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](https://mathoverflow.net/questions/94295/infinite-loop-space-structure-of-bu) . 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
$$P(i)\times X\_{n\_1}\times\cdots \times X\_{n\_i}\to X\_{n\_1+\cdots+ n\_i}$$



Is it possible to construct an action of an $E_\infty$ operad on $BU$ that respects filtration by $BU(n)$?
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.




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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.)


