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It is well-known that $BU^+$ is homotopy equivalent to an infinite loop space where $U$ is the limit of the unitary groups $U(n)$ for $n \rightarrow \infty$ and $+$ denotes Quillen's Plus construction.

On the other hand there is a tool to detect infinite loop spaces by the action from $E\Sigma_p$ ($\Sigma_p$ is the symmetric group).

My question is: how does the map $E\Sigma_p \times_{\Sigma_p} BU^p \rightarrow BU$ looks like. Is it possible to write it down.

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2 Answers 2

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Firstly, no plus construction is needed here. The plus construction is used to kill a perfect subgroup of the fundamental group, but $\pi_1(BU)$ is already trivial.

Next, note that the space $\mathbb{C}^\infty=\bigoplus_{n=0}^\infty\mathbb{C}$ has a natural Hermitian inner product (with respect to which it is not complete). For any Hermitian space $\mathcal{V}$ that is isomorphic to $\mathbb{C}^\infty$, we consider the space $\mathcal{V}\oplus\mathcal{V}$ and its subspaces $\mathcal{V}_L=\mathcal{V}\oplus 0$ and $\mathcal{V}_R=0\oplus\mathcal{V}$. Let $B(\mathcal{V})$ denote the space of subspaces $\mathcal{A}\leq\mathcal{V}\oplus\mathcal{V}$ such that $\mathcal{A}\cap\mathcal{V}_L$ has finite codimension in $\mathcal{A}$, and also the same finite codimension in $\mathcal{V}_L$. To understand this in more detail, suppose we have a subspace $V\leq\mathcal{V}$ with $\dim(V)=n<\infty$, giving a decomposition $$ \mathcal{V}\oplus\mathcal{V} = V_L\oplus (V^\perp)_L \oplus V_R \oplus (V^\perp)_R. $$ We put $$ B(\mathcal{V};V) = \{A\oplus (V^\perp)_L : A\leq V_L\oplus V_R,\; \dim(W) = n\}. $$ We find that $B(\mathcal{V};V)$ is naturally identified with a finite-dimensional Grassmann manifold, so it has a natural compact Hausdorff topology. Moreover, the set $B(\mathcal{V})$ is the colimit of the sets $B(\mathcal{V};V)$, so we give it the colimit topology. One can check that $B(\mathcal{V})$ is then a model for the homotopy type $BU$.

Now suppose we have two Hermitian spaces $\mathcal{V}$ and $\mathcal{W}$ as above, and a linear map $\alpha\colon\mathcal{V}\to\mathcal{W}$ that preserves inner products. (This implies that $\alpha$ is injective, but it need not be surjective.) Given a point $\mathcal{A}=A\oplus(V^\perp)_L\in B(\mathcal{V};V)$, we have a point $$ \mathcal{B} = (\alpha\oplus\alpha)(A) \oplus(\alpha(V)^\perp)_R \in B(\mathcal{W};\alpha(V)). $$ One can check that this does not really depend on the choice of $V$, so we have a well-defined map $\alpha_*\colon B(\mathcal{V})\to B(\mathcal{W})$. One can also check that this is functorial. There are also evident maps $$ B(\mathcal{V})\times B(\mathcal{W}) \to B(\mathcal{V}\oplus\mathcal{W}), $$ making $B$ into a lax monoidal functor.

Now let $E(k)$ denote the space of inner-product preserving linear maps from $\mathcal{V}^k$ to $\mathcal{V}$. This has an evident action of $\Sigma_k$, which is free because inner-product preserving maps are always injective. It is a standard fact that $E(k)$ is also contractible, so it is a model for $E\Sigma_k$.

Now suppose we have elements $\mathcal{A}_1,\dotsc,\mathcal{A}_k\in B(\mathcal{V})$ and a map $\alpha\in E$. We then apply $\alpha_*$ to $\bigoplus_i\mathcal{A}_i$ to get a point $\gamma(\alpha;\mathcal{A}_1,\dotsc,\mathcal{A}_k)\in B(\mathcal{V})$. This construction gives the map $$ \gamma:E(k)\times_{\Sigma_k}B(\mathcal{V})^k\to B(\mathcal{V}) $$ that you need.

Note: An earlier version of this answer said that $\alpha_*(\mathcal{A})$ should just be $(\alpha\oplus\alpha)(\mathcal{A})$, but that is not correct, and in fact $(\alpha\oplus\alpha)(\mathcal{A})$ need not lie in $B(\mathcal{W})$. I thank Jack Smith for pointing out this error.

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  • $\begingroup$ I suppose the questioner intends "Quillen's plus construction" in the sense of "group completion", rather than the correct sense of "construction that kills a perfect subgroup of the fundamental group without changing the homology". $\endgroup$ Apr 17, 2012 at 15:22
  • $\begingroup$ @Charles: perhaps. Then of course the correct statement is that $\mathbb{Z}\times BU$ is the group completion of $\coprod_nBU(n)$. If, as in the question, we take the colimit rather than the coproduct of the spaces $BU(n)$, we get $BU$, which is already group-complete. $\endgroup$ Apr 17, 2012 at 15:28
  • $\begingroup$ Thank you, for making this precise. The proof of Neil Strickland works of course in any case. $\endgroup$
    – berl13
    Apr 17, 2012 at 19:56
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There are a number of equivalent ways of seeing $BU\times Z$ or $BU$ as an infinite loop space. The description that Neil gives is the action of the linear isometries operad $\mathcal{L}$ (complex version) on $BU$. A large number of related spaces have such a structure, as explained in the first section of the first chapter of "$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra", SLN 577 (1977). Morever, the Bott maps are maps of $\mathcal{L}$-spaces, which ties in Bott's original proof that $BU$ is an infinite loop space. The fact that $\coprod_n U(n)$ is a permutative category gives a quite different operad action on $\coprod_n BU(n)$, whose associated infinite loop space is $BU\times \mathbf{Z}$. It is not obvious a priori how to compare the infinite loop structures on these two models. The question is resolved in this and related examples (e.g. $BTop$) in my paper "The spectra associated to $\mathcal{I}$-monoids". Math. Proc. Camb. Phil. Soc. 84(1978), 313--322.

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  • $\begingroup$ That's very interesting, so another question arises for me. Perhaps it leaves the frame of this question. But do you know if there are any standard techniques to compare infinite loop space structures on the same space. More precisely, the situation is the following. One has a space $X$ which is an infinite loop space w.r.t. two infinite loop spaces structures and the corresponding multiplications on its zero component commute. Are there tools to say to which extent the structures differ. I have an example in mind where I am pretty sure that they are not the same. $\endgroup$
    – berl13
    Apr 18, 2012 at 12:12
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    $\begingroup$ Do you really mean ``commute''? In any case, probably the closest thing would be to understand moduli spaces of infinite loop structures (there is relevant spectrum level work of Goerss and Hopkins). By way of example, there is a beautiful but very special case where there is an answer: Adams and Priddy proved that BSU and BSO have unique infinite loop structures (whereas BU and BO do not: $\oplus$ and $\otimes$ give inequivalent infinite loop structures) $\endgroup$
    – Peter May
    Apr 18, 2012 at 18:55

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