Bott periodicity implies that $\Omega(SU)\simeq G(\infty)$. Here, by $G(\infty)$, I mean the direct limit $\underset{m\to \infty}{\lim} G_m(\mathbb{C}^{2m})$ where $G_m(\mathbb{C}^{2m})\subset G_{m+1}(\mathbb{C}^{2m+2})$ by stabilization (or some similar nice model for $BU$). This is the classifying space for $U = \underset{m\to\infty}{\lim} U(m)$, and may be identified with $U/(U\times U)$, where the coordinates of each factor of the product subgroup alternate. From the path space construction, one knows that there is a contractible Serre fibration $\Omega SU \to E \to SU$, where $E$ is the space of paths in $SU$ from the identity. The fibers are each homotopic to $G(\infty)$.

The question I have is whether this may be realized by a contractible fiber bundle $G(\infty)\to E \to SU$ ? Other quasi-fibrations were given by Aguilar and Prieto, and by Behrens. There are also fibrations of this sort constructed using symplectic reduction by Latour and by Giroux, where the fibers are homotopic to $G(\infty)$. What I'm asking for is whether there exists a contractible fiber bundle rather than just a fibration or quasi-fibration?

Presumably such a bundle would arise from a map $SU \to BDiff(G(\infty))$. The isometry group of $G(\infty)$ contains $SU$, so one could ask a fortiori whether there is a map $SU \to BSU$, that is a map $f: SU \to G(\infty)$, such that the induced fiber bundle is contractible? Added clarification: The induced bundle would come from the fiber product of the pull-back of the $U$ bundle $U\to E \to G(\infty)$ with the action of $U$ on $G(\infty)$: $f^{*}(E) \times_{U} G(\infty)$.


2 Answers 2


I'm not completely sure that I understand your notation, so this may be not what you want, but in case it is close enough, I'll have a go. The bit I'm assuming is that $G_m$ is the Grassmannian of $m$-places in $\mathbb{C}^{2m}$. That seems fairly safe, but my brain is refusing to check the homotopy types of everything involved at this hour.

I think that you can do this if you combine a construction of mine from The Co-Riemannian Structure of Smooth Loop Spaces with some bits from Loop Groups by Pressley and Segal, and then check a few details about how stuff holds together in the limit.

The bit that you need from Loop Groups is that the polynomial loop group, $\Omega_{\text{pol}} S U(n)$ acts on the finite restricted Grassmannian $Gr_0(H)$. Let me remind you what that space is: we start with the polarised Hilbert space, $H = L^2(S^1,\mathbb{C}^n)$, polarised as $H = H_+ \oplus H_-$ where $H_+$ are those functions with only positive Fourier coefficients and $H_-$ with strictly negative Fourier coefficients. From this, we define

$$ Gr_0(H) = \{W \subseteq H : \exists k : z^k H_+ \subseteq W \subseteq z^{-k} H_+\} $$

This is the union of $Gr(H_{-k,k})$ where $H_{-k,k} = z^{-k}H_+/z^kH_+$ so this (if I'm reading things aright) is the $G(\infty)$ of your question (Loop Groups, section 7.2).

The next bit that we need is that $\Omega_{\text{pol}} SU(n)$ acts on this space. This is from Theorem 8.3.2 and Proposition 8.3.3 in Loop Groups.

Putting these two together, if we have a principal $\Omega_{\text{pol}} SU(n)$-bundle over a space then we get a $G(\infty)$-fibre bundle over said space.

So now comes the bit from my work. In Section 3.2.3 of The co-Riemannian Structure of Smooth Loop Spaces, I construct a principal $\Omega_{\text{pol}} SU(n)$-bundle over $SU(n)$, with the property that under the obvious inclusion $\Omega_{\text{pol}} SU(n) \to \Omega SU(n)$ then this becomes the bundle coming from the usual path construction (so, although I don't need this in that paper, Bott periodicity implies that it is contractible).

So now we have a fibre bundle with fibre $G(\infty)$ over $SU(n)$. The last bit that you need is to show that under the inclusion $SU(n) \to SU(n+1)$ then these fibre bundles are compatible. The bit where this needs care is in the action of $\Omega_{\text{pol}} SU(n)$ on $Gr_0(H)$. But I think that if you include $L^2(S^1, \mathbb{C}^n)$ in to $L^2(S^1, \mathbb{C}^{n+1})$ at the same time, then you should get a diagram that works. The resulting Grassmannian will be $\bigcup Gr_0(L^2(S^1,\mathbb{C}^n))$ but that will still be $G(\infty)$ (assuming that I've understood the question correctly).

As I said, there's a few ifs and buts here: if I understood the question correctly, and if everything holds together in the limit (but I'm pretty sure that the second "if" is fine), but obviously as I'm less sure about the first if I haven't checked all the details.

  • $\begingroup$ Thanks for the answer, that sounds promising. I'll have to do some reading on Loop groups to understand your answer. $\endgroup$
    – Ian Agol
    Mar 14, 2011 at 19:22

This is a lousy answer, but it gets you part of the way. There is an $E_\infty$ H-space structure on $G(\infty)$ given by direct sum of subspaces of $\mathbb{C}^\infty$. Consequently one is entitled to form the classifying space $B(G(\infty))$.

Now, one can rectify $G(\infty)$ into a group; that is, there is a group $G$ and a homotopy equivalence $G \to G(\infty)$. For instance, one could take $G$ to be the Kan loop group on the singular complex of $B(G(\infty))$, a (very large) group model for $\Omega B(G(\infty)) \simeq G(\infty)$. Then one gets a homotopy equivalence $BG \simeq BG(\infty)$.

Bott periodicity implies that there is a homotopy equivalence $SU \to B(G(\infty))$. Replace this with a map $f: SU \to BG$. The associated principal $G$-bundle $E \to SU$ is contractible, since $f$ is a homotopy equivalence. So you get a contractible fiber bundle over $SU$ with fibre homotopy equivalent to $G(\infty)$.

That's not quite what you want. If you could show that the homotopy equivalence $G \to G(\infty)$ was via an action that mimicked left translation in $G$, then you can form the simultaneous quotient $E \times_G G(\infty)$; this would be precisely the sort of object that you had in mind.

I'm also ignoring the issue about whether the universal bundle $EG \to BG$ is in fact a fibre bundle; $G$ is potentially sufficiently exotic that the standard theorems about Lie groups don't necessarily hold. Again, though, if you could show that $G$ acts on $G(\infty)$ nicely, then you could realize $G$ (as you say above) as a subgroup of $Diff(G(\infty))$, where I believe that the result is known to hold.


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