# Thom isomorphism's effect on module structure of n-oriented spectra

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?

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I took the liberty to edit your question, so as to put backticks only around the problematic latex expression. Please feel free to revert my edit. – Ricardo Andrade Jun 11 '13 at 23:37
No that's perfect, looks much better. Thanks! – Jon Beardsley Jun 12 '13 at 2:13
Jon, as the map $X(n) \to X(n+1)$ is a map of ring spectra, this seems to me like it likely boils down to verifying compatibility of the structure of a ring spectrum on $X(n)$ which the Thom isomorphism. Yes? – Tyler Lawson Jun 13 '13 at 3:15
Yeah, pretty much. I'm currently looking at some oldish papers of Mark Mahowald on Thom spectra which are ring spectra, and specifically Thom spectra coming from loop spaces (hence $A_\infty$-ring spectra) to see how he does this kind of thing. I think I'm close to understanding it (though I've thought that about plenty of things before, and been miserably wrong). – Jon Beardsley Jun 13 '13 at 3:29
@ArtiePrendergast-Smith, that's the reason. Old MathJax had problems with this and with superscripts. The shortcut fix was to use backticks. Apparently, the superscripts have been fixed. But when I put the braces around the single character subscripts, it fixes the display. – Fred Kline Jun 26 '13 at 20:29

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.
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 $\Omega SU(n)\to\Omega SU(n+1)\to BF$).