In Thom's 1952 paper, Thom showed that the Thom class, the Stiefel–Whitney classes, and the Steenrod operations were all related. He used these ideas to prove in the 1954 paper Quelques propriétés globales des variétés differentiables that the cobordism groups could be computed as the homotopy groups of certain Thom spectra $MG$. (The colimit of $n$-iterated suspension of $MG(n)$ is $MG$.)

Let us denote Thom space of the vector bundle $E$ over the base space $B$ as $T(E)=Thom(B,E)$ following Wikipedia notation.

Since Thom class, the Stiefel–Whitney classes, and the Steenrod operations are all related.

Question: Do we have the following relations $$ H^*(MTG,\mathbb{Z}_2)=H^*(BG,\mathbb{Z}_2)U? $$ or $$ H^*(MG,\mathbb{Z}_2)=H^*(BG,\mathbb{Z}_2)U? $$ where $U$ is the Thom class with $Sq^iU=w_i U$? True or false?

How do we show this?

Attempt: The $MG$ and $MTG$ are two related Thom space, related by $$ MG=Thom(BG,V), \quad MTG=Thom(BG,-V), $$ with $V$ the vector bundle.

The Pontryagin-Thom isomorphism provides a relation between the bordism groups of manifolds with (stable) tangential structure and homotopy groups of the Madsen-Tillman spectrum $MTG$. The $MTG$ is a close cousin of the more usual Thom spectrum $MG$ associated to tangential structure $G$.

  • $H^*(MTG,\mathbb{Z}_2)=H^*(BG,\mathbb{Z}_2)U^\prime $ where $U^\prime $ is the Thom class of $-V$ with $Sq^iU^\prime =w_iU^\prime $, this may not be correct.

  • $H^*(MG,\mathbb{Z}_2)=H^*(BG,\mathbb{Z}_2)U$ where $U$ is the Thom class of $V$ with $Sq^iU=w_iU$, $H^*(MTG,\mathbb{Z}_2)=H^*(BG,\mathbb{Z}_2)V$, $UU^\prime =1$, this may be correct?

  • $\begingroup$ What does your notation mean, when you put a Thom class beside a cohomology group? $\endgroup$ – Ryan Budney Aug 11 '18 at 0:50
  • 1
    $\begingroup$ @Ryan, thanks very much for the comment. the Thom class beside a cohomology group means cup product of the elements of the cohomology group with $U$. $\endgroup$ – wonderich Aug 11 '18 at 2:43

More generally, we have

$$H^* Thom(BG,E) \cong H^*(BG) e(E)$$ as modules over the Steenrod algebra, with e(E) the Euler class of $E$. I guess the base space doesn't have to be a classifying space. This is mentioned in the section 3 of Mitchell, Priddy: Stable splittings derived from the Steinberg module, Topology Volume 22, Issue 3, 1983, Pages 285-298 where $G$ is elementary abelian or product of $\Sigma _p$, but as long as we don't want to identify the result with a sub module of a localization, everything works in your case as well. One can prove this, for example, by using the Thom diagonal and Cartan formula, but probably you can read off from Thom's paper.


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