\mathrm{B}\Sigma_n$ which is the disjoing union of classifying spaces of the symmetric groups its group completion has homotopy groups equal to the stable homotopy groups of spheres which shows that quite dramatic things can happen to the homotopy groups upon group completion (all homotopy groups from degree $2$ on of the original space are trivial).

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Thanks for your answer! Ad 1): I see that I can apply this variant of Whitehead if I have a map. Where do I get the map from? Ad 2): I meant homotopy limits. If I haven't miscalculated, group completion should commute with products, at least under suitable flatness conditions. I'm actually interested mostly in a homotopy limit over a cosimplicial diagram. Perhaps one can use here the homology spectral sequence for a cosimplicial space. I need to check. –  Lennart Meier Aug 29 '10 at 10:25
Ad 3): I know that the homotopy groups of group completions in general can go wild (else algebraic K-theory would behave quite differently!). But note that the $\Sigma_n$ are no infinite-loop spaces and therefore do not fit the conditions I posed (which are hopefully clearer after my comments to the question). –  Lennart Meier Aug 29 '10 at 10:26
The map is obtained by obstruction theory (i.e., by induction over a Postnikov tower), see for instance Adams for a more detailed discsussion. I'll have to think about the other two parts. –  Torsten Ekedahl Aug 29 '10 at 11:03
@Torsten: Nice answer. One minor point is that you need to make the assumption that X is homotopy-associative (or possibly something weaker) in order to describe the localization as having homology obtained by tensoring up the base ring. –  Tyler Lawson Aug 29 '10 at 12:26
@Tyler: Thanks you are right. (I could possibly weasel my way out of that by claiming that what is needed is the associativity of the Pontryagin product which seems to be implied by the OP's use of "Pontryagin ring".) –  Torsten Ekedahl Aug 29 '10 at 13:17
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