Thom spectrum of $(\mathrm{Spin}\times_{Z_2} \mathrm{SO}(d))$

Question

$\DeclareMathOperator\MSO{MSO}\DeclareMathOperator\MSpin{MSpin}\DeclareMathOperator\SO{SO}\DeclareMathOperator\BSO{BSO}\DeclareMathOperator\BG{BG}\DeclareMathOperator\BO{BO}\DeclareMathOperator\MG{MG}\DeclareMathOperator\Spin{Spin}\MG(d)$ is the Thom space of the pullback of the vector bundle $V_d$ over $\BO(d)$ along the map $\BG(d) \to \BO(d)$. The colimit of $\Sigma^{-d}\MG(d)$ is $\MG$. The $\MG$ is the Thom spectrum of $G$.

It is known that $M(\Spin\times \SO(d))= \MSpin \wedge \BSO(d)_+$, where $\BSO(d)_+$ is a disjoint union of $\BSO(d)$ with a point.

My question is: What is the Thom spectrum of $(\Spin\times_{Z_2} \SO(d))$?

The hope is to break down the spectrum to familiar one of $\MSpin$. the smash product $\wedge$, the suspension $\Sigma^{-n}$, and perhaps the Thom spectrum $\MSO(d)$.

For $d=4$, $M(\Spin\times_{Z_2}\SO(4))=\MSpin\wedge\Sigma^{-3}\MSO(3)\wedge\Sigma^{-3}\MSO(3)$, which breaks down to the familiar spectra.

For $d=6$, $M(\Spin\times_{Z_2}\SO(6))$, could we do the similar break down?

For $d=8$, $M(\Spin\times_{Z_2}\SO(8))$, could we do the similar break down?

Note that $\SO(6)=\Spin(6)/Z_2=\operatorname{PSU}(4)$. $\SO(8)=\Spin(8)/Z_2$.

EDIT: The $d$ is even for $\SO(d)$ here. Also the modded out $Z_2$ is shared between the identified normal of Spin and SO groups.

• Spin is the tangent structure (tangent spin bundle) of any dimensions of any base Spin-manifold.
• $\SO(d)$ is the principal SO structure and principal SO-bundle.

Answer 1

$\newcommand{\SO}{\mathrm{SO}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\PSO}{\mathrm{PSO}}\newcommand{\Z}{\mathbb Z}$This is a partial answer: I am not very familiar with the representation theory of $\PSO_{2d}$, and I have to make an assumption to get the splitting. I think the assumption is right, but I don't have a proof of it. In any case, once that assumption is in place, the argument is similar to the splitting argument made by Freed-Hopkins (see section 10).

The central extension $$1\to \Z/2\to \SO_{2d}\to\PSO_{2d}\to 1$$ defines a class $c\in H^2(B\PSO_{2d};\Z/2)$. The assumption I have to make is: there is a representation $\rho\colon \PSO_{2d}\to \SO_m$ such that if $V_\rho\to B\PSO_{2d}$ denotes the associated vector bundle, $w_1(V_\rho) = 0$ and $w_2(V_\rho) = c$.

Given that data, we can make the diagram where $\Spin_n\times_{\Z/2}\SO_{2d}\to\SO_n\times\PSO_{2d}$ is the quotient by the common $\Z/2$ and $\SO_n\times\SO_m\to\SO_{n+m}$ is the block diagonal embedding. The composition along the lower left $\Spin_n\times_{\Z/2}\SO_{2d}\to \SO_{n+m}$ kills the central $\Z/2$, and therefore we can choose a lift $\phi\colon \Spin_n\times_{\Z/2}\SO_{2d}\to \Spin_{n+m}$.

Stabilizing in $n$, let $S$ denote a tautological (stable) vector bundle on a classifying space, thought of as a map to $B\mathrm O$. Then there is a commutative diagram It is possible to show that the horizontal arrow is a homotopy equivalence (the proof is essentially the same as the one given by Freed-Hopkins). The data defining a Thom spectrum is a space with a map to $B\mathrm O$, and a homotopy equivalence of that data induces a homotopy equivalence of Thom spectra: $$\mathit{MT}(\Spin\times_{\Z/2}\SO_{2d}) \overset\simeq\longrightarrow \mathit{MTSpin}\wedge (B\PSO_{2d})^{V_\rho - m}.$$ Here $(B\PSO_{2d})^{V-m}$ is the Thom spectrum of the rank-zero stable vector bundle $V_\rho-m\to B\PSO_{2d}$.