[Note]: My question will be a bit long. So, first, thank you for your careful reading, generous comments, helps and answers, in advance!

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}$, enter image description here
    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.
  • $\begingroup$ [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!] $\endgroup$
    – Qfwfq
    Feb 12, 2019 at 19:36
  • $\begingroup$ ok, then, thanks, please upvote too to get us some attention! $\endgroup$
    – wonderich
    Feb 12, 2019 at 20:29
  • 2
    $\begingroup$ @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 $M \otimes_A N$? $\endgroup$ Feb 12, 2019 at 23:13
  • $\begingroup$ @Qfwfq, could you change my notation back, as Theo said, please? thank you! $\endgroup$
    – wonderich
    Feb 13, 2019 at 16:54
  • 1
    $\begingroup$ @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". $\endgroup$
    – Qfwfq
    Feb 13, 2019 at 18:09

1 Answer 1


$\newcommand{\Z}{\mathbb Z}\newcommand{\RP}{\mathbb{RP}}$ 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.

  • $\begingroup$ thanks +1, I am amazed there are experts answering a technical question $\endgroup$
    – wonderich
    Feb 13, 2019 at 0:41
  • $\begingroup$ p.s. I suppose there are still better ways to precise writing down the invariants. $\endgroup$
    – wonderich
    Feb 13, 2019 at 0:45
  • $\begingroup$ 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 $\endgroup$
    – wonderich
    Feb 13, 2019 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .