Skip to main content
6 events
when toggle format what by license comment
May 11, 2020 at 17:42 comment added Lennart Meier @DenisNardin You're right that it would be actually more natural to use the latter because one doesn't have to use the Thom isomorphism. It was more out of habit.
May 11, 2020 at 16:07 comment added Denis Nardin Is there a particular reason why you use $M\mathbb{R}$ instead of $\Sigma^\infty BU_{\mathbb{R}}$ or is that only for convenience of computation?
May 11, 2020 at 11:47 comment added Lennart Meier ...further $M\mathbb{R} \simeq \mathrm{hocolim} \Sigma^{-k-k\sigma} M\mathbb{R}_{k+k\sigma}$ (this is a general fact, see e.g. Section 2 of Hill--Hopkins--Ravenel's Kervaire paper). As $\Phi^{C_2}$ is symmetric monoidal, it also preserves duals and thus takes $\Sigma^{-k-k\sigma}(-) = D(S^{k+k\sigma}) \otimes -$ to $\Sigma^{-k}$ as the $C_2$-fixed points of $S^{k+k\sigma}$ are $S^k$. Thus, we see that the geometric fixed points of $M\mathbb{R}$ are the homotopy colimit over $k$-fold desuspensions of Thom spaces over $BO(n)$, i.e. exactly $MO$.
May 11, 2020 at 11:43 comment added Lennart Meier Sure. $\Phi = \Phi^{C_2}: \mathrm{Sp}^{C_2} \to \mathrm{Sp}$ is the geometric fixed point functor. It is characterized by the following properties: 1) $\Phi^{C_2}\Sigma^{\infty}X = \Sigma^{\infty}X^{C_2}$; 2) $\Phi^{C_2}$ is symmetric monoidal; 3) $\Phi^{C_2}$ commutes with homotopy colimits. If you want to compute the geometric fixed points of $M\mathbb{R}$ you can argue like follows: $M\mathbb{R}_{k+k\sigma}$ is the Thom space of the universal bundle on $BU(n)$ with the complex conjugation action. Its fixed points are the corresponding Thom space over $BO(n)$...
May 11, 2020 at 11:09 comment added მამუკა ჯიბლაძე Could you please add few words to mention what $\Phi$ is? Hu and Kriz do explain it referring to Lewis-May-Steinberger, but...
May 11, 2020 at 8:59 history answered Lennart Meier CC BY-SA 4.0