I am currently trying to wrap my head around all kinds of different definitions of the notion of (generalized) Thom spectrum. My setup is as follows: Suppose I have a commutative (symmetric) ring spectrum $R_{\bullet}$ and a group $G$, which acts on each of the spaces $R_n$ in a way that is compatible with all structure maps and the $\Sigma_n$-action. Moreover, let $P \to X$ be a principal $G$-bundle classified by the map $f \colon X \to BG$.

I can associate a Thom spectrum to this in different ways:

  • I can form the bundle of spectra $\mathcal{R}_n = P \times_{G} R_n \to X$, which has a zero section $\sigma_n \colon X \to \mathcal{R}_n$. The first Thom spectrum is then $Mf^{(1)}_n = \mathcal{R}_n / \sigma_n(X) = r_!\mathcal{R}_n$, where $r \colon X \to \{pt\}$ and $r_!$ is the shriek map as defined by May and Sigurdsson.
  • I can use the two-sided bar construction $B(P, G, R)$ and define $Mf^{(2)}_n = r_!B(P,G,R_n)$ as in Definition 23.5.1 of May and Sigurdsson.
  • I can form the smash product $Mf^{(3)} = \Sigma^{\infty}P_+ \wedge_{\Sigma^{\infty} G_+} R$.
  • I can view $f$ as a map $f \colon BG \to BGL_1(R)$ and use the $\infty$-categorical construction in the work of Ando, Blumberg, Gepner, Hopkins and Rezk to get $Mf^{(4)}$.

What are the precise conditions on $R,G,P$ and $X$ that are needed such that these definitions agree?

Here is what I got so far: I think I understand how $Mf^{(1)}$ and $Mf^{(2)}$ are related (even though I am not sure, if I need any cofibrancy conditions) and it is proven in the paper mentioned in the fourth point that $Mf^{(3)}$ and $Mf^{(4)}$ are equivalent.

Does the bar construction in this case always give a representation of the smash product?


1 Answer 1


The Thom spectrum $Mf^{(1)}$ obtained from the group action of $G$ on $R$ agrees with the (non-derived!) smash product of $\Sigma^{\infty}_+ P$ and $R$ over $\Sigma^{\infty}_+G$. The only difference between this and $Mf^{(3)}$ is that the latter uses the derived smash product. Therefore the two agree if either $R$ or $\Sigma^{\infty}_+P$ is cofibrant as $\Sigma^{\infty}_+G$-module spectrum. This happens for example, if $X$ is a finite CW-complex. Then $P$ is (homeomorphic to) a finite (free) $G$-CW-complex. Then $\Sigma_+^{\infty}P$ is cofibrant, since we can lift $G$-equivariant maps to the base space of $G$-equivariant trivial fibrations.


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