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 $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2}$$, • 2) 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. $$\text{the \eta-invariant of the Dirac operator acting on the 8twisted) Dirac spinor bundle}$$ 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 $$A \cup \eta$$?

• 3) 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 $$B' \in H^2(B \mathbb Z_{4}, \mathbb Z_{2})=\mathbb Z_{2}$$.

• 4) 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!] – Qfwfq Feb 12 at 19:36
• ok, then, thanks, please upvote too to get us some attention! – wonderich Feb 12 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 $M \otimes_A N$? – Theo Johnson-Freyd Feb 12 at 23:13
• @Qfwfq, could you change my notation back, as Theo said, please? thank you! – wonderich Feb 13 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". – Qfwfq Feb 13 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 – wonderich Feb 13 at 0:41
• p.s. I suppose there are still better ways to precise writing down the invariants. – wonderich Feb 13 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 – wonderich Feb 13 at 15:39