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

Let $(X, *)$ be pointed a (1-connected) space, and let $\Omega X$ denote its based loops space. Then, as one knows very well, $\Omega X$ is a group up to homotopy (this includes all the necessary higher homotopies). Let $Y$ be a $\Omega X$-space, i.e. a space, on which $\Omega X$ acts from the left so that all the usual associativity conditions hold (up to all the necessary higher homotopies).

I wonder, if anybody knows, whether there exists a theory of $\Omega X$-equivariant cohomologies of such $Y$? If yes, can you give me a reference?

In particular, I am interested in the values of such an equivariant cohomology in case, when $X$ is a smooth manifold, $Y$ is equal to the fibre of a smooth bundle over $X$,and the action of $\Omega X$ on $Y$ is determined by a connection on $X$.

share|improve this question
add comment

6 Answers 6

As far as I understand, you wish to know the equivariant homology $H_{\Omega X}(Y)$, where $Y \to E \to X$ is a fibration which defines the ''action''. Preferring simple answers, equivariant homology means Borel-equivariant homology, i.e. homology of the Borel construction. But what is $E \Omega X \times_{\Omega X} Y$? It fits into a fibration

$$Y \to E \Omega X \times_{\Omega X} Y \to B \Omega X \simeq X,$$

the last equivalence uses that $X$ is path connected. But if you look at this fibration, you see that it must be the original fibration $Y \to E \to X$. What else should it be? Proving this is an exercise with fibration techniques. So the Borel construction is $E$ and it follows:

''The $\Omega X$-Borel-equivariant homology of $Y$ is the homology of $E$.''

share|improve this answer
add comment

There are a few things to say.

As Johannes Ebert mentioned, it's much easier to thing of "spaces with a $\Omega X$-action" in other terms. Assuming $X$ is path-connected, from the point of view of homotopy theory this is the same as a certain homotopy theory of spaces over $X$: a map $E \to X$ has a homotopy fiber $Y$ with a coherent action of $\Omega X$ and, in the reverse direction, forming the Borel construction gives us the reverse construction. We think of $E \to X$ as the classifying map for a principal $\Omega X$-bundle $Y \to E$, given by the homotopy fiber. So now we just need to determine appropriate equivariant cohomology theories that we can take values in. (Note that these equivariant cohomology theories are going to be analogous with local coefficient systems, rather than with the "genuine" equivariant cohomology theories as in Lewis-May-Steinberger's book or more recent references).

We can either think of the appropriate spectra in a few ways:

  • Spectra equipped with a coherent $\Omega X$-action. It's easier to work with another framework, but the mantras will be that if $Y$ has an $\Omega X$-action and $F$ is a spectrum with an $\Omega X$-action, then the equivariant homology of Y will be the homotopy groups of $Y \wedge_{\Omega X} F$ and the cohomology will be homotopy classes of equivariant maps $[Y,F]_{\Omega X}$.
  • Modules over the spherical group ring $\mathbb{S}[\Omega X]$ that David Ben-Zvi mentioned. This is an $A_\infty$-ring spectrum, or if you use the Moore loop space it's a genuinely associative ring spectrum, and you can talk about its modules (e.g. in the framework of Elmendorf-Kriz-Mandell-May's book or in the paper on symmetric spectra by Hovey-Shipley-Smith). Given our object $Y$ with $\Omega X$-action, its suspension spectrum $\Sigma^\infty Y$ becomes a module over $\mathbb{S}[\Omega X]$. If $F$ is any $\mathbb{S}[\Omega X]$-module, we can determine $\pi_* (\Sigma^\infty Y \wedge_{\mathbb S[\Omega X]} F)$ or $[\Sigma^\infty Y, F]_{\mathbb{S}[\Omega X]}$ in the homotopy category, and these determine the $F$-equivariant homology and cohomology.
  • Bundles of spectra over $X$. See Christopher Douglas' paper "Twisted parametrized stable homotopy theory" on the arXiv. In this case, if we have a bundle of spectra $G \to X$ then we can use spaces of sections $E \to G$ over $X$ instead of working with the loop-space action. Such a bundle itself has a classifying map $X \to BhAut(F)$ from $X$ to a classifying space for the automorphisms of the fiber.
  • A particularly nice example of the previous is when $F$ is some kind of ring (an $A_\infty$ ring spectrum, to be precise), and then we can talk about it as a module over itself; the space of self-equivalences as a module is called $GL_1(F)$ and its classifying space is called $BGL_1(F)$. See numerous papers, old and new, on units of ring spectra, such as Ando-Blumberg-Gepner-Hopkins-Rezk's "Units of ring spectra and Thom spectra" and Ando-Blumberg-Gepner's "Twists of K-theory and TMF", both available on Matthew Ando's webpage.

There are numerous interesting examples of the phenomena you describe. Thom spectra are one example. Twisted $K$-theory is another, and you can describe some version of it by looking at spaces with an action of $\Omega K(\mathbb{Z},3)$.

Hope this helps. It's definitely an area connected to active development.

share|improve this answer
add comment

There are probably many places where this is discussed well, but Tyler is right that the construction you need is available in my paper with Blumberg, Gepner, Hopkins and Rezk, eg on the arxiv.

$\Omega X$ is not a group, but it is a group-like $A_\infty$ space. In the language of $\infty$-categories, it is a group-like $\infty$-groupoid.

Let $G$ be such a thing. In the paper, we discuss what it means to have a $G$-space, and given a $G$-space $X$, we discuss how to construct the analogue of $EG\times_G X$. We do this both from the point of view of group-like $A_\infty$ spaces and from the point of view of $\infty$-categories.

We do not go as far as we might. It's worth pointing out that if you try to extend the notion of "principal G-bundle" to the case that $G$ is a group-like $\infty$-groupoid, what you'd find is that the path fibration over $X$ is precisely a principal $\Omega X$ bundle. Continuing in this vein, you'd find that a Serre fibration is a fiber bundle with structure group $\Omega X$.

In particular, in the setting of $\infty$-categories, you find that Serre fibrations and classical fiber bundles are kind of indistinguishable.

share|improve this answer
    
"Serre fibrations and classical fiber bundles are kind of indistinguishable." ...Except for local triviality, that is. If that is what you (gshar) want, look at Dold's paper on locally homotopy trivial fibrations. –  David Roberts Feb 24 '11 at 5:15
1  
(Hi Matt, welcome to MO!) –  Tyler Lawson Feb 24 '11 at 12:57
add comment

Until the relevant experts arrive, I can mention that one relevant key phrase is "Waldhausen A-theory" (or K-theory of spaces, cf. Waldhausen's paper with this title), which is the K-theory of $\Omega X$-actions on spectra (i.e. modules for the "spectral group algebra" of $\Omega X$, ie its suspension spectrum). We can think of these intuitively as local systems of spectra over $X$ (a notion that's easy to make precise eg in $\infty$-categorical settings). So you're asking about Ext in this category of modules from the trivial local system to a cohomology theory applied to $Y$. One useful reference is

. arXiv:0912.1670 Title: Derived Koszul Duality and Involutions in the Algebraic K-Theory of Spaces Authors: Andrew J. Blumberg, Michael A. Mandell

where they study the relation between this theory of "spectral local systems" and modules over the dual of $X$ (ie the spectral version of cochains on $X$ - you get such a module by taking exactly the Ext from the trivial local system that you're looking for, unless I've gotten my dualities off.)

share|improve this answer
add comment

This is in response to the OP's recent reference request.

It sounds like the reference you seek may be John Milnor's paper

Construction of universal bundles. I. Ann. of Math. (2) 63 (1956), 272–284.

Milnor starts with the Serre path-loop fibration with fibre $\Omega X$ on a based countable simplicial complex $X$. He then replaces continuous paths (loops) with "simplicial paths (loops)". It turns out that the resulting bundle is still universal (meaning the space of based simplicial paths is contractible) and the fibre $\tilde{\Omega} X$, the space of based simplicial loops, is a topological group.

share|improve this answer
add comment

Thank you very much to everybody, who answered me! Here are few observations, concerning your answers.

The answer by Johannes Ebert is the closest to what is now on my mind. His is the answer, which I guessed would and should be right (i.e. The $\Omega X$-Borel-equivariant homology of $Y$ is the homology of $E$), even though it is a bit too simple to spark interest, of course. But at present I just need a reference to a paper in which such construction is dicussed: there are few moments which are not evident for me, e.g. it is not quite clear for me now, what is the proper way to look at the universal space $E\Omega X$, since $\Omega X$ is not a group, but only a homotopy group, etc. (clearly, it should be equivalent to the based paths space, but I need an explicit construction, similar to the bar-resolution, or something like this, which would fit arbitrary good H-space). It seems from what Johannes says, that these are well-known things, so, dear Johannes, if you have a reference to any paper, in which such questions are discussed, please, let me know.

And of course, I would like to have a deeper understanding of such equivariant theories in general, i.e. in the full generality of the local system approach, so, my special thanks to Tyler Lawson and David Ben-Zvi for an extensive list of possible constructions! In effect, my interest is caused by an attempt to understand the topological counterpart of various equivariant theories with coefficients, in the case when the group is replaced by a d.g. Hopf algebra or something equivalent, so this is what I really need!

share|improve this answer
1  
This would work better as a revision to the question. I've merged your user accounts, so you should be able to do it yourself. –  S. Carnahan Feb 24 '11 at 7:42
add comment

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.