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I was inspired by the following algebraic topology orals question:

"Is $S^1$ the loop space of another space?"

This is easy to see if you recognize that $S^1$ is a $K(\mathbb{Z},1)$, and the loop space of any $K(G,n)$ is a $K(G,n-1)$.

I then also remembered that the loop space functor is a functor from pointed topological spaces and continuous maps to the category of H-spaces and continuous homomorphisms. H-spaces being topological spaces that satisfy the axioms of a group up to homotopy (see Spanier, Chapter 1, Section 5).

I have three questions:

  1. Is there a useful criterion for when an H-space is actually a topological group?
  2. Seeing that $S^1$,$S^3$, and $S^7$ are the only spheres that support group structures, it doesn't seem coincidental that $S^1$ is a loop space, because it is in fact an H-space. Since $CP^{\infty}$ is the loop space of $K(Z,3)$ it too is an H-space, but is it known if it is a topological group?
  3. Even if not, is there a way (other than concatenation of loops) to "see" this structure on $CP^{\infty}$?

Thanks!

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For 3., see mathoverflow.net/questions/11117/… For 2, S^7 doesn't support a group structure. I don't THINK it supports an H-space structure either, but that I'm much less sure about. –  Jason DeVito Jan 21 '10 at 1:12
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$S^7$ does support an H-space structure. Give it as the unit octonians, the only thing that fails to hold exactly is associativity, but it holds up to homotopy. The spheres that admit H-space structures are precisely the unit vectors in a normed division algebra. It's a theorem of Adams that these are the only ones. –  Charles Siegel Jan 21 '10 at 1:53
    
Thanks for the clarification! I wasn't sure about associativity (I'm not so sure why I DIDN'T think octonionic multiplication would give it an H-space structure - just shows I STILL have no intuition about the octonians!). –  Jason DeVito Jan 21 '10 at 2:22

4 Answers 4

up vote 11 down vote accepted

Here's a few thoughts on your questions.

  1. See algori's answer. (Incidentally, the "necessity is clear" step is because if $G$ is a topological group then it has a classifying space and then $G \simeq \Omega B G$, hence is homotopy equivalent to a loop space.)

  2. For $CP^\infty$, here's a construction that makes it a topological group. Take the unitary group on a Hilbert space, $U(H)$. This is contractible by Kuiper's theorem (MR0179792). The centre of this group is the circle, $S^1$, acting by diagonal operators. As this is normal, the quotient $PU(H) = U(H)/S^1$ is a topological group. Since $U(H)$ is contractible, this is a $K(\mathbb{Z},2)$ and hence "is" $CP^\infty$.

  3. The group structure on $CP^\infty$ can be viewed in a nice way using the fact that $CP^\infty$ represents $H^2(-,\mathbb{Z})$ and that $H^2(-,\mathbb{Z})$ classifies complex line bundles. Both of these views of $[-,CP^\infty]$ have obvious groups structures: for $H^2(-,\mathbb{Z})$ it is addition whilst for complex line bundles it is given by tensor product (the inverse operation is complex conjugation). (These two group operations are the same operation under the correspondence, by the way). As these are natural operations, they are represented by a group structure on the representing space, making $CP^\infty$ a group object in the $hTop$. That this is the "correct" group structure (aka, that coming from $CP^\infty \simeq K(\mathbb{Z},2) \simeq \Omega K(\mathbb{Z},3)$) is clearest from the characterisation of $CP^\infty$ as the representing space for $H^2(-,\mathbb{Z})$. The equivalence $CP^\infty \simeq \Omega K(\mathbb{Z},3)$ comes from the suspension isomorphism, $H^2(-,\mathbb{Z}) \cong H^3(\Sigma -, \mathbb{Z})$ which is additive and hence preserves the group structures on the representing spaces.

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Here is a partial answer to question 1: the necessary and sufficient condition for a (sufficiently reasonable, say a CW-complex) space to be homotopy equivalent to a topological group is that it should have the homotopy type of a loop space, or, in other words, that it should admit a structure of an $A_\infty$-space. The necessity is clear. On the other hand, if $X=\Omega Y$, then Milnor constructs in "The construction of the universal bundles I", section 3, a group $G(Y)$ with the same homotopy type as $X$. The construction is as follows (we assume $Y$ to be a polyhedron): take the subset of the disjoint union of $Y^n,n\geq 1$ formed by all sequences such that any two consecutive elements are in the same simplex and the first and the last elements are the base point, and take the quotient of this subset with respect to the equivalence relation generated by $(x_1,\ldots,x,x,\ldots x_n)\sim (x_1,\ldots,x,\ldots,x_n)$ and $(x_1,\ldots,x,y,x,\ldots x_n)\sim (x_1,\ldots,x\ldots,x_n)$; the product is the concatenation product.

Here are some remarks:

  1. The above is somewhat (but not completely) similar to what happens when one "strictifies" an $A_\infty$ algebra by taking the cobar construction of the bar construction.

  2. An H-space $X$ can have several non-homotopic products. These are one-to-one with $[X\wedge X,X]$, see e.g. Stasheff, H-spaces from a homotopy point of view, p.11, LNM 161 (which also has useful references to earlier work).

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I really like this answer, but since I asked so many questions, it was hard to choose a unique answer. I appreciate the references! –  Justin Curry Jan 21 '10 at 21:09

I think its nice to see a group structure that is explicitly holomorphic. I hope the following is correct:

Take $\mathbb{C}(t)^{\times}/\mathbb{C}^{\times}$. This has a group structure. Say we consider the product of $\sum_{i=a}^{b} x_{i}t^{i}$ and $\sum_{j=c}^{d}y_{j}t^{j}$. It is $\sum_{k=a+c}^{b+d}(\sum_{\{(i,j)|i+j=k\}}x_{i}y_{j})t^{k}$. This gives some map from $\mathbb{C}\mathbb{P}^{a-b-1}\times \mathbb{C}\mathbb{P}^{c-d-1} \to \mathbb{C}\mathbb{P}^{a+c-b-d-1}$ and the group sturcture on $\mathbb{C}\mathbb{P}^{\infty}$ is the limit of these maps.

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I'm worried this doesn't actually work: Not every element of the field $\mathbb{C}(t)$ can be written as $\sum_{i=a}^b x_i t^i$, and the set of those that do, do not form a group. I don't see a simple way to fix this. –  Alison Miller Sep 5 '10 at 20:05
    
One can also multiply arbitrary elements of the form $\sum_{i=N}^{\infty} x_{i}t^{i}$ for any integer $N$. I was stopping at a finite number because I was trying to show that for some group structure that already exists the multiplication was induced by maps of topological cells. –  Oren Ben-Bassat Oct 27 '10 at 17:49
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maybe I should have used $\mathbb{C}((t))^{\times}$ instead, the formal Laurent series –  Oren Ben-Bassat Oct 27 '10 at 17:56

(Re: 2 & 3)

Since $\operatorname{Sym}^n\mathbb CP^1=\mathbb CP^n$ («by Viète's formulas»), $\mathbb CP^\infty$ is the free abelian monoid generated by $\mathbb CP^1=S^2$ (with some fixed point as a unit). (This is exatly the operation «representing» tensor product of $U(1)$-bundles aka addition in $H^2$ on $[-,\mathbb CP^\infty]=\operatorname{Bun}_{U(1)}(-)=H^2(-;\mathbb Z)$ — cf. other answers.)

Unfortunately, this operation lacks (stict) inverse. But by Dold-Thom theorem $\widetilde{\mathbb Z[S^2]}:=\mathbb Z[S^2]/\mathbb Z[pt]$ (the free abelian group generated by $S^2$ with some (fixed) point as a unit) has homotopy type of $K(\mathbb Z,2)$. Moreover, the natural map $\mathbb CP^\infty=\operatorname{Sym}^\infty(S^2)\to\widetilde{\mathbb Z[S^2]}$ is a homotopy equivalence.

(And any $K(G,n)$ can be made an abelian topological group in analogous way: $\widetilde{G[S^n]}$ has desired homotopy type.)

/* Essentially x-posted from math.SE */

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Another way of phrasing this associative multiplication on $CP^\infty$ is to take it to be the projectivization of the space of polynomials. Multiplication of polynomials descends to a monoid structure on $CP^\infty$. To get inverses, one can take the projectivization of the space of rational functions. –  Ben Wieland Jun 16 '11 at 17:15

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