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It is well known that $BU$ is an infinite loop space, and as such it has an action of an $E_\infty$ operad. An explicit construction of such an action is given, for example, in an answer to this MO question . My question is whether it is possible to construct an action that respects the filtration of $BU$ by $BU(n)$.

Let me spell out the question more explicitly: is it possible to construct a filtered space $X_1 \subset X_2 \subset \cdots X$ such that for each $n$ $X_n\simeq BU(n)$, and there is an action of an $E_\infty$ operad $P$ on $X$ such that operad structure maps like $$P(i)\times X^i\to X$$ restrict to maps $$P(i)\times X_{n_1}\times\cdots \times X_{n_i}\to X_{n_1+\cdots+ n_i}$$

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  • $\begingroup$ Doesn't the multiplicative structure of algebraic K-theory respect the filtration? In this case, shouldn't a similar product (as the one defined by Loday) work? $\endgroup$ – user40276 Aug 26 '16 at 6:01
  • $\begingroup$ I believe the multiplicative structure on $K$-theory is induced by tensor products which changes the rank. $\endgroup$ – Sean Tilson Aug 26 '16 at 9:52
  • $\begingroup$ What happens if you filter the topological category of complex vector spaces by rank and apply Segal's machine? I guess it's not so obvious what the $X_n$'s end up being in this case, though... $\endgroup$ – Dylan Wilson Aug 26 '16 at 12:12
  • $\begingroup$ @DylanWilson Whenever I try any such approach, I get an $E_\infty$ operad acting on $\coprod BU(n)$. This action does respect the obvious filtration. It makes $\coprod BU(n)$ into a homotopy commutative monoid, whose group completion is ${\mathbb Z}\times BU$. However, I am really interested in obtaining a filtered $E_\infty$ structure on $BU$. $\endgroup$ – Gregory Arone Aug 26 '16 at 12:19
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    $\begingroup$ I'm perplexed. We now have two conflicting answers, and both are so beautiful I refuse to disbelieve either one or other. Moreover you accepted one of the answers, so seemingly you find it settled. I believe in such situation it would be better to clarify the status of the second answer. If there is a flaw, imo it must be clearly indicated. $\endgroup$ – მამუკა ჯიბლაძე Aug 30 '16 at 9:32
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Unfortunately there is no such filtration.

At first, this looks very similar (but not as strong as) asking for a map of $E_\infty$ spaces $\coprod BU(n) \to BU$ which would become a splitting map $ku \to bu$ of spectra. We know that doesn't happen, because there's a nontrivial $k$-invariant.

We can look more carefully at how this $k$-invariant works and it leads us to a method to give an actual contradiction. The $k$-invariant is detected by the fact that the generator $1_{ku}$ of $\pi_0 ku$ is annihilated by the Hopf element $\eta$, and the Toda bracket $\langle 2, \eta, 1_{ku}\rangle$ contains the Bott element $\beta$ and does not contain zero. This is spelled out in the following way on the level of $E_\infty$ spaces. Suppose $X$ is an $E_\infty$ space with multiplication $\smile$ and associated spectrum $KX$, and let $\alpha \in \pi_0(X)$ have image $[\alpha] \in \pi_0(KX)$. Up to translating path components, the element $\eta[\alpha] \in \pi_1(KX)$ lifts (up to changing path components) to the element $\alpha \smile_1 \alpha \in \pi_1(X, \alpha \smile \alpha)$. We can find a canonical nullhomotopy of the path composite of $\alpha \smile_1 \alpha$ with itself, expressing the identity $2 \eta [\alpha] = 0$; if we also have a nullhomotopy of $\alpha \smile_1 \alpha$, then we can use this to construct a representative for the bracket. If you carry this out for the basepoint of $BU(1)$ using the standard map $E\Sigma_2 \times_{\Sigma_2} BU(1)^2 \to BU(2)$, you find that you get the generator of $\pi_2 BU(2) = \Bbb Z$. However, in $bu$ this would mean that there was a bracket $\langle 2,\eta, 0\rangle$ which did not contain zero; that would be bad.


Of course, it turns out to be a lot easier to appeal to some machinery. Kochman calculated the Dyer-Lashof operations on $H_* BU = \Bbb F_2[x_1, x_2, \dots]$. His calculations show, for example, that $Q^4 x_1 = x_1^3 + x_1 x_2 + x_3$. As a result, the map $H_6 (E\Sigma_2 \times_{\Sigma_2} BU(1)^2) \to H_6(BU)$ is surjective, but the map $H_6(BU(2)) \to BU$ is not (it is missing $x_1^3$). This means that our operad structure map $P(2) \times X_1 \times X_1 \to X$ could never land in a subobject $X_2$ as you want. (This assumes that I've read and understood correctly in calculating the operations.)

(I actually find Priddy's method for calculating the Dyer-Lashof operations here a little easier than trying to understand Kochman's algorithm: Priddy calculates the Dyer-Lashof operations on $H_*(\coprod BU(n)) = \Bbb F_2[a_0,a_1,\dots]$ and then you can deduce the operations on the generators $x_i = a_i a_0^{-1}$ of $H_* BU$ by the Cartan formula. The inverse screws up the property of preserving homogeneous degree and it's what's messing us up here.)

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  • $\begingroup$ This is great! Stupid question: what's the difference between ku and bu? $\endgroup$ – Dylan Wilson Aug 29 '16 at 12:31
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    $\begingroup$ @Dylan I'm using $ku$ for the connective complex K-theory spectrum and $bu$ for its $0$-connected cover, which have infinite loop spaces $\Bbb Z \times BU$ and $BU$ respectively. (This is not consistent with the older literature, which uses $bu$ for the former.) $\endgroup$ – Tyler Lawson Aug 29 '16 at 13:33
  • $\begingroup$ Am I correct in interpreting the final argument as a proof that this filtration is not even E_2? (Since it looks like you only need Q^4 on a degree 2 class). $\endgroup$ – Dylan Wilson Nov 15 '17 at 2:10
  • $\begingroup$ Whoops, I guess it’s an E_3 operation. Never mind. I wonder if it’s possible to modify the first argument to rule out an E_2 filtration. (This is all motivated from conversations with Jeremy Hahn on his recent preprint with Allen Yuan. They show that the corresponding filtration on \OmegaSU(n) from Mitchell is A_infty, and ask whether it is E_2. I gather that the expectation is that it isn’t, but it’s stated as a question.) $\endgroup$ – Dylan Wilson Nov 15 '17 at 3:26
  • $\begingroup$ @DylanWilson Yeah, since everything is in even degree you're kind of hosed for interesting E_2-algebra operations, because both $Q_1(-)$ and $[-,-]$ take even-degree classes to odd ones. $\endgroup$ – Tyler Lawson Nov 15 '17 at 16:07
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let me give a possibly helpful "answer" by ancient history. Since I don't actually have an answer, these should be comments, but they are too long for that. In Section 1 of Chapter I of "$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra", http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf, I gave a systematic account of the action of the linear isometries operad $\mathcal{L}$ on infinite homogeneous spaces $G/\prod_{1\leq i \leq n} G_i)$ of classical groups. It includes all of the classical groups and, via the usual Grassmannians, all of their classifying spaces. By construction, all of these are colimits related to, but not at all the same as, the one given by the natural filtration you ask about. They are all given by functors $T$ defined on a graded category of finite dimensional (real) inner product spaces, the grading given by dimension; see Definitions 1.6 and 1.8, op cit. So for the real analogue of your question $T$ would send $V$ to $O(V\oplus V)/O(V)\times O(V)$. The $\mathcal{L}$-space associated to $T$ is the colimit $T(\mathbf{R}^{\infty})$ of the $T(V)$. This is not what you ask for, but it is the most geometric action in sight. The construction there was tailored towards proving that all of the Bott maps are $E_{\infty}$ maps, and there might be some alternative packaging that does what you want. You mention the quite different operad acting on $\coprod BU(n)$, which by construction from the permutative category $\coprod U(n)$ behaves as you wish before group completion. Perhaps a comparison between these two different operad actions might be relevant, and I gave that in "The spectra associated to $\mathcal{I}$-monoids", http://www.math.uchicago.edu/~may/PAPERS/24.pdf.

Several comments refer to the multiplicative structure $\otimes$ rather than to the additive structure $\oplus$ you are implicitly asking about. The bipermutative category $\coprod BU(n)$ gives rise to an $E_{\infty}$ ring space structure, but that would lead to a different (and slightly ill-posed) question with your sum of the $n_i's$ replaced by their product.

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Fix your favourite small infinite vector space $V$; note that the space of (ordered) decompositions $f:V \simeq \bigoplus_1^k V$ is contractible, so the space of sets of inclusions $\{ f_j | j = 1,...,k\}$ (i.e., forgetting the order) is a model of $B \mathfrak{S}_k$. To such a set of inclusions $I$ and a choice of suspaces $W_j \subset V$ indexed by $I$, write $\bigoplus_I W_j = \sum f_j W_j$.

Our model of $BU$ will be the subspaces of $V$ of finite corank, and the filtration $BU_n$ will be the subspaces specifically of corank at-most-$n$.

Claim: $BU_n$ is a model of $\mathbf{B}U(n)$.

Idea: I'll describe a blow-up $ X \to BU_n \backslash BU_{n-2}$ and argue both equivalences $X \simeq BU_n \backslash BU_{n-1}$ and $X \simeq BU_n\backslash BU_{n-2}$.

Given a space $W \lt V$ of corank $n-1$ the fiber of the blowup over $W$ consists of all choices of a basis $b :\mathbb{C}[x]\simeq W$ (contractible choice!); let $X$ be the mapping cylinder of the map assigning to $(W;b)$ the subspace $b(x\mathbb{C}[x])$; extend the cylinder map to $p : X \to BU_n\backslash BU_{n-2}$ by $(W;b,t) \mapsto b\left((1-t + tx)\mathbb{C}[x]\right)$

The mapping cylinder of course deformation retracts to its base, $BU_n \backslash BU_{n-1}$; on the other hand, the homotopy specified by $s \mapsto t \mapsto ts$ interpolates $p$ and the map forgetting both $t$ and $b$; which map has contractible (set-theoretic) fibers.

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  • $\begingroup$ I am having difficulties with a couple of steps here. 1. As far as I understand, the space of ordered decompositions is the space of isomorphisms Iso$(V^k, V)$. If so, it is homeomorphic to an infinite unitary group and I don't see why it is contractible. 2. It seems to me that $BU_n$ is the space of subspaces of $V$ of corank exactly $n$, while spaces of corank at most $n$ give you the disjoint union of $BU(i)$ from $1$ to $n$. BTW, I would be really happy to have an answer to this question. So if I am misunderstanding, and someone can explain, I will be grateful. $\endgroup$ – Gregory Arone Aug 27 '16 at 15:01
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    $\begingroup$ Corank is not continuous; one can linearly homotopify Hilbert's Hotel, but one pays for it by leaving the space of Fredholm operators; more importantly, Iso(V,V) is not the stable general linear group; stable U is the operators $g \in Iso(V,V)$ with $g-1$ of finite rank. $\endgroup$ – Jesse C. McKeown Aug 27 '16 at 19:48
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    $\begingroup$ You are right about the group of isomorphisms. Do you have a reference for the model of $BU_n$ as the space of subspaces of corank at most $n$? Thank you! $\endgroup$ – Gregory Arone Aug 27 '16 at 21:52
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    $\begingroup$ I guess the point is supposed to be that the space of subspaces of corank exactly n (which you say is homotopy equivalent to X) models BU(n) because one can build a vector bundle over it where the fibers are V/W? And they you're trying to argue that when you include the subspaces of smaller rank, you're not changing the homotopy type. I guess I see better the strategy now. $\endgroup$ – Dan Ramras Aug 30 '16 at 20:06
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    $\begingroup$ @მამუკაჯიბლაძე I was indeed trying to modify that construction to fit the spec. That union you describe (or something like it) tells you about subspaces of $\Bbb C^\infty$ which are of exact corank n and which, for some $k$, contain all elements $(z_i)_{i \geq 0}$ such that $z_i = 0$ when $i < k$. It doesn't e.g. describe the subspace $\{(z_i) | \sum z_i = 0\}$. $\endgroup$ – Tyler Lawson Sep 2 '16 at 5:16

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