This question is specifically related to the spectra $X(n)$ used in Devinatz, Hopkins and Smith's proof of the nilpotence conjectures, but any general answer in terms of the Thom isomorphism would also be appreciated.


The spectra $X(n)$ are Thom spectra coming from the maps $\Omega SU(n)\to\Omega SU\simeq BU\to BF$. We say that a spectrum $E$ has a complex orientation of degree $n$, or is $n$-oriented, if there is a class $x\in \tilde{E}^2(\mathbb{C}P^n)$ whose restriction to $\tilde{E}^2(\mathbb{C}P^1)\cong\pi\_0(E)$ is 1 (similarly to complex orientations). There is a one-to-one correspondence between $n$-orientations of $E$ and ring-spectrum maps $X(n)\to E$, again just like in the complex oriented case. We know also that there is a Thom isomorphism for such spectra: $t:E\wedge X(n)\overset{\simeq}\to E\wedge \Omega SU(n)\_+$. It's not hard to see that $X(n+1)$ is $n$-oriented, with the orientation coming from the Thomification of the map $\Omega SU(n)\to\Omega SU(n+1)$. Moreover, $X(n+1)$ is an $X(n)$-algebra. My question is whether or not the Thom isomorphism respects the module structure $X(n)\wedge X(n+1)\to X(n+1)$. That is, is there an obvious map $X(n+1)\wedge\Omega SU(n)\_+\to X(n+1)$ and is its precomposition with the Thom isomorphism the same as the given module action?



Thom isomorphism's effect on module structure of n-oriented spectra
Okay, so maybe I'm being naive here, but I think this is probably a naive question. The answer might be a little complicated by the fact that I don't have any good way to drawcommutative diagrams on here. Also, this is all pretty much lifted directly from Mahowald's  [paper](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077313574) on the subject and Eric Peterson's  [notes](http://math.berkeley.edu/~ericp/latex/talks/msri-2013.pdf) on that paper.


Recall the "shear" map for an $H$-space $X$ given by $\sigma: X\times X\to X\times X$, $\sigma(x,y)=(x,x^{-1}y)$. This is a homotopy equivalence, with homotopy inverse given by $(x,y)\mapsto (x,xy)$, I believe. However, notice that if we compose with the multiplication map, we don't get the same maps, i.e. $(x,y)\mapsto (x,x^{-1}y)\mapsto y$ rather than $xy$. Now let's say we've got some map $f:X\to BF$. Then we've got a few maps floating around, $ff:X\times X\overset{\mu}\to X\overset{f}\to BF$ whose Thom spectrum is $Th(f)\wedge Th(f)$ and $f0:X\times X\overset{\sigma}\to X\times X\overset{\mu}\to X\overset{f}\to BF$ whose associated Thom spectrum is $Th(f)\wedge \Sigma^\infty\_+ X$. But since $\sigma$ is an equivalence, we must have that its Thomification is an equivalence, hence $Th(f)\wedge Th(f)\overset{\sim}{\underset{\sigma}\to}Th(f)\wedge \Sigma^\infty\_+X$ is also an equivalence (inducing the Thom isomorphism). But notice that the map $\mu\circ \sigma$ also Thomifies to a map $Th(f)\wedge\Sigma^\infty\_+\to Th(f)$. And basically by drawing out the commutative diagram, you'll see that this "action" of $\Sigma^\infty\_+ X$ on $Th(f)$ is the same thing as going backwards along the Thom isomorphism and then applying the multiplication. 


I hope I'm not saying anything silly here... =P


The general case for a spectrum which is $X(n)$-oriented or whatever, follows from the fact that $\Omega SU(n)\to \Omega SU(n+1)$ is an inclusion (and so we can factor the map $\Omega SU(n)\to BF$ through $\Omega SU(n)\to\Omega SU(n+1)\to BF$). 


