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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 non-abelian group $Sp(1) \cong Spin(3) \cong SU(2)$?

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what does this hat notation mean? – Sean Tilson Nov 16 '11 at 4:57
Well, it looks like $K(\widehat{blah},n)$ is some sort of Eilenberg-MacLane space, but that is really the wrong way to think about it, as $blah$ isn't a discrete group. – David Roberts Nov 16 '11 at 5:05
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 $BSp(1)$. – Torsten Ekedahl Nov 16 '11 at 5:13
$\mathbb CP^{\infty}=K(\mathbb Z,2)$, no? – Will Sawin Nov 16 '11 at 5:13
$\widehat{G}$ 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 non-abelian group, the machinery of K(G,n)s cannot possibly work, which is the point of the question. – Alexander Moll Nov 16 '11 at 6:32

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 $$\pi_i(\mathbb{H}P^\infty)\cong\pi_{i-1}(Sp(1))=\pi_{i-1}(S^3)$$ for each $i$. In particular, $\mathbb{H}P^\infty$ is not an Eilenberg-Mac Lane space.

However, this implies that it 'is' an Eilenberg-Mac Lane space after rationalization, $$\mathbb{H}P^\infty\simeq_{\mathbb{Q}}K(\mathbb{Z},4).$$

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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 $BSp(1)$? – Alexander Moll Nov 16 '11 at 14:18
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? – Mark Grant Nov 16 '11 at 15:07
Also, you could try looking at Chapter 1 of "Algebraic models in Geometry" by Felix, Oprea and Tanré. – Mark Grant Nov 16 '11 at 15:16
My first reaction was Tannaka-Krein 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 pre-existing work on/near this topic... – Alexander Moll Nov 16 '11 at 15:21
This is neat: in each of these three cases, you have $K( \pi_i(G), i+1 )$. – Dave Anderson Nov 16 '11 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 non-canonically 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 Mac-Lane 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 $\pi_{n-1}(G)$. 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.

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Looking at Chern and Stiefel-Whitney classes, it seems like homology or cohomology should be involved more than homotopy groups. Unsure. – Will Sawin Nov 17 '11 at 4:46
If we divide $S_n$ into two copies of $B_n$ glued together at $S_{n-1}$, and all fibrations on $B_{n}$ are trivial, then if we try to glue the fibrations together, the error will be an element of $\pi_{n-1}(G)$. – Will Sawin Nov 17 '11 at 5:09
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). – David Roberts Nov 17 '11 at 5:30
Um, good point. That would be another thing that's not explained. If it's isomorphic to $\pi_{n-1}$, 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 Eilenberg-MacLane spaces for things other than groups. My argument sketch was motivated by this paper: – Will Sawin Nov 17 '11 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.

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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 Eilenberg-Maclane 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 (, Section 2. I guess the same should work for $SU(n)$ when you leave out the $c_1$ factor.

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