# Proving that a space cannot be delooped.

Suppose we have some pointed connected topological space $X$. How can we determine if there exists a space $BX$, called delooping of $X$, such that its space of based loops $\Omega BX$ is homotopy equivalent to $X$? More generally, when can we deloop a space $n$ times?

## Abstract nonsense

There is a general theory answering this question, going back to J.D. Stasheff's 1963 papers "Homotopy Associativity of H-Spaces. I, II" and vastly developed later. The basic answer is that $X$ should carry an action of $A^\infty$ operad, corresponding to composition of loops. This composition is not associative (at least with an obvious definition), but associativity fails in a well-behaved way --- up to coherent higher homotopies. If $X$ is an iterated loop space $X = \Omega^n Y$, then it carries $n$ compatible actions of $A^\infty$, because there are roughly $n$ really different ways to compose two $n$-dimensional spheres in $Y$: a choice of composition corresponds to a map $S^n \to S^n \vee S^n$, which can be obtain by contracting some base-pointed equator on $S^n$, and there are $n$ really different ways to choose it. Composition of $n$-dimensional spheres is also not associative, and different choices of composition do not commute, but all these statements fail up to coherent higher homotopies, encoded in an action of so-called $E_n$ operad on $X$ ($E_1 = A^\infty$). In this case $X$ can be delooped up to $n$ times. An example of explicit construction is given in "The Geometry of Iterated Loop Spaces" of J.P. May: there is a natural right action of $E_n$ operad on the $n$-th suspension functor $\Sigma^n$, so we can consider "derived tensor product" $\Sigma^n \otimes_{E_n} X$. It is a geometric realization of simplicial space $$\Sigma^n X \leftarrow \Sigma^n E_n X \leftleftarrows \Sigma^n E_n E_n X \leftleftarrows \dots$$ (last term shoul have 3, not 2 arrows, but how can you do this with MO TeX?)

## The problems

It's all nice and neat, but how can you determine if a space can have an action of $E_n$? In all examples I'm familiar with this action is supplied with the definition of $X$, because $X$ is defined as an iterated loop space, or a group, or have a natural action of something that is close to $E_n$, like little disks in some manifold $E_M$, or is a representing object for some cohomology theory. Further problem is, even if we find some group structure on $X$, it doesn't necessarily mean that we have captured all structure. For example, a group can be noncommutative, but still admit higher homotopies, assembling into the action of $E_n$ or even $E_\infty$. A simple example is the infinite-dimensional projective unitary group $PU(H)$, where $H$ is some separable Hilbert space. Kuiper's theorem states that $U(H)$ is contractible. Its center is $U(1)$ and we have a fiber sequence $$U(1) \to U(H) \to PU(H)$$ This means that $PU(H)$ is $K(\mathbb{Z},2)$ and an infinity-loop space. While it may be that these higher commutativity homotopies can be written down explicitly, it is not at all obvious that they exist in the first place. If $X$ is an infinity-loop space, we can at least hope that after several iterations of taking loops we will get some periodicity, or an obviously commutative group (naive, but is there much choice?). If $X$ is $n$-loop space for finite $n$, then even this hope is futile.

While I'm not expecting that there is some way, besides finding a certain $E_n$ action, to prove that $X$ is $n$-loop space (but I still would be glad to hear any different methods), I do hope that it is possible to prove that $X$ is not $n$-loop. Possibly there are some homotopic or homological invariants that would be obstructions to finding such a structure. However, I can only think of obvious ones: $\pi_1$ should be abelian. Perhaps some homotopical or homological operations should also vanish for $n$-loop spaces, but I don't know how to describe them.

An iterative check for the existence of $A^\infty$ structure can be performed based on Milnor's theorem. If $X$ has $A^\infty$ action, then it is homotopy equivalent to a topological group $G$, and we get "projective space bundles" $$G \to G^{\ast n} \to B_n$$ Here $B_n$ is some space, and $G^{\ast n}$ is n-th iterated join. For $G=S^1$ or $G=S^3$ this gives standard Hopf bundles over complex and quaternionic projective spaces. A theorem of Adams asserts that such bundles do not exist for spheres of other dimensions (except for $\mathbb{O}P^1$ and $\mathbb{O}P^2$, $G=S^7$). However, his methods seem to rely on $G$ being a sphere, and I am not sure how to prove or disprove the existence of a fibration with fiber $X$ and total space $X^{\ast n}$ for a general space $X$.

## Question

Are there different ways to prove that $X$ is $n$-loop space, besides finding an explicit $E_n$ action? More importantly, how can one prove that such action does not exist? What homological/homotopical invariants and operations can be used to disprove its existence, or prove that some known $E_n$ structure can not be lifted to a $E_{n+k}$ structure?

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Key word is "obstruction theory." See Robinson: arxiv.org/pdf/1301.1572v1.pdf and Goerss-Hopkins: math.northwestern.edu/~pgoerss/spectra/obstruct.pdf For example. –  Dylan Wilson Apr 12 '13 at 15:04
You can also find explicit examples in Stasheff's book on H-spaces. –  Dylan Wilson Apr 12 '13 at 15:04
Every n-fold loop space, by neglect of structure, is an $H_n$ space. This implies that there is an action of some bits of the Dyer-Lashof algebra on its homology. There is something called the nishida relation that tells you how this must relate to the action of the steenrod algebra. I would recommend at looking at Steinberger's chapter of math.uchicago.edu/~may/BOOKS/h_infty.pdf –  Sean Tilson Apr 12 '13 at 16:16
So let me rephrase my last statement, determining when something has an $E_n$ structure is not easy in general, as far as I can tell. I think it is probably the case that you might be able to classify all finite dimensional compact lie groups that are n-fold loop spaces. I know that there is work of Bauer, Pedersen, Notbohm, Grodal, Kitchloo and others that is concerned with when things are finite loop spaces and when they can be manifolds. Try looking at that seciont of the $H_{\infty}$ ring spectra volume, it could be that you can determine something about the module structure of the ... –  Sean Tilson Apr 12 '13 at 18:09
cohomology over the steenrod algebra from the nishida relation. Just pick some example space to start with where you know the product in homology and the action of the steenrod algebra. –  Sean Tilson Apr 12 '13 at 18:10

The general question is indeed hard, but in view of Anton's side question to Dylan about compact Lie groups, I feel compelled to advertise a beautiful old result of John Hubbuck that everyone should know: If $X$ is a connected finite CW complex and a homotopy commutative $H$-space, then $X$ is homotopy equivalent (as an $H$-space) to a torus. The reference is: J.R. Hubbuck. On homotopy commutative H-spaces. Topology 8 1969 119–126. Thus the original question trivializes unless $X$ is infinite dimensional.

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Beautiful indeed! Thank you! –  Anton Fetisov Apr 13 '13 at 20:57

I'll assume for simplicity that $X$ is connected and of finite type.

The most basic cohomological criterion (which I suspect that others have considered too elementary to mention) is that if $X$ is a loop space then $H^*(X;\mathbb{Q})$ has a Hopf algebra structure, and so (by a theorem of Milnor and Moore) it is a tensor product of polynomial algebras and exterior algebras over $\mathbb{Q}$. Thus, if the ring structure of $H^*(X;\mathbb{Q})$ is any more complicated than that, then $X$ cannot be a loop space. Similarly, $H^*(X;\mathbb{Z}/p)$ must be a tensor product of polynomial algebras, exterior algebras, and truncated polynomial algebras of the form $(\mathbb{Z}/p)[x]/x^{p^m}$ for various $m$.

In another MO question I asked about the case $X=\mathbb{H}P^\infty_{(p)}$, which I suspect is not a loop space for any prime $p$, although it satisfies the obvious primary cohomological tests. I still don't know a convincing answer for that case.

VERY LATE UPDATE:

Shortly after I wrote the above answer, Gustavo Granja gave a very nice answer (negative, as expected) to my question about $\mathbb{H}P^\infty_{(p)}$. I only just noticed that I never updated this answer to reflect that, as I obviously should have done.

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