Quaternionic $\text{K}$ -theory

We start by discussing $\text{KQ}$ , the Quaternionic analogue of Atiyah’s Real $\text{K}$ -theory of spaces with an involution $\text{KR}$ [Reference AtiyahAti66], first introduced in [Reference DupontDup69]. Such a $K$ -theoretic invariant has already been used in works on four-dimensional Seiberg–Witten theory [Reference Furuta and KametaniFK05, Reference LiLi06]. As is customary, we will use the capitalized versions Real and Quaternionic when dealing with spaces with involutions.

The key observation is the following: a real vector space can be thought of as a complex vector space $E$ equipped with a complex antilinear map $\jmath _{E}$ squaring to $1_{E}$ , while a quaternionic vector space is a complex vector space $E$ equipped with a complex antilinear map $\jmath _{E}$ squaring to $-1_{E}$ . We will denote the latter by $j$ . This can be of course extended to vector bundles over a space $X$ and gives rise to an alternative viewpoint for the invariants $\text{KO}(X)$ and $\text{KSp}(X)$ . From this description, we see that there are natural product maps on real and quaternionic $K$ -theories

$$\begin{eqnarray}\displaystyle \text{KO}^{i}(X)\otimes \text{KO}^{j}(X) & \rightarrow & \displaystyle \text{KO}^{i+j}(X),\nonumber\\ \displaystyle \text{KO}^{i}(X)\otimes \text{KSp}^{j}(X) & \rightarrow & \displaystyle \text{KSp}^{i+j}(X),\nonumber\\ \displaystyle \text{KSp}^{i}(X)\otimes \text{KSp}^{j}(X) & \rightarrow & \displaystyle \text{KO}^{i+j}(X)\nonumber\end{eqnarray}$$

induced by the tensor product of vector bundles with involution

(1) $$\begin{eqnarray}(V,\jmath _{V})\otimes (W,\jmath _{W})=(V\otimes W,\jmath _{V}\otimes \jmath _{W}).\end{eqnarray}$$

As a consequence, $\text{KO}^{\bullet }(X)$ is a ring and $\text{KSp}^{\bullet }(X)$ a module over it. Recall that for a point $\ast$ , the multiplication by $H-4\in \widetilde{\text{KSp}}(S^{4})=\text{KSp}^{-4}(\ast )$ , where we interpret $S^{4}$ as $\mathbb{HP}^{1}$ and $H$ is the quaternionic tautological bundle, induces an isomorphism

$$\begin{eqnarray}\text{KO}^{i}(\ast )\cong \text{KSp}^{i-4}(\ast )\end{eqnarray}$$

and hence in particular the following version of Bott periodicity holds:

see for example [Reference MilnorMil63, p. 142]. Furthermore, the multiplication map

(2) $$\begin{eqnarray}\text{KSp}^{-5}(\ast )\otimes \text{KO}^{-1}(\ast )\rightarrow \text{KSp}^{-6}(\ast ),\end{eqnarray}$$

in which all involved groups are $\mathbb{Z}/2\mathbb{Z}$ , is an isomorphism [Reference Atiyah, Bott and ShapiroABS64].

Suppose now that $X$ is a compact Hausdorff space equipped with an involution (denoted either by $\unicode[STIX]{x1D70F}$ or $x\mapsto \bar{x}$ ). The key example is the $n$ -dimensional torus $(\mathbb{R}/2\unicode[STIX]{x1D70B}\mathbb{Z})^{n}$ with the involution induced by $x\mapsto -x$ . We will denote this space with involution by $\mathbb{T}^{n}$ or, when the dimension is not important, simply by $\mathbb{T}$ . A Real (respectively Quaternionic) vector bundle $E$ is a complex vector bundle equipped with a complex antilinear map $j$ covering $\unicode[STIX]{x1D70F}$ such that $j^{2}$ is $1_{E}$ (respectively $-1_{E}$ ). The Grothendieck group of Real vector bundles was introduced in [Reference AtiyahAti66] and is denoted by $\text{KR}(X)$ . It is important to remark that it is not the $\mathbb{Z}_{2}$ -equivariant $\text{K}$ -theory of $X$ .

There are natural maps

$$\begin{eqnarray}\text{K}(X)\leftarrow \text{KQ}(X)\rightarrow \text{KSp}(X^{\unicode[STIX]{x1D70F}})\end{eqnarray}$$

given respectively by forgetting the quaternionic structure and restricting to the fixed points of the involution. If $X_{\ast }$ is a space with base point $\ast$ (which we assume to be a fixed point for $\unicode[STIX]{x1D70F}$ ), we can define the reduced Quaternionic $K$ -theory $\widetilde{\text{KQ}}(X_{\ast })$ as the kernel of the augmentation map to

$$\begin{eqnarray}\text{KQ}(X_{\ast })\rightarrow \text{KQ}(\ast ).\end{eqnarray}$$

As usual, we can then extend the definition of $\text{KQ}$ to define a cohomology theory on spaces with or without base points via

$$\begin{eqnarray}\displaystyle \widetilde{\text{KQ}}^{-i}(X_{\ast }) & = & \displaystyle \widetilde{\text{KQ}}(\unicode[STIX]{x1D6F4}^{i}X_{\ast }),\nonumber\\ \displaystyle \text{KQ}^{-i}(X) & = & \displaystyle \widetilde{\text{KQ}}(\unicode[STIX]{x1D6F4}^{i}(X\cup \ast )).\nonumber\end{eqnarray}$$

Here by $\unicode[STIX]{x1D6F4}X_{\ast }$ we denote the reduced suspension

$$\begin{eqnarray}[0,1]\times X_{\ast }/(\{0\}\times X_{\ast }\cup \{1\}\times X_{\ast }\cup [0,1]\times \{\ast \})\end{eqnarray}$$

with the involution induced by $(t,x)\mapsto (t,\unicode[STIX]{x1D70F}(x))$ . We can also define the alternative suspension $\widetilde{\unicode[STIX]{x1D6F4}}X_{\ast }$ with the same underlying space but involution induced by $(t,x)\mapsto (1-t,\unicode[STIX]{x1D70F}(x))$ . The following is the key version of Bott periodicity for this version of $\text{K}$ -theory (compare also with the Real case in [Reference AtiyahAti66]).

Proposition 2 [Reference DupontDup69].

There are canonical isomorphisms

$$\begin{eqnarray}\displaystyle \widetilde{\text{KQ}}^{i}(\unicode[STIX]{x1D6F4}\widetilde{\unicode[STIX]{x1D6F4}}X_{\ast }) & \cong & \displaystyle \widetilde{\text{KQ}}^{i}(X_{\ast }),\nonumber\\ \displaystyle \text{KQ}^{i+8}(X) & \cong & \displaystyle \text{KQ}^{i}(X).\nonumber\end{eqnarray}$$

The first isomorphism implies the isomorphism

$$\begin{eqnarray}\widetilde{\text{KQ}}^{i}(\widetilde{\unicode[STIX]{x1D6F4}}^{j}X_{\ast })\cong \widetilde{\text{KQ}}^{i+j}(X_{\ast }),\end{eqnarray}$$

while the second isomorphism is determined via the product map induced by the tensor product of bundles with involutions (1)

$$\begin{eqnarray}\text{KQ}^{i}(X)\otimes \text{KO}^{j}(\ast )\rightarrow \text{KQ}^{i+j}(X)\end{eqnarray}$$

by multiplication with the generator of $\text{KO}^{8}(\ast )=\mathbb{Z}$ .

As an example, using the above we have

$$\begin{eqnarray}\displaystyle \text{KQ}^{i}(\mathbb{T}^{1}) & = & \displaystyle \text{KQ}^{i-8}(\mathbb{T}^{1})\nonumber\\ \displaystyle & = & \displaystyle \widetilde{\text{KQ}}^{i-1}(\unicode[STIX]{x1D6F4}^{7}(\mathbb{T}^{1}\cup \ast ))=\widetilde{\text{KSp}}^{i-1}(S^{6}\vee S^{7})=\text{KSp}^{i-7}(\ast )\oplus \text{KSp}^{i-8}(\ast ),\nonumber\end{eqnarray}$$

where we used that $\mathbb{T}\wedge X$ is naturally identified with $\widetilde{\unicode[STIX]{x1D6F4}}X$ . In particular, when $k=1$ , we obtain

(3) $$\begin{eqnarray}\text{KQ}^{1}(\mathbb{T}^{1})=\mathbb{Z}/2\mathbb{Z}.\end{eqnarray}$$

More generally, we have the following.

Lemma 1. We have the isomorphism

$$\begin{eqnarray}\text{KQ}^{1}(\mathbb{T}^{n})=\mathbb{Z}_{2}^{a_{n}}\oplus \mathbb{Z}^{b_{k}},\end{eqnarray}$$

where

$$\begin{eqnarray}a_{n}=\bigoplus _{\substack{ k\equiv 1,2\text{mod }8 \\ \quad 1\leqslant k\leqslant n}}\binom{n}{k},\quad b_{n}=\bigoplus _{\substack{ k=3,7\text{ mod }8 \\ \quad 1\leqslant k\leqslant n}}\binom{n}{k}.\end{eqnarray}$$

Proof. From the discussion above, we see that

$$\begin{eqnarray}\text{KQ}^{1}(\mathbb{T}^{n})=\widetilde{\text{KQ}}(\unicode[STIX]{x1D6F4}^{7}(\mathbb{T}^{n}))=\widetilde{\text{KQ}}^{-7}(\mathbb{T}^{n})=\widetilde{\text{KQ}}^{1}(\mathbb{T}^{n}),\end{eqnarray}$$

where we used the fact that $\text{KSp}^{-7}(\ast )=0$ . Recall the basic homotopy equivalence (see for example [Reference HatcherHat02, Proposition 4I.1])

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}(X\times Y)=\unicode[STIX]{x1D6F4}X\vee \unicode[STIX]{x1D6F4}Y\vee \unicode[STIX]{x1D6F4}(X\wedge Y),\end{eqnarray}$$

which also holds at the level of space with involutions. Applying this to $\mathbb{T}^{n}=\mathbb{T}^{1}\times \mathbb{T}^{n-1}$ , by the periodicity theorem we obtain that

$$\begin{eqnarray}\displaystyle \widetilde{\text{KQ}}^{i}(\mathbb{T}^{n}) & = & \displaystyle \widetilde{\text{KQ}}^{i+1}(\unicode[STIX]{x1D6F4}\mathbb{T}^{n})\nonumber\\ \displaystyle & = & \displaystyle \widetilde{\text{KQ}}^{i+1}(\unicode[STIX]{x1D6F4}\mathbb{T}^{1})\oplus \widetilde{\text{KQ}}^{i+1}(\unicode[STIX]{x1D6F4}\mathbb{T}^{n-1})\oplus \widetilde{\text{KQ}}^{i+1}(\unicode[STIX]{x1D6F4}(\mathbb{T}\wedge \mathbb{T}^{n-1}))\nonumber\\ \displaystyle & = & \displaystyle \widetilde{\text{KQ}}^{i}(\mathbb{T}^{1})\oplus \widetilde{\text{KQ}}^{i}(\mathbb{T}^{n-1})\oplus \widetilde{\text{KQ}}^{i+1}(\mathbb{T}^{n-1}).\nonumber\end{eqnarray}$$

From this, it inductively follows that

$$\begin{eqnarray}\widetilde{\text{KQ}}^{i+1}(\mathbb{T}^{n})=\bigoplus _{k=1}^{n}\widetilde{\text{KQ}}^{k+i}(\mathbb{T})^{\oplus \binom{n}{k}}\end{eqnarray}$$

and the result follows. ◻

An index theorem for Quaternionic families

Consider a quaternionic bundle on a compact Riemannian manifold $E\rightarrow M$ with $j$ acting from the right. A Quaternionic family $\unicode[STIX]{x1D701}$ is a continuous family $\{T_{p}\}_{p\in P}$ of first-order elliptic self-adjoint operators

$$\begin{eqnarray}T_{p}:C^{\infty }(E)\rightarrow C^{\infty }(E),\end{eqnarray}$$

where $P$ is a space with involution $p\mapsto \bar{p}$ and we have

$$\begin{eqnarray}T_{p}(s\cdot j)=(T_{\bar{p}}s)\cdot j\quad \text{for every }s\in C^{\infty }(E).\end{eqnarray}$$

To such an object we can associate a topological index and an analytical index as follows (see [Reference Atiyah, Patodi and SingerAPS76], which treats the classical case without involutions, for more details). Let

$$\begin{eqnarray}\unicode[STIX]{x1D70B}:SM\rightarrow M,\end{eqnarray}$$

the projection from the unit cotangent bundle. For each $p\in P$ , the symbol $T_{p}$ defines a self-adjoint automorphism $\unicode[STIX]{x1D70E}_{p}$ of the bundle $\unicode[STIX]{x1D70B}^{\ast }E\rightarrow SM$ . This defines a decomposition $\unicode[STIX]{x1D70B}^{\ast }E=E_{p}^{+}\oplus E_{p}^{-}$ in positive and negative eigenspaces. In particular, we get a vector bundle

$$\begin{eqnarray}E_{\unicode[STIX]{x1D701}}^{+}=\mathop{\bigcup }_{p\in P}E_{p}^{+}\rightarrow SM\times P,\end{eqnarray}$$

which is indeed Quaternionic for the action of $j$ on $E$ and hence a class $[E_{P}^{+}]\in \text{KQ}(SM\times P)$ . The symbol class of $\unicode[STIX]{x1D701}$ is defined to be the image of this class under the coboundary map

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}:\text{KQ}(SM\times P)\rightarrow \text{KQ}^{1}(TM\times P),\end{eqnarray}$$

where $TM$ is the tangent bundle of $M$ . More concretely, this image can be described as follows (see [Reference Atiyah, Patodi and SingerAPS76, Lemma 3.1]). We say that a symbol is positive/negative if $E_{\unicode[STIX]{x1D701}}^{\mp }=0$ . Two self-adjoint symbols are said to be stably equivalent if

$$\begin{eqnarray}\unicode[STIX]{x1D70E}\oplus \unicode[STIX]{x1D6FC}\oplus \unicode[STIX]{x1D6FD}\approx \unicode[STIX]{x1D70E}^{\prime }\oplus \unicode[STIX]{x1D6FC}^{\prime }\oplus \unicode[STIX]{x1D6FD}^{\prime }\end{eqnarray}$$

with $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FC}^{\prime }$ positive and $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FD}^{\prime }$ negative. Then the image of the coboundary map $\unicode[STIX]{x1D6FF}$ is naturally in bijection with stable equivalence classes of self-adjoint symbols. Finally, we define the topological index to be the image of the symbol class under the index map

$$\begin{eqnarray}\text{KQ}^{1}(TM\times P)\rightarrow \text{KQ}^{1}(P);\end{eqnarray}$$

see [Reference Atiyah and SingerAS71a] for more details about the latter.

Fix now a quaternionic separable Hilbert space $(H,j)$ and consider the space $\text{Op}$ of complex linear Fredholm self-adjoint operators which are not essentially definite (see [Reference Atiyah and SingerAS69]). This comes with a natural involution sending the operator $T$ to the operator

$$\begin{eqnarray}v\mapsto -(T(v\cdot j))\cdot j,\end{eqnarray}$$

the fixed points of which are exactly the quaternionic linear operators. With this involution, this is naturally a classifying space for $\text{KQ}^{1}$ , i.e.

(4) $$\begin{eqnarray}\text{KQ}^{1}(P)=[P,\text{Op}]_{\mathbb{ Z}/2\mathbb{Z}},\end{eqnarray}$$

where on the right-hand side we consider the homotopy classes of equivariant maps. The proof of this fact follows quite closely the non-equivariant case and we provide here a brief sketch of it. First of all, one identifies the space $\text{Fred}_{0}$ of Fredholm operators on $H$ of index zero (with the same involution as above) as a classifying space for $\widetilde{\text{KQ}}$ (see for example [Reference AtiyahAti67]). Then one shows that there is an equivariant homotopy equivalence

(5) $$\begin{eqnarray}\text{Op}\approx \widetilde{\unicode[STIX]{x1D6FA}}\,\text{Fred}_{0},\end{eqnarray}$$

where, for a space with involution $(X,\unicode[STIX]{x1D70F})$ , we denote its $\widetilde{\unicode[STIX]{x1D6FA}}X$ loop space (based at a fixed point of $\unicode[STIX]{x1D70F}$ ) equipped with the involution sending a based loop $\unicode[STIX]{x1D6FE}(t)$ for $t\in S^{1}$ to the loop $\unicode[STIX]{x1D70F}(\unicode[STIX]{x1D6FE}(\bar{t}))$ . The operation $\widetilde{\unicode[STIX]{x1D6FA}}$ is the adjoint of $\tilde{\unicode[STIX]{x1D6F4}}$ . To show (5), fix a quaternionic linear operator $T_{0}\in \text{Op}$ and consider for $T\in \text{Op}$ the loop $\unicode[STIX]{x1D6FE}_{T}(t)$ in $\text{Fred}_{0}$ given by

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{T}(t)=\left\{\begin{array}{@{}l@{}}i\cos (t)+\sin (t)T,\quad t\in [0,\unicode[STIX]{x1D70B}],\quad \\ i\cos (t)-\sin (t)T_{0},\quad t\in [\unicode[STIX]{x1D70B},2\unicode[STIX]{x1D70B}],\quad \end{array}\right.\end{eqnarray}$$

which we think of as based at $t=3/2\unicode[STIX]{x1D70B}$ (see Figure 1). As (right) multiplication by $i$ is sent by conjugation to $-i$ , if we consider the involution sending $t$ to $\unicode[STIX]{x1D70B}-t$ $\text{mod}$ $2\unicode[STIX]{x1D70B}$ (and $t=3/2\unicode[STIX]{x1D70B}$ as the base point), we obtain an equivariant map

$$\begin{eqnarray}\text{Op}\rightarrow \widetilde{\unicode[STIX]{x1D6FA}}\,\text{Fred}_{0},\end{eqnarray}$$

where the loop space is based at $T_{0}$ . This map is, up to a simple manipulation, the one shown in [Reference Atiyah and SingerAS69] to be a homotopy equivalence.

Figure 1. The path of operators $\unicode[STIX]{x1D6FE}_{T}(t)$ is equivariant under reflection across the vertical axis in this picture.

By (4), the Quaternionic family of operators $\unicode[STIX]{x1D701}$ parametrized by the space with involution $P$ determines (after performing a standard trick to make them Fredholm) a class in $\text{KQ}^{1}(P)$ called the analytical index. We then have the following.

The proof of the result follows very closely that of the classical index theorem for families of self-adjoint operators (see [Reference Atiyah, Patodi and SingerAPS76], to which we refer for details). For example, using a loop construction as above, one obtains a Quaternionic family of elliptic symbols giving rise to an element in $\text{KQ}^{0}(S^{1}\times X\times TY)$ , and similarly the family of operators induces a class in $\text{KQ}^{0}(S^{1}\times X)$ . From this, one reduces to the usual index theorem (cf. [Reference AtiyahAti66]).

Families of Dirac operators

While the results we will discuss hold more in general for $(8k+3)$ -dimensional spin manifolds, we will focus on the case of three-manifolds. Let $Y$ be a three-manifold equipped with a self-conjugate spin $^{c}$ structure $\mathfrak{s}$ and consider the family of Dirac operators $\{D_{B}\}_{[B]\in \mathbb{T}}$ parametrized by the $b_{1}(Y)$ -dimensional torus $\mathbb{T}$ of flat connections. The torus $\mathbb{T}$ comes with the natural involution by conjugation (or, equivalently, $x\mapsto -x$ ), which has exactly $2^{b_{1}(Y)}$ fixed points corresponding to the spin connections of the elements of $\text{Spin}(\mathfrak{s})$ .

We start by discussing a key example. Suppose that $b_{1}(Y)=1$ . We know from (3) that

$$\begin{eqnarray}\text{KQ}^{1}(\mathbb{T}^{1})=\mathbb{Z}/2\mathbb{Z},\end{eqnarray}$$

so that there are exactly two homotopy classes of Quaternionic families of operators. These can be described explicitly as follows. Given any family over $\mathbb{T}^{1}$ , the operators at $0$ and $\unicode[STIX]{x1D70B}$ are quaternionic, so that in particular their kernel is even dimensional over $\mathbb{C}$ . This implies that the $\text{mod }2$ spectral flow between the two operators is well defined. Indeed, both cases can be realized: in the case of $S^{2}\times S^{1}$ we have even spectral flow, while in the case of the zero surgery on the trefoil we have odd spectral flow. Both cases can be computed explicitly, the first one using a metric with positive scalar curvature, the second using the flat metric (see [Reference Kronheimer and MrowkaKM07, §§36.1 and 37.4]). Hence, the $\text{mod }2$ spectral flow is a complete invariant for families. With this in mind, we prove the following.

Proof. By Proposition 3, the analytical index coincides with the topological index. To compute the latter, we first consider the map

(6) $$\begin{eqnarray}\text{KQ}^{1}(\mathbb{T})\rightarrow \text{K}^{1}(\mathbb{T}),\end{eqnarray}$$

which forgets the involution. This map is injective after tensoring with $\mathbb{Q}$ . This is clear from the computation of the first group (see Lemma 1) and the fact that the statement is true for the natural map $\text{KSp}^{i}(\ast )\rightarrow \text{K}^{i}(\ast )$ . Of course, this map simply sends the family to its index (as a family of self-adjoint operators), which is in turn determined via the Chern character

$$\begin{eqnarray}\text{ch}:\text{K}^{1}(\mathbb{T})\rightarrow H^{\text{odd}}(\mathbb{T};\mathbb{Q})\end{eqnarray}$$

by the triple cup product, as follows from the usual index theorem for families (see [Reference Kronheimer and MrowkaKM07, Lemma 35.1.2] for an explicit computation). An important point here is that $\text{K}^{1}(\mathbb{T})$ is torsion-free, so no information is lost by taking the Chern character. Hence, we only need to understand the torsion of $\text{KQ}^{1}(\mathbb{T})$ , which is according to Lemma 1 a suitable direct sum of $\mathbb{Z}/2\mathbb{Z}$ s.

Pick a basis $\mathbf{x}_{1},\ldots ,\mathbf{x}_{n}$ of $H^{1}(Y,\mathbb{Z})$ . This determines a splitting in one-dimensional factors $\mathbb{T}=\mathbb{T}_{1}\times \cdots \times \mathbb{T}_{n}$ , and (choosing a base spin structure $\mathbf{s}_{0}$ ), we can also use it to identify the set $\text{Spin}(\mathfrak{s})$ with subsets of $\{1,\ldots ,n\}$ . From the computation in Lemma 1, we then know that the summands $\mathbb{Z}/2\mathbb{Z}$ to $\text{KQ}^{1}(\mathbb{T})$ arise from the $\widetilde{\text{KQ}}^{1}$ -groups of wedge summands of the form

$$\begin{eqnarray}\mathbb{T}_{i_{1}}\wedge \cdots \wedge \mathbb{T}_{i_{k}},\end{eqnarray}$$

where $k=1$ or $2$ modulo $8$ . Consider now a standard equivariant loop $\unicode[STIX]{x1D6FE}$ in $\mathbb{T}$ whose fixed points are precisely $\mathbf{s}_{0}$ and the spin structure corresponding to $\{i_{1},\ldots ,i_{k}\}$ . Then the suspension of the inclusion $\unicode[STIX]{x1D6FE}{\hookrightarrow}\mathbb{T}$ factors through the suspension of

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}{\hookrightarrow}\mathbb{T}_{i_{1}}\wedge \cdots \wedge \mathbb{T}_{i_{k}}.\end{eqnarray}$$

We claim that the family over $\mathbb{T}_{i_{1}}\wedge \cdots \wedge \mathbb{T}_{i_{k}}$ is trivial if and only if the family over $\unicode[STIX]{x1D6FE}$ is trivial (recall that both have $\widetilde{\text{KQ}}^{1}$ equal to $\mathbb{Z}/2\mathbb{Z}$ ). Let us first focus on the case $k=2$ . Then we can identify

$$\begin{eqnarray}\mathbb{T}_{1}\wedge \mathbb{T}_{2}=\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE};\end{eqnarray}$$

see Figure 2. The intuition behind the claim is then the following. Any Quaternionic family on $\unicode[STIX]{x1D6FE}$ is trivial once we forget the involution, as the total spectral flow around $\unicode[STIX]{x1D6FE}$ vanishes by equivariance. Hence, we can extend the family to an equivariant family on $\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE}$ by first choosing any extension to the upper hemisphere and then extending it to the lower hemisphere by equivariance. The spectral flow $\text{mod }2$ between the two fixed points only depends on the spectral flow $\text{mod }2$ of $\unicode[STIX]{x1D6FE}$ and determines the corresponding element in $\widetilde{\text{KQ}}^{1}(\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE})=\mathbb{Z}/2\mathbb{Z}$ . In more algebraic terms, we can exploit the sequence of of maps

$$\begin{eqnarray}\widetilde{\text{KO}}(S^{1})\otimes \widetilde{\text{KQ}}^{1}(\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE})\longrightarrow \widetilde{\text{KQ}}^{1}(\unicode[STIX]{x1D6F4}\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE})\cong \widetilde{\text{KQ}}^{1}(\unicode[STIX]{x1D6FE}),\end{eqnarray}$$

where the first map is given by multiplication, while the second map is given by Bott periodicity; see Proposition 2. The composition maps the family over $\mathbb{T}_{1}\wedge \mathbb{T}_{2}$ to the family over $\unicode[STIX]{x1D6FE}$ . Furthermore, it is an isomorphism, as it can be identified with the multiplication

$$\begin{eqnarray}\text{KO}^{-1}(\ast )\otimes \widetilde{\text{KQ}}^{2}(\unicode[STIX]{x1D6FE})\rightarrow \widetilde{\text{KQ}}^{1}(\unicode[STIX]{x1D6FE})\end{eqnarray}$$

and hence with the multiplication map (2). The general case in which $k=1$ or $2$ modulo $8$ follows from the cases $k=1,2$ by identifying

$$\begin{eqnarray}\mathbb{T}_{i_{1}}\wedge \cdots \wedge \mathbb{T}_{i_{k}}=\widetilde{\unicode[STIX]{x1D6F4}}^{k-1}\unicode[STIX]{x1D6FE}\end{eqnarray}$$

and using the $8$ -periodicity of the invariants; see again Proposition 2.

Figure 2. The space $\mathbb{T}_{1}\wedge \mathbb{T}_{2}$ is obtained from the square on the left by quotienting the boundary. The involution in the picture is given by sending $x$ to $-x$ , where the origin is the dot, and the quotient is identified with $\widetilde{\unicode[STIX]{x1D6F4}}\unicode[STIX]{x1D6FE}$ . The result is the two-sphere on the right, where the involution has exactly the two dots as fixed points.

To sum up, the component in each $\mathbb{Z}/2\mathbb{Z}$ -summand of $\text{KQ}^{1}(\mathbb{T})$ can be interpreted in terms of the restriction of the family to a certain equivariant loop $\unicode[STIX]{x1D6FE}$ , which is in turn determined by a certain $\text{mod }2$ spectral flow by the discussion above on the case $b_{1}=1$ .◻

Coupled Morse homology

Given a family of operators $L$ in $S_{\ast }(H:H_{1})$ parametrized by a smooth manifold $Q$ , in [Reference Kronheimer and MrowkaKM07, ch. 33] (and in particular §33.3) the authors constructed the coupled Morse homology $\bar{H}_{\ast }(Q,L)$ , which is a relatively graded module over $\mathbb{F}[U]$ with $U$ of degree $-2$ . We can assume in our context that the family has no spectral flow around loops in $Q$ , so that the grading is absolute. We very quickly recall its definition and refer the reader to [Reference Kronheimer and MrowkaKM07] for more details. Fix a metric and choose a generic Morse function $f$ on $Q$ . After a small perturbation of the family, we can assume that the following genericity assumption holds:

( $\ast$ ) for all critical points $q$ of $f$ , the operator $L_{q}$ has no kernel and simple spectrum.

The chain complex for the coupled Morse homology $\bar{C}_{\ast }(Q,L)$ is generated over $\mathbb{F}$ by the projectivizations of the eigenspaces (which are all one dimensional by ( $\ast$ )) of the operators at the critical points. We then look at equivalence classes of pairs of paths $(\unicode[STIX]{x1D6FE}(t),\unicode[STIX]{x1D719}(t))$ , where $\unicode[STIX]{x1D6FE}$ is a Morse trajectory for $f$ and $\unicode[STIX]{x1D719}$ is a path in the unit sphere of $H$ satisfying a given differential equation. The differential counts the number of these paths in zero-dimensional moduli spaces. The two key properties of this construction are the following.

1. – The coupled Morse homology $\bar{H}_{\ast }(Q,L)$ only depends on the homotopy class of the family $L$ and hence on the corresponding element in $\text{K}^{1}(Q)=[Q,S_{\ast }(H:H_{1})]$ .

2. – If $(Y,\mathfrak{s})$ is a three-manifold equipped with a torsion spin $^{c}$ structure, the result of the construction applied to the family of Dirac operators parametrized by flat connections $\{D_{B}\}_{[B]\in \mathbb{T}}$ is $\overline{\mathit{HM}}_{\ast }(Y,\mathfrak{s})$ .

An analogous construction can be performed if the manifold $Q$ comes with an involution $\unicode[STIX]{x1D70F}$ and the family of operators $L$ is Quaternionic for this involution. We will assume that the fixed points of $\unicode[STIX]{x1D70F}$ are isolated, so that it is locally modeled on $x\mapsto -x$ . The main complication is that now the operators at the fixed points of the action are quaternionic–linear, so the transversality assumption $(\ast )$ cannot be achieved (respecting the involution) as the eigenspaces will always be even dimensional over $\mathbb{C}$ . The problem can be solved, as in [Reference LinLin15], by allowing Morse–Bott singularities of a very specific kind. Indeed, generically the operators at the fixed points of $\unicode[STIX]{x1D70F}$ will have no kernel and two-dimensional eigenspaces: each of these will give rise, after projectivization, to a copy of $S^{2}$ on which the involution $j$ acts as the antipodal map. With this in mind, the construction of $\text{Pin}(2)$ -monopole Floer from [Reference LinLin15] carries over without significant differences (see also [Reference LinLin16] for an introduction): indeed, as we do not have to deal with boundary obstructedness phenomena, the technical details are significantly easier in this case. The output is a version of the chain complex $\bar{C}_{\ast }(Q,L)$ whose homology is $\bar{H}_{\ast }(Q,L)$ which is equipped with a natural chain involution $\unicode[STIX]{x1D70F}$ . We then define $\bar{H}_{\ast }^{\unicode[STIX]{x1D70F}}(Q,L)$ to be the homology of the invariant subcomplex. This is naturally a module over the ring ${\mathcal{R}}$ . We record the main features of this construction in the following result, whose proof follows along the lines of the results contained in [Reference LinLin15] (and is in fact much simpler).

Putting the pieces together, we can finally prove the main result of the present paper.

Proof of Theorem 1.

Using the equivariant coupled Morse homology introduced above, we see that $\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})$ only depends on the homotopy class of the Quaternionic family of operators $\{D_{B}\}_{[B]\in \mathbb{T}}$ , which is by definition the analytic index of the family. Using the index theorem for families (Proposition 3), we showed in Proposition 4 that the analytical index is determined by the triple cup product of $Y$ and the $\text{mod }2$ spectral flow between the Dirac operators corresponding to elements of $\text{Spin}(\mathfrak{s})$ . Hence, we only need to show that the latter is determined by the Rokhlin invariants of the spin structures. This can be seen by looking at the absolute gradings in the chain complex (as introduced in [Reference Kronheimer and MrowkaKM07, §28.3]). Choose a standard equivariant Morse function $f$ on $\mathbb{T}$ so that its $2^{b_{1}(Y)}$ critical points correspond to the elements of $\text{Spin}(\mathfrak{s})$ . Given two spin structures $\mathbf{s}_{0}$ and $\mathbf{s}_{1}$ with corresponding spin connections $B_{0}$ and $B_{1}$ , we have that the zero-dimensional chains in a stable critical submanifold $C_{i}$ over $(B_{i},0)$ have relative grading

$$\begin{eqnarray}\text{ind}_{f}(\mathbf{s}_{0})-\text{ind}_{f}(\mathbf{s}_{1})-2\text{sf}(D_{B_{0}},D_{B_{1}})\text{ mod }4.\end{eqnarray}$$

On the other hand, zero-dimensional chains in $C_{i}$ have absolute grading $-\unicode[STIX]{x1D70E}(W_{i})/4+\text{ind}_{f}(\mathbf{s}_{i})$ modulo $4$ , where $W_{i}$ is any manifold whose boundary is $Y$ on which $\mathbf{s}_{i}$ extends. For this computation we exploit the fact that on a spin four-manifold the Dirac operator is quaternionic, so that its (real) index is divisible by $4$ (see also [Reference LinLin15, ch. 4]). Comparing this with the formula above, we see that the $\text{mod }2$ spectral flow between $D_{B_{0}}$ and $D_{B_{1}}$ is exactly the difference between the Rokhlin invariants of $\mathbf{s}_{0}$ and $\mathbf{s}_{1}$ .◻

Rokhlin invariants as a cubic map

The rest of this section is devoted to understand the structure of the Rokhlin invariants of a pair $(Y,\mathfrak{s})$ and in particular to prove Proposition 1. First, recall some features of spin manifolds in low dimensions (we refer the reader to [Reference KirbyKir89, chs 4 and 5] for a more thorough discussion). Of course,

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{0}^{\text{Spin}}=\mathbb{Z}.\end{eqnarray}$$

In dimension one, the spin cobordism group is

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{1}^{\text{Spin}}=\mathbb{Z}/2\mathbb{Z},\end{eqnarray}$$

the generator being the trivial double cover of the circle (see Figure 3). We will refer to this as the Lie structure and denote it by $\mathbf{s}_{\text{Lie}}$ . In dimension two, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{2}^{\text{Spin}}=\mathbb{Z}/2\mathbb{Z}\end{eqnarray}$$

and the spin cobordism class is determined by the Arf invariant of $(\unicode[STIX]{x1D6F4},\mathbf{s})$ . More precisely, by restricting to loops the spin structure $\mathbf{s}$ determines a map

$$\begin{eqnarray}q:H_{1}(\unicode[STIX]{x1D6F4},\mathbb{Z})\otimes \mathbb{F}\rightarrow \unicode[STIX]{x1D6FA}_{1}^{\text{Spin}}=\mathbb{Z}/2\mathbb{Z},\end{eqnarray}$$

which is a quadratic refinement of the intersection product. The Arf invariant of this quadratic form in $\mathbb{Z}/2\mathbb{Z}$ is then the spin cobordism class of $(\unicode[STIX]{x1D6F4},\mathbf{s})$ .

Figure 3. The two spin structures on the circle. The one on the left corresponds to the non-trivial double cover of the circle and extends to the disk. The one of the right is the Lie structure; it corresponds to the trivial double cover of the circle and does not extend.

In dimension three, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{3}^{\text{Spin}}=0,\end{eqnarray}$$

i.e. every spin manifold is a spin boundary. Consider as an example the three-manifold $Y=S^{1}\times S^{1}\times S^{1}$ . This has eight spin structures, seven with Rokhlin invariant $0$ and one with Rokhlin invariant $1$ . Indeed, a spin structure on $Y$ is a product of spin structures on the circle. If at least one of the factors is not the Lie structure, then it extends to a manifold diffeomorphic to $S^{1}\times S^{1}\times D^{2}$ , which has signature zero. On the other hand, the product of the Lie structures (which we denote by $\mathbf{s}_{0}=\mathbf{s}_{\text{Lie}}\times \mathbf{s}_{\text{Lie}}\times \mathbf{s}_{\text{Lie}}$ ) extends to a manifold with signature $-8$ , namely the complement of a regular fiber of the elliptic fibration $E(1)\rightarrow \mathbb{C}P^{1}$ (see [Reference KirbyKir89, ch. 5]). With this in mind, we can proceed on the proof of Proposition 1.

Proof of Proposition 1.

Consider two spin structures $\mathbf{s}$ and $\mathbf{s}^{\prime }$ in $\text{Spin}(\mathfrak{s})$ so that they differ by an element $\mathbf{x}\in H^{1}(Y;\mathbb{Z})\otimes \mathbb{F}$ . Let $\unicode[STIX]{x1D6F4}\subset Y$ be an oriented surface Poincaré dual to $\mathbf{x}$ . We claim that the difference of their Rokhlin invariants satisfies

$$\begin{eqnarray}\unicode[STIX]{x1D707}(\mathbf{s})-\unicode[STIX]{x1D707}(\mathbf{s}^{\prime })=[\mathbf{s}|_{\unicode[STIX]{x1D6F4}}]\in \unicode[STIX]{x1D6FA}_{2}^{\text{Spin}}=\mathbb{Z}/2\mathbb{Z}.\end{eqnarray}$$

To show this, we construct a suitable spin cobordism between $(Y,\mathbf{s})$ and $(Y,\mathbf{s}^{\prime })$ as follows. Consider the manifold $Y\times [-1,1]$ with fixed spin structures $\mathbf{s}$ and $\mathbf{s}^{\prime }$ at the boundary. Then $\unicode[STIX]{x1D6F4}\times \{0\}$ is a characteristic surface (in a relative sense): there is a spin structure on its complement restricting to $\mathbf{s}$ and $\mathbf{s}^{\prime }$ at the boundaries and which induces the non-trivial element in $\unicode[STIX]{x1D6FA}_{1}^{\text{Spin}}$ on the unit circle of a normal fiber of $\unicode[STIX]{x1D6F4}$ . The proof of this is a direct generalization of the closed case; see [Reference KirbyKir89, §11.2]. Consider now $\unicode[STIX]{x2202}\text{nbhd}(\unicode[STIX]{x1D6F4}\times \{0\})$ , which is naturally identified with $S^{1}\times \unicode[STIX]{x1D6F4}$ . This has the induced spin structure $\mathbf{s}_{\text{Lie}}\times \mathbf{s}|_{\unicode[STIX]{x1D6F4}}$ . Hence, to find a spin cobordism from $(Y,\mathbf{s})$ to $(Y,\mathbf{s}^{\prime })$ , it suffices to find a spin manifold whose boundary is $S^{1}\times \unicode[STIX]{x1D6F4}$ on which this spin structure extends and glue it in. The example of the three-torus discussed above (which is the case $S^{1}\times T^{2}$ ) readily implies that if $\mathbf{s}|_{\unicode[STIX]{x1D6F4}}$ is trivial, then one can find an extension to a manifold with $\unicode[STIX]{x1D70E}=0$ modulo 16 and, if $\mathbf{s}|_{\unicode[STIX]{x1D6F4}}$ is not, then one can find an extension to a manifold with $\unicode[STIX]{x1D70E}=8$ modulo 16. This is because, from the properties of the Arf invariant, writing $\unicode[STIX]{x1D6F4}=\#^{g}T^{2}$ , $\mathbf{s}|_{\unicode[STIX]{x1D6F4}}$ bounds if and only if the number of restrictions to the summands $\mathbf{s}|_{T^{2}}$ that do not bound is even.

To conclude the proof, we need to show that the map

$$\begin{eqnarray}\displaystyle H^{1}(Y;\mathbb{Z})\otimes \mathbb{F} & \rightarrow & \displaystyle \mathbb{Z}/2\mathbb{Z}\nonumber\\ \displaystyle \mathbf{x} & \mapsto & \displaystyle [\mathbf{s}|_{\text{PD}(\mathbf{x})}]\nonumber\end{eqnarray}$$

is cubic. To see this, fix a basis $\mathbf{x}_{1},\ldots ,\mathbf{x}_{n}$ with transverse dual surfaces $\unicode[STIX]{x1D6F4}_{1},\ldots ,\unicode[STIX]{x1D6F4}_{n}$ . Then, if $\mathbf{x}=\sum \unicode[STIX]{x1D706}_{i}\mathbf{x}_{i}$ , and $\unicode[STIX]{x1D6F4}$ is dual to $\mathbf{x}$ , we have

$$\begin{eqnarray}[\mathbf{s}|_{\unicode[STIX]{x1D6F4}}]=\sum \unicode[STIX]{x1D706}_{i}[\mathbf{s}|_{\unicode[STIX]{x1D6F4}_{i}}]+\sum \unicode[STIX]{x1D706}_{i}\unicode[STIX]{x1D706}_{j}[\mathbf{s}|_{\unicode[STIX]{x1D6F4}_{i}\cap \unicode[STIX]{x1D6F4}_{j}}]+\sum \unicode[STIX]{x1D706}_{i}\unicode[STIX]{x1D706}_{j}\unicode[STIX]{x1D706}_{k}[\mathbf{s}|_{\unicode[STIX]{x1D6F4}_{i}\cap \unicode[STIX]{x1D6F4}_{j}\cap \unicode[STIX]{x1D6F4}_{k}}],\end{eqnarray}$$

as can be seen directly by a cut and paste argument (see [Reference KirbyKir89, §11.3]). Finally, the term $[\mathbf{s}|_{\unicode[STIX]{x1D6F4}_{i}\cap \unicode[STIX]{x1D6F4}_{j}\cap \unicode[STIX]{x1D6F4}_{k}}]$ is clearly the triple cup product $\langle \mathbf{x}_{i}\cup \mathbf{x}_{j}\cup \mathbf{x}_{k},[Y]\rangle$ modulo  $2$ .◻

Indeed, the proof shows that not only is the function cubic (with the cubic part determined by the triple cup product), but also that the linear and quadratic coefficients can be determined very explicitly in terms of embedded surfaces. For example, in the case of the three-torus discussed above, fix $\mathbf{s}_{0}$ as the base spin structure. Using this to identify the set of spin structures with $H^{1}(Y;\mathbb{Z})\otimes \mathbb{F}$ , the map $\unicode[STIX]{x1D707}$ is $1$ on any non-zero element. Denoting by $\mathbf{x}_{i}$ the generator of the circle in the $i$ th factor, we see that this is the cubic map

$$\begin{eqnarray}\sum \unicode[STIX]{x1D706}_{i}\mathbf{x}_{i}\mapsto 1+\unicode[STIX]{x1D706}_{1}+\unicode[STIX]{x1D706}_{2}+\unicode[STIX]{x1D706}_{3}+\unicode[STIX]{x1D706}_{1}\unicode[STIX]{x1D706}_{2}+\unicode[STIX]{x1D706}_{2}\unicode[STIX]{x1D706}_{3}+\unicode[STIX]{x1D706}_{1}\unicode[STIX]{x1D706}_{3}+\unicode[STIX]{x1D706}_{1}\unicode[STIX]{x1D706}_{2}\unicode[STIX]{x1D706}_{3}.\end{eqnarray}$$

The coefficients can be interpreted topologically as follows. First consider a surface $\unicode[STIX]{x1D6F4}_{i}$ Poincaré dual to $\mathbf{x}_{i}$ . Then $\mathbf{s}_{0}|_{\unicode[STIX]{x1D6F4}_{i}}$ is a product of Lie structures and hence it has Arf invariant $1$ , which corresponds to the coefficient of $\unicode[STIX]{x1D706}_{i}$ . Furthermore, for $i\neq j$ the intersection $\unicode[STIX]{x1D6F4}_{i}\cap \unicode[STIX]{x1D6F4}_{j}$ has induced the Lie structure and hence it is $1\in \unicode[STIX]{x1D6FA}_{1}^{\text{Spin}}$ . This corresponds to the coefficient of $\unicode[STIX]{x1D706}_{i}\unicode[STIX]{x1D706}_{j}$ .

$\text{Pin}(2)$ -standard manifolds

We say that a three-manifold $Y$ equipped with a self-conjugate spin $^{c}$ structure $\mathfrak{s}$ is $\text{Pin}(2)$ -standard if the triple cup product of $Y$ vanishes and the spin structures in $\text{Spin}(\mathfrak{s})$ all have the same Rokhlin invariant. We claim that in this case we have

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=\tilde{{\mathcal{R}}}\otimes H^{1}(\mathbb{T};\mathbb{F}).\end{eqnarray}$$

Indeed, our main theorem implies that, up to grading shift,

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=\overline{\mathit{HS}}_{\ast }(\#^{b_{1}(Y)}S^{2}\times S^{1},\mathfrak{s}_{0}),\end{eqnarray}$$

where $\mathfrak{s}_{0}$ is the unique torsion spin $^{c}$ structure. The latter can be computed for example using the connected sum spectral sequence (see [Reference LinLin17]). Indeed, we know that

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(S^{2}\times S^{1},\mathfrak{s}_{0})=\tilde{{\mathcal{R}}}\otimes H^{1}(S^{1};\mathbb{F})\end{eqnarray}$$

and, as this is a free module over ${\mathcal{R}}$ , the invariant of the connected sum is simply the tensor product over ${\mathcal{R}}$ of the invariants (as the spectral sequence collapses at the $E^{2}$ page).

Manifolds with $b_{1}=2$

In this case the triple cup product vanishes, so that the invariant is determined by the Rokhlin invariants. If all of them coincide, then the manifold is $\text{Pin}(2)$ -standard, so that the result discussed above holds. For simplicity, we assume that the homology of the manifold does not have $2$ -torsion, so that the reduction $\text{mod }2$ map

$$\begin{eqnarray}H^{1}(Y;\mathbb{Z})\rightarrow H^{1}(Y;\mathbb{F})\end{eqnarray}$$

is surjective. This assumption, the vanishing of the triple cup product, and Poincaré duality imply that the cup product of two basis elements of $H^{1}(Y;\mathbb{Z})$ has to be zero $\text{mod }2$ . Hence, the cubic form from Theorem 1 has to be linear, so that the four Rokhlin invariants coincide in pairs. In particular, in light of Theorem 1, we can compute $\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})$ as the invariant for the manifold obtained by zero surgeries on each component of a split link, one component being a trefoil and one component being unknotted. For this case,

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=({\mathcal{I}}\oplus {\mathcal{I}}\langle 2\rangle )\otimes H^{1}(S^{1};\mathbb{F}),\end{eqnarray}$$

as follows by looking at the connected sum spectral sequence.

Figure 4. Taking the band sum of $n$ copies of the Borromean rings, and doing zero surgery on each of the components, one obtains a three-manifold with $b_{1}=3$ and triple cup product $n$ . Here we have depicted the case $n=2$ .

Manifolds with $b_{1}=3$

Let us assume as above that the homology of the manifold does not have $2$ -torsion. There are several cases to discuss. First of all, if we pick a basis $\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3}\in H^{1}(Y;\mathbb{Z})$ , the value

$$\begin{eqnarray}\langle \mathbf{x}_{1}\cup \mathbf{x}_{2}\cup \mathbf{x}_{3},[Y]\rangle \in \mathbb{Z}\end{eqnarray}$$

is well defined (up to sign). With a little abuse of terminology, we will refer to $m$ as the triple cup product of $Y$ . Recall that examples of three-manifolds with triple cup product $m$ can be provided by the construction in [Reference Ruberman and StrleRS00] by doing surgery on a band sum of $m$ copies of the Borromean rings; see Figure 4.

First of all, we consider the case in which the triple cup product is even. By Poincaré duality, the cup product on $H^{1}(Y;\mathbb{Z})$ vanishes modulo  $2$ . As above, this implies that the Rokhlin function is linear. Hence, either all Rokhlin invariants coincide (in which case the manifold is $\text{Pin}(2)$ -standard) or exactly half of them take one value. We can compute the homology in the latter case as follows. We can write the torus of flat connections as $\mathbb{T}^{1}\times \mathbb{T}^{2}$ in such a way that the spin structures in $\{0\}\times \mathbb{T}^{2}$ and $\{\unicode[STIX]{x1D70B}\}\times \mathbb{T}^{2}$ all have the same Rokhlin invariant. We can consider the two-step filtration coming from the value in the component $\mathbb{T}^{1}$ . The $E^{1}$ page is the direct sum of the equivariant coupled Morse homologies of the families parametrized by $\{0\}\times \mathbb{T}^{2}$ and $\{\unicode[STIX]{x1D70B}\}\times \mathbb{T}^{2}$ , which are $\text{Pin}(2)$ -standard. In particular, we have

$$\begin{eqnarray}E^{1}=\tilde{{\mathcal{R}}}\otimes ((H^{1}(\mathbb{T}^{2};\mathbb{F}))\oplus (H^{1}(\mathbb{T}^{2};\mathbb{F}))\langle 1\rangle ).\end{eqnarray}$$

Here the shift of the second summand comes from the difference of the Rokhlin invariants. Furthermore, each summand $H^{1}(\mathbb{T}^{2};\mathbb{F})$ has a filtration coming from the index of the Morse function on $\mathbb{T}^{2}$ . The $d_{1}$ differential maps the first summand to the second and it also lowers the filtration level on $H^{1}(\mathbb{T}^{2};\mathbb{F})$ . Using the description of the moduli spaces in the case $b_{1}=1$ , the filtration-preserving component is readily computed to be multiplication by $Q^{2}$ (up to grading shift). Hence, we have

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=({\mathcal{I}}\oplus {\mathcal{I}}\langle 2\rangle )\otimes (H^{1}(\mathbb{T}^{2};\mathbb{F})).\end{eqnarray}$$

In the case where the triple cup product is odd, there are again two cases (as can be shown by a direct inspection): either seven spin structures attain one value and the remaining one a different one (as in the case of the three-torus) or five spin structures attain one value and three attain the other. The latter case can be realized from the general example in Figure 4 by tying a knot of Arf invariant $1$ in one of the components. We already see a difference with the even case in usual monopole Floer homology: as shown in [Reference Kronheimer and MrowkaKM07, §35.3] (see in particular the proof of Theorem 35.3.2), in the odd case $\overline{\mathit{HM}}_{\ast }(Y,\mathfrak{s})$ has rank three in each degree, rather than four. In the first of the two possible cases, we will show that

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=(H^{1}(\mathbb{T}^{3})\oplus H^{2}(\mathbb{T}^{3}))\otimes \tilde{{\mathcal{R}}}.\end{eqnarray}$$

An analogous computation was provided in different terms in the case of the three-torus in [Reference LinLin15]. To see this, first recall that the Gysin exact triangle

implies that (if we think of $\overline{\mathit{HS}}_{\ast }$ as a $\mathbb{F}[Q]/Q^{3}$ -module) each cyclic summand of $\overline{\mathit{HS}}_{\ast }$ corresponds to a rank-two subgroup $\mathbb{F}\oplus \mathbb{F}$ of $\overline{\mathit{HM}}_{\ast }$ . Furthermore, if this summand is isomorphic to $\mathbb{F}[Q]/Q^{i}$ , then the generators of this subgroup differ in degree by $i$ .

We look at the spectral sequence from Proposition 6. The $E^{1}$ page is given by

where each column repeats four-periodically. The groups in the $i$ th column correspond to the critical points of index $3-i$ . The spin structure corresponding to the last column has different Rokhlin invariant (so that it is shifted in degree by two). We will use the convention that between two consecutive groups in the same column the map $Q$ has the highest possible rank.

The differential $d_{1}$ goes from one column to the one on its right and, by Proposition 6, the $E^{2}$ page is given by

Because of the module structure, the only possible non-trivial differential $d_{2}$ is the dashed one. Furthermore, as $\overline{\mathit{HM}}_{\ast }$ has rank three in each degree, from the discussion on the Gysin sequence above the differential $d_{3}$ is forced to be the arrow drawn. Now, if $d_{2}$ is not zero, the final result is

Of course, there are no possible extensions as $\mathbb{F}[Q]/Q^{3}$ -modules, so that this is indeed $\overline{\mathit{HS}}_{\ast }$ . On the other hand, this module requires seven generators over $\mathbb{F}[Q]/Q^{3}$ , so that one obtains a contradiction with the computation of $\overline{\mathit{HM}}_{\ast }$ from [Reference Kronheimer and MrowkaKM07] using the Gysin sequence. Hence, $d_{2}$ vanishes and the $E^{\infty }$ page is

Again this requires seven generators over $\mathbb{F}[Q]/Q^{3}$ , but there is now space for a non-trivial extension: this is shown by the dotted arrow and the result follows.

Finally, in the case in which exactly five spin structures have the same Rokhlin invariant, we have

$$\begin{eqnarray}\overline{\mathit{HS}}_{\ast }(Y,\mathfrak{s})=(\text{I}\oplus \text{I}\langle 2\rangle )\otimes H^{1}(\mathbb{T}^{3};\mathbb{F})\end{eqnarray}$$

by an analogous argument. We just point out that this is an example in which the differential $d_{2}$ of the spectral sequence in Proposition 6 is non-zero. Indeed, we can assume after a basis change that the Rokhlin map is

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{1}\mathbf{x}_{1}+\unicode[STIX]{x1D706}_{2}\mathbf{x}_{2}+\unicode[STIX]{x1D706}_{3}\mathbf{x}_{3}\mapsto \unicode[STIX]{x1D706}_{1}+\unicode[STIX]{x1D706}_{2}+\unicode[STIX]{x1D706}_{3}+\unicode[STIX]{x1D706}_{1}\unicode[STIX]{x1D706}_{2}\unicode[STIX]{x1D706}_{3},\end{eqnarray}$$

so that the $E^{1}$ page looks like

repeated as before four-periodically. Here the components in the third column are shifted in light of the Rokhlin invariants. So, the $E^{2}$ page is

and, if we suppose that $d_{2}$ is zero, we see that also $d_{3}$ has to be zero (for degree reasons and because it is a map of ${\mathcal{R}}$ -modules). On the other hand, this group cannot fit in the Gysin exact sequence with $\overline{\mathit{HM}}_{\ast }$ .