[Bott periodicity](http://en.wikipedia.org/wiki/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](http://www.ams.org/mathscinet-getitem?mr=1886093). There are also fibrations of this sort constructed using symplectic reduction by [Latour](http://www.ams.org/mathscinet-getitem?mr=1088270) and by [Giroux](http://dl.dropbox.com/u/8592391/msri10.pdf), 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 $G(\infty)$: $f^{\*}(E) \times\_{U} G(\infty)$. 



Does there exist a contractible fiber bundle with fiber $G(\infty)$ and base $SU(\infty)$?
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](http://www.math.ntnu.no/~stacey/Research/Preprints/coriemloop.html) 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.


