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?

onlyaround the problematic latex expression. Please feel free to revert my edit. $\endgroup$ – Ricardo Andrade Jun 11 '13 at 23:375more comments