Computation of $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$ and Smith isomorphism

question: I am looking for the literature with the result or the computation of Pin- bordism group: $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$. Can someone point out some useful ways to do this or any helpful Refs?

Some helpful background: There is isomorphism between the following Spin and the Pin- bordism group, known as the Smith isomorphism: $$\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)' \to \Omega^{Pin-}_{d}(pt)$$ in particular, the $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$ is not exactly the the usual Spin bordism group $\Omega^{Spin}_{d+1}(B\mathbb{Z}_2)'$, but the reduction, based on a relation: $$\Omega^{Spin}_{d+1}(BG)=\Omega^{Spin}_{d+1}(BG)' \oplus \Omega^{Spin}_{d+1}(pt)$$ where the reduction of the spin bordism group $\Omega^{Spin}_{d+1}(BG)$ to $\Omega^{Spin}_{d+1}(BG)'$ gets rid of the $\Omega^{Spin}_{d+1}(pt)$. This part has something to do with the kernel of the forgetful map to $\Omega^{Spin}_{d+1}(pt)$.

In principle, to compute $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$, we may prove and use the following relations (any comments about this approach):

$$\Omega^{Spin}_{d+1}(B(\mathbb{Z}_2)^2)' \to \Omega^{Pin-}_{d}(B\mathbb{Z}_2)?$$

Some useful info:

$\Omega^{Pin-}_1(pt)=\mathbb{Z}_2, \Omega^{Pin-}_2(pt)=\mathbb{Z}_8, \Omega^{Pin-}_3(pt)=0, \Omega^{Pin-}_4(pt)=0$

$\Omega^{Spin}_1(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_2(B\mathbb{Z}_2)=\mathbb{Z}_2^2, \Omega^{Spin}_3(B\mathbb{Z}_2)=\mathbb{Z}_8, \Omega^{Spin}_4(B\mathbb{Z}_2)=\mathbb{Z}$

This is the reference that I have at hand: Kirby-Taylor, Pin structures on low-dimensional manifolds

I am willing to hear some guidance along this line of thinking, or related issue.

Here's an approach that works up to about dimension 7, outlined by Freed-Hopkins, §10, and explained in more detail by Campbell.

There's a weak equivalence $$\Sigma^{-1} \mathrm{MPin}^-\simeq \mathrm{MSpin}\wedge \mathrm{MTO}_1$$, where $$\mathrm{MTO}_1$$ is a Madsen-Tillmann spectrum, the Thom spectrum of the virtual vector bundle $$(\underline{\mathbb R} - S)\to B\mathrm O_1$$, where $$\underline{\mathbb R}$$ is the trivial line bundle and $$S\to B\mathrm O_1$$ is the tautological line bundle. Hence, to understand $$\Omega_d^{\mathrm{Pin}^-}(B\mathbb Z/2)$$, it suffices to understand the homotopy groups of $$\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2$$.

We'll use the Adams spectral sequence, but there's a key trick that makes it simpler. Let $$\mathcal A(1)$$ denote the subalgebra of the Steenrod algebra generated by $$\mathrm{Sq}^1$$ and $$\mathrm{Sq}^2$$. Then, Anderson, Brown, and Peterson proved that, as $$\mathcal A$$-modules,

$$H^*(\mathrm{MSpin};\mathbb F_2)\cong \mathcal A\otimes_{\mathcal A(1)} (\mathbb F_2\oplus M),$$

where $$M$$ is a graded $$\mathcal A(1)$$-module which is $$0$$ in dimension less than 8.

Thus we can invoke a change-of-rings theorem for the $$E_2$$-page of the Adams spectral sequence for $$\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2$$: using the Adams grading, when $$t -s < 8$$, \begin{align*} E_2^{s,t} &= \mathrm{Ext}_{\mathcal A}^{s,t}(H^*(\mathrm{MSpin}\wedge\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2)\\ &\cong \mathrm{Ext}_{\mathcal A}^{s,t}((A\otimes_{\mathcal A(1)} \mathbb F_2)\otimes H^*(\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2)\\ &\cong \mathrm{Ext}_{\mathcal A(1)}^{s,t}(H^*(\mathrm{MTO_1}\wedge B\mathbb Z/2; \mathbb F_2), \mathbb F_2). \end{align*} Explicitly calculating this is tractable, because $$\mathcal A(1)$$ is small and we're only going up to dimension 7.

• The $$\mathcal A(1)$$-module structure on $$\tilde H^*(B\mathbb Z/2; \mathbb F_2)$$ is standard, and Campbell describes it in Example 3.3 of his paper.
• Campbell also calculates the $$\mathcal A(1)$$-module structure on $$H^*(\mathrm{MTO}_1; \mathbb F_2)$$, and describes the answer in Example 6.6 and Figure 6.4.

A priori, the above method cannot detect torsion away from the prime 2. But $$\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)$$ cannot have any $$p$$-torsion for an odd prime $$p$$: the $$E^2$$-page for the Atiyah-Hirzebruch spectral sequence computing $$\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)_{(p)}$$ is

$$E^2_{q_1,q_2} = H_{q_1}(B\mathbb Z/2; \Omega_{q_2}^{\mathrm{Pin}^-})_{(p)},$$

but the homology of $$B\mathbb Z/2$$ has no $$p$$-torsion when $$p\ne 2$$, so the spectral sequence vanishes. Therefore its $$E^\infty$$-page, the $$p$$-localization of $$\Omega_*^{\mathrm{Pin^-}}(B\mathbb Z/2)$$, is trivial, so the above method suffices (for $$*\le 7$$).

• Is it obvious there isn't any p-torsion in Pin^- bordism of a point itself (the $q_1=0$ line of your spectral sequence)? The OP indicates this is true to degree 4, at least. – Mike Miller Eismeier Oct 12 '18 at 22:45
• @MikeMiller I don't know. It should be about as hard as the analogous fact for spin bordism. – Arun Debray Oct 13 '18 at 1:38
• I'm not sure which version of Pin bordism the article "Pin Cobordism and Related Topics" refers to, but it states that $\overline\Omega_{n+1}^{\text{Spin}}(B\Bbb Z/2) = \Omega_n^{\text{Pin}}$. This would give you the desired result, by precisely the AHSS argument you suggest, and the spin bordism result you mention (which is I think more well-known). – Mike Miller Eismeier Oct 13 '18 at 1:46