Decompose $MT(E(d)\times_{\mathbb Z_2} SU(2))$ as the wedge sum or smash product of spectra

Consider the extension
$$1\to SU(2)\to X\to O\to1,$$

there are 4 possibilities for $$X$$: $$X=O\times SU(2)$$ or $$E\times_{\mathbb{Z}_2}SU(2)$$ or $$Pin^+\times_{\mathbb{Z}_2}SU(2)$$ or $$Pin^-\times_{\mathbb{Z}_2}SU(2)$$ where $$E$$ is defined in Freed-Hopkins's work1 as the colimit of the group $$E(d)$$, the group $$E(d)$$ is defined to be the subgroup of $$O(d)\times\mathbb{Z}_4$$ consisting of the pairs $$(A,j)$$ such that $$\det A=j^2$$, where $$\mathbb{Z}_4=\{\pm1,\pm\sqrt{-1}\}$$ is the multiplicative group of order 4.

Here the notation $$G_1\times_{\mathbb{Z}_2} G_2 :=\frac{G_1\times G_2}{\mathbb{Z}_2}$$ is defined as mod out the common $$\mathbb{Z}_2$$ of $$G_1\times G_2$$.

The question is about computing $$MT(E(d)\times_{\mathbb Z_2} SU(2))$$ and the bordism group $$\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$$.

(1) There is a short exact sequence of groups: $$1\to SO(d)\to E(d)\to\mathbb{Z}_4\to 1$$. So naively, people may suspect that $$MT(E(d)\times_{\mathbb Z_2} SU(2))=MT E(d)\wedge\Sigma^{-3}M SO(3)=MSO(d)\wedge\Sigma^{-2}M\mathbb Z_4\wedge\Sigma^{-3}M SO(3).$$ However, this is likely to be incorrect.

(2) The space $$B(E \times_{\mathbb Z_2}SU(2))$$ sits in a homotopy pullback square: a map $$M \to B(E \times_{\mathbb Z_2}SU(2))$$ is determined by two maps $$M \to BO$$ and $$M\to BSO(3)$$ which correspond to bundles $$TM$$ and $$V_{SO(3)}$$ such that $$w_1(TM)^2=w_2(V_{SO(3)})$$.

To compute the bordism group $$\Omega_d^{E \times_{\mathbb Z_2}SU(2)}$$, we need to know the Madsen-Tillmann spectrum $$MT(E \times_{\mathbb Z_2}SU(2))$$ and decompose it as the wedge sum or smash product of some familiar spectra.

The figure attachment here is my own attempt,

but the map $$f$$ is not a homotopy equivalence. I actually obtain an identification $$\text{Thom(B(E \times_{\mathbb Z_2}SU(2)),-2V)=MT(Pin^+ \times_{\mathbb Z_2}SU(2))}$$ which is already known in 1604.06527 paper, but we need to know $$\text{Thom(B(E \times_{\mathbb Z_2}SU(2)),-V)=MT(E \times_{\mathbb Z_2}SU(2)),}$$

where $$V$$ is the induced virtual bundle of dimension 0 by $$B(E \times_{\mathbb Z_2}SU(2)) \to BO$$.

• Is Thom$$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$$=smash product of Thom$$(B(E \times_{\mathbb Z_2}SU(2)),-V)$$ and Thom$$(B(E \times_{\mathbb Z_2}SU(2)),-V)$$? If so, how to obtain Thom$$(B(E \times_{\mathbb Z_2}SU(2)),-V)$$ as the "square root" of Thom$$(B(E \times_{\mathbb Z_2}SU(2)),-2V)$$?

1 Reflection positivity and invertible topological phases Daniel S. Freed, Michael J. Hopkins, arXiv:1604.06527

Let $$Y$$ be a space, $$V$$ be a virtual bundle of dimension $$0$$ over $$Y$$ (this $$V$$ is your $$-V$$). Then $$Thom(Y,2V)$$ is almost never (except when $$Y$$ is contractible, or something like that) a smash product of the form $$Thom(Y,V)$$ with itself. You can see this by looking at the homology: the homology of the former is more or less isomorphic to that of $$Y$$, whereas the latter is more or less isomorphic to its square.