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This is indeed the natural concatenation of the two double complexes
$D^{+}_{*,*}$ and $D^{-}_{*,*}$.

######  Corollary 5.29.

Let $M$ be a right $A=\Lambda_{R}[u]$-module. Then $U\colon
H^{\infty}_{A}(M)\rightarrow H^{\infty}_{A}(M)$ is an isomorphism.

###### Proof.

$U\colon\operatorname{Tot}^{\Pi,-\infty}(D^{\infty}_{*,*},\partial^{\prime},\partial^{\prime\prime})\rightarrow\operatorname{Tot}^{\Pi,-\infty}(D^{\infty}_{*,*},\partial^{\prime},\partial^{\prime\prime})$
is an isomorphism. ∎

As in the cases $C^{\pm}_{A}$ we will need calculations for $M=R$, $M=R\oplus
R[2]$ and $M=\Lambda_{R}[u]$. This is easily achieved with the help of the
above explicit model.

######  Corollary 5.30.

We have

$H^{\infty}_{A}(A)=0,\;\;H^{\infty}_{A}(R)=R[T,T^{-1}]\;\mbox{ and
}\;H^{\infty}_{A}(R\oplus R[2])=R[S,S^{-1}],$

where $|T|=-4$, $|S|=-2$ and the $R[U]$-action is given by $U\cdot
T^{i}=T^{i+1}$ and $U\cdot S^{i}=S^{i+2}$ for all $i\in\mathbb{Z}$.

We note once again that our notation should not be interpreted to mean that
there is any product structure in $H^{\infty}_{A}(R\oplus R[2])$ or
$H^{\infty}_{A}(R)$.

###  5.4. The Equivariant Instanton Floer Groups

In this section we will give the definition of the various flavors of
equivariant instanton Floer homology. Before we can get to this we have to
recall a few definitions concerning filtered DG modules. Let $M$ be a $DG$
module equipped with an increasing filtration

$\cdots\subset F_{p-1}M\subset F_{p}M\subset F_{p+1}M\subset\cdots\subset M$

by DG submodules. The minimal filtration quotients $F_{p}M/F_{p-1}M$ enter
many times in the following discussion, so we introduce the simpler notation
$\overline{F}_{p}M\coloneqq F_{p}M/F_{p-1}M$. The filtration gives rise to a
homological spectral sequence with

(5.9) $E^{1}_{p,q}=H_{p+q}(\overline{F}_{p}M)\implies H_{p+q}(M)$

(we will elaborate on this in the next section). However, without any
additional information, this spectral sequence may fail to converge in any
sense to the desired target $H(M)$.

The following properties of a filtration are highly desirable in order to
achieve any reasonable convergence properties of the above spectral sequence.
The filtration $\\{F_{p}M\\}_{p}$ of $M$ is said to be [Boa99, Definition 2.1]

  1. (i)

exhaustive if $\operatorname{colim}_{p}F_{p}M=\bigcup_{p}F_{p}M=M$,

  2. (ii)

Hausdorff if $\operatorname{lim}_{p}F_{p}M=\bigcap_{p}F_{p}M=0$ and

  3. (iii)

complete if $\operatorname{Rlim}_{p}F_{p}M=0$.

Here $\operatorname{Rlim}$ is the first (and only nonzero) right derived
functor of $\operatorname{lim}$ (see [Boa99, Section 1] for additional
details). It is important to note that these limits are calculated degreewise.
In particular, if the filtration is degreewise bounded below, that is, for
each degree $n$ there exists $p_{0}(n)$ such that $F_{p}M_{n}=0$ for all
$p\leq p_{0}$, then the filtration is automatically complete Hausdorff.
Similarly, if the filtration is degreewise bounded above then the filtration
is automatically exhaustive.

######  Definition 5.31.

Let $M$ be a DG module equipped with an increasing filtration
$\\{F_{p}M\\}_{p}$ by DG submodules. Then the full completion of
$(M,\\{F_{p}M\\}_{p})$ is defined to be the complex

$\hat{M}\coloneqq\operatorname{lim}_{q}\operatorname{colim}_{p}F_{p}M/F_{q}M$

equipped with the filtration
$F_{p}\hat{M}\coloneqq\operatorname{lim}_{q}F_{p}M/F_{q}M$.

######  Remark 5.32.

The standard completion of a filtered module $M$ is typically defined by
$\operatorname{lim}_{q}M/F_{q}M$ (see for instance [Boa99, 2.7]) and is
naturally filtered by $\\{\operatorname{lim}_{q}F_{p}M/F_{q}M\\}_{p}$. We have
therefore chosen to use the terminology full completion to distinguish it from
this notion. The lemma below demonstrates that the two notions coincide if the
filtration of $M$ is exhaustive.

A map $f\colon M\rightarrow M^{\prime}$ between filtered DG modules is
filtered if $f(F_{p}M)\subset F_{p}M^{\prime}$ for all $p\in\mathbb{Z}$. There
is therefore a category of filtered DG modules and filtered maps. A sequence
$M^{\prime}\rightarrow M\rightarrow M^{\prime\prime}$ of filtered DG modules
and filtered maps is short exact if $F_{p}M^{\prime}\rightarrow
F_{p}M\rightarrow F_{p}M^{\prime\prime}$ is short exact for each
$p\in\mathbb{Z}$. The relevant properties of the full completion are
summarized in the following lemma.

######  Lemma 5.33.

Let $M$ be a filtered DG module. Then the following holds true.

  1. (i)

The filtration $F_{p}\hat{M}=\operatorname{lim}_{q<p}F_{p}M/F_{q}M$ of
$\hat{M}$ is exhaustive and complete Hausdorff.

  2. (ii)

For each $q<p$ there is a natural isomorphism $F_{p}M/F_{q}M\rightarrow
F_{p}\hat{M}/F_{q}\hat{M}$.

  3. (iii)

If the filtration of $M$ is exhaustive, then
$\hat{M}=\operatorname{lim}_{q}M/F_{q}M$.

  4. (iv)

If the filtration of $M$ is complete Hausdorff, then
$\hat{M}=\operatorname{colim}_{p}F_{p}M=\bigcup_{p}F_{p}M$.

The full completion $M\mapsto\hat{M}$ defines a functor from the category of
filtered modules to itself, which preserves short exact sequences.

###### Proof.

Let $M$ be a filtered DG module and set
$N=\operatorname{colim}_{p}F_{p}M=\bigcup_{p}F_{p}M$ filtered by
$F_{p}N=F_{p}M$. Then the full completion of $M$ coincides with the standard
completion of $N$, i.e., $\hat{M}=\operatorname{lim}_{q}N/F_{q}N$. Therefore
part (i) and (ii) follow from the corresponding properties of the standard
completion (see [Boa99, Proposition 2.8]). If the filtration of $M$ is
exhaustive, then $\operatorname{colim}_{p}F_{p}M/F_{q}M=M/F_{q}M$ for each $q$
and part (iii) easily follows. For part (iv)

$\hat{M}=\operatorname{lim}_{q}\operatorname{colim}_{p}F_{p}M/F_{q}M=\operatorname{lim}_{q}N/F_{q}N=N=\operatorname{colim}_{p}F_{p}M.$

It is clear that $M\mapsto\hat{M}$ is functorial with respect to filtered
maps. The exactness of $\hat{M}$ follows from the exactness of sequential
colimits and exactness of the standard completion in the category of
$R$-modules (see for instance [AM69, Corollary 10.3]). ∎

######  Remark 5.34.

If $B$ is a DG algebra and $M$ is a left or right $B$-module equipped with a
filtration by $B$-submodules the definition of the full completion and the
above lemma apply without change. The reason for this is that quotients,
limits and colimits are formed in the category of graded modules, with the
$B$-action and differentials carried along.

Recall from the previous section that for a degreewise $R$-free augmented DG
algebra $A$, four functors $C^{\bullet}_{A}$,
$\bullet\in\\{+,-,(+,tw),\infty\\}$, from the category of right $A$-modules to
the category of left $C^{-}_{A}(R)$-modules were defined. By Theorem 5.6 and
Theorem 5.18 these are all exact and preserve quasi-isomorphisms. Therefore,
if $M$ is a right $A$-module equipped with a filtration $F_{p}M$ by
$A$-submodules, $C^{\bullet}_{A}(M)$ carries a natural filtration
$F_{p}C^{\bullet}_{A}(M)\coloneqq C^{\bullet}_{A}(F_{p}M)$ by
$C^{-}_{A}(R)$-submodules. Now, even if the initial filtration of $M$ is
exhaustive and complete Hausdorff, the resulting filtration of
$C^{\bullet}_{A}(M)$ may fail to satisfy these properties. This motivates the
following definition.

######  Definition 5.35.

Let $A$ be a degreewise $R$-free augmented DG algebra. For
$\bullet\in\\{+,-,(+,tw),\infty\\}$ define a functor $\hat{C}^{\bullet}_{A}$
from the category of filtered right $A$-modules to the category of filtered
left $C^{-}_{A}(R)$-modules by

$\displaystyle\hat{C}^{\bullet}_{A}(M)\coloneqq$
$\displaystyle\operatorname{lim}_{q}\operatorname{colim}_{p}C^{\bullet}_{A}(F_{p}M/F_{q}M)\;\;\mbox{
filtered by }\;$ $\displaystyle F_{p}\hat{C}^{\bullet}_{A}(M)\coloneqq$
$\displaystyle\operatorname{lim}_{q<p}C^{\bullet}_{A}(F_{p}M/F_{q}M)\;\;\;\;\;\;\;\;\;\mbox{
for all}\;p\in\mathbb{Z}.$

In other words, $\hat{C}^{\bullet}_{A}$ is the composition of the natural
extension of $C^{\bullet}_{A}$ to filtered $A$-modules followed by the full
completion functor. We denote the homology of $\hat{C}^{\bullet}_{A}(M)$ by
$\hat{H}^{\bullet}_{A}(M)$.

The various flavors of equivariant instanton Floer homology are obtained by
applying this construction to the framed Floer complex.

######  Definition 5.36.

Let $A=C^{gm}_{*}(\operatorname{SO}(3))$ and let $M=\widetilde{CI}(Y,E)$ be
the framed Floer complex associated with an $\operatorname{SO}(3)$-bundle $E$
over a closed oriented three-manifold $Y$, equipped with the index filtration.
For $\bullet\in\\{+,-,(+,tw),\infty\\}$ define

$CI^{\bullet}(Y,E)\coloneqq\hat{C}^{\bullet}_{A}(M)\;\;\mbox{ and
}\;\;I^{\bullet}(Y,E)\coloneqq H(CI^{\bullet}(Y,E)).$

The groups $I^{\bullet}(Y,E)$, for $\bullet\in\\{+,-,(+,tw),\infty\\}$, are
called the positive, negative, positive twisted and Tate equivariant instanton
Floer homologies of $(Y,E)$, respectively. These homology groups are modules
over $H^{-}_{A}(R)$.

For $A=\Lambda_{R}[u]$ and $M=DCI(Y,E)$ we similarly define the complexes

$DCI^{\bullet}(Y,E)=\hat{C}^{\bullet}_{A}(M).$

######  Remark 5.37.

For $A=C_{*}(\operatorname{SO}(3);R)$ and $M=R$ it holds true that
$H^{-}_{A}(R)\cong H^{-*}(\operatorname{BSO}(3);R)$ (see [Eis19, Lemma A.21]).

The norm map $N_{M}\colon C^{+,tw}_{A}(M)\rightarrow C^{-}_{A}(M)$ is a
natural in $M$ and therefore extends to a natural map
$\hat{N}_{M}\colon\hat{C}^{+,tw}_{A}(M)\rightarrow\hat{C}^{-}_{A}(M)$ for any
filtered module $M$. The following simple lemma ensures that
$\hat{C}^{\infty}_{A}(M)$ is naturally isomorphic to
$\operatorname{Cone}(\hat{N}_{M})$.

######  Lemma 5.38.

Let $A$ be a degreewise $R$-free augmented DG algebra and let $M$ be a
filtered right $A$-module. Then there is a natural (filtered) isomorphism of
$C^{-}_{A}(R)$-modules

$\operatorname{Cone}(\hat{N}_{M})\cong\hat{C}^{\infty}_{A}(M).$

In particular, we have the following exact triangle of $H^{-}_{A}(R)$-modules.

${\hat{H}^{+,tw}_{A}(M)}$${\hat{H}^{-}_{A}(M)}$${\hat{H}^{\infty}_{A}(M)}$$\scriptstyle{H(\hat{N}_{M})}$$\scriptstyle{[0]}$$\scriptstyle{[-1]}$

###### Proof.

For $q<p$ and $n$ there is a natural identification

$\frac{F_{p}C^{\infty}_{A}(M)}{F_{q}C^{\infty}_{A}(M)}_{n}\cong
C^{\infty}_{A}\left(\frac{F_{p}M}{F_{q}M}\right)_{n}=C^{-}_{A}\left(\frac{F_{p}M}{F_{q}M}\right)_{n}\oplus
C^{+,tw}_{A}\left(\frac{F_{p}M}{F_{q}M}\right)_{n-1}.$

In terms of the right hand side the differential is given by

$\left(\begin{array}[]{cc}\partial^{-}_{p,q}&N_{p,q}\\\
0&-\partial^{+,tw}_{p,q}\end{array}\right),$

where $\partial^{-}_{p,q}$ and $\partial^{+,tw}_{p,q}$ denote the differential
in the first and second summand respectively and $N_{p,q}$ denotes the evident
norm map. Applying $\operatorname{lim}_{q}\operatorname{colim}_{p}$ to this
identification, taking into account that this is calculated degreewise with
differentials and $A$-module structure carried along, we obtain
$\hat{C}^{\infty}_{A}(M)_{n}\cong\hat{C}^{-}_{A}(M)_{n}\oplus\hat{C}^{+,tw}_{A}(M)_{n-1}$
in each degree $n$. The differential on the right hand side is given by

$\left(\begin{array}[]{cc}\operatorname{lim}_{q}\operatorname{colim}_{p}\partial^{-}_{p,q}&\operatorname{lim}_{q}\operatorname{colim}_{p}N_{p,q}\\\
0&\operatorname{lim}_{q}\operatorname{colim}_{p}\partial^{+,tw}_{p,q}\end{array}\right).$

Therefore, the right hand side is precisely
$\operatorname{Cone}(\hat{N}_{M})$, since by definition
$\hat{N}_{M}=\operatorname{lim}_{q}\operatorname{colim}_{p}N_{p,q}$. ∎

To verify that $DCI^{\bullet}(Y,E)$ also calculates $I^{\bullet}(Y,E)$ when
$\frac{1}{2}\in R$ we will establish the necessary invariance results for the
functors $\hat{C}^{\bullet}_{A}$. The appropriate tool is the following lemma,
which is a simple consequence of the Eilenberg-Moore comparison theorem of
[Wei94, Theorem 5.5.11]. For a filtered map $f\colon M\rightarrow N$ between
filtered DG modules we use the notation
$\overline{f}_{p}\colon\overline{F}_{p}M\rightarrow\overline{F}_{p}N$ for the
induced maps between the the minimal filtration quotients.

######  Lemma 5.39.

Let $M$ and $N$ be DG modules equipped with exhaustive and complete Hausdorff
filtrations. Then if $f\colon M\rightarrow N$ is a filtered map such that the
induced map
$\overline{f}_{p}\colon\overline{F}_{p}M\rightarrow\overline{F}_{p}N$ is a
quasi-isomorphism for each $p\in\mathbb{Z}$, then $f$ is a quasi-isomorphism.

######  Proposition 5.40.

Let $A$ be a degreewise $R$-free augmented DG algebra. Suppose $f\colon
M\rightarrow N$ is a filtered map of filtered right $A$-modules such that
$\overline{f}_{p}\colon\overline{F}_{p}M\rightarrow\overline{F}_{p}N$ is a
quasi-isomorphism for each $p\in\mathbb{Z}$. Then for
$\bullet\in\\{+,-,(+,tw),\infty\\}$ the induced map

$\hat{f}^{\bullet}\colon\hat{C}^{\bullet}_{A}(M)\rightarrow\hat{C}^{\bullet}_{A}(N)$

is a (filtered) quasi-isomorphism of $C^{-}_{A}(R)$-modules. Furthermore, upon
passage to homology, these isomorphisms are compatible with the exact triangle
of Lemma 5.38.

###### Proof.

Let $\bullet\in\\{+,-,(+,tw),\infty\\}$ and note that the functors
$C^{\bullet}_{A}$ and $\hat{C}^{\bullet}_{A}$ are covariant in $M$. For
simplicity of notation let
$g=\hat{f}^{\bullet}\colon\hat{C}^{\bullet}_{A}(M)\rightarrow\hat{C}^{\bullet}_{A}(N)$
denote the induced filtered map of $C^{-}_{A}(R)$-modules. Then by Lemma 5.33
part (ii) and the exactness of $C^{\bullet}_{A}$, there is for each $p$ a
commutative diagram

${C^{\bullet}_{A}(\overline{F}_{p}M)}$${C^{\bullet}_{A}(\overline{F}_{p}N)}$${\overline{F}_{p}\hat{C}^{\bullet}_{A}(M)}$${\overline{F}_{p}\hat{C}^{\bullet}_{A}(N).}$$\scriptstyle{(\overline{f}_{p})^{\bullet}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\overline{g}_{p}}$

By assumption $\overline{f}_{p}$ is a quasi-isomorphism, so since the functor
$C^{\bullet}_{A}$ preserves quasi-isomorphisms it follows that
$\overline{f}_{p}^{\bullet}$ is a quasi-isomorphism. As the vertical maps are
isomorphisms, we conclude that $\overline{g}_{p}$ is a quasi-isomorphism. By
Lemma 5.33 part (i) the filtrations of $\hat{C}^{\bullet}_{A}(M)$ and
$\hat{C}^{\bullet}_{A}(N)$ are exhaustive and complete Hausdorff. We conclude
that $g=\hat{f}^{\bullet}$ is a quasi-isomorphism using the comparison result;
Lemma 5.39. The lemma applies equally well to the filtered subcomplexes.
Consequently, the restrictions $g_{p}\colon
F_{p}\hat{C}^{\bullet}_{A}(M)\rightarrow F_{p}\hat{C}^{\bullet}(N)$ are also
quasi-isomorphisms for all $p$. This shows that $g$ is in fact a filtered
quasi-isomorphism. ∎

Next we have to establish a similar invariance result for variations in the DG
algebra.

######  Proposition 5.41.

Let $f\colon A\rightarrow B$ be a quasi-isomorphism of degreewise $R$-free
augmented algebras and let $M$ be a filtered $B$-module. Then there is for
each $\bullet\in\\{+,-,(+,tw),\infty\\}$ a natural isomorphism of
$H^{-}_{A}(R)\cong H^{-}_{B}(R)$-modules

$\hat{H}^{\bullet}_{A}(f^{-1}M)\cong\hat{H}^{\bullet}_{B}(M),$

compatible with the exact triangles of Lemma 5.38.

###### Proof.

We first consider the case $\bullet=+$. For each pair $q<p$ there is by
Proposition 5.5 a natural quasi-isomorphism

$f_{*}\colon C^{+}_{A}(F_{p}f^{-1}M/F_{q}f^{-1}M)\rightarrow
C^{+}_{B}(F_{p}M/F_{q}M)$

linear along $f^{*}\colon C^{-}_{B}(R)\rightarrow C^{-}_{A}(R)$. By applying
$\operatorname{lim}_{q}\operatorname{colim}_{p}$ we obtain a filtered map
$g=\hat{f}_{*}\colon\hat{C}^{+}_{A}(f^{-1}M)\rightarrow\hat{C}^{+}_{B}(M)$
linear along $f^{*}:C^{-}_{B}(R)\rightarrow C^{-}_{A}(R)$. By construction,
the following diagram commutes for each $p$.

${C^{+}_{A}(\overline{F}_{p}f^{-1}M)}$${C^{+}_{B}(\overline{F}_{p}M)}$${\overline{F}_{p}\hat{C}^{+}_{A}(f^{-1}M)}$${\overline{F}_{p}\hat{C}^{+}_{B}(M)}$$\scriptstyle{\overline{(f_{*})}_{p}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\overline{g}_{p}}$

As the upper horizontal map is a quasi-isomorphism, so is the lower horizontal
map and we conclude that $g=\hat{f}_{*}$ is a filtered quasi-isomorphism by
Lemma 5.39. For $\bullet=-$ there is for each pair $q<p$ a natural quasi-
isomorphism

$f^{*}\colon C^{-}_{B}(F_{p}M/F_{q}M)\rightarrow
C^{-}_{A}(F_{p}f^{-1}M/F_{q}f^{-1}M).$

By passing to the colimit and limit we obtain a filtered map
$\hat{f}^{*}\colon\hat{C}^{-}_{B}(M)\rightarrow\hat{C}^{-}_{A}(f^{-1}M)$,
linear along $f^{*}\colon C^{-}_{B}(R)\rightarrow C^{-}_{A}(R)$. The same
argument as above shows that this is a filtered quasi-isomorphism.

For the two remaining cases recall that there is a relative functor
$C^{+,tw}_{f}$, see Definition 5.19, from the category of right $B$-modules to
the category of left $C^{-}_{A}(R)$-modules. By Lemma 5.20 this functor is
exact and preserves quasi-isomorphism, so we may promote it to a functor
$\hat{C}^{+,tw}_{f}$ between the corresponding filtered categories as we did
with the others. By passing to the colimit and then the limit in the
commutative diagram of Proposition 5.21 applied to $F_{p}M/F_{q}M$ for all
$q<p$, we obtain a commutative diagram

${\hat{C}^{+,tw}_{B}(M)}$${\hat{C}^{+,tw}_{f}(M)}$${\hat{C}^{+,tw}_{A}(f^{-1}M)}$${\hat{C}^{-}_{B}(M)}$${\hat{C}^{-}_{A}(f^{-1}M)}$${\hat{C}^{-}_{A}(f^{-1}M)}$$\scriptstyle{\hat{\beta}_{M}}$$\scriptstyle{\hat{N}_{M}^{B}}$$\scriptstyle{\hat{N}^{f}_{M}}$$\scriptstyle{\hat{N}^{A}_{M}}$$\scriptstyle{\hat{\alpha}_{M}}$$\scriptstyle{\hat{f}^{*}}$$\scriptstyle{=}$

of filtered DG modules over $C^{-}_{A}(R)$ or $C^{-}_{B}(R)$ as appropriate.
Furthermore, all the maps are filtered and $C^{-}_{A}(R)$-linear,
$C^{-}_{B}(R)$-linear or linear along $f^{*}\colon C^{-}_{B}(R)\rightarrow
C^{-}_{A}(R)$ as appropriate. By the same proposition, as $f\colon
A\rightarrow B$ is a quasi-isomorphism of degreewise free DG algebras, all the
horizontal arrows in the diagram applied to $F_{p}M/F_{q}M$ are quasi-
isomorphisms. This implies, by the now familiar argument, that all the
horizontal arrows in the above diagram are filtered quasi-isomorphisms. In
particular, there is a natural isomorphism
$\hat{H}^{+,tw}_{A}(f^{-1}M)\cong\hat{H}^{+,tw}_{B}(M)$. Moreover, we obtain a
zigzag of quasi-isomorphisms between the cones of the vertical maps

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In view of Lemma 5.38, we obtain an isomorphism
$\hat{H}^{\infty}_{A}(f^{-1}M)\cong\hat{H}^{\infty}_{B}(M)$ linear along
$H^{-}_{A}(R)\cong H^{-}_{B}(R)$ as well. The above diagram also ensures that
these isomorphisms are compatible with the corresponding exact triangles. ∎

The key application of these invariance results is the following.

######  Corollary 5.42.

Suppose that $\frac{1}{2}\in R$. Then for each
$\bullet\in\\{+,-,(+,tw),\infty\\}$ there is an isomorphism
$H(DCI^{\bullet}(Y,E))\cong I^{\bullet}(Y,E)$ of $H^{-}_{A}(R)\cong
R[U]$-modules. These isomorphisms are compatible with the corresponding exact
triangles.

###### Proof.

As $\frac{1}{2}\in R$ there are, according to Proposition 2.9, a quasi-
isomorphism $i\colon\Lambda_{R}[u]\rightarrow
C^{gm}_{*}(\operatorname{SO}(3))$ of DG algebras and a zigzag

${\widetilde{CI}(Y,E)}$${X}$${DCI(Y,E)}$$\scriptstyle{f}$$\scriptstyle{g}$

of filtered objects and filtered $\Lambda_{R}[u]$-homomorphisms such that
$\overline{f}_{p}$ and $\overline{g}_{p}$ are quasi-isomorphism for each
$p\in\mathbb{Z}$. The required result now follows from Propositions 5.40 and
5.41. ∎

For $A=\Lambda_{R}[u]$ we can extend a number of the results from the previous
section.

######  Proposition 5.43.

Let $A=\Lambda_{R}[u]$ and let $M$ be a filtered $A$-module. Then there is a
filtered quasi-isomorphism
$\hat{f}_{M}\colon\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$ of
degree $3$ and a (filtered) homotopy
$\hat{s}_{M}\colon\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$ such
that $\hat{f}_{M}U-U\hat{f}_{M}=\partial\hat{s}_{M}-\hat{s}_{M}\partial$ for
$U\in R[U]\cong C^{-}_{A}(R)$. Moreover, there is a (filtered) quasi-
isomorphism

$\operatorname{Cone}(\hat{N}_{M}\circ\hat{f})\rightarrow\operatorname{Cone}(\hat{N}_{M})\cong\hat{C}^{\infty}_{A}(M),$

which is $R[U]$-linear up to homotopy, where the source is given the
$R[U]$-module structure determined by the homotopy
$\hat{t}_{M}=\hat{N}_{M}\circ\hat{s}_{M}$ as in Lemma 5.25. Furthermore, the
resulting isomorphism in homology is compatible with the associated exact
triangles.

###### Proof.

Recall from Lemma 5.24 that there is for each right $A$-module $N$ a quasi-
isomorphism $f_{N}\colon C^{+}_{A}(N)\rightarrow C^{+,tw}_{A}(N)$ of degree
$3$ and a homotopy $s_{N}\colon C^{+}_{A}(N)\rightarrow C^{+,tw}_{A}(N)$ such
that $f_{N}U-Uf_{N}=\partial s_{N}-s_{N}\partial$. Let $f_{p,q}$ and $s_{p,q}$
denote the maps obtained for $N=F_{p}M/F_{q}M$ for $q<p$. By naturality we may
pass to the colimit over $p$ and then to the limit over $q$ to obtain maps
$\hat{f}_{M}\colon\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$ and
$\hat{s}_{M}\colon\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$. As
$f_{p,q}U-Uf_{p,q}=s_{p,q}\partial-\partial s_{p,q}$ is valid for each pair
$q<p$, it follows that
$\hat{f}_{M}U-U\hat{f}_{M}=\partial\hat{s}_{M}-\hat{s}_{M}\partial$. The fact
that $\hat{f}_{M}$ is a quasi-isomorphism is proved as before using Lemma
5.39.

The final statement follows by applying
$\operatorname{lim}_{q}\operatorname{colim}_{p}$ to the diagram (5.8) in the
proof of Theorem 5.28 for $F_{p}M/F_{q}M$ and another application of Lemma
5.39. It is clear from this construction that one obtains an isomorphism
between the corresponding exact triangles in homology. ∎

Combining the above proposition with Corollary 5.42 we obtain the following.

######  Corollary 5.44.

Suppose that $\frac{1}{2}\in R$. Then there is a degree $3$ isomorphism
$I^{+}(Y,E)\cong I^{+,tw}(Y,E)$ of $R[U]$-modules and there is an exact
triangle of $R[U]$-modules

${I^{+}(Y,E)}$${I^{-}(Y,E)}$${I^{\infty}(Y,E)}$,$\scriptstyle{[3]}$$\scriptstyle{[0]}$$\scriptstyle{[-4]}$

where the numbers specify the degrees of the maps.

We will now restrict ourselves to the cases of interest, namely,

$(A,M)=(C^{gm}_{*}(\operatorname{SO}(3)),\widetilde{CI}(Y,E))\;\;\mbox{ or
}\;\;(A,M)=(\Lambda_{R}[u],DCI(Y,E)),$

where $M$ is equipped with the index filtration. Recall from Definition 2.8
and Definition 2.4 that $M$ was obtained from a multicomplex
$(M_{*,*},\\{\partial^{r}\\}_{r=0}^{5})$, and that the index filtration was
defined to be the column filtration, i.e., $F_{p}M_{n}=\bigoplus_{s\leq
p}M_{s,n-s}$. In both cases it holds true that $M_{s,t}=0$ for all $t>4$ and
$t<0$ (for $M=DCI(Y,E)$ this is true for $t\geq 4$). This implies that the
filtration is degreewise finite and therefore exhaustive and complete
Hausdorff. Explicitly,

(5.10) $F_{p}M_{n}=\left\\{\begin{array}[]{cl}M_{n}&\mbox{ for all }p\geq n\\\
0&\mbox{ for all }p\leq n-5\end{array}\right.$

The following lemma indicates why it is desirable to introduce the full
completions of the various functors.

######  Lemma 5.45.

The filtration of $C^{+}_{A}(M)$, $F_{p}C^{+}_{A}(M)=C^{+}_{A}(F_{p}M)$, is
degreewise bounded above, exhaustive and Hausdorff. The filtration of
$C^{-}_{A}(M)$ is degreewise bounded below and complete Hausdorff. If $M\neq
0$ the filtration of $C^{+}_{A}(M)$ fails to be complete and the filtration of
$C^{-}_{A}(M)$ fails to be exhaustive. Therefore,

$\hat{C}^{+}_{A}(M)=\operatorname{lim}_{q}C^{+}_{A}(M/F_{q}M)\;\;\mbox{ and
}\;\;\hat{C}^{-}_{A}(M)=\operatorname{colim}_{p}C^{-}_{A}(F_{p}M).$

###### Proof.

We will use the underlying bigrading of the multicomplex $M_{*,*}$. As a
graded module (not DG) we have $C^{+}_{A}(M)=B(M,A,R)\cong M\otimes BA$.
Therefore, in degree $n$

(5.11) $C^{+}_{A}(M)_{n}=\bigoplus_{u+r=n}M_{u}\otimes
BA_{r}=\bigoplus_{s+t+r=n}M_{s,t}\otimes BA_{r},$

and $F_{p}C^{+}_{A}(M)$ is obtained by imposing the condition $s\leq p$ in the
above right hand direct sum. First, notice that $A$ and hence $BA$ are
supported in nonnegative degrees. Then, as $M_{s,t}=0$ for $t<0$, it follows
$F_{p}C^{+}_{A}(M)_{n}=C^{+}_{A}(M)_{n}$ for $p\geq n$. This shows that the
filtration is degreewise bounded above and hence exhaustive. It also follows
from the above description that the filtration is Hausdorff. On the other hand

(5.12)
$\lim_{p}C^{+}_{A}(M/F_{p}M)_{n}\cong\lim_{p}\bigoplus_{\begin{subarray}{c}s+t+r=n\\\
s>p\end{subarray}}M_{s,t}\otimes BA_{r}\cong\prod_{s+t+r=n}M_{s,t}\otimes
BA_{r}.$

This does not coincide with the direct sum in equation (5.11) if $M\neq 0$,
due to the periodicity $M_{s,t}\cong M_{s+8,t}$ for all $s,t$. Thus,
$C^{+}_{A}(M)$ fails to be complete if $M\neq 0$.

Next, there is an isomorphism
$C^{-}_{A}(M)=\operatorname{Hom}_{A}(EA,M)\cong\operatorname{Hom}(BA,M)$ of
graded modules. In a fixed degree $n$, we find

(5.13)
$C^{-}_{A}(M)_{n}=\operatorname{Hom}(BA,M)_{n}=\prod_{s+t-r=n}\operatorname{Hom}(BA_{r},M_{s,t})$

and as above $F_{p}C^{-}_{A}(M)$ is obtained by imposing the condition $s\leq
p$ in the right hand product. For the factor
$\operatorname{Hom}(BA_{r},M_{s,t})$ to be nonzero it is necessary that $r\geq
0$ and $0\leq t\leq 4$, and therefore $s=n+r-t\geq n-4$. Hence,
$F_{p}C^{-}_{A}(M)_{n}=0$ for $p<n-4$ showing that the filtration is
degreewise bounded below, and thus complete Hausdorff. Moreover, the above
also implies that the right hand product in (5.13) subject to the condition
$s\leq p$ is finite and therefore coincides with the direct sum. Consequently,
$\operatorname{colim}_{p}F_{p}C^{-}_{A}(M)$ is given by

(5.14) $\operatorname{colim}_{p}\bigoplus_{\begin{subarray}{c}s+t-r=n\\\ s\leq
p\end{subarray}}\operatorname{Hom}(BA_{r},M_{s,t})=\bigoplus_{s+t-r=n}\operatorname{Hom}(BA_{r},M_{s,t}),$

which does not coincide with the product in (5.13) as long as $M\neq 0$. This
shows that the filtration fails to be exhaustive if $M\neq 0$.

The description of $\hat{C}^{+}_{A}(M)$ and $\hat{C}^{-}_{A}(M)$ given in the
statement is now a consequence of Lemma 5.33 part (iii) and (iv). ∎

######  Remark 5.46.

Miller [Eis19, Appendix A] defines completed versions of $C^{\pm}_{A}$ using
what he names the completed bar construction and the finitely supported cobar
construction. He defines the completed bar construction, $\hat{B}(M,A,N)$, to
be the completion of $B(M,A,N)$ with respect to the filtration by internal
degree, while the finitely supported cobar construction
$c\hat{B}(N,A,M)\subset\operatorname{Hom}_{A}(B(N,A,A),M)$ is the subcomplex
consisting of those functionals that vanish on $B_{p,q}(N,A,A)$ for all
sufficiently large $p$. The above lemma, in particular equations (5.12) and
(5.14), verifies that our $\hat{C}^{+}_{A}(M)$ and $\hat{C}^{-}_{A}(M)$
coincide with Miller’s $\hat{B}(M,A,R)$ and $c\hat{B}(R,A,M)$ for
$M=\widetilde{CI}(Y,E)$ and $M=DCI(Y,E)$ with $A$ as appropriate.

The final part of this section is devoted to extending the concrete models of
Proposition 5.8 and Theorem 5.28 to the complex $DCI^{\bullet}(Y,E)$ for
$\bullet\in\\{+,-,\infty\\}$. We will therefore assume that $\frac{1}{2}\in R$
throughout.

######  Theorem 5.47.

Let $A=\Lambda_{R}[u]$ and let
$(D^{\pm}_{*,*},\partial^{\prime},\partial^{\prime\prime})$ be the double
complexes associated with $M=DCI(Y,E)$ in Proposition 5.8. Then there are
isomorphisms of DG $R[U]$-modules

$DCI^{+}(Y,E)\cong\operatorname{Tot}^{\Pi}(D^{+}_{*,*},\partial^{\prime},\partial^{\prime\prime})\;\;\mbox{
and
}\;\;DCI^{-}(Y,E)\cong\operatorname{Tot}^{\oplus}(D^{-}_{*,*},\partial^{\prime},\partial^{\prime\prime}).$

In particular, there are identifications

$DCI^{+}(Y)_{n}\cong\prod_{p\geq 0}M_{n-4p}\;\;\mbox{ and
}\;\;DCI^{-}(Y)_{n}\cong\bigoplus_{p\geq 0}M_{n+4p}$

for each $n\in\mathbb{Z}$. An element $x\in DCI^{\pm}(Y)_{n}$ can therefore be
expressed by a sequence $x=(m_{n+4p})_{p}$ where we require $p\leq 0$ in the
$+$ case and $p\geq 0$ and almost all $m_{n+4p}=0$ in the $-$ case. The
differential is in both cases given by

$\partial x=(\partial_{M}m_{n+4p}-(-1)^{n}m_{n+4(p-1)}u)_{p}$

and the action of $R[U]$ is determined by $U\cdot x=(m_{n+4(p+1)})_{p}$.

###### Proof.

The isomorphism
$C^{+}_{A}(M)\cong\operatorname{Tot}^{\oplus}(D^{+}_{*,*},\partial^{\prime},\partial^{\prime\prime})$
of Proposition 5.8 yields in degree $n$

$C^{+}_{A}(M)_{n}\cong\bigoplus_{s\geq 0}M_{n-4s}=\bigoplus_{s\geq
0,t}M_{t,n-4s-t}$

and $F_{p}C^{+}_{A}(M)$ is obtained by imposing the condition $t\leq p$ in the
right hand sum. Now the calculation of the limit
$\operatorname{lim}_{p}C^{+}_{A}(M)_{n}/F_{p}C^{+}_{A}(M)_{n}$ goes through as
in equation (5.12) of the previous proof to show that
$\hat{C}^{+}_{A}(M)_{n}\cong\prod_{s\geq 0}M_{n-4s}$.

The other case is analogous and the calculation of the relevant colimit
proceeds as in equation (5.14) of the previous proof. ∎

Note that the only difference between the above theorem and Proposition 5.8 is
that we have interchanged the type of totalization we apply to the double
complexes.

We also obtain the following explicit model computing $I^{\infty}(Y,E)$.

######  Theorem 5.48.

Let $A=\Lambda_{R}[u]$, let
$(D^{\pm}_{*,*},\partial^{\prime},\partial^{\prime\prime})$ and
$(D^{\infty}_{*,*},\partial^{\prime},\partial^{\prime\prime})$ be the double
complexes associated with $M=DCI(Y,E)$ in the above theorem and Theorem 5.28,
respectively. Let
$\nu,\psi\colon\operatorname{Tot}^{\Pi}(D^{+},\partial^{\prime},\partial^{\prime\prime})\rightarrow\operatorname{Tot}^{\oplus}(D^{-},\partial^{\prime},\partial^{\prime\prime})$
of degree $3$ and $0$ respectively, be given degreewise by the compositions

${\prod_{s\geq 0}M_{n-4s}}$${M_{n}}$${M_{n+3}}$${\bigoplus_{t\geq
0}M_{n+3+4t}}$${\prod_{s\geq 0}M_{n-4s}}$${M_{n}}$${\bigoplus_{t\geq
0}M_{n+4t},}$$\scriptstyle{\pi}$$\scriptstyle{(-1)^{n}u}$$\scriptstyle{\iota}$$\scriptstyle{\pi}$$\scriptstyle{\iota}$

where $\pi$ and $\iota$ denote the projection and inclusion, respectively.
Then $\nu$ is a chain map satisfying $\nu U-U\nu=\partial\psi-\psi\partial$.
Define $\operatorname{Cone}(\nu)$ with the $R[U]$-module structure adjusted by
the homotopy $\psi$ as in Lemma 5.25. Then there is an isomorphism
$R[U]$-modules
$\operatorname{Tot}^{\Pi,\infty}(D^{\infty}_{*,*},\partial^{\prime},\partial^{\prime\prime})\cong\operatorname{Cone}(\nu)$
and a filtered quasi-isomorphism between this complex and $DCI^{\infty}(Y,E)$,
which is $R[U]$-linear up to homotopy. Moreover, this quasi-isomorphism is
compatible with the corresponding exact triangles in homology.

###### Proof.

By Proposition 5.43 there is a filtered quasi-isomorphism
$\hat{f}_{M}\colon\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$ of
degree $3$ and a filtered homotopy
$\hat{s}_{M}:\hat{C}^{+}_{A}(M)\rightarrow\hat{C}^{+,tw}_{A}(M)$ such that
$\hat{f}_{M}U-U\hat{f}_{M}=\partial\hat{s}_{M}-\hat{s}_{M}\partial$.
Furthermore, there is an induced filtered quasi-isomorphism
$\operatorname{Cone}(\hat{N}_{M}\circ\hat{f}_{M})\rightarrow\hat{C}^{\infty}_{A}(M)$,
which is $R[U]$-linear up to homotopy when the cone is given the adjusted
$R[U]$-module structure determined by the homotopy
$\hat{t}_{M}=\hat{N}_{M}\circ\hat{s}_{M}$. In terms of the explicit models of
the above theorem $\hat{N}_{M}\circ\hat{f}_{M}$ and
$\hat{N}_{M}\circ\hat{s}_{M}$ are precisely $\nu$ and $\psi$ as defined in the
statement. The verification of this is done just as in Lemma 5.24. The fact
that
$\operatorname{Cone}(\nu)\cong\operatorname{Tot}^{\Pi,\infty}(D^{\infty}_{*,*},\partial^{\prime},\partial^{\prime\prime})$
is proved just as in Theorem 5.28. ∎

######  Remark 5.49.

The proofs of Theorem 5.47 and Theorem 5.48 go through without any essential
change with $M=\widetilde{CI}(Y,E)$, regarded as a $\Lambda_{R}[u]$-module, in
place of $M=DCI(Y,E)$.

We may extract the following noteworthy consequences, showing in particular
that the framed homology groups $\widetilde{I}(Y,E)\coloneqq H(DCI(Y,E))\cong
H(\widetilde{CI}(Y,E))$ may be recovered from the $R[U]$-action on
$I^{\pm}(Y,E)$ up to extension.

######  Corollary 5.50.

There are long exact sequences

${\tilde{I}_{n}(Y,E)}$${I^{+}(Y,E)_{n}}$${I^{+}(Y,E)_{n-4}}$${\tilde{I}(Y,E)_{n-1}}$${\tilde{I}(Y,E)_{n-3}}$${I^{-}(Y,E)}$${I^{-}(Y,E)_{n-4}}$${\tilde{I}(Y,E)_{n-4}.}$$\scriptstyle{U}$$\scriptstyle{U}$

Moreover, $U:I^{\infty}(Y,E)\rightarrow I^{\infty}(Y,E)$ is an isomorphism.

###### Proof.

Let $M=DCI(Y,E)$ and $A=\Lambda_{R}[u]$. In terms of the explicit models for
$DCI^{\pm}(Y,E)=\hat{C}^{\pm}_{A}(M)$ the action of $U\in R[U]$ are in degree
$n$ given by the natural shifts

$U\colon\prod_{s\geq 0}M_{n-4s}\rightarrow\prod_{s\geq 0}M_{n-4-4s}\;\;\mbox{
and }\;\;U\colon\bigoplus_{s\geq 0}M_{n+4s}\rightarrow\bigoplus_{s\geq
0}M_{n-4+4s},$

respectively. Explicitly, $(x_{s})_{s\geq 0}\mapsto(x_{s+1})_{s\geq 0}$ in the
first case and $(y_{s})_{s\geq 0}\mapsto(0,y_{0},y_{1},\cdots)$ in the second.
From this we deduce that the first map is surjective with kernel
$D^{+}_{0,*}\cong M$ and the second is injective with cokernel isomorphic to
$D^{-}_{0,*}\cong M$. There are therefore short exact sequences of chain
complexes

${0}$${M}$${D^{+}_{*}}$${D^{+}_{*}[4]}$${0}$${0}$${D^{-}_{*}[-4]}$${D^{-}_{*}}$${M}$${0}$$\scriptstyle{U}$$\scriptstyle{U}$

giving rise to the stated long exact sequences in homology. The final
statement is a consequence of the fact that the shift

$U\colon
D^{\infty}_{n}=\prod_{s\to\infty}M_{n-4s}\rightarrow\prod_{s\to\infty}M_{n-4-4s}=D^{\infty}_{n-4}$

is an isomorphism in each degree $n$ and therefore also an isomorphism upon
passage to homology. ∎

######  Remark 5.51.

The fact that $U\colon I^{\infty}(Y,E)\rightarrow I^{\infty}(Y,E)$ is an
isomorphism when $\frac{1}{2}\in R$ was first proved using a spectral sequence
argument in [Eis19].

###  5.5. The Index Spectral Sequences

For $\bullet\in\\{+,-,(+,tw),\infty\\}$ the complexes $CI^{\bullet}(Y,E)$ and
$DCI^{\bullet}(Y,E)$ introduced in Definition 5.36 come equipped with
filtrations and therefore give rise to spectral sequences. These will be
called index spectral sequences as the filtrations are induced by the index
filtrations of $\widetilde{CI}(Y,E)$ and $DCI(Y,E)$. The purpose of this
section is to recall some theory on spectral sequences and then to establish
the basic properties of the index spectral sequences. We will mainly follow
Boardman’s paper [Boa99], but we will stick to homological notation. All of
this will be very useful for our calculations in the next section.

Recall that an (unrolled) exact couple $(A,E,i,j,k)$ consists of two sequences
of graded modules $(A_{s})_{s}$ and $(E_{s})_{s}$, indexed over
$s\in\mathbb{Z}$, and homomorphisms $i=(i_{s}:A_{s-1}\rightarrow A_{s})_{s}$,
$j=(j_{s}:A_{s}\rightarrow E_{s})_{s}$ and $k=(k_{s}:E_{s}\rightarrow
A_{s-1})_{s}$ such that the sequence

${A_{s-1}}$${A_{s}}$${E_{s}}$${A_{s-1}}$${A_{s}}$$\scriptstyle{i_{s}}$$\scriptstyle{j_{s}}$$\scriptstyle{k_{s}}$$\scriptstyle{i_{s}}$

is exact for each $s$. We will generally omit the lower index on the maps,
letting the source and target specify the map in question. In our situation
the degrees of $(i,j,k)$ will be $(0,0,-1)$. For each $r\geq 1$ and $s$ set

$\displaystyle Z^{r}_{s}$ $\displaystyle\coloneqq
k^{-1}\operatorname{Im}[i^{(r-1)}:A_{s-r}\rightarrow A_{s-1}]$ $\displaystyle
B^{r}_{s}$ $\displaystyle\coloneqq
j\operatorname{Ker}[i^{(r-1)}:A_{s}\rightarrow A_{s+r-1}]$ $\displaystyle
E^{r}_{s}$ $\displaystyle\coloneqq Z^{r}_{s}/B^{r}_{s},$

where $i^{(r)}$ denotes the $r$-fold composition $i\circ\cdots\circ i$ for
$r\geq 1$ and $i^{(0)}$ is the identity. Furthermore, put
$Z^{\infty}_{s}\coloneqq\bigcap_{r}Z^{r}_{s}$,
$B^{\infty}_{s}\coloneqq\bigcup_{r}B^{r}_{s}$ and $E^{\infty}_{s}\coloneqq
Z^{\infty}_{s}/B^{\infty}_{s}$. For the later convergence theory it is also
important to introduce the derived limit
$RE^{\infty}_{s}\coloneqq\operatorname{Rlim}_{s}Z^{r}_{s}$. There are
differentials $d^{r}\colon E^{r}_{s}\rightarrow E^{r}_{s-r}$ for each $s$ and
$r\geq 1$, defined in the following way. Let $x\in Z^{r}_{s}$ represent the
class $[x]\in E^{r}_{s}$. Choose $y\in A_{s-r}$ such that $i^{(r-1)}(y)=k(x)$.
Then $d^{r}([x])=[j(y)]\in E^{r}_{s-r}$. The situation is illustrated in the
following diagram

(5.15) ${y\in A_{s-r}}$${k(x)\in A_{s-1}}$${j(y)\in E_{s-r}}$${x\in
E_{s}.}$$\scriptstyle{i^{(r-1)}}$$\scriptstyle{j}$$\scriptstyle{k}$

One may then show that $d^{r}$ is well-defined, that $d^{r}\circ d^{r}=0$ and
that

$E^{r+1}_{s}\cong\operatorname{Ker}(d^{r}\colon E_{s}^{r}\rightarrow
E^{r}_{s-r})/\operatorname{Im}(d^{r}\colon E^{r}_{s+r}\rightarrow E^{r}_{s})$

for each $r\geq 1$ and $s$. In other words, $(E^{r},d^{r})_{r\geq 1}$ is a
spectral sequence. We call it the spectral sequence associated with the exact
couple $(A,E,i,j,k)$. Every exact couple we will meet is covered by the
following example (see [Boa99, Section 9]).

######  Example 5.52.

Let $C$ be a DG module equipped with an increasing filtration
$\\{F_{s}C\\}_{s}$ by DG submodules. For each $s\in\mathbb{Z}$ let
$A_{s}=H(F_{s}C)$ and $E_{s}=H(F_{s}C/F_{s-1}C)$. Define $i\colon
A_{s-1}\rightarrow A_{s}$, $j\colon A_{s}\rightarrow E_{s}$ and $k\colon
E_{s}\rightarrow A_{s-1}$ to be the maps induced by the short exact sequence

${0}$${F_{s-1}C}$${F_{s}C}$${F_{s}C/F_{s-1}C}$${0}$

upon passage to homology. In particular, $k$ is the connecting homomorphism of
degree $-1$. Then $(A,E,i,j,k)$ is an exact couple. The spectral sequence
associated with this exact couple is the spectral sequence mentioned in (5.9).
If $B$ is a DG algebra and $C$ is a $B$-module filtered by $B$-submodules,
then $A_{s}$, $E_{s}$ carry the structure of graded $H(B)$-modules, and the
maps $i,j,k$ are $H(B)$-linear. Moreover, this structure carries in a natural
way over to the whole spectral sequence.

Returning to the situation of an exact couple $(A,E,i,j,k)$ put

(5.16)
$A_{\infty}\coloneqq\operatorname{colim}_{s}A_{s},\;\;A_{-\infty}\coloneqq\operatorname{lim}_{s}A_{s}\;\;\mbox{
and }\;\;RA_{-\infty}\coloneqq\operatorname{Rlim}_{s}A_{s}.$

For us the relevant target for the spectral sequence is the colimit
$A_{\infty}$, filtered by
$F_{s}A_{\infty}\coloneqq\operatorname{Im}(A_{s}\rightarrow A_{\infty})$. For
each $s\in\mathbb{Z}$ there is a natural inclusion $\phi_{s}\colon
F_{s}A_{\infty}/F_{s-1}A_{\infty}\rightarrow E^{\infty}_{s}$ given by the
composition

(5.17)
${F_{s}A_{\infty}/F_{s-1}A_{\infty}}$${\operatorname{Im}j/B^{\infty}_{s}}$${Z^{\infty}_{s}/B^{\infty}_{s}=E^{\infty}_{s}.}$$\scriptstyle{\cong}$

Here, the first isomorphism is given by sending $[x]\in
F_{s}A_{\infty}/F_{s-1}A_{\infty}$, represented by $x\in F_{s}A_{\infty}$, to
$[j(y)]\in\operatorname{Im}j/B^{\infty}_{s}$, where $y\in A_{s}$ is a lift of
$x$ along the surjection $A_{s}\rightarrow F_{s}A_{\infty}$, and the second
map is induced by the inclusion $\operatorname{Im}j=\operatorname{Ker}k\subset
Z^{\infty}_{s}$ (see [Boa99, Lemma 5.6] and its proof).

######  Definition 5.53.

The spectral sequence $(E^{r},d^{r})_{r\geq 1}$ associated with the exact
couple $(A,E,i,j,k)$ converges strongly to the colimit $A_{\infty}$ if the
filtration $F_{s}A_{\infty}=\operatorname{Im}(A_{s}\rightarrow A_{\infty})$ of
$A_{\infty}$ is exhaustive and complete Hausdorff and the map $\phi_{s}\colon
F_{s}A_{\infty}/F_{s-1}A_{\infty}\rightarrow E^{\infty}_{s}$ is an
isomorphisms for each $s\in\mathbb{Z}$.

Strong convergence is the ideal form of convergence. In theory, if we can can
calculate $E^{\infty}$ we would know the subquotients
$F_{s}A_{\infty}/F_{s-1}A_{\infty}$ and provided we can solve the extension
problems to determine $F_{s}A_{\infty}/F_{t}A_{\infty}$ for each pair $s>t$,
we can recover the target group as

$A_{\infty}\cong\operatorname{lim}_{s}\operatorname{colim}_{t}F_{t}A_{\infty}/F_{s}A_{\infty}.$

######  Definition 5.54.

[Boa99, Definition 5.10] Let $(A,E,i,j,k)$ be an exact couple with associated
spectral sequence $(E^{r},d^{r})_{r\geq 1}$. If $RA_{-\infty}=0$ and
$A_{-\infty}=0$ (see (5.16)), then the spectral sequence is said to converge
conditionally to the colimit $A_{\infty}$

In our situation of a filtered DG module and the corresponding exact couple
described in Example 5.52 we have the following result.

######  Theorem 5.55.

[Boa99, Theorem 9.2] Let $C$ be a DG module equipped with an increasing
filtration $F_{s}C$ by DG submodules. Let $(A_{s},E_{s},i,j,k)$ be the
associated exact couple with corresponding spectral sequence
$(E^{r},d^{r})_{r\geq 1}$. If the filtration of $C$ is exhaustive and complete
Hausdorff, then the spectral sequence $(E^{r},d^{r})_{r\geq 1}$ converges
conditionally to the colimit $A_{\infty}=H(C)$.

In the presence of conditional convergence there are two remaining
obstructions to achieving strong convergence; namely, the vanishing of the
group $RE^{\infty}$ and another group $W$. We refer to [Boa99, Lemma 8.5] for
the general definition of $W$, but we note that if the exact couple is
constructed from a filtered complex, then $W$ is isomorphic to the kernel of
the canonical interchange map (see [HR19])

$\kappa\colon\operatorname{colim}_{p}\operatorname{lim}_{q}H(F_{p}C/F_{q}C)\rightarrow\operatorname{lim}_{q}\operatorname{colim}_{p}H(F_{p}C/F_{q}C).$

We then have the following convergence theorem.

######  Theorem 5.56.

[Boa99, Theorem 8.2] Let $(A,E,i,j,k)$ be an exact couple and assume that the
associated spectral sequence $(E^{r},d^{r})_{r\geq 1}$ converges conditionally
to the colimit $A_{\infty}$. Then if $W=0$ and $RE^{\infty}=0$ the spectral
converges strongly to the colimit $A_{\infty}$.

The following result gives a few standard criteria for the vanishing of
$RE^{\infty}$ and $W$ in the situation of a filtered complex.

######  Proposition 5.57.

Let $C$ be a filtered complex with associated exact couple $(A,E,i,j,k)$ and
spectral sequence $(E^{r},d^{r})_{r\geq 1}$.

  1. (i)

If the filtration of $C$ is degreewise bounded below, then $RE^{\infty}=0$ and
$W=0$.

  2. (ii)

If the filtration of $C$ is degreewise bounded above then $W=0$.

  3. (iii)

If the spectral sequence degenerates at some finite stage; that is,
$E^{r_{0}}=E^{r_{0}+1}=\cdots=E^{\infty}$ for some finite $r_{0}\geq 1$, then
$RE^{\infty}=0$ and $W=0$.

We may now apply this theory to the filtered complexes $CI^{\bullet}(Y,E)$ for
$\bullet\in\\{+,-,(+,tw),\infty\\}$. This is a good point to explain that the
mod $8$ periodicity of $\widetilde{CI}(Y,E)$ carries over to the complexes
$CI^{\bullet}(Y,E)$. Let $M=\widetilde{CI}(Y,E)$ and let
$A=C^{gm}_{*}(\operatorname{SO}(3))$ or $A=\Lambda_{R}[u]$. Then the
periodicity may be expressed as an isomorphism $M\cong M[8]$ of $A$-modules.
Moreover, this isomorphism is compatible with the filtration in the sense that
$F_{p+8}M\cong(F_{p}M)[8]$ for all $p\in\mathbb{Z}$. There is therefore an
induced isomorphism

$\displaystyle CI^{\bullet}(Y,E)$
$\displaystyle=\;\operatorname{lim}_{q}\operatorname{colim}_{p}C^{\bullet}_{A}(F_{p+8}M/F_{q+8}M)$
$\displaystyle\cong\;\operatorname{lim}_{q}\operatorname{colim}_{p}C^{\bullet}_{A}(F_{p}M/F_{q}M)[8]=CI^{\bullet}(Y,E)[8].$

This isomorphism is seen to be filtered in the same sense:
$F_{p}CI^{\bullet}(Y,E)_{n}\cong F_{p+8}CI^{\bullet}(Y,E)_{n+8}$ for all $p$
and $n$.

######  Theorem 5.58.

Let $A=C_{*}^{gm}(\operatorname{SO}(3))$. There is a conditionally convergent
spectral sequence of $H^{-}_{A}(R)$-modules

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv
s}H^{\bullet}_{\operatorname{SO}(3)}(\alpha)_{t}\implies
I^{\bullet}(Y,E)_{s+t}$

for each $\bullet\in\\{+,-,(+,tw),\infty\\}$, where

$H^{\bullet}_{\operatorname{SO}(3)}(\alpha)\coloneqq
H^{\bullet}_{A}(C^{gm}_{*}(\alpha))$

for each critical orbit $\alpha\in\mathcal{C}$. Moreover, the spectral
sequences are periodic in the sense that there are isomorphisms
$E_{s,t}^{r}\cong E_{s+8,t}^{r}$ for all $s,t,r$. These isomorphisms commute
with the differentials and the $H^{-}_{A}(R)$-action and they are compatible
with the target in the sense that

${\overline{F}_{s}I^{\bullet}_{s+t}}$${\overline{F}_{s+8}I^{\bullet}_{s+t+8}}$${E^{\infty}_{s,t}}$${E^{\infty}_{s+8,t}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$

commutes for all $s,t$.

The spectral sequence for $I^{-}(Y,E)$ converges strongly, while the spectral
sequence for $I^{+}(Y,E)$ converges strongly provided $RE^{\infty}=0$.

###### Proof.

Let $M=\widetilde{CI}(Y,E)$ equipped with the index filtration. For each
$\bullet\in\\{+,-,(+,tw),\infty\\}$ the complex
$CI^{\bullet}(Y,E)=\hat{C}^{\bullet}_{A}(M)$ carries by construction an
exhaustive and complete Hausdorff filtration with minimal filtration quotients

$\overline{F}_{p}\hat{C}^{\bullet}_{A}(M)\cong
C^{\bullet}_{A}(\overline{F}_{p}M).$

There is therefore by Theorem 5.55 a conditionally convergent spectral
sequence

$E^{1}_{s,t}=H^{\bullet}_{A}(\overline{F}_{s}M)_{s+t}\implies\hat{H}^{\bullet}_{A}(M)_{s+t}=I^{\bullet}(Y,E)_{s+t}.$

From the definition of $\widetilde{CI}(Y,E)$ (Definition 2.4) we may identify

$\overline{F}_{s}M_{s+t}=\bigoplus_{j(\alpha)\equiv s}C^{gm}_{t}(\alpha)$

so that $E^{1}_{s,t}\cong\bigoplus_{j(\alpha)\equiv
s}H_{\operatorname{SO}(3)}(\alpha)_{t}$ in the notation of the statement.

Next, the periodicity isomorphisms $F_{p}\hat{C}^{\bullet}_{A}(M)_{n}\cong
F_{p+8}\hat{C}^{\bullet}_{A}(M)_{n+8}$, compatible with the differentials and
the $C^{-}_{A}(R)$-structure, induce a morphism of the associated exact
couple. That is, there are isomorphisms $A_{s}\cong A_{s+8}$, $E_{s}\cong
E_{s+8}$ each $s$, compatible with the structure maps. It is then a
straightforward exercise to verify that this gives rise to the periodicity in
the spectral sequence, compatible with the target, as stated.

Finally, for the convergence we know by Lemma 5.45 that the filtration of
$\hat{C}^{+}_{A}(M)$ is degreewise bounded above and that the filtration of
$\hat{C}^{-}_{A}(M)$ is degreewise bounded below. This implies by Proposition
5.57 and Theorem 5.56 that the spectral sequence associated with $I^{-}(Y,E)$
converges strongly and that the spectral sequence associated with $I^{+}(Y,E)$
converges strongly provided $RE^{\infty}=0$. ∎

Provided $\frac{1}{2}\in R$, these spectral sequences can equally well be
constructed from $M=DCI(Y,E)$ equipped with the index filtration and
$A=\Lambda_{R}[u]$. This is what we will assume when we calculate these groups
for binary polyhedral spaces in the next section. Furthermore, in this setting
the calculation of the groups

$H^{\bullet}_{\operatorname{SO}(3)}(C^{gm}_{*}(\alpha))\cong
H^{\bullet}_{\Lambda_{R}[u]}(H_{*}(\alpha))$

for each $\bullet\in\\{+,-,(+,tw),\infty\\}$, are contained in Lemma 5.7,
Corollary 5.9 and Corollary 5.30. From these calculations it becomes clear
that the spectral sequences for $I^{+}$ and $I^{-}$ are contained in the upper
and lower half-planes respectively, while the spectral sequence for
$I^{\infty}$ is a whole-plane spectral sequence.

##  6\. Calculations for Binary Polyhedral Spaces

We will from this point require that $2\in R$ is invertible. The universal
example to have in mind is $R=\mathbb{Z}[\frac{1}{2}]$. Our main aim is to
give explicit calculations of the $H^{-}_{\Lambda_{R}[u]}(R)=R[U]$-modules

$I^{+}(\overline{Y}_{\Gamma},E),\;\;I^{-}(\overline{Y}_{\Gamma},E)\;\;\mbox{
and }\;\;I^{\infty}(\overline{Y}_{\Gamma},E)$

for each finite subgroup $\Gamma\subset\operatorname{SU}(2)$, where
$E\rightarrow\overline{Y}_{\Gamma}$ is the trivial
$\operatorname{SU}(2)$-bundle. To simplify the notation slightly we will omit
the reference to the bundle in the notation from this point. In the final part
we also include the calculations of $I^{\bullet}(Y_{\Gamma})$ for
$\bullet\in\\{+,-,\infty\\}$, omitting some details.

The two results Theorem 4.21 and Proposition 4.25 give a concrete description
of the Donaldson model $\operatorname{DCI}(\overline{Y}_{\Gamma})$ in all
cases. The key tools needed in the calculations are the index spectral
sequence in combination with the explicit models for $DCI^{\pm}$ of Theorem
5.8. In the $I^{+}$ case, we will see that the index spectral sequence
immediately degenerates, and we are left with solving an extension problem in
the category of $R[U]$-modules. In the $I^{-}$ case there are a number of
differentials in the index spectral sequence, but we will in each case verify
that the spectral sequence degenerates on some finite page $E^{r_{0}}$. The
extension problems in this case are neatly solved by observing that
$E^{\infty}_{s}$ is a free $R[U]$-module for all $s$.

In [Eis19, Corollary 8.7] Miller calculates $I^{\infty}(Y)$ for all rational
homology spheres. This, of course, includes $\overline{Y}_{\Gamma}$, so our
calculation of this group should only be regarded as a verification of his
result. In addition, we observe that the norm map vanishes in homology so that
the exact triangle relating the three groups splits up into short exact
sequences.

Throughout this section we let $A=\Lambda_{R}[u]$ with $|u|=3$. Given a finite
subgroup $\Gamma\subset\operatorname{SU}(2)$ we will always write
$\mathcal{C}$ for the set of critical orbits, or equivalently, the set of
$1$-dimensional quaternionic representations of $\Gamma$. Recall that for each
$\alpha\in\mathcal{C}$ we have fixed generators $b_{\alpha}\in H_{0}(\alpha)$,
$t_{\alpha}=b_{\alpha}\cdot u\in H_{3}(\alpha)$ if $\alpha$ is irreducible and
$t_{\alpha}\in H_{2}(\alpha)$ if $\alpha$ is reducible.

###  6.1. The Case $I^{+}$

In this section we will calculate $I^{+}(\overline{Y}_{\Gamma})$ as an
$R[U]$-module for all finite subgroups $\Gamma\subset\operatorname{SU}(2)$.
Even though our arguments do not explicitly rely on the index spectral
sequence, we note that the motivation for the upcoming arguments is the
following observation.

######  Lemma 6.1.

Let $M=DCI(\overline{Y}_{\Gamma})$ for a finite subgroup
$\Gamma\subset\operatorname{SU}(2)$. Then the index spectral sequence

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H_{A}^{+}(\alpha)_{t}\implies
I^{+}(\overline{Y}_{\Gamma})_{s+t}$

degenerates at the $E^{1}$ page, that is, all differentials are trivial and
$E^{1}=E^{\infty}$.

###### Proof.

From the calculations in Lemma 5.7 and Corollary 5.9 we see that
$H^{+}_{A}(\alpha)$ vanishes in odd degrees for all types of orbits $\alpha$.
Then, as $j(\alpha)\equiv 0\ (\mathrm{mod}\ 4)$ for all
$\alpha\in\mathcal{C}$, it follows that $E^{1}_{s,t}=0$ if $s$ or $t$ is odd.
Since the differential $d^{r}$ has bidegree $(-r,r-1)$ we conclude that
$d^{r}=0$ for all $r\geq 1$ as required. ∎

From this one may determine $I^{+}(\overline{Y}_{\Gamma})$ as an $R$-module.
However, the structure as an $R[U]$-module is more subtle and this is the
reason we spell out our argument more directly in terms of the filtration of
the chain complex $DCI^{+}(\overline{Y}_{\Gamma})$.

Let $\Gamma\subset\operatorname{SU}(2)$ be a finite subgroup and let
$M=DCI(\overline{Y}_{\Gamma})$. Let $\mathcal{C}^{irr}\subset\mathcal{C}$
denote the subset of free critical orbits or equivalently irreducible flat
connections. Define $M^{irr}\subset M$ to be the submodule generated by
$\mathcal{C}^{irr}$, i.e.,

$M^{irr}_{s,t}=\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{C}^{irr}\\\
j(\alpha)\equiv s\end{subarray}}H_{t}(\alpha)\;\;\mbox{ and
}\;\;M^{irr}_{n}=\bigoplus_{s+t=n}M^{irr}_{s,t}.$

The following lemma is the key tool in resolving our extension problems in the
category of $R[U]$-modules.

######  Lemma 6.2.

There is an $R$-linear map $\psi\colon M\rightarrow M$ of degree $-4$
satisfying

  1. (i)

$\psi(x)\cdot u=(-1)^{|x|}\partial_{M}x$ for all $x\in M$,

  2. (ii)

$\operatorname{Im}(\psi)\subset M^{irr}$,

  3. (iii)

$\psi(\operatorname{Im}(u))=0$ and

  4. (iv)

$\psi(\operatorname{Im}\partial_{M})=0$.

###### Proof.

By Theorem 4.21 the only possibly nontrivial differentials in $M$ are
$\partial_{M}\colon M_{4s}\rightarrow M_{4s-1}$ for $s\in\mathbb{Z}$. The same
result also shows that $u\colon M_{4s-4}^{irr}\rightarrow
M_{4s-1}^{irr}=M_{4s-1}$ is an isomorphism for each $s$. We may therefore
define $\psi\colon M_{4s}\rightarrow M_{4s-4}^{irr}\subset M_{4s-4}$ by
$u^{-1}\circ\partial_{M}$ and set $\psi=0\colon M_{t}\rightarrow M_{t-4}$ in
all other degrees. The relation $\psi(x)\cdot u=(-1)^{|x|}\partial_{M}x$ is
then satisfied for all $x\in M$ and by construction
$\operatorname{Im}(\psi)\subset M^{irr}$. For (iii) and (iv) it is then
sufficient to note that $\operatorname{Im}(u)$ and
$\operatorname{Im}(\partial_{M})\subset M^{irr}$ are supported in the degrees
$M_{4s-1}$ for $s\in\mathbb{Z}$. ∎

######  Remark 6.3.

It is not difficult to show that the conditions (i) and (ii) determine the map
$\psi$ uniquely, but we will not use this explicitly.

Recall from Corollary 5.47 that for $M=DCI(\overline{Y}_{\Gamma})$ there is an
identification $DCI^{+}(\overline{Y}_{\Gamma})_{n}\cong\prod_{p\geq
0}M_{n-4p}$ for each $n$. Moreover, the differential is given by

$\partial(m_{n-4p})_{p\geq
0}=(\partial_{M}m_{n-4p}-(-1)^{n}m_{n-4(p+1)}u)_{p\geq 0}.$

The point of the map $\psi$ should now be clear; given $m\in M_{n}$ the
element $(\psi^{p}(x))_{p\geq 0}\in\prod_{p\geq 0}M_{n-4p}$ gives an explicit
extension of $m$ to a cycle in $DCI^{+}(Y)$.

Using the description of the differentials in Proposition 4.21 we note that
$\psi$ is given on a generator $b_{\alpha}\in H_{0}(\alpha)\subset M_{4p}$ by

(6.1)
$\psi(b_{\alpha})=u^{-1}\left(\sum_{\beta\in\mathcal{C}^{irr}}n_{\beta\alpha}t_{\beta}\right)=\sum_{\beta\in\mathcal{C}^{irr}}n_{\beta\alpha}b_{\beta}$

(the integers $n_{\beta\alpha}$ were defined in Definition 4.20).

Recall from Lemma 5.45 that

$DCI^{+}(\overline{Y}_{\Gamma})=\hat{C}_{A}^{+}(M)=\operatorname{lim}_{p}C^{+}_{A}(M/F_{p}M),$

filtered by
$F_{p}\hat{C}_{A}^{+}(M)=\hat{C}_{A}^{+}(F_{p}M)=\operatorname{lim}_{q<p}C^{+}_{A}(F_{p}M/F_{q}M)$.
The concrete model given in Corollary 5.47 applies equally well to these
subcomplexes. Recall that we use the notation
$\overline{F}_{p}M=F_{p}M/F_{p-1}M$.

######  Lemma 6.4.

Let $M=DCI(\overline{Y}_{\Gamma})$ be equipped with the index filtration. Then
$F_{4p}M=F_{4p+t}M$ for all $p\in\mathbb{Z}$ and $0\leq t\leq 3$. Moreover,
there are identifications for $s\in\mathbb{Z}$ and $0\leq t\leq 3$

(6.4) $\displaystyle C^{+}_{A}(\overline{F}_{4p}M)_{4s+t}$
$\displaystyle\cong\left\\{\begin{array}[]{cc}M_{4p+t}&\mbox{ for }s\geq p\\\
0&\mbox{ for }s<p\end{array}\right.$ (6.7)
$\displaystyle\hat{C}^{+}_{A}(F_{4p}M)_{4s+t}$
$\displaystyle\cong\left\\{\begin{array}[]{cc}\prod_{q\geq
0}M_{4(p-q)+t}&\mbox{ for }s\geq p\\\ \prod_{q\geq 0}M_{4(s-q)+t}&\mbox{ for
}s<p.\end{array}\right.$

Define $\zeta\colon
C^{+}_{A}(\overline{F}_{4p}M)\rightarrow\hat{C}^{+}_{A}(F_{4p}M)$ for $x\in
M_{4p+t}\cong C^{+}(\overline{F}_{4p}M)_{4s+t}$, $s\geq p$, by

$\zeta(x)=(\psi^{q}(x))_{q\geq 0}\in\prod_{q\geq
0}M_{4(p-q)+t}=\hat{C}^{+}_{A}(F_{4p}M),$

where $\psi$ is the map of Lemma 6.2, and otherwise to be zero. Then $\zeta$
is a chain map splitting the exact sequence of DG $R$-modules (i.e.,
$\pi\circ\zeta=1$)

${0}$${\hat{C}^{+}_{A}(F_{4p-1}M)}$${\hat{C}^{+}_{A}(F_{4p}M)}$${C^{+}_{A}(\overline{F}_{4p}M)}$${0.}$$\scriptstyle{i}$$\scriptstyle{\pi}$$\scriptstyle{\zeta}$

###### Proof.

The fact that $F_{4p+t}M=F_{4p}M$ for $0\leq t\leq 3$ follows from the
structure theorem 4.21. From the same result one concludes that
$F_{4p}M_{n}=M_{n}$ for $n\leq 4p+3$ and $F_{4p}M_{n}=0$ for $n\geq 4(p+1)$.
Therefore, $\overline{F}_{4p}M_{n}$ coincides with $M_{n}$ for $4p\leq n\leq
4p+3$ and vanishes otherwise. Using this, the given degreewise formulas for
$C^{+}_{A}(\overline{F}_{4p}M)$ and $\hat{C}^{+}_{A}(F_{4p}M)$ follow from
Proposition 5.8 and Corollary 5.47, respectively.

Next we have to verify that $\zeta$ is a chain map. The differential in
$C^{+}_{A}(\overline{F}_{4p}M)$ is given by $u\colon M_{4p}\rightarrow
M_{4p+3}$ in degree $4s$ for $s>p$ and is otherwise $0$. The differential in
$\hat{C}^{+}_{A}(F_{4p}M)$ is given in degree $4s+t$ by

$\partial((m_{4(s-q)+t})_{q\geq
0})=(\partial_{M}m_{4(s-q)+t}-(-1)^{t}m_{4(s-q-1)+t}u)_{q\geq 0},$

where if $s>p$ it is assumed that $m_{4(s-q)+t}=0$ for $q\leq s-p$. In view of
this formula and the fact that $\psi(x)\cdot u=(-1)^{|x|}\partial_{M}x$ for
all $x\in M$ by Lemma 6.2, we see that for $x\in
M_{4p+t}=C^{+}_{A}(\overline{F}_{4p}M)_{4s+t}$, $s\geq p$,

$\partial\circ\zeta(x)=\partial(\psi^{q}x)_{q\geq
0}=\left\\{\begin{array}[]{cl}0&\mbox{ if }(s,t)=(p,0)\mbox{ or }1\leq t\leq
3\\\ (x\cdot u,0,0,\cdots)&\mbox{ if }s>p,t=0\end{array}\right.$

In the first case we also have $\partial x=0$ and hence
$\zeta\circ\partial(x)=0$, while in the second case we have $\partial x=x\cdot
u$ so that $\zeta\circ\partial(x)=(\psi^{q}(x\cdot u))_{q\geq 0}=(x\cdot
u,0,0,\cdots)$ as $\psi$ vanishes on $\operatorname{Im}u$ by Lemma 6.2. In all
other degrees $\zeta$ vanishes, so we may conclude from the above that $\zeta$
is a chain map.

Finally, the projection $\pi\colon\hat{C}^{+}_{A}(F_{4p}M)\rightarrow
C^{+}_{A}(\overline{F}_{4p}M)$ is in terms of the formulas of (6.4) given by
the projection $\prod_{q\geq 0}M_{4(p-q)+t}\rightarrow M_{4p+t}$ in degree
$4s+t$ for $s\geq p$ and otherwise $0$. From this description it is clear that
$\pi\circ\zeta=1$ and the proof is complete. ∎

We need one final lemma describing the interaction between the map $\zeta$ and
the action of $R[U]$.

######  Lemma 6.5.

In the situation of the above lemma the map $\zeta\colon
C^{+}_{A}(\overline{F}_{4p}M)\rightarrow\hat{C}^{+}_{A}(F_{4p}M)$ satisfies
$\zeta(Ux)=U\zeta(x)$ for each $x$ with $|x|\neq 4p$. Moreover, in degree $4p$
we have the following commutative diagram

${C^{+}_{A}(\overline{F}_{4p}M)_{4p}}$${\hat{C}^{+}_{A}(F_{4p}M)_{4p}}$${\hat{C}^{+}_{A}(F_{4p}M)_{4(p-1)}}$${C^{+}_{A}(\overline{F}_{4(p-1)}M)_{4(p-1)}}$${\hat{C}^{+}_{A}(F_{4(p-1)}M)_{4(p-1)}.}$$\scriptstyle{\zeta}$$\scriptstyle{\psi}$$\scriptstyle{U}$$\scriptstyle{=}$$\scriptstyle{\zeta}$

###### Proof.

In degree $n$ for $n<4p$ the map $\zeta$ vanishes, so the first assertion is
clear in this case. In degree $4s+t$ with $s>p$ and $0\leq t\leq 3$ we have,
using the formulas of Lemma 6.4,

$C^{+}_{A}(\overline{F}_{4p}M)_{4s+t}=M_{4p+t}\;\;\mbox{ and
}\;\;\hat{C}^{+}_{A}(F_{4p}M)_{4s+t}=\prod_{q\geq 0}M_{4(p-q)+t}.$

It then follows from Proposition 5.8 and Corollary 5.47 that the $U$ action is
given by the identity in these degrees. Therefore, $U\circ\zeta=\zeta\circ U$
in this case as well.

For $x\in M_{4p+t}=C^{+}_{A}(\overline{F}_{4p}M)_{4p+t}$, where $0\leq t\leq
3$, we have $Ux=0$, while

$U(\zeta(x))=U((\psi^{q}x)_{q\geq 0})=(\psi^{q+1}x)_{q\geq 0}=(\psi^{q}(\psi
x))_{q\geq 0}=\zeta(\psi(x)),$

where we regard $\psi x\in
M_{4(p-1)+t}=C^{+}_{A}(\overline{F}_{4(p-1)}M)_{4(p-1)+t}$. This shows that
the given diagram commutes in degree $4p+t$ for $0\leq t\leq 3$. However, in
the proof of Lemma 6.2 it was shown that $\psi\colon M_{4p+t}\rightarrow
M_{4(p-1)+t}$ vanishes for $1\leq t\leq 3$. Therefore $\zeta\circ
U=U\circ\zeta$ in degree $4p+t$ for $1\leq t\leq 3$ as well. This completes
the proof. ∎

To describe $I^{+}(\overline{Y}_{\Gamma})$ and later
$I^{-}(\overline{Y}_{\Gamma})$, it is convenient to introduce the following
definition.

######  Definition 6.6.

For an $R[U]$-module $X$ define the mod $8$ periodic $R[U]$-modules
$X^{\Pi,8}$ and $X^{\oplus,8}$ degreewise by

$X^{\Pi,8}_{n}=\prod_{s\in\mathbb{Z}}X_{n+8s}\;\;\mbox{ and
}\;\;X^{\oplus,8}_{n}=\bigoplus_{s\in\mathbb{Z}}X_{n+8s}.$

The maps $U\colon X^{\Pi,8}_{n}\rightarrow X^{\Pi,8}_{n-4}$ and $U\colon
X^{\oplus,8}_{n}\rightarrow X^{\oplus,8}_{n-4}$ are defined to be the product
and direct sum over the maps $U:X_{n+8s}\rightarrow X_{n-4+8s}$ for
$s\in\mathbb{Z}$, respectively.

Let $\Gamma\subset\operatorname{SU}(2)$ be a finite subgroup. Given
$\alpha\in\mathcal{C}$ we will use the shorthand notation
$H^{+}_{A}(\alpha)\coloneqq H^{+}_{A}(H_{*}(\alpha))$. Write
$\mathcal{C}=\mathcal{C}^{irr}\cup\mathcal{C}^{red}\cup\mathcal{C}^{f.red}$
for the decomposition of the critical orbits into irreducible, reducibles and
fully reducibles. For each $\eta\in\mathcal{C}^{f.red}$,
$\lambda\in\mathcal{C}^{red}$ and $\alpha\in\mathcal{C}^{irr}$ introduce
variables $V_{\eta}$, $W_{\lambda}$ and a generator $g_{\alpha}$ so that

$H^{+}_{A}(\eta)=R[V_{\eta}],\;\;\;\;H^{+}_{A}(\lambda)=R[W_{\lambda}]\;\;\mbox{
and }\;\;H^{+}_{A}(\alpha)=R\cdot g_{\alpha},$

where $|V_{\eta}|=4$, $|W_{\lambda}|=2$ and $|g_{\alpha}|=0$. This is
justified by the calculations of Lemma 5.7 and Corollary 5.9. In the following
we regard the relative grading $j:\mathcal{C}\rightarrow\mathbb{Z}/8$ as
taking values in $\\{0,4\\}$.

######  Theorem 6.7.

The positive equivariant instanton Floer homology
$I^{+}(\overline{Y}_{\Gamma})$ associated with the trivial
$\operatorname{SU}(2)$-bundle over $\overline{Y}_{\Gamma}$ is given by

$\left[\left(\bigoplus_{\alpha\in\mathcal{C}^{irr}}R\cdot
g_{\alpha}[j(\alpha)]\right)\oplus\left(\bigoplus_{\lambda\in\mathcal{C}^{red}}R[W_{\lambda}][j(\lambda)]\right)\oplus\left(\bigoplus_{\eta\in\mathcal{C}^{f.red}}R[V_{\eta}][j(\eta)]\right)\right]^{\Pi,8}.$

The $R[U]$-module structure is determined by

$\displaystyle U\cdot V^{p}_{\eta}$
$\displaystyle=\left\\{\begin{array}[]{cc}V^{p-1}_{\eta}&\mbox{ for }p>0\\\
\sum_{\rho\in C^{irr}}n_{\rho\eta}g_{\rho}&\mbox{ for }p=0\end{array}\right.$
$\displaystyle U\cdot W_{\lambda}^{p}$
$\displaystyle=\left\\{\begin{array}[]{cc}W^{p-2}_{\lambda}&\mbox{ for }p>0\\\
\sum_{\rho\in C^{irr}}n_{\rho\lambda}g_{\rho}&\mbox{ for
}p=0\end{array}\right.$ $\displaystyle U\cdot g_{\alpha}$
$\displaystyle=\sum_{\rho\in C^{irr}}n_{\rho\alpha}g_{\rho},$

where we interpret $W_{\beta}^{-1}=0$ and the integers $n_{\rho_{1}\rho_{2}}$
are given in Proposition 4.25 for each pair
$(\rho_{1},\rho_{2})\in\mathcal{C}$.

###### Proof.

Let as usual $M=DCI(\overline{Y}_{\Gamma})$ be equipped with the index
filtration. By Lemma 6.4 we have $F_{4p}M=F_{4p+t}M$ for $0\leq t\leq 3$, and
the map $\zeta\colon
C^{+}_{A}(\overline{F}_{4p}M)\rightarrow\hat{C}^{+}_{A}(F_{4p}M)$ along with
the inclusion
$i\colon\hat{C}^{+}_{A}(F_{4(p-1)}M)\rightarrow\hat{C}^{+}_{A}(F_{4p}M)$
define a chain isomorphism

$i+\zeta:\hat{C}^{+}_{A}(F_{4(p-1)}M)\oplus
C^{+}_{A}(\overline{F}_{4p}M)\cong\hat{C}^{+}_{A}(F_{4p}M)$

of $DG$ $R$-modules for each $p$. By induction we obtain isomorphisms

$\hat{C}^{+}_{A}(F_{4p}M)\cong\hat{C}^{+}_{A}(F_{4(p-r-1)}M)\oplus\left(\bigoplus_{q=0}^{r}C^{+}_{A}(\overline{F}_{4(p-q)}M)\right)$

for each $p$ and $r\geq 1$. Since the filtration
$F_{p}\hat{C}^{+}_{A}(M)=\hat{C}^{+}_{A}(F_{p}M)$ is degreewise bounded above
and complete Hausdorff, it follows by first passing to the limit over $r$ and
then to the colimit over $p$ that we obtain an isomorphism of DG $R$-modules

$\hat{C}^{+}_{A}(M)\cong\prod_{p\in\mathbb{Z}}C^{+}_{A}(\overline{F}_{4p}M).$

Hence, as products commutes with homology,

(6.8)
$I^{+}(\overline{Y}_{\Gamma})=\hat{H}^{+}_{A}(M)\cong\prod_{p\in\mathbb{Z}}H^{+}_{A}(\overline{F}_{4p}M).$

Write $\mathcal{C}=\mathcal{C}^{0}\cup\mathcal{C}^{1}$ where
$\mathcal{C}^{i}=\\{\alpha\in\mathcal{C}:j(\alpha)\equiv 4i\ (\mathrm{mod}\
8)\\}$ for $i=0,1$. Then

$H^{+}_{A}(\overline{F}_{4p}M)=\left\\{\begin{array}[]{cc}\bigoplus_{\alpha\in\mathcal{C}^{0}}H^{+}_{A}(\alpha)[4p]&\mbox{
if }p\equiv 0\ (\mathrm{mod}\ 2)\\\
\bigoplus_{\alpha\in\mathcal{C}^{1}}H^{+}_{A}(\alpha)[4p]&\mbox{ if }p\equiv
1\ (\mathrm{mod}\ 2),\end{array}\right.$

and we may simplify the expression in (6.8)

(6.9)
$\prod_{p\in\mathbb{Z}}H^{+}_{A}(\overline{F}_{4p}M)=\prod_{p\in\mathbb{Z}}\bigoplus_{\alpha\in\mathcal{C}}H^{+}_{A}(\alpha)[8p+j(\alpha)]=\left(\bigoplus_{\alpha\in\mathcal{C}}H^{+}_{A}(\alpha)[j(\alpha)]\right)^{\Pi,8}.$

By replacing each $H^{+}_{A}(\alpha)$ with $R[V_{\alpha}]$, $R[W_{\alpha}]$ or
$R\cdot g_{\alpha}$ according to whether $\alpha$ is fully reducible,
reducible or irreducible we obtain the additive statement of the theorem.

To finish the proof we need to determine the action of $U$. Returning to the
notation of equation (6.8), for a fixed degree $n$, we have
$I^{+}(\overline{Y}_{\Gamma})_{n}=\hat{H}^{+}_{A}(F_{4p}M)_{n}$ for each $p$
with $4p>n$. This is a consequence of the formula (6.4) given in Lemma 6.4.
Fix the minimal $p$ with $4p>n$ and consider the commutative diagram

${I^{+}(\overline{Y}_{\Gamma})_{n}}$${\hat{H}^{+}_{A}(F_{4p}M)_{n}}$${\prod_{q:4q\leq
n}H^{+}_{A}(\overline{F}_{4q}M)_{n}}$${I^{+}(\overline{Y}_{\Gamma})_{n-4}}$${\hat{H}^{+}_{A}(F_{4p}M)_{n-4}}$${\prod_{q:4q\leq
n}H^{+}_{A}(\overline{F}_{4(q-1)}M)_{n-4}.}$$\scriptstyle{U}$$\scriptstyle{U}$$\scriptstyle{\cong}$$\scriptstyle{U^{\prime}}$$\scriptstyle{\phi}$$\scriptstyle{\cong}$$\scriptstyle{\phi}$

Here, $\phi$ is the isomorphism obtained by taking the limit, as $j\to\infty$,
over the maps

$([x_{q}])_{j\leq q\leq n/4}\mapsto\sum_{j\leq q\leq n/4}[\zeta(x_{q})],$

where $x_{q}\in C^{+}_{A}(\overline{F}_{4q}M)$ represents $[x_{q}]\in
H^{+}_{A}(\overline{F}_{4q}M)$ and $U^{\prime}$ is the map that forces the
right rectangle to commute. Our task is to determine $U^{\prime}$. By Lemma
6.5 we have for $y\in H^{+}_{A}(\overline{F}_{4q}M)_{n}$ that $H(\zeta)\circ
U(y)=U\circ H(\zeta)(y)$ provided $n>4q$, while if $n=4q$, and $y=[x]$ then
$U\circ H(\zeta)(y)=[\psi x]$. It follows that $U^{\prime}=\prod U$ if $n$ is
not divisible by $4$. Otherwise, if $4|n$ so that $n=4(p-1)$, then
$U^{\prime}=\tau+\prod U$, where $\tau=([x_{q}])_{q\leq p-1}=[\psi(x_{p-1})]$.
Therefore, in the description

$I^{+}(\overline{Y}_{\Gamma})=\prod_{p\in\mathbb{Z}}\bigoplus_{\alpha\in\mathcal{C}}H^{+}_{A}(\alpha)[8p+j(\alpha)]$

of (6.9), the action of $U$ is given by taking the product over the internal
$R[U]$-module structure of each factor $H^{+}_{A}(\alpha)$ and adding the
correction term $\tau$. For each $p$, $\tau$ only affects the terms
$H^{+}_{A}(\alpha)[8p+j(\alpha)]_{8p+j(\alpha)}=H^{+}_{A}(\alpha)_{0}$. In
terms of our generators this correspond to $V^{0}_{\alpha}$ if $\alpha$ is
fully reducible, $W_{\alpha}^{0}$ if $\alpha$ is reducible and $g_{\alpha}$ if
$\alpha$ is irreducible. Now, we may require that these generators correspond
to $b_{\alpha}\in H_{0}(\alpha)\subset M_{4q}\cong
C^{+}_{A}(\overline{F}_{4q}M)=H^{+}_{A}(\overline{F}_{4q}M)_{4q}$ under the
identifications of Lemma 6.4. Therefore, using the explicit formula (6.1) for
$\psi$, we obtain

$U\cdot V^{0}_{\eta}=\sum_{\rho\in\mathcal{C}^{irr}}n_{\rho\eta}g_{\rho}$

and similarly for the other types of generators. This completes the proof. ∎

######  Example 6.8.

For $Y=\overline{Y}_{O^{*}}$ there are two irreducibles $\alpha$, $\beta$, the
trivial connection $\theta$ and a fully reducible $\eta$. The grading is given
by $j(\theta)=j(\beta)=0$ and $j(\alpha)=j(\eta)=4$. Furthermore,
$n_{\alpha\beta}=n_{\beta\alpha}=3$ and $n_{\alpha\theta}=n_{\beta\eta}=1$.
This is obtained from Proposition 4.25. The above theorem gives

$I^{+}(Y)=\left(R[V_{\theta}]\oplus R\cdot g_{\beta}\oplus
R[V_{\eta}][4]\oplus R\cdot g_{\alpha}[4]\right)^{\Pi,8}.$

This means that

$I^{+}(Y)_{n}=\left\\{\begin{array}[]{cc}R\\{g_{\beta}\\}\oplus\prod_{i\geq
0}R\\{V_{\theta}^{2i},V_{\eta}^{2i-1}\\}&n\equiv 0\ (\mathrm{mod}\ 8)\\\
R\\{g_{\alpha}\\}\oplus\prod_{i\geq
0}R\\{V_{\theta}^{2i+1},V_{\eta}^{2i}\\}&n\equiv 4\ (\mathrm{mod}\ 8)\\\
0&\mbox{ otherwise. }\end{array}\right.$

For $x=(r_{2i}V^{2i}_{\eta})_{i\geq 0}\in I^{+}(Y)_{4}$, $r_{2i}\in R$, we
have $U\cdot x=r_{0}g_{\alpha}+(r_{2i}V^{2i-1}_{\eta})_{i\geq 1}\in
I^{+}(Y)_{0}$. For $x=rg_{\alpha}$, $r\in R$, we have $U\cdot x=3rg_{\beta}$.

###  6.2. The Case $I^{-}$

In this section we will calculate $I^{-}(\overline{Y}_{\Gamma})$ for all
finite subgroups $\Gamma\subset\operatorname{SU}(2)$. In contrast to the
calculations in the previous section, there will be a number of nontrivial
differentials in the index spectral sequence. Nevertheless, the spectral
sequence will still stabilize after a finite number of steps in each case.

######  Lemma 6.9.

Let $\Gamma\subset\operatorname{SU}(2)$ be a finite subgroup. Then the only
possibly nontrivial differentials in the index spectral sequence

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H^{-}_{A}(\alpha)_{t}\implies
I^{-}(\overline{Y}_{\Gamma})_{s+t}$

are $d^{4r}\colon E^{4r}_{4(s+r),-4(r-1)}\rightarrow E^{4r}_{4s,3}$ for $s$
and $r\geq 1$.

###### Proof.

As $j(\alpha)\equiv 0\mod{4}$ for each $\alpha\in\mathcal{C}$, we have
$E^{1}_{s,*}=0$ unless $4|s$. By Lemma 5.7 and Corollary 5.9 the $R[U]$-module
$H^{-}_{A}(\alpha)$ is given by $R[U]$ with $|U|=-4$, $R[Z][2]$ with $|Z|=-2$
or $R[3]$ depending on whether $\alpha$ is fully reducible, reducible or
irreducible, respectively. This means that $E^{1}_{4s,t}=0$ for $t\geq 4$ and
odd $t\leq 2$. Since the differential $d^{r}$ has bidegree $(-r,r-1)$ we
deduce that it can only be nonzero if it lands in the bidegree $(4s,3)$ for
some $s$. The only such differentials that also begin in a possibly nonzero
group are $d^{4r}\colon E^{4r}_{4(s-r),-4(r-1)}\rightarrow E^{4r}_{4s,3}$. ∎

Our first task will be to give an explicit formula for the nontrivial
differentials. Fix a finite subgroup $\Gamma\subset\operatorname{SU}(2)$ and
let $M=DCI(\overline{Y}_{\Gamma})$ be equipped with the index filtration.
Recall that by Lemma 5.45

$DCI^{-}(\overline{Y}_{\Gamma})=\operatorname{colim}_{p}C^{-}_{A}(F_{p}M)=\bigcup_{p}C^{-}_{A}(F_{p}M).$

In the following we will use the fact from Lemma 6.4 that $F_{4p+t}M=F_{4p}M$
for all $p\in\mathbb{Z}$ and $0\leq t\leq 3$ without further mention.

######  Lemma 6.10.

For all $p,s\in\mathbb{Z}$ and $0\leq t\leq 3$ we have

$\displaystyle C^{-}_{A}(F_{4p}M)_{4s+t}$
$\displaystyle\cong\left\\{\begin{array}[]{cc}\bigoplus_{0\leq q\leq
p-s}M_{4(s+q)+t}&\mbox{ if }s\leq p\\\ 0&\mbox{ if }s>p\end{array}\right.$
$\displaystyle C^{-}_{A}(\overline{F}_{4p}M)_{4s+t}$
$\displaystyle\cong\left\\{\begin{array}[]{cc}M_{4p+t}&\mbox{ if }s\leq p\\\
0&\mbox{ if }s>p\end{array}\right.$

The differential in $C^{-}_{A}(\overline{F}_{4p}M)$ is given by $u\colon
M_{4p}\rightarrow M_{4p+3}$ in degree $4s$ with $s\leq p$, and vanishes
otherwise. The differential in $C^{-}_{A}(F_{4p}M)$ is given by

$\partial((x_{q})_{0\leq q\leq
p-s})=(\partial_{M}x_{q}-(-1)^{t}x_{q-1}u)_{0\leq q\leq p-s+1}$

for $(x_{q})_{0\leq q\leq p-s}\in\bigoplus_{0\leq q\leq
p-s}M_{4(s+q)+t}=C^{-}_{A}(F_{4p}M)_{4s+t}$ where $s\leq p$, and we interpret
$x_{-1}=0=x_{p-s+1}$.

###### Proof.

This is a simple consequence of Proposition 5.8 using the fact that
$F_{4p}M_{n}=M_{n}$ for $n\leq 4p+3$ and $F_{4p}M_{n}=0$ otherwise. ∎

For each $\eta\in\mathcal{C}^{f.red}$, $\lambda\in\mathcal{C}^{red}$ and
$\alpha\in\mathcal{C}^{irr}$ we introduce variables $U_{\eta},Z_{\lambda}$ and
a generator $h_{\alpha}$. Then, according to Lemma 5.7 and Corollary 5.9,

$H^{-}_{A}(\eta)=R[U_{\eta}],\;\;\;\;H^{-}_{A}(\lambda)=R[Z_{\lambda}][2]\;\;\mbox{
and }\;\;H^{-}_{A}(\alpha)=R\cdot h_{\alpha}$

where $|U_{\eta}|=-4$, $|Z_{\lambda}|=-2$ and $|h_{\alpha}|=3$.

######  Lemma 6.11.

Let $\\{E^{r},d^{r}\\}_{r\geq 1}$ be the index spectral sequence with

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H^{-}_{A}(\alpha)_{t}\implies
I^{-}_{s+t}(\overline{Y}_{\Gamma}).$

Then, for each $r\geq 1$ and $s$, $E^{4r}_{4s,-4(r-1)}=E^{1}_{4s,-4(r-1)}$ is
a free $R$-module with generators

$\\{U^{r-1}_{\eta}:\eta\in\mathcal{C}^{f.red}\ni j(\eta)\equiv
4s\\}\cup\\{Z^{2r-1}_{\lambda}:\lambda\in\mathcal{C}^{red}\ni j(\lambda)\equiv
4s\\}$

and $E^{1}_{4s,3}$ is a free $R$-module with generators
$\\{h_{\alpha}:\alpha\in\mathcal{C}^{irr}\ni j(\alpha)\equiv 4s\\}$. Let
$p_{r}:E^{1}_{4s,3}\rightarrow E^{4r}_{4s,3}$ denote the natural surjection.
Then in terms of these generators the differential $d^{r}\colon
E^{4r}_{4(s+r),-4(r-1)}\rightarrow E^{4r}_{4s,3}$ is determined by

$\displaystyle d^{4r}(U^{r-1}_{\eta})$
$\displaystyle=p_{r}\left(\sum_{(\alpha_{1},\cdots,\alpha_{r})}n_{\alpha_{1}\eta}\left(\prod_{i=1}^{r-1}n_{\alpha_{i+1}\alpha_{i}}\right)\cdot
h_{\alpha_{r}}\right)$ $\displaystyle d^{4r}(W^{2r-1}_{\lambda})$
$\displaystyle=p_{r}\left(\sum_{(\alpha_{1},\cdots,\alpha_{r})}n_{\alpha_{1}\lambda}\left(\prod_{i=1}^{r-1}n_{\alpha_{i+1}\alpha_{i}}\right)\cdot
h_{\alpha_{r}}\right),$

where the sums are taken over tuples
$(\alpha_{1},\cdots,\alpha_{r})\in(\mathcal{C}^{irr})^{r}$ for which
$(\eta,\alpha_{1},\cdots,\alpha_{r})$, respectively
$(\lambda,\alpha_{1},\cdots,\alpha_{r})$, form a path in the graph
$\mathcal{S}_{\Gamma}$ (see Definition 3.9).

###### Proof.

Before we introduce the explicit generators we will work out a formula for the
differentials in terms of the complex $M$. By Lemma 6.9 we have
$E^{4r}_{4(s+r),-4(r-1)}=E^{1}_{4(s+r),-4(r-1)}=H^{-}_{A}(\overline{F}_{4(s+r)}M)_{4(s+1)}$
and by Lemma 6.10 the latter group may be identified with
$\operatorname{Ker}(u)\subset M_{4(s+r)}$. Recall from diagram (5.15) that the
differential $d^{4r}$ is defined by first applying the connecting homomorphism
$\delta\colon H^{-}_{A}(\overline{F}_{4(s+r)}M)\rightarrow
H^{-}_{A}(F_{4(s+r)-1}M)$, then lifting along the map
$H^{-}_{A}(F_{4s}M)\rightarrow H^{-}_{A}(F_{4(s+r)-1}M)$ and then pushing down
along $H^{-}_{A}(F_{4s}M)\rightarrow H^{-}_{A}(\overline{F}_{4s}M)$. On the
chain level we have the following diagram

${C^{-}_{A}(\overline{F}_{4(s+r)}M)_{4(s+1)}}$${C^{-}_{A}(F_{4(s+r)-1}M)_{4s+3}}$${C^{-}_{A}(F_{4s}M)_{4s+3}}$${M_{4(s+r)}}$${\bigoplus_{q=0}^{r-1}M_{4(s+q)+3}}$${M_{4s+3}.}$$\scriptstyle{\cong}$$\scriptstyle{\delta}$$\scriptstyle{\cong}$$\scriptstyle{\iota}$$\scriptstyle{\cong}$$\scriptstyle{\delta}$$\scriptstyle{\iota}$

The arrows correspond to the connecting homomorphism $\delta$ are dashed as
they are not well-defined before passing to homology. Given
$x\in\operatorname{Ker}u\subset M_{4(s+r)}$ the element $\delta(x)$ is
represented on the chain level by $(0,0,\cdots,0,\partial_{M}x)$. As $\iota$
is the inclusion of the first summand, this element does not lift along
$\iota$. Let $\psi:M\rightarrow M$ be the map of Lemma 6.2 and let

$y\coloneqq(\psi^{r-q}x)_{q=1}^{r-1}\in\bigoplus_{q=1}^{r-1}M_{4(s+q)}\cong
C^{-}_{A}(F_{4(s+r-1)}M)_{4(s+1)}.$

Then $\partial(y)=(\partial_{M}\psi^{r-1}x,0,0,\cdots,0,-\partial_{M}x)$,
which implies that
$\iota[\partial_{M}\psi^{r-1}x]=[(0,\cdots,0,\partial_{M}x)]$ in homology. By
Lemma 6.10 we have the following sequence of isomorphisms

$M_{4s+3}\cong C^{-}_{A}(F_{4s}M)_{4s+3}\cong
C^{-}_{A}(\overline{F}_{4s}M)_{4s+3}\cong
H^{-}_{A}(\overline{F}_{4s}M)_{4s+3}\cong E^{1}_{4s,3}.$

Moreover, according to Lemma 6.9, there are no outgoing differentials from
$E^{r}_{4s,3}$ for $r\geq 1$. We conclude that $d^{4r}(x)$ is given by the
image of $\partial_{M}\psi^{r-1}x$ along the surjection
$E^{1}_{4s,3}\rightarrow E^{4r}_{4s,3}$ for each $x\in
E^{4r}_{4(s+r),-4(r-1)}\subset M_{4(s+r)}$. In other words, we have the
following commutative diagram

(6.10)
${E^{4r}_{4(s+r),-4(r-1)}}$${E^{1}_{4(s+r),-4(r-1)}}$${\operatorname{Ker}(u)\subset
M_{4(s+r)}}$${E^{4r}_{4s,3}}$${E^{1}_{4s,3}}$${M_{4s+3}.}$$\scriptstyle{d^{4r}}$$\scriptstyle{=}$$\scriptstyle{\cong}$$\scriptstyle{\partial_{M}\circ\psi^{(r-1)}}$$\scriptstyle{p_{r}}$$\scriptstyle{\cong}$

Now, $\operatorname{Ker}u\subset M_{4(s+r)}$ is freely generated by
$b_{\eta}\in H_{0}(\eta)$, $b_{\lambda}\in H_{0}(\lambda)$ for
$\eta\in\mathcal{C}^{f.red}$ and $\lambda\in\mathcal{C}^{red}$ with
$j(\eta)\equiv j(\lambda)\equiv 4(s+r)$. It is important to note that under
the top horizontal identification in (6.10) we may, and will, require that
these correspond to the generators $U^{r-1}_{\eta}$ and $W^{2r-1}_{\lambda}$
in $E^{1}_{4(s+r),-4(r-1)}$. Similarly, $M_{4s+3}$ is freely generated by
$t_{\alpha}=b_{\alpha}\cdot u\in H_{3}(\alpha)$ for
$\alpha\in\mathcal{C}^{irr}$ with $j(\alpha)\equiv 4s$, and under the lower
horizontal isomorphism we require that these correspond to the generators
$h_{\alpha}\in E^{1}_{4s,3}$. Recall from equation (6.1) that

$\psi(b_{\eta})=u^{-1}\partial_{M}(b_{\eta})=\sum_{\alpha\in\mathcal{C}^{irr}}u^{-1}(n_{\alpha\eta}t_{\alpha})=\sum_{\alpha\in\mathcal{C}^{irr}}n_{\alpha\eta}b_{\alpha}.$

and similarly for $\psi(b_{\lambda})$. Taking into account the fact that
$n_{\alpha\beta}=0$ whenever $\alpha$ and $\beta$ are not adjacent in the
graph $\mathcal{S}_{\Gamma}$, we obtain

$\partial_{M}\psi^{r-1}(b_{\eta})=\sum_{(\alpha_{1},\cdots,\alpha_{r})}(n_{\alpha_{1}\eta}n_{\alpha_{2}\alpha_{2}}\cdots
n_{\alpha_{r}\alpha_{r-1}})\cdot t_{\alpha_{r}},$

where the sum is taken over
$(\alpha_{1},\cdots,\alpha_{r})\in(\mathcal{C}^{irr})^{r}$ for which
$(\eta,\alpha_{1},\cdots,\alpha_{r})$ forms an edge path in
$\operatorname{S}_{\Gamma}$. An analogous formula holds for $b_{\lambda}$, so
in view of diagram (6.10) the proof is complete. ∎

By combining the above lemma with Theorem 4.25 we obtain complete control over
all the differentials in the spectral sequences. The following result takes
care of all the extension problems we will meet.

######  Lemma 6.12.

Let $\Gamma\subset\operatorname{SU}(2)$ be a finite subgroup and let
$(E^{r},d^{r})_{r\geq 1}$ be the index spectral sequence with

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H^{-}_{A}(\alpha)_{t}\implies
I^{-}(\overline{Y}_{\Gamma})_{s+t}.$

Assume that $E^{\infty}_{s}$ is a free $R[U]$-module for $s=0,4$. Then there
is an isomorphism of $R[U]$-modules

$I^{-}(\overline{Y}_{\Gamma})\cong(E^{\infty}_{0}\oplus
E^{\infty}_{4})^{\oplus,8}.$

###### Proof.

Recall that the index spectral sequence is periodic in the sense that
$E^{r}_{s,t}\cong E^{r}_{s+8,t}$, or equivalently $E^{r}_{s}[8]\cong
E^{r}_{s+8}$, for all $s,t\in\mathbb{Z}$ and $r\in\mathbb{N}\cup\\{\infty\\}$.
Therefore, the assumption that $E_{0}^{\infty}$ and $E_{4}^{\infty}$ are free
$R[U]$-modules implies that $E^{\infty}_{s}$ is a free $R[U]$-module for each
$s\in\mathbb{Z}$, as $E^{1}_{s}=0$ for all $s$ not divisible by $4$.

Write $I=I^{-}(\overline{Y}_{\Gamma})$. By Theorem 5.58 $(E^{r},d^{r})_{r\geq
1}$ is a spectral sequence of $R[U]$-modules that converges strongly to $I$.
In other words, $I$ carries an exhaustive and complete Hausdorff filtration
$\\{F_{s}I\\}_{s}$ of $R[U]$-submodules and there is an isomorphism
$E^{\infty}_{s}\cong\overline{F}_{s}I$ of $R[U]$-modules for each
$s\in\mathbb{Z}$. In view of the fact that $E^{\infty}_{s}$ is free over
$R[U]$, the short exact sequence

${0}$${F_{s-1}I}$${F_{s}I}$${\overline{F}_{s}I\cong
E^{\infty}_{s}}$${0}$$\scriptstyle{\iota_{s}}$$\scriptstyle{\pi_{s}}$

splits in the category of $R[U]$-modules for each $s\in\mathbb{Z}$. Hence,
$F_{s-1}I\oplus\overline{F}_{s}I\cong F_{s}I$ and inductively

$F_{s-1}I\oplus\left(\bigoplus_{t=0}^{r}\overline{F}_{s+t}I\right)\cong
F_{s+r}I$

for $s\in\mathbb{Z}$ and $r\geq 1$. These are all isomorphisms of
$R[U]$-modules. By Lemma 6.10 we have $E^{1}_{s,t}=0$ for all $s$ and $t\geq
4$. Therefore, for fixed $n\in\mathbb{Z}$, $(E^{1}_{s})_{n}=E^{1}_{s,n-s}=0$
for all $s\leq n-4=:s_{0}$. Hence, $(\overline{F}_{s}I)_{n}=0$ for all $s\leq
s_{0}$. Since the filtration of $I$ is complete Hausdorff it follows that
$(F_{s}I)_{n}=0$ for all $s\leq s_{0}$. Consequently, using the above
isomorphisms, we obtain

$I_{n}=\operatorname{colim}_{s\geq
s_{0}}(F_{s}I)_{n}\cong\operatorname{colim}_{s\geq
s_{0}}\bigoplus_{i=s_{0}}^{s}(\overline{F}_{s}I)_{n}\cong\bigoplus_{s}(E^{\infty}_{s})_{n}$

for each $n\in\mathbb{Z}$. These isomorphisms piece together to an isomorphism
$I\cong\bigoplus_{s}E^{\infty}_{s}$ of $R[U]$-modules. Finally, we may exploit
the periodicity of $E^{\infty}$ and the fact that $E^{\infty}_{s}=0$ unless
$4|s$ to simplify:

$I\cong\bigoplus_{s}E^{\infty}_{s}\cong\bigoplus_{s}(E^{\infty}_{8s}\oplus
E^{\infty}_{8s+4})\cong\bigoplus_{s}(E^{\infty}_{0}[8s]\oplus
E^{\infty}_{4}[8s])=(E^{\infty}_{0}\oplus E^{\infty}_{4})^{\oplus,8}$

and the proof is complete. ∎

It is not possible to give a uniform result for the calculation of
$I^{-}(\overline{Y}_{\Gamma})$ for all the subgroups
$\Gamma\subset\operatorname{SU}(2)$ simultaneously. We will therefore treat
the simplest cases $\Gamma=I^{*},O^{*},T^{*},C_{m}$ first and treat the more
involved case of binary dihedral groups afterwards.

We have the following table over the flat connections extracted from Appendix
$A$. The notation is compatible with Proposition 4.25.

$\begin{array}[]{|c|c|c|c|}\hline\cr\Gamma&\mathcal{C}^{f.red}&\mathcal{C}^{irr}&\mathcal{C}^{red}\\\
\hline\cr C_{2m}&\theta,\eta&{}\hfil&\lambda_{1},\cdots,\lambda_{m-1}\\\
\hline\cr C_{2m+1}&\theta&{}\hfil&\lambda_{1},\cdots,\lambda_{m}\\\ \hline\cr
I^{*}&\theta&\alpha,\beta&{}\hfil\\\ \hline\cr
O^{*}&\theta,\eta&\alpha,\beta&{}\hfil\\\ \hline\cr
T^{*}&\theta&\alpha&\lambda\\\ \hline\cr\end{array}$

######  Theorem 6.13.

The negative equivariant instanton Floer homology associated with the trivial
$\operatorname{SU}(2)$-bundle over $\overline{Y}_{\Gamma}$ for
$\Gamma=C_{2m},C_{2m+1}$ is given by

$\displaystyle I^{-}(\overline{Y}_{C_{2m}})$
$\displaystyle=\left((R[U_{\theta}]\oplus
R[U_{\eta}][4m]\oplus\bigoplus_{i=1}^{m-1}R[Z_{\lambda_{i}}][4i+2]\right)^{\oplus,8}$
$\displaystyle I^{-}(\overline{Y}_{C_{2m+1}})$
$\displaystyle=\left(R[U_{\theta}]\oplus\bigoplus_{i=1}^{m}R[Z_{\lambda_{i}}][4i+2]\right)^{\oplus,8},$

while for $\Gamma=I^{*},O^{*},T^{*}$ we have
$I^{-}(\overline{Y}_{\Gamma})=(X_{\Gamma})^{\oplus,8}$ where $X_{\Gamma}$ is
the $R[U]$-submodule of $P_{\Gamma}$ generated by $G_{\Gamma}$ specified in
the following table.

$\begin{array}[]{|c|c|c|}\hline\cr\Gamma&P_{\Gamma}&G_{\Gamma}\\\ \hline\cr
I^{*}&R[U_{\theta}]&\\{U_{\theta}^{2}\\}\\\ \hline\cr
O^{*}&R[U_{\theta}]\oplus R[U_{\eta}][4]&\\{U_{\theta}^{1},U_{\eta}^{1}\\}\\\
\hline\cr T^{*}&R[U_{\theta}]\oplus
R[Z_{\lambda}][2]&\\{U_{\theta}^{1},Z_{\lambda}^{0},3U_{\theta}^{0}-Z^{1}_{\lambda}\\}\\\
\hline\cr\end{array}$

###### Proof.

For $\Gamma=I^{*},O^{*},T^{*},C_{n}$ we will write
$\mathcal{C}=\mathcal{C}_{\Gamma}$ for the set of flat connections and
$(E^{r},d^{r})_{r\geq 1}$ will denote the corresponding index spectral
sequence with

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H^{-}_{A}(\alpha)_{t}\implies
I^{-}(\overline{Y}_{\Gamma})_{s+t}.$

In each case we will calculate $E^{\infty}$ and observe that $E^{\infty}_{0}$
and $E^{\infty}_{4}$ are free $R[U]$-modules. By Lemma 6.12 this is enough to
conclude that $I^{-}(\overline{Y}_{\Gamma})=(E_{0}^{\infty}\oplus
E_{4}^{\infty})^{\oplus,8}$. In the following we will make consistent use of
Lemma 6.9, Theorem 4.25 and Lemma 6.11 that together determine all the
differentials explicitly.

$\mathbf{C_{n}}$. As $\mathcal{C}^{irr}=\emptyset$ there are no nontrivial
differentials in the spectral sequence so that $E^{1}=E^{\infty}$. For $n=2m$
we have $\mathcal{C}=\\{\theta,\eta,\lambda_{1},\cdots,\lambda_{m-1}\\}$ with
$\theta$, $\eta$ fully reducible and the $\lambda_{i}$ reducible. The grading
is given by $j(\lambda_{i})\equiv 4i$ and $j(\eta)\equiv 4m$. This gives

$E^{\infty}_{0}\oplus E^{\infty}_{4}=R[U_{\theta}]\oplus
R[U_{\eta}][j(\eta)]\oplus\left(\bigoplus_{i=1}^{m-1}R[Z_{\lambda_{i}}][j(\lambda_{i})+2]\right).$

For $n=2m+1$ we have $\mathcal{C}=\\{\theta,\lambda_{1},\cdots,\lambda_{m}\\}$
with $\theta$ fully reducible and the $\lambda_{i}$ reducible. The grading is
given by $j(\lambda_{i})\equiv 4i$. Hence,

$E^{\infty}_{0}\oplus
E^{\infty}_{4}=R[U_{\theta}]\oplus\left(\bigoplus_{i=1}^{m}R[Z_{\lambda_{i}}][j(\lambda_{i})+2]\right).$

In both cases these are free $R[U]$-modules ($R[Z_{\lambda}]$ is freely
generated by $\\{Z_{\lambda}^{0},Z_{\lambda}^{1}\\}$). The stated results are
obtained by applying $X\mapsto X^{\oplus,8}$ to the above formulas for
$E_{0}^{\infty}\oplus E^{\infty}_{4}$.

$\mathbf{I^{*}}$. In this case we have $\mathcal{C}=\\{\theta,\alpha,\beta\\}$
with $\alpha$ and $\beta$ irreducible. The grading is given by
$j(\theta)=j(\beta)=0$ and $j(\alpha)=4$. The first nontrivial differential in
the index spectral sequence is $d^{4}\colon
E^{4}_{8s,0}=R\\{U^{0}_{\theta}\\}\rightarrow
R\\{h_{\alpha}\\}=E^{4}_{8s-4,3}$ and is given by
$d^{4}(U^{0}_{\theta})=n_{\alpha\eta}h_{\alpha}=h_{\alpha}$. Therefore,
$E^{\infty}_{8s+4,3}=0=E^{\infty}_{8s,0}$. The next differential is
$d^{8}\colon E^{8}_{8s,-4}=R\\{U_{\theta}^{1}\\}\rightarrow
R\\{h_{\beta}\\}=E^{8}_{8(s-1),3}$ and is given by
$d^{8}(U^{1}_{\theta})=n_{\beta\alpha}n_{\alpha\eta}h_{\beta}=4h_{\beta}$. As
$2\in R$ is invertible, this is an isomorphism and we conclude that
$E^{\infty}_{8s,3}=0$ and $E^{\infty}_{8s,-4}=0$. There are no more nontrivial
differentials in the spectral sequence. Therefore, $E^{\infty}_{4}=0$ and
$E^{\infty}_{0}\subset R[U_{\theta}]$ is the free $R[U]$-submodule generated
by $U_{\theta}^{2}$ as required.

$\mathbf{O^{*}}$. Here we have $\mathcal{C}=\\{\theta,\eta,\alpha,\beta\\}$
with $\theta$, $\eta$ fully reducible and $\alpha,\beta$ irreducible. The
grading is $j(\theta)=j(\beta)=0$ and $j(\alpha)=j(\eta)=4$. The first
nontrivial differentials are

$\displaystyle d^{4}\colon$ $\displaystyle
E^{4}_{8s+4,0}=R\\{U^{0}_{\eta}\\}\rightarrow R\\{h_{\beta}\\}=E^{4}_{8s,3}$
$\displaystyle d^{4}\colon$ $\displaystyle
E^{4}_{8s,0}=R\\{U^{0}_{\theta}\\}\rightarrow
R\\{h_{\alpha}\\}=E^{4}_{8s-4,3}.$

These are both isomorphisms as $n_{\beta\eta}=1=n_{\alpha\theta}$. This
implies that $E^{\infty}_{8s,3}=E^{5}_{8s,3}=0$ and
$E^{\infty}_{8s-4,3}=E^{5}_{8s-4,3}=0$. There can therefore be no more
nontrivial differentials. Hence, $E^{\infty}_{0}\subset R[U_{\theta}]$ and
$E^{\infty}_{4}\subset R[U_{\eta}][4]$ are the free $R[U]$-submodules
generated by $U_{\theta}^{1}$ and $U_{\eta}^{1}$, respectively.

$\mathbf{T^{*}}$. In this case $\mathcal{C}=\\{\theta,\alpha,\lambda\\}$ with
$\theta$ fully reducible, $\alpha$ irreducible and $\lambda$ reducible. The
grading is $j(\theta)=j(\lambda)=0$ and $j(\alpha)=4$. The first nontrivial
differential is $d^{4}\colon
E^{4}_{8s,0}=R\\{Z_{\lambda}^{1},U_{\theta}^{0}\\}\rightarrow
R\\{h_{\alpha}\\}=E^{4}_{8s-4,3}$ and is determined by
$d^{4}(U^{0}_{\theta})=n_{\alpha\theta}h_{\alpha}=h_{\alpha}$ and
$d^{4}(Z^{1}_{\lambda})=n_{\alpha\lambda}h_{\alpha}=3h_{\alpha}$. This map is
surjective with kernel $R\\{3U^{0}_{\theta}-Z^{1}_{\lambda}\\}$. Therefore,
$E^{\infty}_{8s-4,3}=E^{4}_{8s-4,3}=0$ and there are no more nontrivial
differentials. We find $E^{\infty}_{4}=0$,
$(E^{\infty}_{0})_{0}=R\\{3U^{0}_{\theta}-Z^{1}_{\lambda}\\}$ and
$(E^{\infty}_{0})_{n}=(R[U_{\theta}]\oplus R[Z_{\lambda}][2])_{n}$ in all
other degrees $n$. It now suffices to observe that $E^{\infty}_{0}$ is indeed
the free submodule of $R[U_{\theta}]\oplus R[Z_{\lambda}][2]$ generated by
$\\{Z_{\lambda}^{0},3U^{0}_{\theta}-Z^{1}_{\lambda},U_{\theta}^{1}\\}$. This
completes the final case and the proof. ∎

We will now consider the binary dihedral groups $D_{m}^{*}$. It is necessary
to partition the calculations into cases depending on the residue of $m$ mod
$4$. We have the following table over the flat connections

$\begin{array}[]{|c|c|c|c|}\hline\cr{}\hfil&\mathcal{C}^{f.red}&\mathcal{C}^{irr}&\mathcal{C}^{red}\\\
\hline\cr
D_{4n}^{*}&\theta,\eta_{1},\eta_{2},\eta_{3}&\alpha_{1},\cdots,\alpha_{2n}&{}\hfil\\\
\hline\cr D_{4n+1}^{*}&\theta,\eta&\alpha_{1},\cdots,\alpha_{2n}&\lambda\\\
\hline\cr
D_{4n+2}^{*}&\theta,\eta_{1},\eta_{2},\eta_{3}&\alpha_{1},\cdots,\alpha_{2n+1}&{}\hfil\\\
\hline\cr D_{4n+3}^{*}&\theta,\eta&\alpha_{1},\cdots,\alpha_{2n+1}&\lambda\\\
\hline\cr\end{array}$

in agreement with Appendix $A$ and Proposition 4.25. In all cases the grading
of the irreducibles are given by $j(\alpha_{i})\equiv 4i\mod{8}$. For
$D^{*}_{4n+2}$ and $D^{*}_{4n+3}$ all the fully reducibles and reducibles
satisfy $j(\rho)=0$. For $D^{*}_{4n}$ we have $j(\theta)=j(\eta_{1})=0$,
$j(\eta_{2})=j(\eta_{3})=4$, while for $D^{*}_{4n+1}$ we have
$j(\theta)=j(\eta)=0$ and $j(\lambda)=4$. For the convenience of the reader we
include the relevant diagrams for $S_{D^{*}_{m}}$ from Proposition 4.25.

$\mathcal{S}_{D^{*}_{2m}}$$\alpha_{1}$$\alpha_{2}$$\alpha_{m-1}$$\alpha_{m}$$\theta$$\eta_{1}$$\eta_{2}$$\eta_{3}$$(2|2)$$(2|2)$$1$$1$$1$$1$

$\mathcal{S}_{D^{*}_{2m+1}}$$\alpha_{1}$$\alpha_{2}$$\alpha_{m-1}$$\alpha_{m}$$\lambda$$\theta$$\eta$$(2|2)$$(2|2)$$2$$1$$1$

######  Theorem 6.14.

The negative instanton Floer homology associated with the trivial
$\operatorname{SU}(2)$-bundle over $\overline{Y}_{\Gamma}$ for
$\Gamma=D^{*}_{m}$ is given by
$I^{-}(\overline{Y}_{D_{m}^{*}})=(X_{m})^{\oplus,8}$, where $X_{m}$ is the
$R[U]$-submodule of $P_{m}$ generated by $G_{m}$ specified in the following
tables.

$\begin{array}[]{|c|c|}\hline\cr m&P_{m}\\\ \hline\cr 4n&R[U_{\theta}]\oplus
R[U_{\eta_{1}}]\oplus R[U_{\eta_{2}}][4]\oplus R[U_{\eta_{3}}][4]\\\ \hline\cr
4n+1&R[U_{\theta}]\oplus R[U_{\eta}]\oplus R[Z_{\lambda}][6]\\\ \hline\cr
4n+2&R[U_{\theta}]\oplus R[U_{\eta_{1}}]\oplus R[U_{\eta_{2}}]\oplus
R[U_{\eta_{3}}]\\\ \hline\cr 4n+3&R[U_{\theta}]\oplus R[U_{\eta}]\oplus
R[Z_{\lambda}][2]\\\ \hline\cr\end{array}$

$\begin{array}[]{|c|c|}\hline\cr m&G_{m}\\\ \hline\cr
4n&\\{U^{0}_{\theta}-U_{\eta_{1}}^{0},U^{0}_{\eta_{2}}-U^{0}_{\eta_{3}},U_{\theta}^{n},U_{\eta_{2}}^{n}\\}\\\
\hline\cr
4n+1&\\{U_{\theta}^{0}-U_{\eta}^{0},Z^{0}_{\lambda},U_{\theta}^{n},Z^{2n+1}_{\lambda}\\}\\\
\hline\cr
4n+2&\\{U^{0}_{\theta}-U_{\eta_{1}}^{0},U^{0}_{\eta_{2}}-U^{0}_{\eta_{3}},U_{\theta}^{n}-U_{\eta_{2}}^{n},U_{\theta}^{n+1}\\}\\\
\hline\cr
4n+3&\\{U^{0}_{\theta}-U^{0}_{\eta},Z^{0}_{\lambda},2U^{n}_{\theta}-Z_{\lambda}^{2n+1},U_{\theta}^{n+1}\\}\\\
\hline\cr\end{array}$

###### Proof.

Observe first that the modules $P_{m}$ in the above table coincides with the
$R[U]$-submodule of $E^{1}_{0}\oplus E^{1}_{4}$ in the index spectral sequence
generated by the reducibles and fully reducibles. Furthermore, the submodule
generated by $G_{m}$ is free in each case. Therefore, following the same
procedure as in the proof of Theorem 6.13, it will be sufficient to show that
$E^{\infty}_{0}\oplus E^{\infty}_{4}$ is the $R[U]$-submodule of $P_{m}$
generated by $G_{m}$ in each case.

For $m=4n,4n+1$ we have

$E^{1}_{8s,3}=R\\{h_{\alpha_{2}},h_{\alpha_{4}},\cdots,h_{\alpha_{2n}}\\}\mbox{
and
}E^{1}_{8s+4,3}=R\\{h_{\alpha_{1}},h_{\alpha_{3}},\cdots,h_{\alpha_{2n-1}}\\}$

for each $s\in\mathbb{Z}$. For $m=4n+2,4n+3$ we have the same formulas except
that one generator $h_{\alpha_{2n+1}}$ is adjoined to the latter group. To
simplify the notation put $h_{i}\coloneqq h_{\alpha_{i}}$ for $1\leq i\leq
2n+1$. For each $r\geq 1$ let $p_{r}\colon
E^{1}_{4s,3}=E^{4}_{4s,3}\rightarrow E^{4r}_{4s,3}$ denote the natural
surjection.

Consider first the cases $m=4n,4n+1$. To unify the notation slightly we write
$(U_{1},U_{2},U_{3},U_{4})=(U_{\theta},U_{\eta_{1}},U_{\eta_{2}},U_{\eta_{3}})$
for $m=4n$ and $(U_{1},U_{2},Z)=(U_{\theta},U_{\eta},Z_{\lambda})$ for
$m=4n+1$. We claim that for $1\leq r\leq n$ the differentials

$\displaystyle f^{r}\coloneqq d^{4r}\colon$ $\displaystyle
E^{4r}_{8s,-4(r-1)}\rightarrow E^{4r}_{8s-4r,3}$ $\displaystyle g^{r}\coloneqq
d^{4r}\colon E^{4r}_{8s+4,-4(r-1)}\rightarrow E^{4r}_{8s-4(r-1),3}$

are given by the formulas

$\displaystyle f^{r}(U_{1}^{r-1})$
$\displaystyle=f^{r}(U_{2}^{r-1})=2^{r-1}p_{r}(h_{r})$ $\displaystyle
g^{r}(U_{3}^{r-1})$
$\displaystyle=g^{r}(U_{4}^{r-1})=2^{r-1}p_{r}(h_{2n-r+1})$ $\displaystyle
g^{r}(Z^{2r-1})$ $\displaystyle=2^{r}p_{r}(h_{2n-r+1}).$

We verify this by induction on $r$. Let $1\leq r\leq n-1$ and assume that the
statement is true for all $i$ with $1\leq i<r$. This implies in particular
that $p_{r}(h_{i})=0$ for $1\leq i<r$ and $2n-r+1<i\leq 2n$. By Lemma 6.11 we
have

$f^{r}(U_{1}^{r-1})=p_{r}\left(\sum_{(\beta_{1},\cdots,\beta_{r})}(n_{\beta_{1}\theta}n_{\beta_{2}\beta_{1}}\cdots
n_{\beta_{r}\beta_{r-1}})\cdot h_{\beta_{r}}\right),$

where the sum runs over all
$(\beta_{1},\cdots,\beta_{r})\in(\mathcal{C}^{irr})^{r}$ for which
$(\theta,\beta_{1},\cdots\beta_{r})$ forms an edge path in
$\mathcal{S}_{\Gamma}$. As such a path has length $r$, it must, in view of the
graphs $\mathcal{S}_{D_{m}^{*}}$ shown above, terminate at some vertex
$\alpha_{i}$ with $i\leq r$. By the inductive hypothesis
$p_{r}(h_{\alpha_{i}})=0$ for $i<r$, so the only nonzero term in the formula
corresponds to the path $(\alpha_{1},\alpha_{2},\cdots,\alpha_{r})$. As
$n_{\alpha_{i+1}\alpha_{i}}=2$ for each $i$ and $n_{\alpha_{1}\theta}=1$ we
conclude that $f^{r}(U_{1}^{r-1})=2^{r-1}p_{r}(h_{r})$. The other formulas
follow by essentially identical arguments, we only note that the additional
factor $2$ picked up in the last formula follows from the fact that
$n_{\alpha_{2n}\lambda}=2$. This completes the inductive step and the claim is
verified.

The above formulas for the differentials imply, as $2\in R$ is invertible,
that the generators $h_{i}$, $1\leq i\leq 2n$, of $E^{1}_{8s,3}$ and
$E^{1}_{8s+4,3}$ are killed off two by two until we reach $E^{4n+1}_{8s,3}=0$,
$E^{4n+1}_{8s+4,3}=0$. Therefore, by Lemma 6.9, $E^{4n+1}=E^{\infty}$.
Moreover, we find
$E^{\infty}_{8s,-4(r-1)}=\operatorname{Ker}(f^{r})=R\\{(U_{1}^{r-1}-U_{2}^{r-1})\\}$
for $1\leq r\leq n$, and similarly

$E^{\infty}_{8s+4,-4(r-1)}=\left\\{\begin{array}[]{cl}R\\{(U_{3}^{r-1}-U_{4}^{r-1})\\}&\mbox{
for }m=4n\\\ 0&\mbox{ for }m=4n+1\end{array}\right.$

for $1\leq r\leq n$. In all other degrees we have
$E^{1}_{s,t}=E^{\infty}_{s,t}$. We therefore see that $E^{\infty}_{0}$ is
freely generated as an $R[U]$-module by
$\\{(U_{1}^{0}-U_{2}^{0}),U_{1}^{n}\\}$ for both $m=4n$ and $m=4n+1$, while
$E^{0}_{4}$ is freely generated by $\\{(U_{3}^{0}-U_{4}^{0}),U_{3}^{n}\\}$ for
$m=4n$ and by $\\{Z_{\lambda}^{0},Z_{\lambda}^{2n+1}\\}$ for $m=4n+1$. We have
thus completed the cases $m=4n,4n+1$.

Consider now the cases $m=4n+2,4n+3$. As earlier we write
$(U_{1},U_{2},U_{3},U_{4})=(U_{\theta},U_{\eta_{1}},U_{\eta_{2}},U_{\eta_{3}})$
for $m=4n+2$ and $(U_{1},U_{2},Z)=(U_{\theta},U_{\eta},Z_{\lambda})$. Here
$U^{r-1}_{i},Z^{2r-1}\in E^{1}_{8s,-4(r-1)}$ for each $i$ and $r\geq 1$. For
$1\leq r\leq n+1$ the differentials

$d^{4r}\colon E^{4r}_{8s,-4(r-1)}\rightarrow E^{4r}_{8s-4r,3}$

are given by the formulas

$\displaystyle d^{4r}(U_{1}^{r-1})$
$\displaystyle=d^{4r}(U_{2}^{r-1})=2^{r-1}p_{r}(h_{r})$ $\displaystyle
d^{4r}(U_{3}^{r-1})$
$\displaystyle=d^{4r}(U_{4}^{r-1})=2^{r-1}p_{r}(h_{2n+2-r})$ $\displaystyle
d^{4r}(Z^{2r-1})$ $\displaystyle=2^{r}p_{r}(h_{2n+2-r}).$

This may be verified by induction in exactly the same way as in the cases
$n=4m,4m+1$. We conclude that $E^{4n+5}_{8s,3}=0=E^{4n+5}_{8s+4,3}$ and hence
$E^{4n+5}=E^{\infty}$. In addition, for $0\leq r\leq n-1$ we obtain

$E^{\infty}_{8s,-4r}=\operatorname{Ker}(d^{4(r+1)})=\left\\{\begin{array}[]{ll}R\\{(U_{1}^{r}-U_{2}^{r}),(U_{3}^{r}-U_{4}^{r})\\}&\mbox{if
}m=4n+2\\\ R\\{(U_{1}^{r}-U_{2}^{r})\\}&\mbox{if }m=4n+3\end{array}\right.$

Furthermore, for $m=4n+2$ we have
$d^{4(n+1)}(U_{i}^{n})=2^{n}p_{n+1}(h_{n+1})$ for each $1\leq i\leq 4$, and
for $m=4n+3$ we have

$2d^{4(n+1)}(U_{i}^{n})=d^{4(n+1)}(Z^{2n+1})=2^{n+1}p_{n+1}(h_{2n+1})$

for $i=1,2$. From this we deduce that

$E^{\infty}_{8s,-4n}=\left\\{\begin{array}[]{ll}R\\{(U_{1}^{n}-U_{2}^{n}),(U_{3}^{n}-U_{4}^{n}),(U_{1}^{n}-U_{3}^{n})\\}&\mbox{if
}m=4n+2\\\ R\\{(U_{1}^{n}-U_{2}^{n}),(2U_{1}^{n+1}-Z^{2n+1})\\}&\mbox{if
}m=4n+3\end{array}\right.$

In all other degrees we have $E^{1}_{s,t}=E^{\infty}_{s,t}$. For $m=4n+2$ we
conclude that $E^{\infty}_{0}$ is freely generated by
$\\{U_{1}-U_{2},U_{3}-U_{4},U_{1}^{n}-U_{3}^{n},U_{1}^{n+1}\\}$ and for
$m=4n+3$ we conclude that $E^{\infty}_{0}$ is freely generated by
$((U_{1}^{0}-U_{2}^{0}),Z_{\lambda}^{0},2U_{1}^{n}-Z^{2n+1},U_{1}^{n+1}\\}$.
In both cases $E^{\infty}_{4}=0$. This completes the final two cases and hence
the proof. ∎

###  6.3. The Case $I^{\infty}$

This is the simplest calculation. Let $\Gamma\subset\operatorname{SU}(2)$ be a
finite subgroup and write as usual $M=DCI(\overline{Y}_{\Gamma})$ and
$\mathcal{C}$ for the set of flat connections. Introduce a variable $T_{\eta}$
for each $\eta\in\mathcal{C}^{f.red}$ and a variable $S_{\lambda}$ for each
$\lambda\in\mathcal{C}^{red}$, where $|T_{\eta}|=-4$, $|S_{\lambda}|=-2$. Then
according to Corollary 5.30 we have $H^{\infty}_{A}(\alpha)=0$ for $\alpha$
irreducible, $H^{\infty}_{A}(\lambda)=R[S_{\lambda},S_{\lambda}^{-1}]$ for
$\lambda$ reducible and $H^{\infty}_{A}(\eta)=R[T_{\eta},T_{\eta}^{-1}]$ for
$\eta$ fully reducible. The $R[U]$-module structure is given by $U\cdot
T^{i}_{\eta}=T^{i+1}_{\eta}$ and $U\cdot S^{i}_{\lambda}=S^{i-2}_{\lambda}$
for each $i\in\mathbb{Z}$.

######  Definition 6.15.

For an $R[U]$-module $X$ we define the mod $8$ periodic $R[U]$-module
$X^{\Pi_{\infty},8}$ degreewise by

$X^{\Pi_{\infty},8}_{n}=\prod_{s\to\infty}X_{n+8s}$

and define $U\colon X^{\Pi_{\infty},8}_{n}\rightarrow
X^{\Pi_{\infty},8}_{n-4}$ to be the product over $U\colon X_{n+8s}\rightarrow
X_{n-4+8s}$ for $s\in\mathbb{Z}$.

######  Theorem 6.16.

The Tate equivariant instanton Floer homology associated with the trivial
$\operatorname{SU}(2)$-bundle over $\overline{Y}_{\Gamma}$ is given by

$I^{\infty}(\overline{Y}_{\Gamma})=\left[\left(\bigoplus_{\eta\in\mathcal{C}^{f.red}}R[T_{\eta},T_{\eta}^{-1}][j(\eta)]\right)\oplus\left(\bigoplus_{\lambda\in\mathcal{C}^{red}}R[S_{\lambda},S_{\lambda}^{-1}][j(\lambda)]\right)\right]^{\Pi_{\infty},8}.$

###### Proof.

Let $(E^{r},d^{r})_{r\geq 1}$ be the index spectral sequence with

$E^{1}_{s,t}=\bigoplus_{j(\alpha)\equiv s}H^{\infty}_{A}(\alpha)_{t}\implies
I^{\infty}(\overline{Y}_{\Gamma})_{s+t}.$

From the fact that $j(\alpha)\equiv 0\ (\mathrm{mod}\ 4)$ for each
$\alpha\in\mathcal{C}$ and that $H^{\infty}_{A}(\alpha)$ is concentrated in
even degrees for all types of orbits, we conclude that $E^{1}_{s,t}=0$ for all
$(s,t)$ with $s$ odd or $t$ odd. It follows that there are no nontrivial
differentials in the spectral sequence so that $E^{1}=E^{\infty}$. This
implies by Proposition 5.57 part (iii) and Theorem 5.55 that the spectral
sequence converges strongly.

Write $I=I^{\infty}(\overline{Y}_{\Gamma})$. To resolve the extension problems
observe first that $E^{1}_{4s}$ is a (shifted) direct sum of $R[U]$-modules
$R[T_{\eta},T_{\eta}^{-1}]$ and $R[S_{\lambda},S_{\lambda}^{-1}]$. These
contain free $R[U]$-submodules $R[T_{\eta}]$ and $R[S_{\lambda}]$ (freely
generated by $\\{S^{0}_{\lambda},S^{1}_{\lambda}\\}$), respectively, and the
whole module is the localization of these free submodules in the
multiplicatively closed subset $\\{U^{i}\\}_{i\geq 0}\subset R[U]$. By
Corollary 5.50 $U\colon I\rightarrow I$ is an isomorphism, and the same is
valid for the $R[U]$-submodule $F_{s}I$ for each $s\in\mathbb{Z}$. We may
therefore for each $s$ construct a splitting of the short exact sequence

${0}$${F_{s-1}I}$${F_{s}I}$${\overline{F}_{s}I\cong
E^{\infty}_{s}}$${0}$$\scriptstyle{\xi_{s}}$

by defining $\xi$ to be a section on the free submodule and then extending to
the whole module using the universal property of localization. By induction we
obtain an isomorphism

$F_{s}I/F_{t}I\cong\bigoplus_{q=t+1}^{s}E^{\infty}_{q}$

of $R[U]$-modules for each pair $s>t$. Since the filtration of $I$ is
exhaustive and complete Hausdorff we obtain, exploiting the periodicity,

$I\cong\operatorname{lim}_{t}\operatorname{colim}_{s}F_{s}I/F_{t}I\cong\prod_{s\to-\infty}E^{\infty}_{s}\cong\prod_{s\to-\infty}(E^{\infty}_{0}[-8s]\oplus
E^{\infty}_{4}[-8s])$

and the final term is precisely $(E^{\infty}_{0}\oplus
E^{\infty}_{4})^{\Pi_{\infty},8}=(E^{1}_{0}\oplus
E^{1}_{4})^{\Pi_{\infty},8}$. The proof is completed by observing that

$E^{1}_{0}\oplus
E^{1}_{4}=\left(\bigoplus_{\eta\in\mathcal{C}^{f.red}}R[T_{\eta},T_{\eta}^{-1}][j(\eta)]\right)\oplus\left(\bigoplus_{\lambda\in\mathcal{C}^{red}}R[S_{\lambda},S_{\lambda}^{-1}][j(\lambda)]\right).$

∎

######  Proposition 6.17.

For each finite subgroup $\Gamma\subset\operatorname{SU}(2)$ the homology norm
map

$H(N)\colon I^{+}(\overline{Y}_{\Gamma})_{n}\rightarrow
I^{-}(\overline{Y}_{\Gamma})_{n+3}$

vanishes. In particular, the exact triangle of $R[U]$-modules of Corollary
5.44 splits into a short exact sequence

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###### Proof.

The calculations of Theorem 6.7, Theorem 6.13 and Theorem 6.14 show that
$I^{+}(\overline{Y}_{\Gamma})$ and $I^{-}(\overline{Y}_{\Gamma})$ are
concentrated in even degrees. Therefore, as the norm map has degree $3$, it
must vanish. ∎

###  6.4. Calculations for $Y_{\Gamma}$

In this section we explain, omitting some details, the necessary modifications
needed to calculate the equivariant instanton Floer groups for $Y_{\Gamma}$,
that is, $S^{3}/\Gamma$ equipped with the standard orientation inherited from
$S^{3}$. The tools needed to handle this are contained in [Eis19, Theorem
7.10]. First, the given result states that there is an isomorphism
$I^{\infty}(Y_{\Gamma})\cong I^{\infty}(\overline{Y}_{\Gamma})$. Second, in
the proof it is verified that $DCI(Y_{\Gamma})\cong
DCI(\overline{Y}_{\Gamma})^{\vee}$ and from this it is not hard to express
$DCI(Y_{\Gamma})$ as the totalization of a multicomplex.

######  Lemma 6.18.

Let $i\colon\mathcal{C}\rightarrow\mathbb{Z}/8$ denote the grading for
$Y_{\Gamma}$ and let $j\colon\mathcal{C}\rightarrow\mathbb{Z}/8$ denote the
grading for $\overline{Y}_{\Gamma}$. Then

$i(\alpha)=\left\\{\begin{array}[]{ll}j(\alpha)-3&\mbox{ if
}\alpha\in\mathcal{C}^{irr}\\\ j(\alpha)-2&\mbox{ if
}\alpha\in\mathcal{C}^{red}\\\ j(\alpha)&\mbox{ if
}\alpha\in\mathcal{C}^{f.red}\end{array}\right.$

Moreover,
$DCI(Y_{\Gamma})=\operatorname{Tot}^{oplus}(DCI(Y_{\Gamma})_{*,*},\\{\partial^{r}\\}_{r=1}^{4})$
where

$DCI(Y_{\Gamma})_{s,t}=\bigoplus_{i(\alpha)\equiv s}H_{t}(\alpha),$

$\partial^{2}=0$ and $\partial^{1},\partial^{3},\partial^{4}$ are given for
$b_{\alpha}\in H_{0}(\alpha)\subset DCI(Y_{\Gamma})_{4s+1,0}$,
$\alpha\in\mathcal{C}^{irr}$, by

$\displaystyle\partial^{1}(b_{\alpha})$
$\displaystyle=\sum_{\eta\in\mathcal{C}^{f.red}}n_{\alpha\eta}b_{\eta}$
$\displaystyle\partial^{3}(b_{\alpha})$
$\displaystyle=\sum_{\lambda\in\mathcal{C}^{red}}n_{\alpha\lambda}t_{\lambda}$
$\displaystyle\partial^{4}(b_{\alpha})$
$\displaystyle=\sum_{\beta\in\mathcal{C}^{irr}}n_{\alpha\beta}t_{\beta}.$

and vanish otherwise.

The patterns seen in the calculations of $I^{\pm}(\overline{Y}_{\Gamma})$ are
reversed for $I^{\pm}(Y_{\Gamma})$. Compare the following with Lemma 6.1 and
Lemma 6.9.

######  Lemma 6.19.

Let $(E^{r},d^{r})_{r\geq 1}$ be the index spectral sequence with

$E^{1}_{s,t}=\bigoplus_{i(\alpha)\equiv s}H^{-}_{A}(\alpha)_{t}\implies
I^{-}(Y_{\Gamma})_{s+t}.$

Then the spectral sequence immediately degenerates, i.e., $E^{1}=E^{\infty}$.

Let $(E^{r},d^{r})_{r\geq 1}$ be the index spectral sequence with

$E^{1}_{s,t}=\bigoplus_{i(\alpha)\equiv s}H^{+}_{A}(\alpha)_{t}\implies
I^{+}(Y_{\Gamma})_{s+t}.$

Then the only possibly nontrivial differentials are $d^{2r+1}\colon
E^{2r+1}_{4s+1,0}\rightarrow E^{2r+1}_{4s-2r,2r}$ for $r\geq 0$ and
$s\in\mathbb{Z}$.

###### Proof.

In the first case recall that $H^{-}_{A}(\alpha)=R\cdot h_{\alpha}$ with
$|h_{\alpha}|=3$ for $\alpha\in\mathcal{C}^{irr}$,
$H^{-}_{A}(\lambda)=R[Z_{\lambda}][2]$, $|Z_{\lambda}|=-2$ for
$\lambda\in\mathcal{C}^{red}$ and $H^{-}_{A}(\eta)=R[U_{\eta}]$,
$|U_{\eta}|=-4$ for $\eta\in\mathcal{C}^{f.red}$. By the above lemma
$i(\rho)\equiv 0\ (\mathrm{mod}\ 2)$ for each
$\rho\in\mathcal{C}^{ref}\cup\mathcal{C}^{f.red}$ and $i(\alpha)\equiv 1\
(\mathrm{mod}\ 4)$ for each $\alpha\in\mathcal{C}^{irr}$. Therefore
$E^{1}_{s,t}$ can only be nonzero if the total degree $s+t$ is even. As each
differential $d^{r}$ reduces the total degree by one, they must all vanish and
the first assertion follows.

In the second case, $H^{+}_{A}(\eta)=R[V_{\eta}]$, $|V_{\eta}|=4$,
$H^{+}_{A}(\lambda)=R[W_{\lambda}]$, $|W_{\lambda}|=2$ and
$H^{+}_{A}(\alpha)=R\cdot g_{\alpha}$, $|g_{\alpha}|=0$, where
$\eta\in\mathcal{C}^{f.red}$, $\lambda\in\mathcal{C}^{red}$ and
$\alpha\in\mathcal{C}^{irr}$. Due to the indexing mentioned above the only
possibly nontrivial module $E^{1}_{s,t}$ with $s+t$ odd are $E^{1}_{4s+1,0}$.