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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.