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 $*\le 7$).

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    $\begingroup$ 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. $\endgroup$ – Mike Miller Oct 12 '18 at 22:45
  • $\begingroup$ @MikeMiller I don't know. It should be about as hard as the analogous fact for spin bordism. $\endgroup$ – Arun Debray Oct 13 '18 at 1:38
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    $\begingroup$ 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). $\endgroup$ – Mike Miller Oct 13 '18 at 1:46

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