# Twisted spin bordism invariants in 5 dimensions

The spin $$G$$-bordism invariant can be twisted in the way that the Spin($$d$$) group of the $$d$$-manifold and the $$G$$ group can be combined and mod out any shared normal subgroup.

For example, we can consider $$Spin(d) \times_N G \equiv \frac{Spin(d) \times G }{N}$$.

Here we denote $$G_1 \times_N G_2 \equiv \frac{G_1 \times G_2 }{N}$$ in general.

I am trying to understand a particular twisted spin bordism invariant in 5 dimensions, based on a computation of Adams spectral sequence.

I have obtained that 5d bordism groups as

$$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{2}}= 0,$$ $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{4}} =\mathbb Z_{16},$$ $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2},$$

My question: What are the above $$\mathbb Z_{32}$$ generator, $$\mathbb Z_{16}$$ generator, and $$\mathbb Z_{2}$$ generator as

(1) the 5d topological terms, and

(2) its 5d manifold generators?

What I have done which leads to a tentative clue for an answer:

1. I find Adams $$\mathcal A$$ module structure has the following for $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{4}} =\mathbb Z_{16}$$ and $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2}$$, 1. The $$\mathbb Z_{16}$$ in $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{4}} =\mathbb Z_{16}$$ is similar to the $$\Omega_4^{Pin^+}(pt) =\mathbb Z_{16}$$, which is well-known to be generated by a 4d $$\eta$$-invariant that can be obtained from a Dirac spinor in 4d. $$\text{the \eta-invariant of the Dirac operator acting on the (twisted) Dirac spinor bundle}$$ Thus $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{4}} =\mathbb Z_{16}$$ may be generated by $$A \cup \eta?$$ except that there is only an $$A \in H^1(B \mathbb Z_2,\mathbb Z_2)=\mathbb Z_2$$, there is no $${\mathbb Z_{16}}$$ class of $$A \cup \eta$$?
1. The $$\mathbb Z_{2}$$ in $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2}$$ may be generated by $$A' B' \cup \text{Arf}$$ where $$A' \in H^1(B \mathbb Z_{4}, \mathbb Z_{2})=\mathbb Z_{2}$$ and $$B' \in H^2(B \mathbb Z_{4}, \mathbb Z_{2})=\mathbb Z_{2}$$.
1. The $$\mathbb Z_{32}$$ in $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2}$$ may be related to the earlier $$\mathbb Z_{16}$$ and has something to do with the Postnikov square.
• [uhm, just changed the notation for "twisted" product $G\times^N H=(G\times H)/N$ cause when I see $G\times_N H$ I tend to think of a fibered product; of course if you don't like the notation change you can revert it back!] Feb 12 '19 at 19:36
• ok, then, thanks, please upvote too to get us some attention! Feb 12 '19 at 20:29
• @Qfwfq I've certainly seen the notation $G \times_N H$ for this, and I have not seen $G \times^N H$. Although I agree the latter is more consistent with other established notation, the former I think is more standard... I think it is coined from a balanced tensor product $M \otimes_A N$? Feb 12 '19 at 23:13
• @Qfwfq, could you change my notation back, as Theo said, please? thank you! Feb 13 '19 at 16:54
• @wonderich: done. In general, to ruturn to a previous version of the question all you have to do is click on the "edited [time] ago" link on the bottom of the OP (just on the left of your icon) and select the version you want to roll back to and click "rollback". Feb 13 '19 at 18:09

$$\newcommand{\Z}{\mathbb Z}\newcommand{\RP}{\mathbb{RP}}$$ Let $$G_1 := \mathrm{Spin}_5\times_{\Z/2}\Z/4$$. The group $$\Omega_5^{G_1}$$ is discussed by Tachikawa-Yonekura, §3.1 and §3.4. Given a 5-manifold $$M$$ with a $$G_1$$-structure given by the principal $$G_1$$-bundle $$P\to M$$, there's a principal $$\Z/2$$-bundle $$Q := P\times_{G_1} \Z/2\to M$$. If $$N\subset M$$ is a representative of the Poincaré dual to $$w_1(Q)\in H^1(M;\Z/2)$$, then $$N$$ acquires a pin$$+$$-structure. Then:

• The map $$\Omega_5^{G_1}\to\Omega_4^{\mathrm{Pin}^+}\cong\Z/16$$ sending $$M$$ to the bordism class of $$N$$ is an isomorphism; in particular, one can realize the isomorphism $$\Omega_4^{\mathrm{Pin}^+}\to\Z/16$$ as an $$\eta$$-invariant, and putting this together gives a complete bordism invariant for $$\Omega_5^{G_1}$$.
• A manifold generator of $$\Omega_5^{G_1}$$ is $$\RP^5$$ with a $$G_1$$-structure such that $$Q$$ is the nontrivial principal $$\Z/2$$-bundle; then an embedded $$\RP^4\subset\RP^5$$ represents the Poincaré dual of $$w_1(Q)$$ and $$\RP^4$$ acquires a pin$$+$$-structure representing $$1$$ or $$-1$$ in $$\Omega_4^{\mathrm{Pin}^+}\cong\Z/16$$.

Let $$G_2 := \mathrm{Spin}_5\times_{\Z/2}\Z/8$$. Then $$\Omega_5^{G_2}$$ is discussed by Hsieh, §2.2. In particular:

• Given an element $$R$$ of the representation ring of $$\Z/8$$, Hsieh shows how to define an exponentiated $$\eta$$-invariant $$\eta_R\colon\Omega_5^{G_2}\to\mathrm U_1$$, and proves that these are complete invariants, i.e. there is some combination of $$\eta$$-invariants for some of these representations realizing the maps $$\Omega_5^{G_2}\to\Z/32$$ and $$\Omega_5^{G_2}\to\Z/2$$ that you're interested in.
• Lens spaces, with various $$G_2$$-structures, generate $$\Omega_5^{G_2}$$.

Given $$R$$ as above and a lens space $$L(m, \vec a)$$ with $$G_2$$-structure, Hsieh provides a formula (equation (2.35) in the paper) for $$\eta_R(L(m, \vec a))$$; this should make it possible to explicitly determine which $$\eta$$-invariants and which lens spaces you're looking for.

• thanks +1, I am amazed there are experts answering a technical question Feb 13 '19 at 0:41
• p.s. I suppose there are still better ways to precise writing down the invariants. Feb 13 '19 at 0:45
• Thanks, I accept it as a useful comment/answer. However, a more advanced question based on more detailed thought is given here: mathoverflow.net/q/323152/27004 Feb 13 '19 at 15:39