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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
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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
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@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

1 Answer 1

up vote 1 down vote accepted

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 on the subject and Eric Peterson's notes 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 $\Omega SU(n)\to\Omega SU(n+1)\to BF$).

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