$e^{K}_{G}(V_{j}(1))=(1-\tau^{-j}\xi^{-1})^{a_{j}}$, where $\xi$ denote the
standard $1$-dimensional representation of $S^{1}$. Hence
$K^{*}_{G}(\mathbb{P}(V_{j}))\cong R(G)[\xi,\xi^{-1}]/(e^{K}_{G}(V_{j}(1)))$.
Hence

$S^{-1}K^{*}_{G}(\mathbb{P}(V_{j}))\cong\frac{\mathbb{Q}(\omega)[\xi,\xi^{-1}]}{(1-\tau^{-j}\xi^{-1})^{a_{j}})}.$

Let $\mathbb{C}_{j}$ denote the $1$-dimensional representation of $G$ where
$g$ acts as multiplication by $\omega^{j}$. Then $N_{j}$ is a direct sum of
line bundles of the form $\mathbb{C}_{i}(1)$ where $i\neq j$. So it suffices
to show that $e^{K}_{G}(\mathbb{C}_{i}(1))=1-\omega^{-i}\xi^{-1}$ is
invertible. We have

$\displaystyle 1-\omega^{-i}\xi^{-1}$
$\displaystyle=1-\omega^{-(i-j)}+\omega^{-(i-j)}(1-\omega^{-j}\xi^{-1})$
$\displaystyle=(1-\omega^{-(i-j)})(1-u)$

where $u=(1-\omega^{i-j})^{-1}(1-\omega^{-j}\xi^{-1})$. Then $u^{a_{j}}=0$ in
$S^{-1}K^{*}_{G}(\mathbb{P}(V_{j}))$ because
$0=e^{K}_{G}(V_{j}(1))=(1-\omega^{-j}\xi^{-1})^{a_{j}}$. Therefore

$(1-\omega^{-i}\xi^{-1})^{-1}=(1-\omega^{-(i-j)})^{-1}(1+u+u^{2}+\cdots),$

where there are only finitely many terms on the right since $u$ is nilpotent.
∎

Equipped now with the $K$-theoretic equivalents of Lemmas 6.2 and 6.4 and
using the localisation theorem in $K$-theory, we obtain the $K$-theoretic
equivalent of Theorem 6.5:

######  Theorem 6.7.

Let $f$ be a $G$-monopole map. Assume that $\mathbb{R}\oplus H^{+}$ can be
given a $G$-invariant complex structure which we use to $K$-orient $H^{+}$.
Then

$\chi(SW^{K,\phi}_{G,f}(\theta),g)=\chi(e^{K}_{G}(H^{+}/(H^{+})^{g}),g)\sum_{j=0}^{n-1}\chi(SW^{K,\phi}_{G,f^{s_{j}}}(e^{K}_{G_{\mathfrak{s}}}(D/D_{j})^{-1}\theta,g).$

###  6.3. Free actions revisited

In this section we will apply Theorem 6.7 to the case of free actions. Let $X$
be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let $G$ be a
group which acts smoothly, orientation preservingly and freely on $X$. Assume
that $b_{+}(X)^{G}=b_{+}(X/G)>0$. Let $\mathfrak{s}$ be a $G$-invariant spinc-
structure and assume that $d(X,\mathfrak{s})=0$. Since
$d(X,\mathfrak{s})=2d-b_{+}(X)-1=0$, this implies that $b_{+}(X)$ is odd.

In the case of free actions with $d(X,\mathfrak{s})=0$, we have a method of
$K$-orienting $\mathbb{R}\oplus H^{+}(X/H)$ for every subgroup $H$. First,
from the proof of Proposition 5.13 we see that $\mathbb{R}\oplus H^{+}(X)$ is
isomorphic to several copies of the real regular representation. In fact,
since $d(X,\mathfrak{s})=0$, there are $2d_{0}$ copies, where $d_{0}=d/|G|$.
Therefore, $\mathbb{R}\oplus H^{+}(X)$ can be (non-canonically) equipped with
a $G$-invariant complex structure $I$ for which $\mathbb{R}\oplus H^{+}(X)$ is
isomorphic as a complex representation to $d_{0}$ copies of the complex
regular representation. Fix a choice of such a complex structure. We use this
to define a $K$-orientation on $\mathbb{R}\oplus H^{+}(X)$. Then for any
$H\subseteq G$, we have an isomorphism $\mathbb{R}\oplus
H^{+}(X/H)\cong(\mathbb{R}\oplus H^{+}(X))^{H}$. Moreover, $(\mathbb{R}\oplus
H^{+}(X))^{H}$ is a complex subspace of $\mathbb{R}\oplus H^{+}(X)$ and hence
it has an induced complex structure and a $K$-orientation.

######  Theorem 6.8.

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let $G$
be a finite group which acts smoothly and freely on $X$. Let $\mathfrak{s}$ be
a spinc-structure whose isomorphism class is fixed by $G$ and suppose that
$d(X,\mathfrak{s})\geq 0$. Assume that $b_{+}(X)^{G}>0$ and if
$b_{+}(X)^{G}=1$ then fix a chamber $\phi$.

Fix a choice of $G$-invariant complex structure on $\mathbb{R}\oplus H^{+}(X)$
for which $\mathbb{R}\oplus H^{+}(X)$ is isomorphic to a sum of copies of the
complex regular represention and use this to $K$-orient $\mathbb{R}\oplus
H^{+}(X/H)$ for every $H\subseteq G$, as described above. Then we have

$\sum_{g\in G}\sum_{s^{\prime}}SW(X/\langle
g\rangle,\mathfrak{s}^{\prime})=0\;({\rm mod}\;|G|)$

where the second sum is over spinc-structures on $X/\langle g\rangle$ whose
pullback to $X$ is isomorphic to $\mathfrak{s}$. Equivalently, we have

$\sum_{(H)\text{
cyclic}}\phi(|H|)|G/N_{G}H|\sum_{\mathfrak{s}^{\prime}|q_{H}^{*}(\mathfrak{s}^{\prime})\cong\mathfrak{s}}SW(X/H,\mathfrak{s}^{\prime})=0\;({\rm
mod}\;|G|)$

where the first sum is over conjugacy classes of cyclic subgroups of $G$ and
the second sum is over spinc-structures on $X/H$ whose pullback to $X$ is
isomorphic to $\mathfrak{s}$. If $b_{+}(X/H)=1$ then
$SW(X/H,\mathfrak{s}^{\prime})$ is defined using the chamber $\phi$.

###### Proof.

Let $f$ be the $G$-equivariant Bauer–Furuta monopole map of
$(X,\mathfrak{s})$. Let $M=SW^{K}_{G,f}(1)\in R(G)$. The restriction of $M$ to
$R(1)\cong\mathbb{Z}$, which is the rank of $M$, is just the ordinary
($K$-theoretic) Seiberg–Witten invariant of $(X,\mathfrak{s})$. Since
$d(X,\mathfrak{s})=0$, it follows that this equals the usual Seiberg–Witten
invariant $SW(X,\mathfrak{s})$ (see [4, §6]). Thus
$\chi(M,1)=SW(X,\mathfrak{s})$.

Let $g\in G$ have order $n$. Set $H=\langle g\rangle$. Since $H$ is cyclic, we
can choose a splitting $H_{\mathfrak{s}}\cong S^{1}\times H$. Then as in
Section 6.2, let $s_{j}$ denote the splitting given by
$s_{j}(g)=(\omega^{-j},g)$, $\omega=e^{2\pi i/n}$. Let $\mathbb{C}_{j}$ be the
$1$-dimensional representation of $H$ where $g$ acts as multiplication by
$\omega^{j}$. Use the splitting $s_{0}$ to regard $D$ as a representation of
$H$. Since the action is free, we have

(6.1) $D\cong\bigoplus_{j=0}^{n-1}\mathbb{C}_{j}^{d/n}.$

Recall that we have fixed a complex structure on $\mathbb{R}\oplus H^{+}(X)$
for which $\mathbb{R}\oplus H^{+}(X)$ is isomorphic to a sum of copies of the
regular representation of $G$. Restricting to $H$, we therefore have that
$\mathbb{R}\oplus H^{+}(X)$ as a representation of $H$ is isomorphic to a sum
of copies of the regular representation. Hence

(6.2) $H^{+}/(H^{+})^{H}\cong\bigoplus_{j=1}^{n-1}\mathbb{C}_{j}^{d/n}.$

Consider the reduced monopole map $f^{s_{j}}$. Its $H$-equivariant
$K$-theoretic Seiberg–Witten invariant takes the form of a map of
$R(H)$-modules:

$SW^{K}_{H,f^{s_{j}}}:R(H)[\xi,\xi^{-1}]\to R(H).$

We have $R(H)\cong\mathbb{Z}[t]/(t^{n}-1)$, where $t=[\mathbb{C}_{1}]$. By
Proposition 6.1, it follows that
$SW^{K}_{H,f^{s_{j}}}(1)=SW(X/H,\mathfrak{s}_{s_{j}})$. Under the splitting
$s_{j}$, $s_{j}^{*}(\xi)=t^{-j}$, hence $s_{j}^{*}(t^{j}\xi)=1$. This implies
that $SW^{K}_{H,f^{s_{j}}}(t^{j}\xi)=SW(X/H,\mathfrak{s}_{s_{j}})$, hence
$SW^{K}_{H,f^{s_{j}}}(\xi)=t^{-j}SW(X/H,\mathfrak{s}_{s_{j}})$ and more
generally $SW^{K}_{H,f^{s_{j}}}(\xi^{m})=t^{-mj}SW(X/H,\mathfrak{s}_{s_{j}})$
for any $m$. Therefore,

$\chi(SW^{K}_{H,f^{s_{j}}}(\xi^{m}),g)=\omega^{-mj}SW(X/H,\mathfrak{s}_{s_{j}}).$

Theorem 6.7 gives

(6.3)
$\chi(M,g)=\chi(e^{K}_{H}(H^{+}/(H^{+})^{H}),g)\sum_{j=0}^{n-1}\chi(SW^{K}_{H,f^{s_{j}}}(e^{K}_{H_{\mathfrak{s}}}(D/D_{j})^{-1}),g).$

From Equation (6.1) we have

$e^{K}_{H_{\mathfrak{s}}}(D/D_{j})=\prod_{i\neq
j}(1-\omega^{-i}\xi^{-1})^{d/n}$

and hence

$\displaystyle\chi(SW^{K}_{H,f^{s_{j}}}(e^{K}_{H_{\mathfrak{s}}}(D/D_{j})^{-1},g)$
$\displaystyle=\left(\prod_{i\neq
j}(1-\omega^{j-i})^{-d/n}\right)SW(X/H,\mathfrak{s}_{s_{j}})$
$\displaystyle=\left(\prod_{i=1}^{n-1}(1-\omega^{j})\right)^{-d/n}SW(X/H,\mathfrak{s}_{s_{j}}).$

From Equation (6.2) we have

$e^{K}_{H}(H^{+}/(H^{+})^{H})=\prod_{i=1}^{n-1}(1-t^{-i})^{d/n},$

hence

$\chi(e^{K}_{H}(H^{+}/(H^{+})^{H}),g)=\left(\prod_{i=1}^{n-1}(1-\omega^{-i})\right)^{d/n}.$

Substituting into Equation (6.3) gives

$\chi(M,g)=\sum_{j=0}^{n-1}SW(X/H,\mathfrak{s}_{s_{j}}).$

Now since $\sum_{g\in G}\chi(M,g)=0\;({\rm mod}\;|G|)$, we have shown that

$\sum_{g\in G}\sum_{s^{\prime}}SW(X/\langle
g\rangle,\mathfrak{s}^{\prime})=0\;({\rm mod}\;|G|).$

∎

###  6.4. A constraint on smooth group actions

######  Theorem 6.9.

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let $G$
be a finite group acting on $X$ by orientation preserving diffeomorphisms. Let
$\mathfrak{s}$ be a $G$-invariant spinc-structure. Assume that
$b_{+}(X)^{G}>0$ and that $d(X,\mathfrak{s})=0$, hence $b_{+}(X)$ is odd.
Suppose also that $\mathbb{R}\oplus H^{+}(X)$ can be equipped with a
$G$-invariant complex structure. Suppose that $SW(X,\mathfrak{s})\neq 0\;({\rm
mod}\;|G|)$. Then for some non-trivial cyclic subgroup $\\{1\\}\neq H\subseteq
G$ and some splitting $s:H\to G_{\mathfrak{s}}$, we have
$2\,dim_{\mathbb{C}}(D^{sH})>dim_{\mathbb{R}}(H^{+}(X)^{H})$.

###### Proof.

Let $f$ be the $G$-equivariant Bauer–Furuta monopole map of
$(X,\mathfrak{s})$. Let $M=SW^{K}_{G,f}(1)\in R(G)$. Then
$\chi(M,1)=SW(X,\mathfrak{s})$. The complex structure on $\mathbb{R}\oplus
H^{+}(X)$ yields a complex structure on $\mathbb{R}\oplus H^{+}(X)^{H}$ for
every $H\subseteq G$ and hence a $K$-orientation.

Let $g\in G$ have order $n$. Theorem 6.7 gives

$\chi(M,g)=\chi(e^{K}_{H}(H^{+}/(H^{+})^{H}),g)\sum_{s}\chi(SW^{K,\phi}_{H,f^{s}}(e^{K}_{G_{\mathfrak{s}}}(D/D^{sH})^{-1}\theta,g)$

where the sum is over splittings $s:H\to H_{\mathfrak{s}}$. Now since $s(H)$
acts trivially on the domain and codomain of $f^{s}$, there are no
obstructions to achieving equivariant transversality. The expected dimension
of the moduli space of $sH$-invariant solutions to the Seiberg–Witten
equations is $2\,dim_{\mathbb{C}}(D^{sH})-dim_{\mathbb{R}}(H^{+}(X)^{H})-1$.
If this is negative then $SW^{K}_{H,f^{s}}=0$.

Now we re-write the congruence $\sum_{g\in G}\chi(M,g)=0\;({\rm mod}\;|G|)$ as

$SW(X,\mathfrak{s})=-\sum_{g\neq 1}\chi(M,g)\;({\rm mod}\;|G|).$

Hence if $SW(X,\mathfrak{s})\neq 0\;({\rm mod}\;|G|)$ then $\chi(M,g)\neq 0$
for some $g\in G\setminus\\{1\\}$ and thus
$2\,dim_{\mathbb{C}}(D^{sH})>dim_{\mathbb{R}}(H^{+}(X)^{H})$ for some
splitting $s$ of $H$. ∎

###  6.5. Divisibility conditions

Let $f:S^{V,U}\to S^{V^{\prime},U^{\prime}}$ be an ordinary monopole map over
a point, $V=\mathbb{C}^{a},V^{\prime}=\mathbb{C}^{a^{\prime}}$,
$U=\mathbb{R}^{b},U^{\prime}=\mathbb{R}^{b^{\prime}}$, $d=a-a^{\prime}$,
$b_{+}=b^{\prime}-b$. Assume that $b_{+}\geq 1$ and that $2d-b_{+}-1=2m$ is
even and non-negative. Choose an orientation on $H^{+}$ and choose a chamber
if $b_{+}=1$. Then the abstract Seiberg–Witten invariant of $f$ is defined and
given by $SW_{f}(x^{m})\in H^{0}(pt;\mathbb{Z})=\mathbb{Z}$. Let us denote it
by $SW(f)$ for simplicity. If $f$ is the Bauer–Furuta monopole map of
$(X,\mathfrak{s})$, then $SW(f)=SW(X,\mathfrak{s})$ is the usual
Seiberg–Witten invariant.

######  Theorem 6.10.

Let $f:S^{V,U}\to S^{V^{\prime},U^{\prime}}$ be an ordinary monopole map over
a point and let $p$ be a prime. If $SW(f)\neq 0\;({\rm mod}\;p)$, then
$(b_{+}-1)/2$ is divisible by $p^{e+1}$ whenever
$p^{e}\leq\left\lfloor\frac{m}{p-1}\right\rfloor$, where $2d-b_{+}-1=2m$. In
particular, if $SW(f)\neq 0\;({\rm mod}\;p)$ and $b_{+}\neq 1\;({\rm
mod}\;2p)$ then $m\leq p-2$.

###### Proof.

We give the proof in the case $p$ is odd. The case $p=2$ is similar. We will
make use of the Steenrod powers $P^{i}$. Using the Borel model, the Steenrod
powers can be defined on $S^{1}$-equivariant cohomology with
$\mathbb{Z}_{p}$-coefficients. Let $W$ be an $S^{1}$-equivariant vector bundle
on $B$ with Thom class $\tau_{B}$. Then

(6.4) $P^{j}(\tau_{W})=q_{j}(W)\tau_{W}$

for some characteristic class $q_{j}(W)\in
H^{2(p-1)j}_{S^{1}}(B;\mathbb{Z}_{p})$. Setting $q_{t}=q_{0}+tq_{1}+\cdots$,
one finds [20, Chapter 19] that if $W$ is complex with Chern roots
$\\{a_{i}\\}$ then:

$q_{t}(W)=\prod_{i}(1+ta_{i}^{p-1}).$

Setting $P_{t}=P^{0}+tP^{1}+\cdots$, we have
$P_{t}(\tau_{W})=q_{t}(W)\tau_{W}$.

Considering $\mathbb{C}$ with the standard $S^{1}$-action and $\mathbb{R}$
with the trivial $S^{1}$-action as $S^{1}$-equivariant vector bundles over
$B=pt$, we have

$q_{t}(\mathbb{C})=1+tx^{p-1},\quad q_{t}(\mathbb{R})=1.$

Since $deg(x)=2$, we have $P^{1}(x)=x^{p}$ and $P^{j}(x)=0$ for $j>1$. Hence
$P_{t}(x)=(1+tx^{p-1})x$. Consider
$f^{*}:H^{*}_{S^{1}}(S^{V^{\prime},U^{\prime}};S^{U};\mathbb{Z}_{p})\to
H^{*}_{S^{1}}(S^{V,U};S^{U};\mathbb{Z}_{p})$. Suspending if necessary, we can
assume $U$ is even-dimensional. Then
$f^{*}(\tau^{\phi}_{V^{\prime},U^{\prime}})=\delta\tau_{U}\eta^{\phi}$, where
$\eta^{\phi}=SW(f)x^{a-1-m}$. Applying $P_{t}$ to both sides of

$SW(f)\delta\tau_{U}x^{a-1-m}=f^{*}(\tau^{\phi}_{V^{\prime},U^{\prime}})$

gives

$\displaystyle(1+tx^{p-1})^{a-1-m}SW(f)\delta\tau_{U}x^{a-1-m}$
$\displaystyle=(1+tx^{p-1})^{a^{\prime}}f^{*}(\tau^{\phi}_{V^{\prime},U^{\prime}})$
$\displaystyle=(1+tx^{p-1})^{a^{\prime}}SW(f)\delta\tau_{U}x^{a-1-m}.$

(To prove this we used that Equation (6.4) also holds for the refined Thom
class $\tau^{\phi}_{V^{\prime},U^{\prime}}$. This follows by a straightforward
extension of the usual proof. We also used that $P_{t}$ commutes with the
coboundary operator $\delta$).

If $SW(f)\neq 0\;({\rm mod}\;p)$, then the above equation reduces to

$(1+tx^{p-1})^{a-1-m}x^{a-1-m}=(1+tx^{p-1})^{a^{\prime}}x^{a-1-m}.$

This is an equality in
$H^{*}(\mathbb{P}(V);\mathbb{Z}_{p})[[t]]\cong\mathbb{Z}_{p}[x][[t]]/(x^{a})$.
Multiplying both sides by
$(1+tx^{p-1})^{-a^{\prime}}=1-a^{\prime}tx^{p-1}+\cdots$ gives

$(1+tx^{p-1})^{d-1-m}x^{a-1-m}=x^{a-1-m}\text{ in
}\mathbb{Z}_{p}[x][[t]]/(x^{a}).$

Expanding on the left, we get

$\binom{d-1-m}{j}=0\;({\rm mod}\;p)\text{ whenever }1\leq j\leq m/(p-1).$

Let $p^{e}\leq\lfloor m/(p-1)\rfloor$. Setting $j=p^{u}$ with $u=0,1,\dots,e$
we get

$\binom{d-1-m}{p^{u}}=0\;({\rm mod}\;p)\text{ for }0\leq u\leq e.$

This implies that $d-1-m$ is divisible by $p^{e+1}$. Noting that
$2m=2d-b_{+}-1$, we have $d-m-1=(b_{+}-1)/2$, and hence $(b_{+}-1)/2$ is
divisible by $p^{e+1}$. ∎

######  Remark 6.11.

The conclusion that $m\leq p-2$ when $SW(f)\neq 0\;({\rm mod}\;p)$ and
$b_{+}\neq 1\;({\rm mod}\;2p)$ was also shown in [17, Corollary 1.5].

###  6.6. $\mathbb{Z}_{p}$-actions

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let
$G=\mathbb{Z}_{p}=\langle g\rangle$ act on $X$ by orientation preserving
diffeomorphism, where $p$ is a prime. For convenience we will work with
$\mathbb{Z}_{p}$-coefficients. For odd $p$ we have
$H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z}_{p})\cong\mathbb{Z}_{p}[u,v]/(u^{2})$
where $deg(u)=1$, $deg(v)=2$. For $p=2$ we have
$H^{*}_{\mathbb{Z}_{2}}(pt;\mathbb{Z}_{2})\cong\mathbb{Z}_{2}[u]$ where
$deg(u)=2$. In this case we set $v=u^{2}$ and sometimes write $u=v^{1/2}$.

Suppose that $H^{+}(X)^{G}\neq 0$ and let $\phi$ be a chamber. Let
$\mathfrak{s}$ be a $G$-invariant spinc-structure. Fix a trivialisation
$G_{\mathfrak{s}}\cong S^{1}\times G$. This amounts to choosing a lift of
$\mathbb{Z}_{p}$ to the spinor bundles associated to $\mathfrak{s}$. Then the
equivariant Seiberg–Witten invariants
$SW_{\mathbb{Z}_{p},X,\mathfrak{s}}^{\phi}:H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z}_{p})[x]\to
H^{*-d(X,\mathfrak{s})}_{\mathbb{Z}_{p}}(pt;\mathbb{Z}_{p})$ are defined. Let
$\mathbb{C}_{j}$ be the $1$-dimensional complex representation of $G$ where
$g$ acts as multiplication by $\omega^{j}$, $\omega=e^{2\pi i/p}$. Then
$D=\bigoplus_{j=0}^{p-1}\mathbb{C}_{j}^{d_{j}}$ for some
$d_{0},\dots,d_{p-1}$. The weights $d_{j}$ can be computed using the $G$-spin
theorem. Let $b_{0}$ be the dimension of $H^{+}(X)^{G}$. To each
$j\in\mathbb{Z}_{p}$ we obtain a splitting $s_{j}:\mathbb{Z}_{p}\to
S^{1}\times\mathbb{Z}_{p}$ given by $s_{j}(g)=(\omega^{-j},g)$. Let $f$ denote
the $G$-monopole map associated to $(X,\mathfrak{s})$ and $f^{s_{j}}$ the
reduced monopole map. The expected dimension of the moduli space for
$f^{s_{j}}$ is $2d_{j}-b_{0}-1$. Let $\delta_{j}=d_{j}-(b_{0}+1)/2$. If this
is integral and non-negative then we obtain a reduced Seiberg–Witten invariant
$\overline{SW}^{s_{j},\phi}_{G,X,\mathfrak{s}}=SW_{f^{s_{j}}}^{\phi}(x^{\delta_{j}})\in\mathbb{Z}$.
If $\delta_{j}$ is non-integral or negative, then we set
$\overline{SW}^{s_{j},\phi}_{G,X,\mathfrak{s}}=0$. Recall that each splitting
$s_{j}$ defines an isomorphism
$\psi_{s_{j}}:H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z}_{p})\to
H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})[x]$. If we use $s_{0}$ to identify
$H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z}_{p})$ with
$H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})[x]$ then $\psi_{s_{j}}$ is given by
$\psi_{s_{j}}(x)=x-jv$, $\psi_{s_{j}}(v)=v$. Since $s_{j}(G)$ acts trivially
on the domain and codomain of $f_{s_{j}}$, we have that
$SW^{\phi}_{G,f^{s_{j}}}(\psi_{s_{j}}^{-1}(x^{m}))=\overline{SW}^{s_{j},\phi}_{G,X,\mathfrak{s}}$
if $m=\delta_{j}$ and is zero otherwise. Then since
$\psi_{s_{j}}^{-1}(x)=x+jv$, it follows that for any
$\theta\in\mathbb{Z}_{p}[x,v]$,
$SW_{G,f^{s_{j}}}^{\phi}(\theta)=c_{j}\overline{SW}^{s_{j},\phi}_{G,X,\mathfrak{s}}$,
where $c_{j}$ is the coefficient of $(x+jv)^{\delta_{j}}$ when $\theta$ is
written as a polynomial in $x+jv$ with coefficients in $\mathbb{Z}_{p}[v]$.

Let $k=\mathbb{Z}_{p}(v)$ be the ring of rational functions in $v$ with
coefficients in $\mathbb{Z}_{p}$. Let $k(x)$ be the ring of rational functions
in $x$ with coefficients in $k$. For any $a\in k$ we have a natural
homomorphism $k(x)\to k((x-a))$ from $k(x)$ into the ring $k((x-a))$ of formal
Laurent series in $x-a$ with coefficients in $k$ and with finite polar part.
If $f\in k(x)$ then the image of $f$ in $k((x-a))$ can be uniquely written as
$f=\sum_{j=-n}^{\infty}c_{j}(x-a)^{j}$, $c_{j}\in k$. We refer to $c_{j}$ as
the coefficient of $(x-a)^{j}$ in the Laurent expansion of $f$ at $a$. For any
$j\in\mathbb{Z}_{p}$ and integers $n,n_{0},n_{1},\dots,n_{p-1}$, define
$c_{j}(n;n_{0},\dots,n_{p-1})$ to be the coefficient of $(x+jv)^{n}$ in the
Laurent expansion of $\prod_{i=0}^{p-1}(x+iv)^{n_{i}}$. It follows that

$c_{j}(n;n_{0},\dots,n_{p-1})=\left(\sum_{k_{i}}\prod_{i|i\neq
j}\binom{n_{i}}{k_{i}}(i-j)^{n_{i}-k_{i}}\right)v^{\sum_{i}n_{i}-n}$

where the sum is over non-negative integers
$k_{0},\dots,\hat{k}_{j},\dots,k_{p-1}$ such that
$k_{0}+\cdots+\hat{k}_{j}+\cdots+k_{p-1}=n-n_{j}$. We will extend the
definition of $c_{j}(n;n_{0},\dots,n_{p-1})$ to non-integer values of $n$ by
setting $c_{j}(n;n_{0},\dots,n_{p-1})=0$ if $n$ is non-integral.

######  Theorem 6.12.

For any non-negative integers $m_{0},\dots,m_{p-1}$, we have the following
equality in $H^{*}(\mathbb{Z}_{p};\mathbb{Z}_{p})$:

$\displaystyle
SW^{\phi}_{\mathbb{Z}_{p},X,\mathfrak{s}}(x^{m_{0}}(x+v)^{m_{1}}\cdots(x+(p-1)v)^{m_{p-1}})$
$\displaystyle\quad\quad=e_{\mathbb{Z}_{p}}(H^{+}/(H^{+})^{g})\sum_{j=0}^{p-1}c_{j}\left(-\dfrac{(b_{0}+1)}{2};m_{0}-d_{0},\dots,m_{p-1}-d_{p-1}\right)\overline{SW}^{s_{j},\phi}_{G,X,\mathfrak{s}}.$

###### Proof.

By Theorem 6.5, we have

$SW^{\phi}_{G,f}(\theta)=e_{G}(H^{+}/(H^{+})^{g})\sum_{j=0}^{p-1}SW^{\phi}_{G,f^{s_{j}}}(e_{G_{\mathfrak{s}}}(D/D_{j})^{-1}\theta).$

If $\theta=\prod_{i=0}^{p-1}(x+iv)^{m_{i}}$, then

$e_{G_{\mathfrak{s}}}(D/D_{j})^{-1}\theta=\prod_{i=0}^{p-1}(x+iv)^{m_{i}}\prod_{i|i\neq
j}(x+jv)^{-d_{i}}.$

Hence $SW^{\phi}_{G,f^{s_{j}}}(e_{G_{\mathfrak{s}}}(D/D_{j})^{-1}\theta)$
equals $SW^{G,s_{j},\phi}_{X,\mathfrak{s}}$ times the coefficient of
$(x+jv)^{\delta_{j}}$ in $\prod_{i=0}^{p-1}(x+iv)^{m_{i}}\prod_{i|i\neq
j}(x+jv)^{-d_{i}}$. But $\delta_{j}=d_{j}-(b_{0}+1)/2$, so this is
$\alpha_{j}$ times the coefficient of $(x+jv)^{-(b_{0}+1)/2}$ in
$\prod_{i=0}^{p-1}(x+iv)^{m_{i}-d_{i}}$, which is
$c_{j}(-(b_{0}+1)/2;m_{0}-d_{0},\dots,m_{p-1}-d_{p-1})$. ∎

For instance, when $p=2$, we get:

$SW_{\mathbb{Z}_{2},X,\mathfrak{s}}^{\phi}(x^{m_{0}}(x+v)^{m_{1}})=\left(\binom{m_{1}-d_{1}}{\delta_{0}-m_{0}}\overline{SW}^{s_{0},\phi}_{G,X,\mathfrak{s}}+\binom{m_{0}-d_{0}}{\delta_{1}-m_{1}}\overline{SW}^{s_{1},\phi}_{G,X,\mathfrak{s}}\right)v^{m_{0}+m_{1}-\delta}$

where we set $\binom{a}{b}=0$ if $b$ is non-integral or negative.

##  7\. Kähler actions

Suppose $(X,I)$ is a compact complex surface with $b_{1}(X)=0$ and suppose
that $G$ is a finite group which acts on $X$ by biholomorphism. Since
$b_{1}(X)=0$, $X$ admits a Kähler metric. Furthermore, the set of Kähler
metrics is convex so by averaging we can find a $G$-invariant Kähler metric
$g$ with associated Kähler form $\omega$. The Hodge decomposition yields an
equality $H^{+}(X)=\mathbb{R}[\omega]\oplus Re(H^{2,0}(X))$ and hence
$H^{+}(X)\cong\mathbb{R}\oplus H^{0}(X,K_{X})$ where $K_{X}$ denotes the
canonical bundle of $X$. Since $H^{0}(X,K_{X})$ is a complex vector space,
this provides a distinguished orientation on $H^{+}(X)$ given by
$\\{\omega,e_{1},Ie_{1},\dots,e_{n},Ie_{n}\\}$, where $e_{1},\dots,e_{n}$ is a
complex basis for $H^{0}(X,K_{X})$. We also have a distinguished chamber given
by setting $\phi=[\omega]$. We call this the Kähler chamber and we call
$-\phi$ the anti-Kähler chamber. Since
$H^{+}(X)^{G}\cong\mathbb{R}[\omega]\oplus H^{0}(X,K_{X})^{G}$, we see that
$dim(H^{+}(X)^{G})\geq 1$ and equals $1$ only if $H^{0}(X,K_{X})^{G}=0$.

The complex structure on $X$ defines a spinc-structure $\mathfrak{s}_{can}$,
the canonical spinc-structure with spinor bundles
$S^{\pm}=\wedge^{0,ev/odd}T^{*}X$. Any other spinc-structure is of the form
$\mathfrak{s}_{L}=L\otimes\mathfrak{s}_{can}$ for some complex line bundle $L$
and we have $c(\mathfrak{s}_{L})=2c_{1}(L)-c_{1}(K_{X})$. The spinor bundles
for $\mathfrak{s}_{L}$ are given by $S^{\pm}_{L}=L\otimes S^{\pm}$. The charge
conjugate of $\mathfrak{s}_{L}$ is $\mathfrak{s}_{K_{X}^{*}L}$. A spinc-
structure $\mathfrak{s}_{L}$ is preserved by $G$ if and only $G$ preserves the
isomorphism class of $L$. Furthermore, since $G$ has a canonical lift to
$S^{\pm}$ we see that lifts of $G$ to $S^{\pm}_{L}$ correspond to lifts of $G$
to $L$.

Let $L$ be a complex line bundle whose isomorphism class is preserved by $G$.
To compute the equivariant Seiberg–Witten invariants of
$(X,\mathfrak{s}_{L})$, we first consider the Seiberg–Witten equations with
respect to a $2$-form perturbation of the form $\eta=i\lambda\omega$, where
$\lambda$ is a sufficiently large positive real number. Note that it suffices
to consider only the Kähler chamber since charge conjugation exchanges the
Kähler and anti-Kähler chambers. We follow the approach of [27, Chapter 12].
Let $\mathcal{M}=\mathcal{M}(X,L,g,\eta)$ denote the moduli space of solutions
to the $\eta$-perturbed Seiberg–Witten equations on $X$ with respect to the
metric $g$ and spinc-structure $\mathfrak{s}_{L}$. By [27, Proposition 12.23],
we have a natural bijection between $\mathcal{M}$ and the space of effective
divisors on $X$ representing $c_{1}(L)$. If the image of $c_{1}(L)$ in
$H^{2}(X;\mathbb{R})$ is not of type $(1,1)$, then there are no divisors
representing $c_{1}(L)$ and hence $\mathcal{M}$ is empty. On the other hand if
$c_{1}(L)$ is of type $(1,1)$, then $L$ admits a holomorphic structure. Since
$b_{1}(X)=0$, the holomorphic structure is unique up to isomorphism and so we
can regard $L$ as a fixed holomorphic line bundle. Let $V^{i}=H^{i}(X,L)$
denote the cohomology groups of $L$ and let $h^{i}(L)$ denote the dimension of
$V^{i}$. The $V^{i}$ are representations of $G_{\mathfrak{s}_{L}}$ where the
$S^{1}$-subgroup acts by scalar multiplication. Effective divisors
representing $c_{1}(L)$ are in bijection with non-zero holomorphic sections of
$L$ up to scale, so we have an identification
$\mathcal{M}\cong\mathbb{P}(V^{0})$. Under this identification the action of
$G$ on $\mathcal{M}$ corresponds to the action on $\mathbb{P}(V^{0})$ induced
by the action of $G_{\mathfrak{s}_{L}}$ on $V^{0}$.

Although the moduli space $\mathcal{M}$ is a smooth manifold, the perturbation
$\eta=i\lambda\omega$ is typically not a regular perturbation. That is, the
moduli space $\mathcal{M}$ is typically not cut out transversally and its
dimension, $2(h^{0}(L)-1)$, is typically larger than the expected dimension
which is $2(h^{0}(L)-1)-2(h^{1}(L)-h^{2}(L)+h^{2,0}(X))$. To compute the
equivariant Seiberg–Witten equations in this setting we will make use of the
technique of obstruction bundles [13], [27, §12.9]. The failure of
$\mathcal{M}$ to be cut out transversally is measured by a bundle
$Obs\to\mathcal{M}$ called the obstruction bundle. The fibres of $Obs$ are the
cokernels of the linearisation of the Seiberg–Witten equations. As shown in
[27, §12.9], we have an exact sequence of bundles on $\mathcal{M}$:

(7.1) $0\to\widetilde{V}^{1}\to Obs\to
H^{2}(X,\mathcal{O})\to\widetilde{V}^{2}\to 0,$

where $\widetilde{V}^{i}$ denotes the vector bundle over
$\mathbb{P}(V^{0})=S(V^{0})/S^{1}$ given by
$\widetilde{V}^{i}=V^{i}\times_{S^{1}}S(V^{0})$ and $H^{2}(X,\mathcal{O})$ is
to be regarded as a trivial vector bundle on $\mathbb{P}(V^{0})$. In the
presence of a $G$-action, the obstruction bundle $Obs$ is a $G$-equivariant
vector bundle and the above sequence is an exact sequence of $G$-equivariant
vector bundles on $\mathcal{M}$.

######  Lemma 7.1.

Assume that $G_{\mathfrak{s}}$ is split. Then for any $\theta\in
H^{*}_{G_{\mathfrak{s}_{\scalebox{0.7}{$\scriptscriptstyle
L$}}}}\\!(pt;\mathbb{Z})$, we have

$SW_{G}^{\omega}(\theta)=(\pi_{\mathbb{P}(V^{0})})_{*}(e_{G}(Obs)\theta)$

where $\theta$ is regarded as an element of
$H^{*}_{G}(\mathbb{P}(V^{0});\mathbb{Z})$ by pulling it back to $S(V^{0})$ and
using $H^{*}_{G_{\mathfrak{s}_{\scalebox{0.7}{$\scriptscriptstyle
L$}}}}\\!(S(V^{0});\mathbb{Z})\cong H^{*}_{G}(\mathbb{P}(V^{0});\mathbb{Z})$.

###### Proof.

In the non-equivariant setting, this follows from the technique of obstruction
bundles. We need to extend the result to the equivariant setting. If $Obs$
admits a $G$-invariant section whhich is tranverse to the zero section, then
the usual argument can be carried out equivariantly. However in general an
equivariant vector bundle need not admit invariant sections which are
transverse to the zero section. To get around this problem we will work with
families instead of equivariantly.

For any compact smooth manifold $B$ and any principal $G$-bundle $P\to B$, we
have an associated family $E=P\times_{G}X$. Since the Kähler structure of $X$
is preserved by $G$, the family $E$ is a Kähler family in the sense that the
fibres are equipped with a smoothly varying Kähler structure. We will use this
structure to evaluate the families Seiberg–Witten invariants
$SW_{G,E,\mathfrak{s}_{L}}^{\omega}(\theta)$. The families moduli space is
$\mathcal{M}_{E}=P\times_{G}\mathcal{M}$. It is not cut out transversally, but
the obstruction bundle technique can be applied. The obstruction bundle
$Obs_{E}$ for this family is just the associated vector bundle
$Obs_{E}=P\times_{G}Obs$ and therefore we have

$SW_{G,E,\mathfrak{s}_{L}}^{\omega}(\theta)=(\pi_{*})(e(Obs_{E})\theta)$

where $\pi$ is the projection map $\pi:\mathcal{M}_{E}\to B$. It follows
immediately that

$SW_{G,E,\mathfrak{s}_{L}}^{\omega}(\theta)=(f_{P})^{*}((\pi_{\mathbb{P}(V^{0})})_{*}(e_{G}(Obs)\theta))$

where $f_{P}:B\to BG$ is the classifying map for $P$. Then by Theorem 4.2, it
follows that
$SW_{G}^{\omega}(\theta)=(\pi_{\mathbb{P}(V^{0})})_{*}(e_{G}(Obs)\theta)$. ∎

We turn to the computation of $e_{G}(Obs)$. Since $Obs$ is a complex vector
bundle of rank $r=h^{1}(L)-h^{2}(L)+h^{2,0}(X)$, we have
$e_{G}(Obs)=c_{r,G}(Obs)$, where $c_{r,G}$ denotes the $r$-th equivariant
Chern class. Let $c_{G}=1+c_{1,G}+\cdots$ denote the total equivariant Chern
class and $s_{G}=1+s_{1,G}+\cdots$ the total equivariant Segre class. From
(7.1) we have

$c_{G}(Obs)=c_{G}(\widetilde{V}^{1})s_{G}(\widetilde{V}^{2})c_{G}(H^{2}(X,\mathcal{O}))$

and hence

$e_{G}(Obs)=\sum_{\begin{subarray}{c}i+j+k=r\\\ i,j,k\geq
0\end{subarray}}c_{i,G}(\widetilde{V}^{1})s_{j,G}(\widetilde{V}^{2})c_{k,G}(H^{2}(X,\mathcal{O})).$

Assume now that $G_{\mathfrak{s}_{\scalebox{0.7}{$\scriptscriptstyle L$}}}\\!$
is a split extension. Fix a splitting
$G_{\mathfrak{s}_{\scalebox{0.7}{$\scriptscriptstyle L$}}}\\!\cong S^{1}\times
G$. A choice of splitting amounts to a choice of lift of the $G$-action to
$L$. This makes $V^{i}$ into representations of $G$ and then
$\widetilde{V}^{i}\cong V^{i}\otimes\mathcal{O}_{V^{0}}(1)$. The equivariant
Chern and Segre classes of $\widetilde{V}^{i}$ can now be expressed in terms
of the Chern and Segre classes of $V^{i}$ and
$x=c_{1,G}(\mathcal{O}_{V^{0}}(1))$ using the following identities for a
vector bundle $E$ of rank $r$ and a line bundle $N$:

$c_{j}(E\otimes N)=\sum_{l=0}^{j}c_{l}(E)c_{1}(N)^{j-l}\binom{r-l}{j-l},\quad
s_{j}(E\otimes N)=\sum_{l=0}^{j}s_{l}(E)c_{1}(N)^{j-l}\binom{-r-l}{j-l}.$

The same result also applies more generally when $E$ is a virtual vector
bundle. We can simplify the expression for $e_{G}(Obs)$ by writing it in terms
of the virtual bundle $\widetilde{V}^{1}-\widetilde{V}^{2}$, namely

$e_{G}(Obs)=\sum_{\begin{subarray}{c}i+j=r\\\ i,j\geq
0\end{subarray}}c_{i,G}(\widetilde{V}^{1}-\widetilde{V}^{2})c_{j,G}(H^{2}(X,\mathcal{O})).$

We also have $H^{*}_{G_{\mathfrak{s}_{\scalebox{0.7}{$\scriptscriptstyle
L$}}}}\\!(pt;\mathbb{Z})\cong H^{*}_{G}(pt;\mathbb{Z})[x]$ and so it suffices
to compute $SW_{G}^{\omega}(x^{m})$ for each $m\geq 0$. We also have that
$(\pi_{\mathbb{P}(V^{0})})_{*}(x^{j})=s_{j-(d-1),G}(V^{0})$ where
$d=h^{0}(L)-h^{1}(L)+h^{2}(L)$. Putting it all together, we have

$\displaystyle
SW_{G}^{\omega}(x^{m})=(\pi_{\mathbb{P}(V^{0})})_{*}\left(\sum_{\begin{subarray}{c}i+j=r\\\
i,j\geq
0\end{subarray}}c_{i,G}(\widetilde{V}^{1}-\widetilde{V}^{2})c_{j,G}(H^{2}(X,\mathcal{O}))x^{m}\right)$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=r\\\ i,j\geq
0\end{subarray}}\sum_{l=0}^{i}\binom{h^{1}(L)-h^{2}(L)-l}{i-l}c_{l,G}(V^{1}-V^{2})c_{j,G}(H^{2}(X,\mathcal{O}))(\pi_{\mathbb{P}(V^{0})})_{*}(x^{m+i-l})$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=r\\\ i,j\geq
0\end{subarray}}\sum_{l=0}^{i}\binom{h^{1}(L)-h^{2}(L)-l}{i-l}s_{i-l+m-(h^{0}(L)-1),G}(V^{0})c_{l,G}(V^{1}-V^{2})c_{j,G}(H^{2}(X,\mathcal{O})).$

######  Theorem 7.2.

Let $X$ be a compact complex surface with $b_{1}(X)=0$. Let $G$ be a finite
group that acts on $X$ by biholomorphisms. Let $L$ be a $G$-equivariant line
bundle. If $L$ is not holomorphic, or if $L$ is holomorphic and $h^{0}(L)=0$,
then $SW^{\omega}_{G,X,\mathfrak{s}_{L}}=0$. If $L$ is holomorphic and
$h^{0}(L)>0$, then

$\displaystyle SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=r\\\ i,j\geq
0\end{subarray}}\sum_{l=0}^{i}\binom{h^{1}(L)-h^{2}(L)-l}{i-l}s_{i-l+m-(h^{0}(L)-1),G}(V^{0})c_{l,G}(V^{1}-V^{2})c_{j,G}(H^{2}(X,\mathcal{O}))$

where $V^{i}=H^{i}(X,L)$, $d=h^{0}(L)-h^{1}(L)+h^{2}(L)$,
$r=h^{1}(L)-h^{2}(L)+h^{2,0}(X)$.

The expression for $SW_{G,X,\mathfrak{s}_{L}}^{\omega}$ given by Theorem 7.2
gives a formula for $SW_{G,X,\mathfrak{s}_{L}}^{\omega}$ purely in terms of
the representations $V^{0},V^{1},V^{2}$ and $H^{2}(X,\mathcal{O})$ and hence
purely in terms of the complex geometry of $X$. Next, we restrict to the case
$b_{+}(X)=1$.

######  Theorem 7.3.

Let $X$ be a compact complex surface with $b_{1}(X)=0$ and $b_{+}(X)=1$. Let
$G$ be a finite group that acts on $X$ by biholomorphisms. Let $L$ be a
$G$-equivariant line bundle. Let $D=V^{0}-V^{1}+V^{2}$ and
$d=h^{0}(L)-h^{1}(L)+h^{2}(L)$.

  * (1)

If $h^{0}(L)>0$, then

$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=s_{m-(d-1)}(D),\quad
SW_{G,X,\mathfrak{s}_{L}}^{-\omega}(x^{m})=0.$

  * (2)

If $h^{2}(L)>0$, then

$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=0,\quad
SW_{G,X,\mathfrak{s}_{L}}^{-\omega}(x^{m})=-s_{m-(d-1)}(D).$

  * (3)

If $h^{0}(L)=h^{2}(L)=0$, then $SW_{G,X,\mathfrak{s}_{L}}^{\pm\omega}=0$.

###### Proof.

Part (3) is clear, so it remains to prove (1) and (2). Since $b_{+}(X)=1$, we
have $H^{2}(X,\mathcal{O})\cong H^{0}(X,K_{X})^{*}=0$. If
$h^{0}(L),h^{2}(L)>0$, then there are non-zero holomorphic sections
$\alpha,\beta$ of $L$ and $K^{*}_{X}L$. But then $\alpha\beta$ is a non-zero
section of $K_{X}$, which is impossible. This means that at most one of
$h^{0}(L),h^{2}(L)$ is non-zero. Consider the case that $h^{0}(L)>0$,
$h^{2}(L)=0$. The case $h^{0}(L)=0$, $h^{2}(L)>0$ will follow from charge
conjugation symmetry. If $h^{0}(L)>0$, then
$SW_{G,X,\mathfrak{s}_{L}}^{-\omega}=0$ and hence the wall-crossing formula
gives $SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=s_{m-(d-1)}(D)$. ∎

Next, we consider the case that $X$ is a K3 surface:

######  Theorem 7.4.

Let $X$ be a complex K3 surface and let $G$ be a finite group which acts on
$X$ by biholomorphisms. Let $L$ be a $G$-equivariant line bundle.

  * (1)

If $c_{1}(L)$ is not $(1,1)$ then $SW_{G,X,\mathfrak{s}_{L}}^{\pm\omega}=0$.

  * (2)

If $c_{1}(L)$ is $(1,1)$, $c_{1}(L)\neq 0$ and $G$ acts trivially on $K_{X}$,
then there is only one chamber and $SW_{G,X,\mathfrak{s}_{L}}=0$.

  * (3)

If $c_{1}(L)=0$ and $G$ acts trivially on $K_{X}$, then there is only one
chamber and $SW_{G,X,\mathfrak{s}_{L}}(x^{m})=s_{1,G}(L)^{m}$.

  * (4)

If $c_{1}(L)=0$ and $G$ acts non-trivially on $K_{X}$, then
$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=s_{1,G}(L)^{m}$,
$SW_{G,X,\mathfrak{s}_{L}}^{-\omega}=(s_{1,G}(L)-s_{1,G}(K_{X}))^{m}$.

  * (5)

If $c_{1}(L)$ is $(1,1)$, $c_{1}(L)\neq 0$ and $G$ acts non-trivially on
$K_{X}$, then there are two chambers. If $h^{0}(L)>0$, then

$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=e_{G}(H^{0}(X,K_{X}))s_{m-(d-1)}(D),\quad
SW_{G,X,\mathfrak{s}_{L}}^{-\omega}(x^{m})=0.$

If $h^{0}(L)=0$, then

$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=0,\quad
SW_{G,X,\mathfrak{s}_{L}}^{-\omega}(x^{m})=-e_{G}(H^{0}(X,K_{X}))s_{m-(d-1)}(D).$

###### Proof.

Case (1) is clear, so we can assume that $c_{1}(L)$ is $(1,1)$.

Suppose that $h^{0}(L)$ and $h^{2}(L)$ are non-zero. If $\alpha\in
H^{0}(X,L),\beta\in H^{0}(X,L^{*})$ are non-zero, then $\alpha\beta\in
H^{0}(X,\mathcal{O})$ is non-zero and hence non-vanishing. This means that
$\alpha,\beta$ are non-vanishing and hence $c_{1}(L)=0$. This means that if
$c_{1}(L)\neq 0$ then $h^{0}(L)$ or $h^{2}(L)$ is zero. Now if the action of
$G$ on $K_{X}$ is trivial, then $H^{+}(X)^{G}=H^{+}(X)$ and there is only one
chamber. Then it follows that $SW_{G,X,\mathfrak{s}_{L}}=0$ since $h^{0}(L)=0$
or $h^{2}(L)=0$. This proves (2). If $c_{1}(L)\neq 0$ and $G$ acts non-
trivially on $K_{X}$ then there are two chambers, but we have that either
$h^{0}(L)=0$, in which case $SW_{G,X,\mathfrak{s}_{L}}^{\omega}=0$, or
$h^{2}(L)=0$, in which case $SW_{G,X,\mathfrak{s}_{L}}^{-\omega}=0$. Then the
invariants for the other chamber are given by the wall-crossing formula. This
proves (5).

It remains to prove (3) and (4). Hence we assume that $c_{1}(L)=0$. This means
that $h^{0}(L)=h^{2}(L)=1$, $h^{1}(L)=0$, $d=2$, $r=0$. So Theorem 7.2
simplifies to

$SW_{G,X,\mathfrak{s}_{L}}^{\omega}(x^{m})=\binom{-1}{0}s_{m,G}(V^{0})=s_{m,G}(V^{0}).$

But $V^{0}=H^{0}(X,L)\cong L$ is $1$-dimensional, hence
$s_{m,G}(V^{0})=s_{1,G}(L)^{m}$. This proves (3). Finally in case (4) the
formula for $SW_{X,G,\mathfrak{s}_{L}}^{-\omega}$ follows from the formula for
$SW_{G,X,\mathfrak{s}_{L}}^{\omega}$ and charge conjugation, or alternatively
from the wall-crossing formula. ∎

##  8\. Gluing formulas

In this section we prove gluing formulas for the equivariant Seiberg–Witten
invariants of equivariant connected sums.

Let $f:S^{V,U}\to S^{V^{\prime},U^{\prime}}$ be a $G$-monopole map. Define the
$G_{\mathfrak{s}}$-equivariant degree $deg_{G_{\mathfrak{s}}}(f)\in
H^{b_{+}-2d}_{G_{\mathfrak{s}}}(pt;\mathbb{Z}_{w})$ by
$f^{*}(\tau_{V^{\prime},U^{\prime}})=deg_{G_{\mathfrak{s}}}(f)\tau_{V,U}$. The
following result is an extension of [5, Theorem 6.1] to the case where
$G_{\mathfrak{s}}$ is not necessarily split.

######  Proposition 8.1.

Let $f:S^{V,U}\to S^{V^{\prime},U^{\prime}}$ be a $G$-monopole map. If
$e(H^{+})=0$, then $deg_{G_{\mathfrak{s}}}(f)=0$. If $e(H^{+})\neq 0$, then
$d\leq 0$ and $deg_{G_{\mathfrak{s}}}(f)=e(H^{+})s_{G_{\mathfrak{s}},-d}(D)$.
Furthermore in this case we have that
$e_{G}(H^{+})s_{G_{\mathfrak{s}},j}(D)=0$ for $j>-d$.

###### Proof.

Consider the commutative square

$\textstyle{S^{V,U}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{S^{V^{\prime},U^{\prime}}}$$\textstyle{S^{U}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\scriptstyle{\iota}$$\textstyle{S^{U^{\prime}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$

where $\iota:S^{U}\to S^{U^{\prime}}$ is given by inclusion. We have

$i^{*}f^{*}(\tau_{V^{\prime},U^{\prime}})=i^{*}(\tau_{V,U}deg_{G_{\mathfrak{s}}}(f))=\tau_{U}e_{G_{\mathfrak{s}}}(V)deg_{G_{\mathfrak{s}}}(f).$

On the other hand,

$\iota^{*}i^{*}(\tau_{V^{\prime},U^{\prime}})=\iota^{*}(\tau_{U^{\prime}}e_{G_{\mathfrak{s}}}(V^{\prime}))=\tau_{U}e(H^{+})e_{G_{\mathfrak{s}}}(V^{\prime}).$

Since $f\circ i=i\circ\iota$, we may equate these two expressions, giving

$e_{G_{\mathfrak{s}}}(V)deg_{G_{\mathfrak{s}}}(f)=e(H^{+})e_{G_{\mathfrak{s}}}(V^{\prime}).$

Using a spectral sequence argument it can be shown that
$e_{G_{\mathfrak{s}}}(V)$ is not a zero divisor and so the above equation
uniquely determines $deg_{G_{\mathfrak{s}}}(f)$. However, it seems difficult
to obtain an explicit formula for $deg_{G_{\mathfrak{s}}}(f)$ from this
equation. We can remedy this by considering the larger group
$\widehat{G}_{\mathfrak{s}}=S^{1}\times G_{\mathfrak{s}}$. We let
$\widehat{G}_{\mathfrak{s}}$ act on $S^{V,U}$ and $S^{V^{\prime},U^{\prime}}$
through the homomorphism $S^{1}\times G_{\mathfrak{s}}\to G_{\mathfrak{s}}$
given by $(u,g)\mapsto ug$. Then $f$ can be regarded as a
$\widehat{G}_{\mathfrak{s}}$-equivariant map. Repeating the above computation
for this larger group gives

(8.1)
$e_{\widehat{G}_{\mathfrak{s}}}(V)deg_{\widehat{G}_{\mathfrak{s}}}(f)=e(H^{+})e_{\widehat{G}_{\mathfrak{s}}}(V^{\prime}).$

Let $\mathbb{C}_{1}$ be the $1$-dimensional representation of $S^{1}\times
G_{\mathfrak{s}}$ where the first factor acts by scalar multiplication and the
second factor acts trivially. Then
$H^{*}_{\widehat{G}_{\mathfrak{s}}}(pt;\mathbb{Z}_{w})\cong
H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z}_{w})[y]$ where
$y=c_{1}(\mathbb{C}_{1})$. Equation (8.1) can be expanded as

$(y^{a}+c_{G,1}(V)y^{a-1}+\cdots+c_{G,a}(V))deg_{\widehat{G}_{\mathfrak{s}}}(f)=e(H^{+})(y^{a^{\prime}}+c_{G,1}(V^{\prime})y^{a-1}+\cdots+c_{G,a^{\prime}}(V^{\prime})).$

If $e(H^{+})=0$, then it follows that $deg_{\widehat{G}_{\mathfrak{s}}}(f)=0$.
Suppose instead that $e(H^{+})\neq 0$. Then we can write
$deg_{\widehat{G}_{\mathfrak{s}}}(f)=f_{0}y^{k}+f_{1}y^{k-1}+\cdots+f_{k}$ for
some $k\geq 0$, where $f_{0}\neq 0$. Expanding the left hand side of the above
equation we see that $a^{\prime}=k+a$, hence $d=a-a^{\prime}=-k\leq 0$. Now
consider the above equation in the group of formal Laurent series of the form
$\sum_{j=-\infty}^{n}c_{n}y^{n}$ where $n$ is any integer and $c_{j}\in
H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z}_{w})$. This group is a module over the
ring $H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z})[[y^{-1}]]$ of formal power
series in $y$ with coefficients in $H^{*}_{G_{\mathfrak{s}}}(pt;\mathbb{Z})$.
Multiplying both sides by $1+y^{-1}s_{G,1}(V)+y^{-2}s_{G,2}(V)+\cdots$ we
obtain

$y^{a}(f_{0}y^{-d}+f_{1}y^{-d-1}+\cdots+f_{k})=e(H^{+})(y^{a^{\prime}}+s_{G,1}(D)y^{a^{\prime}-1}+s_{G,2}(D)y^{a^{\prime}-2}+\cdots).$

Equating coefficients we see that $f_{j}=e(H^{+})s_{G,j}(D)$ for $j\leq-d$ and
$e(H^{+})s_{G,j}(D)=0$ for $j>-d$. Thus

$deg_{\widehat{G}_{\mathfrak{s}}}(f)=e(H^{+})(y^{-d}+s_{G,1}(D)y^{-d-1}+\cdots+s_{G,-d}(D)).$

Restricting to $G_{\mathfrak{s}}\subset\widehat{G}_{\mathfrak{s}}$ amounts to
setting $y=0$ in the above equality, giving

$deg_{G_{\mathfrak{s}}}(f)=e(H^{+})s_{G,-d}(D).$

∎

###  8.1. Abstract gluing formula

We first prove a general gluing formula for the abtract Seiberg–Witten
invariants of a smash product of monopole maps. Let $G$ be a finite group and
$G_{\mathfrak{s}}$ an $S^{1}$-central extension of $G$. Suppose that we have
two $G$-monopole maps

$f_{i}:S^{V_{i},U_{i}}\to S^{V^{\prime}_{i},U^{\prime}_{i}},\quad i=1,2$

over $B=pt$ and assume $f_{1},f_{2}$ are $G_{\mathfrak{s}}$-equivariant for
the same central extension of $G$. Then the smash product

$f=f_{1}\wedge f_{2}:S^{V,U}\to S^{V^{\prime},U^{\prime}}$

is again $G_{\mathfrak{s}}$-equivariant, where $V=V_{1}\oplus V_{2}$,
$U=U_{1}\oplus U_{2}$, $V^{\prime}=V^{\prime}_{1}\oplus V^{\prime}_{2}$,
$U^{\prime}=U^{\prime}_{1}\oplus U^{\prime}_{2}$. As usual we write
$D=V-V^{\prime}$, $H^{+}=U^{\prime}-U$. Similarly define $D_{i},H^{+}_{i}$ for
$i=1,2$. Since $(H^{+})^{G}=(H^{+}_{1})^{G}\oplus(H^{+}_{2})^{G}$, we see that
$f$ admits a chamber if and only if at least one of $f_{1}$ and $f_{2}$ admits
a chamber. Suppose that this is the case and let $\phi=(\phi_{1},\phi_{2})$ be
a chamber. Thus either $\phi_{1}\neq 0$ or $\phi_{2}\neq 0$.

Our orientation conventions are as follows. We orient $H^{+}$ according to
$H^{+}=H^{+}_{1}\oplus H^{+}_{2}$, orient $U$ according to $U=U_{1}\oplus
U_{2}$ and orient $U^{\prime}$ according to $U^{\prime}=U\oplus H^{+}$. Note
that this is different from taking $U^{\prime}$ to be $U^{\prime}_{1}\oplus
U^{\prime}_{2}$ by a factor of $(-1)^{b_{2}(b_{+,1})}$, where $b_{i}$ denotes
the rank of $U_{i}$ and $b_{+,i}$ the rank of $H^{+}_{i}$.

######  Theorem 8.2.

If $(H^{+}_{1})^{G},(H^{+}_{2})^{G}$ are both non-zero, then the
Seiberg–Witten invariants of $f$ are all zero. If $(H^{+}_{1})^{G}\neq 0$,
$(H^{+}_{2})^{G}=0$, then

$SW_{G,f}^{\phi}(\theta)=SW_{G,f_{1}}^{\phi_{1}}(e_{G}(H^{+}_{2})s_{G_{\mathfrak{s}},-d_{2}}(D_{2})\theta).$

where $s_{G_{\mathfrak{s}},-d_{2}}$ is taken to be zero if $d_{2}>0$. In
particular, if $G_{\mathfrak{s}}\cong S^{1}\times G$ is the trivial extension
then

$SW_{G,f}^{\phi}(\theta)=\sum_{k+l=-d_{2}}s_{G,l}(D_{2})SW_{G,f_{1}}^{\phi_{1}}(x^{k}e_{G}(H^{+}_{2})\theta).$

###### Proof.

Without loss of generality we can assume $(H^{+}_{1})^{G}\neq 0$. If we also
have $(H^{+}_{2})^{G}\neq 0$, then $dim((H^{+})^{G})>1$ and so
$SW_{G,f}^{\phi}$ is independent of the choice of chamber. Hence without loss
of generality we can assume that $\phi=(\phi_{1},0)$, where $\phi_{1}\neq 0$.
We have that
$SW_{G,f}^{\phi}(\theta)=(\pi_{\mathbb{P}(V)})_{*}(\eta^{\phi}\theta)$ where
$f^{*}(\tau^{\phi}_{V^{\prime},U^{\prime}})=(-1)^{b}\delta\tau_{U}\eta^{\phi}$.
Similarly
$SW_{G,f_{1}}^{\phi_{1}}(\theta)=(\pi_{\mathbb{P}(V_{1})})_{*}(\eta^{\phi_{1}}_{1}\theta)$
where
$f^{*}(\tau^{\phi_{1}}_{V^{\prime}_{1},U^{\prime}_{1}})=(-1)^{b_{1}}\delta\tau_{U_{1}}\eta^{\phi_{1}}_{1}$.
Recall that the equivariant degree $deg_{G_{\mathfrak{s}}}(f_{2})$ of $f_{2}$
is defined by
$f_{2}^{*}(\tau_{V^{\prime}_{2},U^{\prime}_{2}})=\tau_{V_{2},U_{2}}deg_{G_{\mathfrak{s}}}(f_{2})$.

Consider the external cup product

$H^{i}_{G_{\mathfrak{s}}}(S^{V^{\prime}_{1},U^{\prime}_{1}},S^{U_{1}};\mathbb{Z})\times
H^{j}_{G_{\mathfrak{s}}}(S^{V^{\prime}_{2},U^{\prime}_{2}},pt;\mathbb{Z})\to
H^{i+j}_{G_{\mathfrak{s}}}(S^{V^{\prime},U^{\prime}},S^{U};\mathbb{Z}).$

Using this cup product we can write

$\tau^{\phi}_{V^{\prime},U^{\prime}}=(-1)^{b_{+,1}b_{2}}\tau^{\phi_{1}}_{V^{\prime}_{1},U^{\prime}_{2}}\tau_{V^{\prime}_{2},U^{\prime}_{2}}.$

The sign factor $(-1)^{b_{+,1}b_{2}}$ arises because our orientation
conventions described above. From this, we find

$\displaystyle(-1)^{b}\delta\tau_{U}\eta^{\phi}$
$\displaystyle=f^{*}(\tau^{\phi}_{V^{\prime},U^{\prime}})$
$\displaystyle=(-1)^{b_{+,1}b_{2}}(f_{1}\wedge
f_{2})^{*}(\tau^{\phi_{1}}_{V^{\prime}_{1},U^{\prime}_{1}}\tau_{V^{\prime}_{2},U^{\prime}_{2}})$
$\displaystyle=(-1)^{b_{+,1}b_{2}}f_{1}^{*}(\tau^{\phi_{1}}_{V^{\prime}_{1},U^{\prime}_{1}})f_{2}^{*}(\tau_{V^{\prime}_{2},U^{\prime}_{2}})$
$\displaystyle=(-1)^{b_{+,1}b_{2}+b_{1}}\delta\tau_{U_{1}}\eta^{\phi_{1}}_{1}\tau_{V_{2},U_{2}}deg_{G_{\mathfrak{s}}}(f_{2})$
$\displaystyle=(-1)^{b_{+,1}b_{2}+b_{1}+(b_{+,1}+1)b_{2}}\delta\tau_{U}e_{G_{\mathfrak{s}}}(V_{2})\eta^{\phi_{1}}_{1}deg_{G_{\mathfrak{s}}}(f_{2})$
$\displaystyle=(-1)^{b}\delta\tau_{U}e_{G_{\mathfrak{s}}}(V_{2})\eta^{\phi_{1}}_{1}deg(f_{2}).$

Hence

(8.2)
$\eta^{\phi}=e_{G_{\mathfrak{s}}}(V_{2})\eta^{\phi_{1}}_{1}deg_{G_{\mathfrak{s}}}(f_{2}).$

Let $\iota_{1}:\mathbb{P}(V_{1})\to\mathbb{P}(V_{2})$ be the inclusion map.
Since $(\iota_{1})_{*}(1)=e_{G_{\mathfrak{s}}}(V_{2})$, we have

$\displaystyle SW_{G,f}^{\phi}(\theta)$
$\displaystyle=(\pi_{\mathbb{P}(V)})_{*}(\eta^{\phi}\theta)$
$\displaystyle=(\pi_{\mathbb{P}(V)})_{*}(e_{G_{\mathfrak{s}}}(V_{2})\eta^{\phi_{1}}_{1}deg_{G_{\mathfrak{s}}}(f_{2})\theta)$
$\displaystyle=(\pi_{\mathbb{P}(V)})_{*}(\iota_{1})_{*}(\eta^{\phi_{1}}_{1}deg_{G_{\mathfrak{s}}}(f_{2})\theta)$
$\displaystyle=(\pi_{\mathbb{P}(V_{1})})_{*}(\eta^{\phi_{1}}_{1}deg_{G_{\mathfrak{s}}}(f_{2})\theta)$
$\displaystyle=SW^{\phi_{1}}_{G,f_{1}}(deg_{G_{\mathfrak{s}}}(f_{2})\theta).$

Now if $(H^{+}_{2})^{G}\neq 0$, then $e(H^{+}_{2})=0$ and
$deg_{G_{\mathfrak{s}}}(f_{2})=0$ by Proposition 8.1. Hence in this case
$SW_{G,f}^{\phi}$ vanishes. If $(H^{+}_{2})^{G}=0$, then $d_{2}\leq 0$ and
Proposition 8.1 gives
$deg_{G_{\mathfrak{s}}}(f_{2})=e(H^{+}_{2})s_{G_{\mathfrak{s}},-d_{2}}(D_{2})$,
which gives

$SW_{G,f}^{\phi}(\theta)=SW_{G,f_{1}}^{\phi_{1}}(e_{G}(H^{+}_{2})s_{G_{\mathfrak{s}},-d_{2}}(D_{2})\theta).$

Lastly if $G_{\mathfrak{s}}\cong S^{1}\times G$ is the trivial extension, then

$s_{G_{\mathfrak{s}},-d_{2}}(D_{2})=\sum_{k=0}^{-d_{2}}x^{k}s_{G,-d_{2}-k}(D_{2})$

and hence

$SW_{G,f}^{\phi}(\theta)=\sum_{k+l=-d_{2}}s_{G,l}(D_{2})SW_{G,f_{1}}^{\phi_{1}}(x^{k}e_{G}(H^{+}_{2})\theta).$

∎

###  8.2. Equivariant connected sum

Let $X,Y$ be compact, oriented, smooth $4$-manifolds. Let $G$ be a finite
group acting smoothly and orientation preservingly on $X$. Let $H$ be a
subgroup of $G$ and assume that $H$ acts smoothly and orientation preservingly
on $Y$. We will construct an action of $G$ on the connected sum of $X$ with
$|G/H|$ copies of $Y$. For this we need to assume that $X$ and $Y$ contain
points $x,y$ whose stabiliser groups are $H$. We also need that
$T_{x}X,T_{y}Y$ are isomorphic as real representations of $H$ via an
orientation reversing isomorphism. Choose such an isomorphism $\psi:T_{x}X\to
T_{y}Y$. Then we can form the $G$-equivariant connected sum
$X\\#ind^{G}_{H}(Y)$ as follows. Here $ind^{G}_{H}(Y)$ is the disjoint union
of $G/H$ copies of $Y$ which is made into a $G$-space by taking
$ind^{G}_{H}(Y)=G\times_{H}Y$. We remove from $X$ and $ind^{G}_{H}(Y)$
equivariant tubular neighbourhoods $\nu(Gx),\nu(Gy)$ of the orbits $Gx,Gy$ and
identify the boundaries of $X\setminus\nu(Gx)$,
$ind^{G}_{H}(Y)\setminus\nu(Gy)$ by the orientation reversing $G$-equivariant
isomorphism $\partial\nu(Gx)\to\partial\nu(Gy)$ determined by $\psi$. We note
that the construction of $X\\#ind^{G}_{H}(Y)$ depends on the choice of points
$x,y$ and the isomorphism $\psi$. The underlying smooth manifold of
$X\\#ind^{G}_{H}(Y)$ is $X\\#|G/H|Y$, the connected sum of $X$ with $|G/H|$
copies of $Y$. Note also that if $b_{1}(X)=b_{1}(Y)=0$, then
$b_{1}(X\\#ind^{G}_{H}(Y))=0$.

Suppose that $\mathfrak{s}_{X}$ is a $G$-invariant spinc-structure on $X$ and
$\mathfrak{s}_{Y}$ is a $H$-invariant spinc-structure on $Y$. Let $S^{\pm}$
denote the spinor bundles corresponding to $\mathfrak{s}_{Y}$. Since $H$ fixes
$y$, we see that $H_{\mathfrak{s}_{Y}}$ lifts to the fibres $S^{\pm}_{y}$.
Hence the extension class of $H_{\mathfrak{s}_{Y}}$ is given by
$Sq_{3}^{\mathbb{Z}}(T_{y}Y)\in H^{3}_{H}(pt;\mathbb{Z})$, the third integral
Stiefel–Whitney class of $T_{y}Y$. Similarly, since $H$ fixes $x$, we see that
the restriction $G_{\mathfrak{s}_{X}}|_{H}$ of the extension
$G_{\mathfrak{s}_{X}}$ to $H$ has extension class
$Sq_{3}^{\mathbb{Z}}(T_{x}X)$. Since $T_{x}X$ and $T_{y}Y$ are orientation
reversingly isomorphic, it follows that the extension classes agree (reversing
orientation has the effect of exchanging the roles of positve and negative
spinors). Therefore $H_{\mathfrak{s}_{Y}}$ and $G_{\mathfrak{s}_{X}}|_{H}$ are
isomorphic as extensions of $H$. The isomorphism $\psi:T_{x}X\to T_{y}Y$
determines the isomorphism $G_{\mathfrak{s}_{X}}|_{H}\to
H_{\mathfrak{s}_{Y}}$. We can induce a $G$-invariant spinc-structure
$ind^{G}_{H}(\mathfrak{s}_{Y})$ on $ind^{G}_{H}(Y)$ as follows. We take the
spinor bundles of $ind^{G}_{H}(\mathfrak{s}_{Y})$ to be given by
$G_{\mathfrak{s}_{X}}\times_{H_{\mathfrak{s}_{Y}}}S^{\pm}$. The underlying
isomorphism class of $ind^{G}_{H}(\mathfrak{s}_{Y})$ is simply given by
equipping each connected component of $ind^{G}_{H}(Y)$ with a copy of
$\mathfrak{s}_{Y}$. However our construction makes it clear that the extension
of $G$ determined by $ind^{G}_{H}(\mathfrak{s}_{Y})$ is isomorphic to
$G_{\mathfrak{s}_{X}}$. It follows that the spinc-structures
$\mathfrak{s}_{X}$ and $ind^{G}_{H}(\mathfrak{s}_{Y})$ can be identified
equivariantly on the boundaries of $X\setminus\nu(Gx)$ and
$ind^{G}_{H}(Y)\setminus\nu(Gx)$ to form a spinc-structure
$\mathfrak{s}_{X}\\#ind^{G}_{H}(\mathfrak{s}_{Y})$ on
$X\\#ind^{G}_{H}(\mathfrak{s}_{Y})$.

To summarise, given a $G$-invariant spinc-structure $\mathfrak{s}_{X}$ on $X$
and a $H$-invariant spinc-structure $\mathfrak{s}_{Y}$ on $Y$, the choice of
an orientation reversing isomorphism $\psi:T_{x}X\to T_{y}Y$ of
representations of $H$ allows us to construct a $G$-invariant spinc-structure
$\mathfrak{s}=\mathfrak{s}_{X}\\#ind^{H}_{G}(\mathfrak{s}_{Y})$ on
$X\\#ind^{G}_{H}(Y)$ and moreover the extensions $G_{\mathfrak{s}}$ and
$G_{\mathfrak{s}_{X}}$ are isomorphic.

Suppose $f:S^{V,U}\to S^{V^{\prime},U^{\prime}}$ is an $H$-equivariant
monopole map with respect to some extension $H_{\mathfrak{s}}$. Suppose
$H\subseteq G$ and $H_{\mathfrak{s}}=G_{\mathfrak{s}}|_{H}$ for some extension
$G_{\mathfrak{s}}$. We will define an induced monopole map

$ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(f):S^{ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(V),ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(U)}\to
S^{ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(V^{\prime}),ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(U^{\prime})}$

as follows. Choose representatives $g_{1},\dots,g_{n}$ for the left cosets of
$H$ in $G$ and choose lifts $\tilde{g}_{i}$ of $g_{i}$ to $G_{\mathfrak{s}}$.
Then $\tilde{g}_{1},\dots,\tilde{g}_{n}$ are representatives for the left
cosets of $H_{\mathfrak{s}}$ in $G_{\mathfrak{s}}$. For any representation $W$
of $H_{\mathfrak{s}}$,
$ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(W)=\mathbb{R}[G_{\mathfrak{s}}]\otimes_{\mathbb{R}[H_{\mathfrak{s}}]}W$.
Regard $S^{W}$ as $W\cup\\{\infty\\}$ and similarly
$S^{ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(W)}=ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(W)\cup\\{\infty\\}$.
Then $ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(f)$ is defined by

$ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(f)(\sum_{i}g_{i}\otimes
w_{i})=\begin{cases}\sum_{i}g_{i}\otimes f(w_{i})&\text{if
}f(w_{i})\neq\infty\text{ for all }i,\\\ \infty&\text{otherwise}.\end{cases}$

This is well-defined because $f$ is $H_{\mathfrak{s}}$-equivariant.

######  Theorem 8.3.

Let $X,Y$ be compact, oriented, smooth $4$-manifolds with
$b_{1}(X)=b_{1}(Y)=0$. Let $G$ be a finite group acting smoothly and
orientation preservingly on $X$. Let $H$ be a subgroup of $G$ and suppose that
$H$ acts smoothly and orientation preservingly on $Y$. Suppose that there is
an $x\in X$ and $y\in Y$ with stabiliser group $H$ and an orientation
reversing isomorphism $\psi:T_{x}X\to T_{y}Y$ of representations of $H$.
Suppose that $\mathfrak{s}_{X}$ is a $G$-invariant spinc-structure on $X$ and
$\mathfrak{s}_{Y}$ is a $H$-invariant spinc-structure on $Y$. Set
$Z=X\\#ind^{G}_{H}(Y)$,
$\mathfrak{s}=\mathfrak{s}_{X}\\#ind^{G}_{H}(\mathfrak{s}_{Y})$. Then:

  * (1)

If $H^{+}(X)^{G},H^{+}(Y)^{H}$ are both non-zero then the equivariant
Seiberg–Witten invariants of $Z$ vanish.

  * (2)

If $H^{+}(X)^{G}\neq 0$ and $H^{+}(Y)^{H}=0$, then for any chamber $\phi\in
H^{+}(X)^{G}\setminus\\{0\\}$ we have

$SW^{\phi}_{G,Z,\mathfrak{s}}(\theta)=SW^{\phi}_{G,X,\mathfrak{s}_{X}}(e(ind^{G}_{H}(H^{+}(Y)))s_{G_{\mathfrak{s}},-d_{Y}}(ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(D_{Y}))\theta).$

###### Proof.

Let $f_{X},f_{Y},f_{Z}$ denote the equivariant monopole maps for $X,Y$ and
$Z$. Bauer’s connected sum formula [9] extends easily to the $G$-equivariant
setting. Then by a straightforward extension of [29, Theorem 3.1], we see that
$f_{Z}=f_{X}\wedge ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(f_{Y})$.

Set $f=f_{Z}$, $f_{1}=f_{X}$,
$f_{2}=ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(f_{Y})$. We use the notation
$H^{+},D,H^{+}_{i},D_{i}$ as in Section 8.1. Then $H^{+}_{1}=H^{+}(X)$,
$H^{+}_{2}=ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(H^{+}(Y))$,
$(H^{+}_{1})^{G}=H^{+}(X)^{G}$, $(H^{+}_{2})^{G}=H^{+}(Y)^{H}$. Hence if
$H^{+}(X)^{G},H^{+}(Y)^{H}$ are both non-zero, then the Seiberg–Witten
invariants of $Z$ vanish by Theorem 8.2.

Now suppose $H^{+}(X)^{G}\neq 0$ and $H^{+}(Y)^{H}=0$. Theorem 8.2 gives

$SW^{\phi}_{G,Z,\mathfrak{s}}(\theta)=SW^{\phi}_{G,X,\mathfrak{s}}(e(H^{+}_{2})s_{G_{\mathfrak{s}},-d_{Y}}(D_{2})\theta).$

The result follows, since
$H^{+}_{2}=ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(H^{+}(Y))$ and
$D_{2}=ind^{G_{\mathfrak{s}}}_{H_{\mathfrak{s}}}(D_{Y})$. ∎

We consider now the case where $H^{+}(X)^{G}=0$ and $H^{+}(Y)^{H}\neq 0$. Here
it is more difficult to obtain a general formula for the cohomological
invariants so we restrict to the case $G=\mathbb{Z}_{p}$ and $H=1$. Let
$\mathbb{C}_{j}$ denote the $1$-dimensional complex representation of
$\mathbb{Z}_{p}=\langle g\rangle$ where $g$ acts as multiplication by
$\omega^{j}$, $\omega=e^{2\pi i/p}$. When $p$ is odd we may choose the
orientation on $ind^{G}_{1}(H^{+}(Y))$ such that
$ind^{G}_{1}(H^{+}(Y))/H^{+}(Y)\cong\bigoplus_{j=1}^{(p-1)/2}\mathbb{C}_{j}^{b_{+}(Y)}$.

Recall that $H^{*}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})\cong\mathbb{Z}[v]/(pv)$,
$deg(v)=2$. When $p=2$ to avoid orientability issues we use
$\mathbb{Z}_{2}$-coefficients. We have
$H^{*}_{\mathbb{Z}_{2}}(pt;\mathbb{Z}_{2})\cong\mathbb{Z}_{2}[u]$, $deg(u)=1$.
In this case we set $v=u^{2}$ and we will also denote $u$ by $v^{1/2}$.

######  Theorem 8.4.

Let $X,Y$ be compact, oriented, smooth $4$-manifolds with
$b_{1}(X)=b_{1}(Y)=0$. Let $G=\mathbb{Z}_{p}$ where $p$ is prime act smoothly
and orientation preservingly on $X$. Suppose that $\mathfrak{s}_{X}$ is a
$G$-invariant spinc-structure on $X$ and $\mathfrak{s}_{Y}$ is a spinc-
structure on $Y$ with $d(Y,\mathfrak{s}_{Y})=2d_{Y}-b_{+}(Y)-1=0$. Set
$Z=X\\#ind^{G}_{1}(Y)$,
$\mathfrak{s}=\mathfrak{s}_{X}\\#ind^{G}_{1}(\mathfrak{s}_{Y})$. Suppose that
$H^{+}(X)^{G}=0$ and $b_{+}(Y)>0$. Then

$SW_{G,Z,\mathfrak{s}}^{\phi}(x^{m})=(-1)^{d_{Y}+1}h{\sum_{l}}^{\prime}e(H^{+}(X))s_{-d_{X}-l}(D_{X})SW(Y,\mathfrak{s}_{Y},\phi_{Y})v^{m+l-(p-1)/2}$

where the sum ${\sum_{l}}^{\prime}$ is over $l$ such that $0\leq l\leq-d_{X}$,
$m+l>0$, $m+l=0\;({\rm mod}\;p-1)$ and $h=\prod_{j=1}^{(p-1)/2}j^{b_{+}(Y)}$
for $p\neq 2$, $h=1$ for $p=2$.

###### Proof.

We give the proof for $p\neq 2$. The case $p=2$ is similar. Let
$f_{X},f_{Y},f_{Z}$ denote the monopole maps for $X,Y,Z$. We have that
$f_{Z}=ind^{G}_{1}(f_{Y})\wedge f_{X}$. Theorem 8.2 gives

$SW_{G,Z,\mathfrak{s}}^{\phi}(x^{m})=SW_{G,ind^{G}_{1}(f_{Y})}^{\phi_{Y}}(e_{G}(H^{+}(X))s_{G_{\mathfrak{s}},-d_{X}}(D_{X})x^{m}).$

Next we will compute $SW_{G,ind^{G}_{1}(f_{Y})}^{\phi_{Y}}$ using Theorem 6.5.
It is easy to see that for each splitting $s_{j}:\mathbb{Z}_{p}\to
S^{1}\times\mathbb{Z}_{p}$, we have $(ind^{G}_{1}(f_{Y}))^{s_{j}}=f_{Y}$.
Furthermore,
$ind^{G}_{1}(D_{Y})\cong\bigoplus_{j=0}^{p-1}\mathbb{C}_{j}^{d_{Y}}$. As
explained above we can orient $ind^{G}_{1}(H^{+}(Y))$ in such a way that
$ind^{G}_{1}(H^{+}(Y))/H^{+}(Y)\cong\bigoplus_{j=1}^{(p-1)/2}\mathbb{C}_{j}^{b_{+}(Y)}$.
Then

$\displaystyle SW^{\phi_{Y}}_{G,ind^{G}_{1}(f_{Y})}(x^{m})$
$\displaystyle=e_{G}(ind^{G}_{1}(H^{+}(Y))/H^{+}(Y))\sum_{j=0}^{p-1}SW^{\phi_{Y}}_{G,f_{Y}}(e_{S^{1}\times
G}(D/D_{j})^{-1}(x+jv)^{m})$
$\displaystyle=\prod_{r=1}^{(p-1)/2}(rv)^{b_{+}(Y)}\sum_{j=0}^{p-1}SW^{\phi_{Y}}_{G,f_{Y}}\left(\psi_{s_{j}}^{-1}\left(\prod_{k=1}^{p-1}(x+kv)^{-d_{Y}}(x-jv)^{m}\right)\right)$

where $\psi_{s_{j}}$ was defined in Section 6.6. Since
$d(Y,\mathfrak{s}_{Y})=0$, we have
$SW^{\phi_{Y}}_{G,f_{Y}}(1)=SW(Y,\mathfrak{s}_{Y},\phi_{Y})$ and
$SW^{\phi_{Y}}_{G,f_{Y}}(\psi_{s_{j}}^{-1}(x^{m}))=0$ for $m>0$. Hence

$\displaystyle SW^{\phi_{Y}}_{G,ind^{G}_{1}(f_{Y})}(x^{m})$
$\displaystyle=(-1)^{d_{Y}}\prod_{r=1}^{(p-1)/2}(rv)^{b_{+}(Y)}SW(Y,\mathfrak{s}_{Y},\phi_{Y})v^{m-d_{Y}(p-1)}\sum_{j=0}^{p-1}(-j)^{m}$
$\displaystyle=(-1)^{d_{Y}}h\,SW(Y,\mathfrak{s}_{Y},\phi_{Y})v^{m-(p-1)/2}\sum_{j=0}^{p-1}j^{m}$

where $h=\prod_{j=1}^{(p-1)/2}j^{b_{+}(Y)}$. But $\sum_{j=0}^{p-1}j^{m}$
equals $-1$ mod $p$ if $m>0$ and $(p-1)$ divides $m$ and is zero otherwise.
Hence

$SW^{\phi_{Y}}_{G,ind^{G}_{1}(f_{Y})}(x^{m})=\begin{cases}(-1)^{d_{Y}+1}h\,SW(Y,\mathfrak{s}_{Y},\phi_{Y})&m>0,m=0\;({\rm
mod}\;p-1),\\\ 0&\text{otherwise}\end{cases}$

Then using

$s_{G_{\mathfrak{s}},-d_{X}}(D_{X})=\sum_{l=0}^{-d_{X}}x^{l}s_{-d_{X}-l}(D_{X}),$

we obtain

$SW_{G,Z,\mathfrak{s}}^{\phi}(x^{m})=(-1)^{d_{Y}+1}h{\sum_{l}}^{\prime}e(H^{+}(X))s_{-d_{X}-l}(D_{X})SW(Y,\mathfrak{s}_{Y},\phi_{Y})v^{m+l-(p-1)/2}$

where the sum is over $l$ with $0\leq l\leq-d_{X}$, $m+l>0$ and $m+l=0\;({\rm
mod}\;p-1)$. ∎

The following corollary follows immediately from Theorem 8.4.

######  Corollary 8.5.

Let $p$ be a prime. Let $X$ be a compact, oriented, smooth $4$-manifold with
$b_{1}(X)=0$ and $b_{+}(X)>0$. Let $\mathfrak{s}$ be a spinc-structure on $X$
such that $d(X,\mathfrak{s})=0$. If $b_{+}(X)=1$ then fix a chamber. Let
$G=\mathbb{Z}_{p}$ act on the connected sum $\\#pX$ of $p$ copies of $X$ by
cyclically permuting the summands (to be precise, we take the equivariant
connected sum $S^{4}\\#pX$ where $\mathbb{Z}_{p}$ acts on $S^{4}$ by
rotation). Then

$SW_{G,\\#pX,\\#p\mathfrak{s}}(x^{m})=\begin{cases}(-1)^{d_{X}+1}hSW(X,\mathfrak{s}_{X})v^{m-(p-1)/2}&m>0,m=0\;({\rm
mod}\;p-1),\\\ 0&\text{otherwise}\end{cases}$

where $h=\prod_{j=1}^{(p-1)/2}j^{b_{+}(X)}$ for $p\neq 2$, $h=1$ for $p=2$.

##  9\. Some examples

###  9.1. Constraints on group actions

Consider the case where $G=\mathbb{Z}_{p}$ for a prime $p$. We use notation
from Section 6.6. Combining the divisibility condition Theorem 6.10 with
Theorem 6.12, we obtain the following constraint on smooth
$\mathbb{Z}_{p}$-actions:

######  Theorem 9.1.

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let
$G=\mathbb{Z}_{p}$ act smoothly on $X$ and let $\mathfrak{s}$ be a spinc-
structure preserved by $G$. If $SW(X,\mathfrak{s})\neq 0\;({\rm mod}\;p)$ and
$b_{0}\neq 1\;({\rm mod}\;2p)$ then there exists an $i$ such that $0\leq
2d_{i}-b_{0}-1\leq 2(p-2)$.

###### Proof.

If the condition $0\leq 2d_{i}-b_{0}-1\leq 2(p-2)$ is not satisfied then
Theorem 6.10 implies that $\overline{SW}^{s_{i}}_{G,X,\mathfrak{s}}=0\;({\rm
mod}\;p)$. If this holds for all $i$ then Theorem 6.12 implies that
$SW_{G,X,\mathfrak{s}}=0\;({\rm mod}\;p)$ and hence
$SW(X,\mathfrak{s})=0\;({\rm mod}\;p)$. ∎

######  Corollary 9.2.

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$. Let
$G=\mathbb{Z}_{2}$ act smoothly on $X$ and let $\mathfrak{s}$ be a spinc-
structure preserved by $G$. If $SW(X,\mathfrak{s})$ is odd and $b_{0}\neq
1\;({\rm mod}\;4)$ then there exists an $i$ such $2d_{i}=b_{0}+1$.

######  Example 9.3.

Let $K$ denote a $K3$ surface given as a degree $3$ cyclic branched cover of
$\mathbb{CP}^{1}\times\mathbb{CP}^{1}$ branched over a smooth curve $\Sigma$
of bi-degree $(3,3)$. This gives an action of $G=\mathbb{Z}_{3}$ on $K$ with
fixed point set a surface of genus $4$ and self-intersection $6$. Similarly we
can realise $4(S^{2}\times S^{2})$ as the branched triple cover of an
unknotted surface in $S^{4}$ of genus $2$ [1]. This gives an action of $G$ on
$4(S^{2}\times S^{2})$ with fixed point set a surface of genus $2$ and self-
intersection zero. Now consider the equivariant connected sum
$X_{0}=4(S^{2}\times S^{2})\\#5K$ of $4(S^{2}\times S^{2})$ and five copies of
$K$ with the given $\mathbb{Z}_{3}$-action. This gives a
$\mathbb{Z}_{3}$-action on $X_{0}$ with fixed point set a single surface of
genus $22$ and self-intersection $30$. The $4$-manifold $X_{0}$ is
homeomorphic to the elliptic surface $X=E(10)$, hence the
$\mathbb{Z}_{3}$-action on $X_{0}$ also defines a continuous, locally linear
$\mathbb{Z}_{3}$-action on $X$. On the other hand we will use Theorem 9.1 to
show that there is no smooth $\mathbb{Z}_{3}$-action on $X$ with the same
fixed point data. Let $\mathfrak{s}$ denote the unique spin structure on $X$.
Then $SW(X,\mathfrak{s})=\binom{8}{4}=1\;({\rm mod}\;3)$. A smooth
$\mathbb{Z}_{3}$-action on $X$ with fixed point set a surface of genus $22$
and self-intersection $30$ will have $b_{0}=5$ by the $G$-signature theorem
and $d_{0}=0$, $d_{1}=d_{2}=5$ by the $G$-spin theorem. But this contradicts
Theorem 9.1 which requires $3\leq d_{i}\leq 4$ for some $i$. So such an action
does not exist.

###  9.2. Exotic group actions

The gluing formula gives a method of contructing group actions which are
homeomorphic but not diffeomorphic. Let $X_{1},X_{2}$ be compact oriented
smooth $4$-manifolds with $b_{1}=0$ and $b_{+}>1$. Assume that there is a
homeomorphism $\varphi:X_{1}\to X_{2}$, but that $X_{1}$, $X_{2}$ have
different mod $p$ Seiberg–Witten invariants for a prime $p$, so in particular
they are not diffeomorphic. More precisely we will assume the following. For
simplicity assume $H_{1}(X_{i};\mathbb{Z})=0$ for $i=1,2$ so that spinc-
structures can be identified with characteristics. Then we will require that
there does not exist an isometry $\psi:H^{2}(X_{1};\mathbb{Z})\to
H^{2}(X_{2};\mathbb{Z})$ for which
$SW(X_{1},\mathfrak{s}_{1})=SW(X_{2},\psi(\mathfrak{s}_{1}))\;({\rm mod}\;p)$
for every spinc-structure $\mathfrak{s}_{1}$ with
$d(X_{1},\mathfrak{s}_{1})=0$ and for some choice of orientations of
$H^{+}(X_{i})$.

Given $g>0$, let $Y_{g}$ denote the degree $p$ cyclic branched cover of an
unknotted embedded surface in $S^{4}$ of genus $g$. Then $Y_{p}$ is
diffeomorphic to $\\#^{g(p-1)}S^{2}\times S^{2}$. The branched covering
construction defines an action of $G=\mathbb{Z}_{p}$ on $Y_{g}$. Moreover
$H^{2}(Y_{g};\mathbb{Z})^{G}=0$, in particular $H^{+}(Y_{g})^{G}=0$. By [14]
there exists a $k>0$ such that $X_{1}\\#k(S^{2}\times S^{2})$ and
$X_{1}\\#k(S^{2}\times S^{2})$ are diffeomorphic. This also implies that
$pX_{1}\\#k(S^{2}\times S^{2})$ and $pX_{2}\\#k(S^{2}\times S^{2})$ are
diffeomorphic. Thus if $g(p-1)\geq k$ then we get two $\mathbb{Z}_{p}$-actions
on $X=pX_{1}\\#Y_{g}\cong pX_{2}\\#Y_{g}$. The first is obtained by taking
$Y_{g}$ with the $\mathbb{Z}_{p}$-action described above and attaching $p$
copies of $X_{1}$ which are permuted by the $\mathbb{Z}_{p}$-action. The
second action on $X$ is given by the same construction but with $X_{2}$ in
place of $X_{1}$. These two actions are equivariantly homeomorphic since we
can apply the homeomorphism $\varphi:X_{1}\to X_{2}$ to each copy of $X_{1}$.
On the other hand they are not equivarianly diffeomorphic. To see this, note
that since $H^{2}(Y_{g};\mathbb{Z})^{G}=0$, one finds that the $G$-invariant
spinc-structures on $X_{i}\\#Y_{g}$ are precisely those of the form
$ind^{G}_{1}(\mathfrak{s}_{i})\\#\mathfrak{s}$ where $\mathfrak{s}_{i}$ is a
spinc-structure on $X_{i}$ and $\mathfrak{s}$ is the unique spin structure on
$Y_{g}$. If $d(X_{i},\mathfrak{s}_{i})=0$ then Theorem 8.4 gives

$SW_{G,X,ind^{G}(\mathfrak{s}_{i})\\#\mathfrak{s}}(x^{p-1})=(-1)^{g(p-1)/2}SW(X_{i},\mathfrak{s}_{i})v^{(p-1)/2}\;({\rm
mod}\;p).$

Our assumption that $X_{1},X_{2}$ have different mod $p$ Seiberg–Witten
invariants then implies that the two different $G$-actions have different
equivariant Seiberg–Witten invariants, so they are not diffeomorphic.

###  9.3. Obstructions to enlarging group actions

Let $G$ be a finite group and $H$ a subgroup of $G$. The compatibility of the
equivariant Seiberg–Witten invariants with the restriction map from
$G$-equivariant cohomology (or $K$-theory) to $H$-equivariant implies a
condition for a smooth $H$-action to extend to $G$. This follows immediately
from Theorem 3.1 (2), but we restate it here from the perspective of extending
a group action.

######  Proposition 9.4.

Let $X$ be a compact, oriented smooth $4$-manifold with $b_{1}(X)=0$. Suppose
that a finite group $H$ acts on $X$ by orientation preserving diffeomorphisms
and suppose that $H^{+}(X)^{H}\neq 0$. Let $\mathfrak{s}$ be a $H$-invariant
spinc-structure and let $\phi\in H^{+}(X)^{H}\setminus\\{0\\}$ be a chamber.
Let $G$ be a finite group containing $H$. If the $H$-action on $X$ extends to
a smooth, orientation preserving action of $G$ which fixes $\mathfrak{s}$ and
$\phi$, then $H_{\mathfrak{s}}$ is the restriction to $H$ of an $S^{1}$
central extension $G_{\mathfrak{s}}\to G$ and for every $\theta$ in the image
of the restriction map $H^{*}_{G_{\mathfrak{s}}}(pt;A)\to
H^{*}_{H_{\mathfrak{s}}}(pt;A)$, we have that
$SW_{H,X,\mathfrak{s}}^{\phi}(\theta)$ is in the image of
$H^{*}_{G}(pt;A_{w})\to H^{*}_{H}(pt;A_{w})$.

Furthermore, if $b_{+}(X)$ is odd and a $H$-equivariant spinc-struture
$\mathfrak{o}$ on $H^{+}(X)$ is given which can be lifted to a $G$-equivariant
spinc-structure, then for every $\theta$ in the image of
$R(G_{\mathfrak{s}})\to R(H_{\mathfrak{s}})$, we have that
$SW_{H,X,\mathfrak{s}}^{\phi,K}(\theta)$ is in the image of $R(G)\to R(H)$.

######  Example 9.5.

Let us consider Proposition 9.4 in the case that
$H=\mathbb{Z}_{p}=\langle\sigma\;|\;\sigma^{p}\rangle$ is cyclic of odd prime
order and
$G=D_{p}=\langle\sigma,\tau\;|\;\sigma^{p},\tau^{2},(\tau\sigma)^{2}\rangle$
is the dihedral group of order $2p$. Both $G$ and $H$ have no non-trivial
$S^{1}$ central extensions. Let $w\in
H^{1}_{D_{p}}(pt;\mathbb{Z}_{2})\cong\mathbb{Z}_{2}$ be the unique non-trivial
element. Recall that
$H^{2k}_{\mathbb{Z}_{p}}(pt;\mathbb{Z})\cong\mathbb{Z}_{p}$ for every $k>0$.
On the other hand a simple calculation shows that
$H^{2k}_{D_{p}}(pt;\mathbb{Z})=0$ for odd $k$ and
$H^{2k}_{D_{p}}(pt;\mathbb{Z}_{w})=0$ for even $k>0$. Also
$R(\mathbb{Z}_{p})\cong\mathbb{Z}[t]/(t^{p}-1)$ and the image of $R(D_{p})\to
R(\mathbb{Z}_{p})$ is the subring generated by $t+t^{-1}$. From this we obtain
non-trivial conditions for a smooth $\mathbb{Z}_{p}$-action on $X$ to extend
to $D_{p}$, where $X$ is a compact, oriented smooth $4$-manifold with
$b_{1}(X)=0$ and $H$ acts smoothly and orientation preservingly. Suppose
$\mathfrak{s}$ is a $H$-invariant spinc-structure on $X$ and $\phi$ is a
$H$-invariant chamber. If the $\mathbb{Z}_{p}$-action extends to a smooth
orientation preserving action of $D_{p}$ which preserves $\mathfrak{s}$ and
$\phi$. Assume $b_{+}(X)$ is odd, so $d(X,\mathfrak{s})$ is even. Then the
condition is that $SW^{\phi}_{\mathbb{Z}_{p},X,\mathfrak{s}}(x^{m})=0$
whenever $2m-d(X,\mathfrak{s})$ is positive and equals $2$ mod $4$ if $\sigma$
preserves orientation on $H^{+}$, or equals $0$ mod $4$ if $\sigma$ reverses
orientation on $H^{+}$.

Consider for instance the case that $X=\\#pY$ is the connected sum of $p$
copies of a $4$-manifold $Y$ and $\sigma$ cyclically permutes the summands.
Then by Corollary 8.5,
$SW_{\mathbb{Z}_{p},\\#pY,\\#p\mathfrak{s}_{Y}}^{\phi}(x^{p-1})$ is a non-zero
multiple of $SW(Y,\mathfrak{s}_{Y},\phi)v^{(p-1)/2}$. We obtain the following
non-existence result: if $SW(Y,\mathfrak{s}_{Y},\phi)\neq 0\;({\rm mod}\;p)$
then there does not exist an extension of the $\mathbb{Z}_{p}$-action on $X$
to a smooth action of $D_{p}$ which fixes $\\#p\mathfrak{s}_{Y}$ and $\phi$
and for which $\tau$ preserves orientation on $H^{+}(X)$ if $p=3\;({\rm
mod}\;4)$ or reverses orientation on $H^{+}(X)$ if $p=1\;({\rm mod}\;4)$.

###  9.4. Obstructions to equivariant connected sum decompositions

Let $X$ be a compact, oriented, smooth $4$-manifold with $b_{1}(X)=0$.
Consider a smooth $G$-action on $X$. If $X$ can be written as an equivariant
connected sum $X=X_{1}\\#X_{2}$ where $b_{+}(X_{1})^{G},b_{+}(X_{2})^{G}>0$,
then the equivariant Seiberg–Witten invariants vanish by Theorem 8.3. This can
be used to limit the possible ways in which $X$ can be an equivariant
connected sum.

######  Example 9.6.

We will construct two smooth $\mathbb{Z}_{3}$-actions on
$X=5\mathbb{CP}^{2}\\#23\overline{\mathbb{CP}}^{2}$ with the same fixed point
data. One of these actions will decompose as an equivariant connected sum
$X=X_{1}\\#X_{2}$ with $b_{+}(X_{1})^{G},b_{+}(X_{2})^{G}>0$, the other will
not decompose in this way. The actions on $X$ will be constructed from the
following:

  * (1)

Let $K$ be the Fermat quartic
$\\{[z_{0},z_{1},z_{2},z_{3}]\in\mathbb{CP}^{3}\;|\;z_{0}^{4}+z_{1}^{4}+z_{2}^{4}+z_{3}^{4}=0\\}$
with $\mathbb{Z}_{3}$-action given by
$[z_{0},z_{1},z_{2},z_{3}]\mapsto[z_{0},z_{2},z_{3},z_{1}]$.

  * (2)

Let $\mathbb{CP}^{2}_{(a)}$ denote $\mathbb{CP}^{2}$ with
$\mathbb{Z}_{3}$-action $[z_{0},z_{1},z_{2}]\mapsto[z_{0},\omega
z_{1},\omega^{a}z_{2}]$, where $\omega=e^{2\pi i/3}$.

  * (3)

Let $2(S^{2}\times S^{2})$ be the branched triple cover of an unknotted torus
in $S^{4}$.

The action of $\mathbb{Z}_{3}$ on the tangent space of an isolated fixed point
will be orientation preservingly isomorphic to
$(z_{1},z_{2})\mapsto(\omega^{-1}z_{1},\omega^{a}z_{2})$, where $a$ is either
$1$ or $-1$. Let $n_{\pm}$ denote the number of isolated fixed points of type
$a=\pm 1$. Then $K$ has $(n_{+},n_{-})=(6,0)$, $\mathbb{CP}^{2}_{(1)}$ has
$(n_{+},n_{-})=(0,1)$, $\mathbb{CP}^{2}_{(2)}$ has $(n_{+},n_{-})=(3,0)$ and
$2(S^{2}\times S^{2})$ has $(n_{+},n_{-})=0$.

Let $Y$ be the equivariant sum
$Y=\overline{\mathbb{CP}}^{2}_{(2)}\\#\overline{\mathbb{CP}}^{2}_{(1)}\\#2(S^{2}\times
S^{2})$, where the first and second summands are connected along isolated
fixed points and the second and third summands are connected along non-
isolated fixed points. Then $X$ is diffeomorphic to $K\\#Y$ and so we obtain a
$\mathbb{Z}_{3}$-action on $X$ by considering $K\\#Y$ as an equivariant
connected sum. We have that $b_{+}(Y)^{G}=0$ and if we equip $2(S^{2}\times
S^{2})$ with its unique spin structure and equip each
$\overline{\mathbb{CP}}^{2}$ summand of $Y$ with a spinc-structure satisfying
$c(\mathfrak{s})^{2}=-1$, then we obtain an invariant spinc-structure
$\mathfrak{s}_{Y}$ on $Y$ for which $d_{Y}=0$. If $\mathfrak{s}_{K}$ denotes
the unique spin structure on $K$, then Theorem 8.3 gives

$SW_{\mathbb{Z}_{3},K\\#Y,\mathfrak{s}_{K}\\#\mathfrak{s}_{Y}}(1)=SW_{\mathbb{Z}_{3},K,\mathfrak{s}_{K}}(1)=SW(K,\mathfrak{s}_{K})=1.$

So $K\\#Y$ has a non-zero equivariant Seiberg–Witten invariant and can’t be
obtained as an equivariant connected sum of the form $X_{1}\\#X_{2}$ with
$b_{+}(X_{1})^{G},b_{+}(X_{2})^{G}>0$.

Next let $K^{\prime}$ be defined as follows. First take the equivariant
connected sum
$K^{\prime}_{0}=3\mathbb{CP}^{2}_{(2)}\\#\overline{\mathbb{CP}}^{2}_{(2)}$
where the three $\mathbb{CP}^{2}_{(2)}$ summands are attached to the three
isolated fixed points of $\overline{\mathbb{CP}}^{2}_{(2)}$. Then let
$K^{\prime}=K^{\prime}_{0}\\#18\overline{\mathbb{CP}}^{2}$, where the
$\mathbb{Z}_{3}$-action permutes the $18$ copies of
$\overline{\mathbb{CP}}^{2}$ in six $3$-cycles. It is easily seen that the
$\mathbb{Z}_{3}$-action on $K^{\prime}$ has the same fixed point data as $K$,
namely $(n_{+},n_{-})=(6,0)$ and no non-isolated fixed points. Hence the
$\mathbb{Z}_{3}$-actions on $K\\#Y$ and $K^{\prime}\\#Y$ have the same fixed
points as well. Moreover, $K^{\prime}\\#Y$ is diffeomorphic to $X$ so this
gives another $\mathbb{Z}_{3}$-action on $X$ with the same fixed point data.
Unlike the first action this one can be decomposed into an equivariant
connected sum $X_{1}\\#X_{2}$ with $b_{+}(X_{1})^{G},b_{+}(X_{2})^{G}>0$. This
is clear because each of the three $\mathbb{CP}^{2}_{(2)}$-summands in
$K^{\prime}_{0}$ have $b_{+}^{G}=1$.

###  9.5. Obstructions to equivariant positive scalar curvature

The $4$-manifolds of the form $a\mathbb{CP}^{2}\\#b\overline{\mathbb{CP}}^{2}$
admit metrics of positive scalar curvature. On the other hand the equivariant
Seiberg–Witten invariants can be used to find many examples of actions on such
manifolds for which there is no invariant metric of positive scalar curvature.
Let $X$ be a simply-connected $4$-manifold with $b_{+}(X)>1$ on which
$\mathbb{Z}_{p}$ acts smoothly. Assume the action has a non-isolated fixed
point and that $X$ has a non-zero mod $p$ Seiberg–Witten invariant. We assume
$X$ is not spin (if necessary we can replace $X$ by a blow up to achieve
this). Then for some $g>0$ we have that $X\\#g(p-1)(S^{2}\times S^{2})$ is
diffeomorphic to $a\mathbb{CP}^{2}\\#b\overline{\mathbb{CP}}^{2}$, where
$a=b_{+}(X)+g(p-1)$, $b=b_{-}(X)+g(p-1)$. Now recall that
$Y_{g}=g(p-1)(S^{2}\times S^{2})$ can be realised as the degree $p$ cyclic
branched cover of an unknotted surface in $S^{4}$ of genus $g$. We have
$b_{+}(Y_{g})^{\mathbb{Z}_{p}}=0$ and then $X\\#Y_{g}$ has non-zero
equivariant Seiberg–Witten invariants by Theorem 8.3. This gives a smooth
$\mathbb{Z}_{p}$-action on $a\mathbb{CP}^{2}\\#b\overline{\mathbb{CP}}^{2}$
which does not admit an invariant positive scalar curvature metric.

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