Given that $$\mathbb{R}P^{\infty} = B O(1) = K(\widehat{O(1)}, 1)$$ $$\mathbb{C} P^{\infty} = B U(1) = K( \widehat{U(1)}, 2)$$ is there any way to make sense of $$\mathbb{H}P^{\infty} = B Sp(1)$$ in a similar manner using the representation theory of the nonabelian group $Sp(1) \cong Spin(3) \cong SU(2)$?

2$\begingroup$ what does this hat notation mean? $\endgroup$– Sean TilsonNov 16, 2011 at 4:57

1$\begingroup$ Well, it looks like $K(\widehat{blah},n)$ is some sort of EilenbergMacLane space, but that is really the wrong way to think about it, as $blah$ isn't a discrete group. $\endgroup$– David Roberts ♦Nov 16, 2011 at 5:05

4$\begingroup$ In any case, disregarding the $K$notation, the result is true and for the same reason the others are; $\mathbb{H}P^\infty$ is the quotient of $S^\infty$ by $Sp(1)$. The action is free and $S^\infty$ is contractible so (more or less) by definition the quotient is a $BSp(1)$. $\endgroup$– Torsten EkedahlNov 16, 2011 at 5:13

1$\begingroup$ $\mathbb CP^{\infty}=K(\mathbb Z,2)$, no? $\endgroup$– Will SawinNov 16, 2011 at 5:13

11$\begingroup$ $\widehat{G}$ denotes the dual group of a locally compact abelian group. In particular, $\widehat{G} \cong G$ for finite groups (but not canonically), and you may recall $\widehat{U(1)} \cong \mathbb{Z}$ from Fourier series. For a nonabelian group, the machinery of K(G,n)s cannot possibly work, which is the point of the question. $\endgroup$– Alexander MollNov 16, 2011 at 6:32
4 Answers
There is a principal bundle
$$Sp(1)\to S^{4n+3} \to \mathbb{H}P^n$$
for each $n$, which on passing to the limit shows that $$\pi_i(\mathbb{H}P^\infty)\cong\pi_{i1}(Sp(1))=\pi_{i1}(S^3)$$
for each $i$. In particular, $\mathbb{H}P^\infty$ is not an EilenbergMac Lane space.
However, this implies that it 'is' an EilenbergMac Lane space after rationalization, $$\mathbb{H}P^\infty\simeq_{\mathbb{Q}}K(\mathbb{Z},4).$$

1$\begingroup$ Thanks for the response  I'm certainly glad to see a 4. I'm not familiar with rational homotopy theory, but I'd like to ask: is there some way of understanding the rationalization of BG in terms of the representation theory of G? How can one explain this $\mathbb{Z}$ in the result for $BSp(1)$? $\endgroup$ Nov 16, 2011 at 14:18

$\begingroup$ Well I'm not familiar with representation theory, but perhaps there is some way to make sense of $\widehat{G}$ when $G$ is not abelian? A few ideas: one could replace continuous homomorphisms from $G$ to $S^1$ with ones to $S^3$, the unit quaternions? One could look at the group of continuous homomorphisms $Sp(1)$ to $S^1$ after rationalization? Maybe this is some appropriate notion of the dual group tensored with the rationals? $\endgroup$ Nov 16, 2011 at 15:07

1$\begingroup$ Also, you could try looking at Chapter 1 of "Algebraic models in Geometry" by Felix, Oprea and Tanré. $\endgroup$ Nov 16, 2011 at 15:16

$\begingroup$ My first reaction was TannakaKrein dualtiy, but I feel like there should be something more concrete for SU(2), probably its Langlands dual group since it's reductive. I'd be surprised if there wasn't any preexisting work on/near this topic... $\endgroup$ Nov 16, 2011 at 15:21

2$\begingroup$ This is neat: in each of these three cases, you have $K( \pi_i(G), i+1 )$. $\endgroup$ Nov 16, 2011 at 23:25
I claim that, in the equivalences you stated, duality has nothing to do with it. Specifically, if your viewpoint were correct, then for finite $G$, $BG(1)$ would be noncanonically isomorphic to $K(G,1)$. In fact, I claim that they are canonically isomorphic. Furthermore, the $\mathbb Z$ in $\mathbb CP^{\infty}$ is canonically isomporphic to the fundamental group, not the character group, of $U(1)$.
Argument: Since both these spaces represent functors, it suffices to consider the underlying functors. Eilenberg MacLane spaces correspond to cohomology functors. It is easy to prove using Cech cohomology that cohomology with coefficients in $G$ naturally classifies principal $G$bundles. This, of course, is exactly what the classifying space classifies  not dual to what the classifying space classifies.
$\mathbb CP^{\infty}$: There is an exact sequence $0\to\mathbb Z \to \mathbb C^+ \to \mathbb C^\times\to 0$, giving a map $H^1(X,\mathbb C^\times)\to H^2(X,\mathbb Z)$. The image is discrete while the kernel, a quotient of $H^1(X,\mathbb C^{+})$, is connected, so the map is exactly the quotient by the connected component of the identity.
$H^1(X,\mathbb C^\times)$ classifies principal $\mathbb C^\times$ spaces. Continuously moving the bundle around in it corresponds to continuously deforming the bundle. These bundles up to derivation are exactly what $BU(1)$ classifies.
EDIT: Idea/sketch for a general proof of this equivalence: Let $G_n$ be the group of principal $G$bundles on $S^{n}$. Then for some reason this should be equivalent to $\pi_{n1}(G)$. Now, the values everywhere of good functors on the category of CW complexes (specifically, representable ones) depend only on the values they take on spheres. So suppose a group had only one nontrivial $G_n$. The principal $G$spaces functor would then be equivalent to $H^n(X,G_n)$, giving an equivalence of classifying spaces.

$\begingroup$ Looking at Chern and StiefelWhitney classes, it seems like homology or cohomology should be involved more than homotopy groups. Unsure. $\endgroup$ Nov 17, 2011 at 4:46

$\begingroup$ If we divide $S_n$ into two copies of $B_n$ glued together at $S_{n1}$, and all fibrations on $B_{n}$ are trivial, then if we try to glue the fibrations together, the error will be an element of $\pi_{n1}(G)$. $\endgroup$ Nov 17, 2011 at 5:09

$\begingroup$ Why is $G_n$ a group? For nonabelian $G$ this is not true in general (modulo the fact you need to pass to isomorphism classes). $\endgroup$– David Roberts ♦Nov 17, 2011 at 5:30

$\begingroup$ Um, good point. That would be another thing that's not explained. If it's isomorphic to $\pi_{n1}$, then, that's why it's a group. Other than that, I don't know. Alternatively, it might not matter too much. We can construct versions of EilenbergMacLane spaces for things other than groups. My argument sketch was motivated by this paper: jstor.org/stable/1970209?seq=2 $\endgroup$ Nov 17, 2011 at 6:06
Any group $G$ has a classifying space $BG$. It can be a finite group, an infinite discrete group a Lie group or any topological group. The construction is always the same, find a contractible space, usually called $EG$, with a free continuous action by $G$. Then $BG$ is the quotient $EG/G$. One then has a fibration (in fact a principal bundle) $$G\to EG \to BG.$$ The long exact sequence in homotopy groups gives an isomorphism of $\pi_i G$ and $\pi_{i+1}BG$ (homotopy groups of $EG$ are all $0$). When $G$ is discrete, this gives $\pi_1BG=G$ and $BG$ is aspherical (no higher homotopy groups). If $G$ is $S^1$, then, by definition, $BS^1=CP^{\infty}$. Moreover $\pi_1 S^1 = \pi_2 CP^{\infty}=\mathbb Z$ are the only nontrivial groups of $S^1$ and $CP^{\infty}$. So $CP^{\infty} = K(Z,2)$. The $3$sphere is a Lie group, a.k.a SU(2), but has lots of nontrivial homotopy groups. So $BS^3$ is not $K(\mathbb Z,4)$ even though $\pi_4 BS^3 = \pi_3 S^3 = \mathbb Z$.
For $n\ge2$, $K(A,n)$ is defined only when $A$ is a abelian group.
This is not really an answer to the question posed but seems to be of relevance to people interested in the question (and is directly related to the case $BSU(2)\cong_{\mathbb{Q}}K(\mathbb{Z},4)$ mentioned by Mark Grant). There is a sequence of groups for which the classifying spaces are rationally products of EilenbergMaclane spaces: namely $BU(n)$. The $i$th Chern class is an element of $H^{2i}(X,\mathbb{Z})$ and hence can be thought of as homotopy class of map to a $K(\mathbb{Z},2i)$ space. Therefore you get a map
$$c_1\times\cdots\times c_n\colon BU(n)\to \prod_{i=1}^nK(\mathbb{Z},2i)$$
which turns out to be a rational homotopy equivalence. I learned this trick from Atiyah & Bott (http://www.jstor.org/stable/10.2307/37156), Section 2. I guess the same should work for $SU(n)$ when you leave out the $c_1$ factor.