$(\varPsi\varPsi^{*})_{00}=(\psi\psi^{*})_{0}\otimes\left(\begin{array}[]{cc}\frac{1}{2}&0\\\
0&-\frac{1}{2}\end{array}\right)$

Notice that $\det(\mathfrak{s}_{L})=\det(S\otimes L)\simeq L^{\otimes 2}$, so
in the notation of Kronheimer and Mrowka, $B^{t}\equiv B_{L}^{\otimes 2}$ is
the induced connection on $\det(\mathfrak{s}_{L})$. In particular,
$F_{B^{t}}=2F_{B_{L}}$ and so the previous equation can be rewritten as

$\frac{1}{2}*F_{B^{t}}+\rho^{-1}(\psi\psi^{*})_{0}=\frac{1}{2}*F_{B^{\det}}$

This agrees with the conventions used in [55, Eq 4.11]. ∎

It is very important to notice that when the Seiberg-Witten monopoles appear
as fixed points of the $S^{1}$ action, the equations they solve are
“automatically” perturbed by the term $F_{B^{\det}}$. These perturbations can
have dramatic consequences, the first one being the following:

######  Lemma 39.

Suppose that $(B_{L},\psi)$ solves the perturbed Seiberg-Witten equations
(27), where $B_{L}\in\mathcal{A}(L)$ is a $U(1)$ connection, and
$\psi\in\varGamma(S\otimes L)$. Then $\psi$ vanishes identically if and only
if $c_{1}(E)\in H^{2}(Y;\mathbb{Z})$ is an even class, i.e,
$w_{2}(\mathfrak{su}(E))$ vanishes. In other words, if $c_{1}(E)$ is an odd
class, then there are no “reducible” solutions to the equations (27).

###### Proof.

Observe that a solution of the form $(B_{L},0)$ must solve the equation

$F_{B^{\det}}=2F_{B_{L}}$

and by Chern-Weil theory we conclude that

$c_{1}(E)=2c_{1}(L)$

from which the assertion in the lemma follows immediately. ∎

######  Remark 40.

Another consequences of these perturbations is that in general the symmetry
between the solutions to the Seiberg-Witten equations for $\mathfrak{s}$ and
its conjugate spin-c structure $\bar{\mathfrak{s}}$ is lost. This is analogous
to a comment Donaldson makes in [28, Section 6].

In fact, if define

$\omega\equiv\frac{1}{4}F_{B^{\det}}$

then $\omega$ is a harmonic representative of $-\frac{\pi i}{2}c_{1}(E)\in
H^{2}(Y;i\mathbb{R})$ and using $F_{B^{t}}=2F_{B_{L}}$, the equations 27 can
be rewritten as [55, Eq 29.2]

(28) $\begin{cases}\frac{1}{2}\rho(F_{B^{t}}-4\omega)-(\psi\psi^{*})_{0}=0\\\
D_{B}\psi=0\end{cases}$

In particular, if $c_{1}(E)$ is an odd class, then the term $\omega$
corresponds to a _non-balanced perturbation_ , in the terminology of
Kronheimer and Mrowka [55, Definition 29.1.1]. A consequence of working with a
non-balanced perturbation is that the definition of the monopole Floer
homology groups $HM(Y,\mathfrak{s},\omega)$ require the choice of a local
coefficient system in order to make sense of the differential, since in
general there are monotonicity issues when defining these groups. Some choices
of local coefficient systems are described in [55, Section 30.2].

It remains a daunting challenge to try to define a suitable version of $SO(3)$
_monopole Floer homology,_ because of the presence of reducibles (i.e, abelian
$U(1)$ monopoles and flat connections).

As a historical analogy, before the construction of monopole Floer homology in
full generality by Kronheimer and Mrowka [55], the simplest case where it
could be defined was for three manifolds with $b_{1}(Y)>0$ and for non-torsion
spin-c structures, since in this case reducibles could be avoided by using
suitable perturbations. The difficult case was that of torsion spin-c
structures, which was solved thanks to the blow-up model introduced by
Kronheimer and Mrowka.

Likewise, we can think of defining $SO(3)$ monopole Floer homology as trying
to define monopole Floer homology in the torsion case, which begs the question
of whether there is an analogue of the non-torsion case where reducibles can
be avoided.

Such an analogue may be provided by a construction of an $U(2)$ _monopole
Floer homology._ We explain the basic ideas behind this construction: assuming
all the technical hurdles can be solved, it can be regarded as giving rise to
a family of Floer homologies, indexed by subintervals of the real line. So in
a sense this would be the Floer-theoretic analogue of the typical wall-
crossing phenomenon for the various moduli spaces which appear in Algebraic
Geometry.

In this case we no longer fix a reference connection on $\det E$, so we are
working with the enlarged configuration space

$\mathcal{C}_{U(2)}(Y,\mathfrak{t})=\mathcal{A}(E)\times\varGamma(S\otimes E)$

Suppose that $\omega$ is a harmonic representative of $c_{1}(E)$ [before it
was chosen to be a representative of $-\frac{\pi i}{2}c_{1}(E)$].

The $U(2)$ monopole equations $SW_{U(2)}(Y,\mathfrak{t},\tau)$ are equations
for a pair $(B,\varPsi)$ which satisfy

(29) $U(2)\text{ monopole equations
}SW_{U(2)}(Y,\mathfrak{t},\tau)\;\;\;\begin{cases}*F_{B}+2\rho^{-1}(\varPsi\varPsi^{*})_{0}+i\tau(*\omega)I_{E}=0\\\
D_{B}\varPsi=0\end{cases}$

The main differences with the $SO(3)$ monopole equations (26) are the
following:

  1. (1)

As mentioned before, we are no longer fixing a reference connection on $\det
E$.

  2. (2)

Since we dropped the reference connection, we now need an equation which
involves the entire curvature $F_{B}$ of $B$, instead of just its traceless
part. Subsequently, $(\varPsi\varPsi^{*})_{0}$ denotes the component of
$\varPsi\varPsi^{*}$ which is traceless only on the “spinorial” component,
that is,
$(\varPsi\varPsi^{*})_{0}\in\varGamma(\text{End}_{0}S\otimes\text{End}E)$.

  3. (3)

Notice that the traceless part of the curvature equation in (29) gives us back
the $SO(3)$ monopole equations (26).

It is not difficult to identify the “critical values” of $\tau$, that is,
those values where the flat connections and Seiberg-Witten solutions can
appear.

######  Lemma 41.

a) Suppose that $(B,\varPsi\equiv 0)$ solves the $U(2)$ monopole equations
(29). Then $\tau=\pi$ and $B$ is a flat $U(2)$ connection. If we assume
$c_{1}(E)$ is odd, then $B$ is irreducible, i.e, not compatible with a
decomposition $E=L_{1}\oplus L_{2}$, $B=B_{1}\oplus B_{2}$.

b) Suppose that $(B=B_{1}\oplus B_{2},\varPsi=\psi\oplus 0)$ solves the
equation (29), where $E=L_{1}\oplus L_{2}$ is compatible with the reduction of
$B$. Then $\tau=2\pi n$ for $n$ an arbitrary integer, and $(B_{1},\psi)$
solves the perturbed Seiberg-Witten equations (27) for the spin-c structure
$S\otimes L_{1}$ with perturbation term $F_{B_{2}}$. Moreover, $c_{1}(L_{1})$
and $c_{1}(L_{2})$ are both integer multiples of $c_{1}(E)$, i.e, their Chern
classes belong to the ray spanned by $c_{1}(E)$ inside the integer lattice
$H^{2}(Y;\mathbb{Z})$.

###### Proof.

a) This is a Chern-Weil argument once again. Taking traces we find that
$\text{tr}(F_{B})=-2i\tau\omega$ and therefore $\tau=\pi$. If $B$ were
compatible with a reduction then

$F_{B}=\left(\begin{array}[]{cc}F_{B_{1}}&0\\\
0&F_{B_{2}}\end{array}\right)=-i\tau\omega I_{E}$

we would conclude that $F_{B_{1}}=F_{B_{2}}$. Therefore $L_{1}\simeq L_{2}$
and $c_{1}(E)$ must represent an even class, which cannot be true by
assumption.

b) In this case the equations become

$\begin{cases}*F_{B_{1}}+2\rho^{-1}(\psi\psi^{*})_{0}+i\tau(*\omega)=0\\\
F_{B_{2}}=-i\tau\omega\\\ D_{B_{1}}\psi=0\end{cases}$

Substituting the second equation into the first one we recognize the perturbed
Seiberg Witten equation 27

$*F_{B_{1}}+2\rho^{-1}(\psi\psi^{*})_{0}-*F_{B_{2}}=0$

By Chern-Weil theory the second equation says that $-2\pi ic_{1}(L_{2})=-i\tau
c_{1}(E)$, so $\tau$ is of the form $\tau=2\pi n$, for $n\in\mathbb{Z}$ an
integer. Since

$c_{1}(E)=c_{1}(L_{1}\oplus
L_{2})=c_{1}(L_{1})+c_{1}(L_{2})=c_{1}(L_{1})+nc_{1}(E)$

this forces $c_{1}(L_{1})$ to be

$c_{1}(L_{1})=(1-n)c_{1}(E)$

∎

######  Remark 42.

a) Notice the second part of the previous lemma implies that the only monopole
Floer groups that the $U(2)$ monopole Floer equations can “see” are those
whose spin-c structure satisfy $c_{1}(\mathfrak{s})=c_{1}(S\otimes
L_{1})=2(1-n)c_{1}(E)$, for $n$ an arbitrary integer. In particular, notice
that all spin-c structures which are torsion can appear in this picture. Also,
in the specific case that $Y$ is an integer homology $S^{1}\times S^{2}$, and
we choose $c_{1}(E)$ as a generator of $H^{2}(Y;\mathbb{Z})\simeq\mathbb{Z}$,
then all spin-c structures can arise in this case as well.

b) Implicitly we are also saying that solutions to equation 29 of the form
$(B=B_{1}\oplus B_{2},\varPhi=\varPsi_{1}\oplus\varPsi_{2})$, $E=L_{1}\oplus
L_{2}$ , do not arise unless one of the spinors $\varPsi_{1}$ or $\varPsi_{2}$
vanish identically. An easy way to see that otherwise by taking the traceless
part of the curvature equation, we would obtain a pair $(B=B_{1}\oplus
B_{2},\varPhi=\varPsi_{1}\oplus\varPsi_{2})$ which satisfies the $SO(3)$
monopole equations 26 where none of the spinors is identically zero. However,
[35, Lemma 5.22] shows that this cannot be the case. A quick look at the
statement of Feehan and Leness shows that they stated this already for a
_perturbed_ version of the $SO(3)$ monopole equations, which had to satisfy
specific properties so that this would not occur. Therefore, this already
shows one needs to be very careful about which properties will continue to
hold after adding (holonomy) perturbations.

Having introduced the $U(2)$ monopole equations, it makes sense to state the
following conjecture:

######  Conjecture 43.

Suppose that $Y$ is a closed oriented 3-manifold, and $E$ is an $U(2)$ bundle
which satisfies the admissibility condition that $w=c_{1}(E)\in
H^{2}(Y;\mathbb{Z})$ is a non-torsion odd class. Choose a spin-c structure
$\mathfrak{s}$ and consider the bundle $V=S\otimes E$ representing the spin-u
structure $\mathfrak{t}$, where $(S,\rho)$ is a spinor bundle representing
$\mathfrak{s}$.

Then for $\tau\neq\pi,2\pi n$ for $n\in\mathbb{Z}$, it is possible to define a
Seiberg-Witten $U(2)$ monopole Floer homology
$HM_{U(2)}(Y,\mathfrak{t},\tau)$. Some expected features of these groups are
the following:

a) The chain complex of $HM_{U(2)}(Y,\mathfrak{t},\tau)$ will be built out of
a suitably perturbed version of the equations 29. These solutions can be
identified with the critical points of a perturbed $U(2)$ Chern-Simons-Dirac
functional $CS_{\mathfrak{t},\tau}$, morally a Morse function on the space
$\mathcal{B}(Y,\mathfrak{t})=\left(\mathcal{A}(E)\times\varGamma(V)\right)/\mathcal{G}(E)$.

b) The differential involves the count of gradient flow lines mod
$\mathbb{R}$\- translations of $CS_{\mathfrak{t},\tau}$ as in every other
Floer homology. In general, there will be monotonicity issues [a bound on the
dimension of flow lines will not imply a uniform energy bound on the
corresponding moduli spaces], so the groups $HM_{U(2)}(Y,\mathfrak{t},\tau)$
will require a Novikov ring (local coefficient system) so that the
differential is well defined.

c) The groups will be $\mathbb{Z}/2\mathbb{Z}$ graded in general.

d) Write $\mathbb{R}\backslash\\{\pi,2\pi n\mid
n\in\mathbb{Z}\\}=\bigcup_{i}I_{i}$ as a disjoint union of open intervals. If
$\tau,\tau^{\prime}$ belong to the same “chamber” $I_{i}$, then
$HM_{U(2)}(Y,\mathfrak{t},\tau)\simeq
HM_{U(2)}(Y,\mathfrak{t},\tau^{\prime})$, i.e, the groups are isomorphic to
each other as long as one does not cross a “wall”.

Clearly behind this conjecture there is a great deal of work that will require
verification, but it could serve as a toy model before a general construction
of $SO(3)$ monopole Floer homology, which would prove even more difficult in
general.

##  7\. $SO(3)$ Monopoles on $S^{1}\times\varSigma$: Framed Monopole Homology

Now we want to understand the solutions to the $SO(3)$ monopole equations on
$Y=S^{1}\times\varSigma$ in terms of the solutions to the $SO(3)$ vortex
equations on $\varSigma$.

Before doing this, it is useful to understand how the spin-c structures on $Y$
are related to the spin-c structures on $\varSigma$. More details can be found
in [68, Section 3]. Throughout our discussion we will use a product metric on
$S^{1}\times\varSigma$.

Suppose $\mathfrak{s}_{Y}$ is a spin-c structure on $Y$. Since
$H^{2}(S^{1}\times\varSigma;\mathbb{Z})$ has no 2-torsion, we can specify
$\mathfrak{s}_{Y}$ by its element $c_{1}(\mathfrak{s}_{Y})\in
H^{2}(S^{1}\times\varSigma;\mathbb{Z})$. Let $S_{Y}$ denote a spinor bundle
representing $\mathfrak{s}_{Y}$ and $L_{Y}=\det S_{Y}$ the corresponding line
bundle.

If we think of $S^{1}\times\varSigma$ as $[0,1]\times\varSigma$ with the
boundaries identified, then $L_{Y}$ is constructed as the pullback under the
projection $[0,1]\times\varSigma\rightarrow\varSigma$ of a line bundle
$L_{\varSigma}$ by gluing along the boundaries with an isomorphism
$u\in\text{Map}(\varSigma,S^{1})$. Then

$c_{1}(\mathfrak{s}_{Y})=c_{1}(L_{Y})=c_{1}(L_{\varSigma})+[S^{1}]\cup[u]$

where $[u]$ is the class of $u$ in $[\varSigma;S^{1}]\simeq
H^{1}(\varSigma;\mathbb{Z})$. The spin-c structure $\mathfrak{s}_{Y}$ induces
a spin-c structure $\mathfrak{s}_{\varSigma}$ with determinant line bundle
$L_{\varSigma}$. If $S_{\varSigma}$ is a spinor bundle for
$\mathfrak{s}_{\varSigma}$ , we can assume that it is of the form

$S_{\varSigma}=(\mathbb{C}\oplus
K_{\varSigma}^{-1})\otimes(L_{\varSigma}\otimes K_{\varSigma})^{1/2}$

where $K_{\varSigma}=\varOmega^{1,0}(\varSigma)$ is the canonical bundle of
$\varSigma$ , and $(L_{\varSigma}\otimes K_{\varSigma})^{1/2}$ denotes a line
bundle whose square is the line bundle $L_{\varSigma}\otimes K_{\varSigma}$.
In particular, if $L_{\varSigma}\simeq\mathbb{C}$ is the trivial line bundle,
then we obtain the canonical spin-c structure on $\varSigma$

$S_{can}\simeq K_{\varSigma}^{1/2}\oplus K_{\varSigma}^{-1/2}$

Notice that an implicit choice of _spin_ structure on $\varSigma$ was made so
that we could find a square root of the canonical bundle, so we will assume a
fixed choice once and for all. Moreover, if $[u]=0$, then $c_{1}(L_{Y})=0$
which means that we can regard the torsion spin-c structure on $Y$ as being of
the form

$S_{tor}=K_{\varSigma}^{1/2}\oplus K_{\varSigma}^{-1/2}$

Now suppose that $E$ is a $U(2)$ bundle on $\varSigma$. We can then consider
the spin-u structure

$V=(K_{\varSigma}^{1/2}\oplus K_{\varSigma}^{-1/2})\otimes E=(\mathbb{C}\oplus
K_{\varSigma}^{-1})\otimes(K_{\varSigma}^{1/2}\otimes E)$

Therefore, it is more elegant to reabsorb $K_{\varSigma}^{1/2}$ into the
bundle $E$ we are considering and work with the canonical spin-u structure
222Notice that $\mathbb{C}\oplus K_{\varSigma}^{-1}$ is not a torsion spin-c
structure in general, but this does not make a big difference for our
arguments, since $\text{ad}(K_{\varSigma}^{1/2}\otimes E)\simeq\text{ad}(E)$.
Moreover, the Chern connection is being implicitly used on
$K_{\varSigma}^{-1}$ , and the trivial connection on the $\mathbb{C}$ factor.

$V_{can}=(\mathbb{C}\oplus K_{\varSigma}^{-1})\otimes E$

It suffices to define the Clifford map $\rho$ on $S=\mathbb{C}\oplus
K_{\varSigma}^{-1}$ . First of all, notice that

$\rho(d\theta)^{2}=-|d\theta|_{g}^{2}1_{S}=-1_{S}$

so $\rho(d\theta)$ is an endomorphism of $S$ which squares to $-1$. We will
thus identify $\rho(d\theta)$ with the matrix

$\rho(d\theta)=\left(\begin{array}[]{cc}i&0\\\ 0&-i\end{array}\right)$

Decompose a one form $\eta\in T^{*}(S^{1}\times\varSigma)$ as

$\eta=td\theta+\varpi$

where $\varpi$ is identified with the pullback of a $(0,1)$ form on
$\varSigma$. Notice that our conventions are such that

$|\eta|_{g_{Y}}=t^{2}+2|\varpi|_{g_{\varSigma}}^{2}$

and that we are regarding $\varOmega^{0,1}(\varSigma)$ only as a _real_ vector
space, so that we have the isomorphism
$\varOmega^{0,1}(\varSigma)\simeq\varOmega^{1}(\varSigma)$.

Finally, we define for
$(\alpha,\beta)\in\mathbb{C}\oplus\pi^{*}K_{\varSigma}^{-1}$ the Clifford map

$\rho(\eta)\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)=\left(\begin{array}[]{c}it\alpha-\sqrt{2}\left\langle\beta,\varpi\right\rangle\\\
\sqrt{2}\alpha\varpi-it\beta\end{array}\right)$

Notice that $K_{\varSigma}^{-1}$ is being interpreted as a _complex_ vector
space, and that $\left\langle\beta,\varpi\right\rangle$ refers to the
hermitian inner product on $\varOmega^{0,1}(\varSigma)$, complex linear in the
first entry. Slightly abusing notations we can write this as

$\rho(\eta)=\left(\begin{array}[]{cc}i\left\langle
d\theta,\eta\right\rangle&-\sqrt{2}\varpi^{*}\\\ \sqrt{2}\varpi&-i\left\langle
d\theta,\eta\right\rangle\end{array}\right)$

It is also useful to have local expressions for these formulas. Consider a
coframe $e^{1},e^{2}$ of $T^{*}\varSigma$ and define

$\begin{cases}\epsilon=\frac{1}{\sqrt{2}}(e^{1}+ie^{2})\\\
\bar{\epsilon}=\frac{1}{\sqrt{2}}(e^{1}-ie^{2})\end{cases}$

Under an almost complex structure $J$ of $\varSigma$, we have that $J(e^{1})$
is mapped to $e^{2}$, thus

$J(\epsilon)=\frac{1}{\sqrt{2}}(e^{2}-ie^{1})=-\frac{i}{\sqrt{2}}(e^{1}+ie^{2})=-i\epsilon$

In other words, $\epsilon$ is of type $(1,0)$. Likewise, $\bar{\epsilon}$ is
of type $(0,1)$ since $J(\bar{\epsilon})=i\bar{\epsilon}$. It is an easy check
that

$\displaystyle\rho(\epsilon)(\alpha,\beta)=(-\sqrt{2}\left\langle\beta,\bar{\epsilon}\right\rangle,0)$
$\displaystyle\iff\left(\begin{array}[]{cc}0&-\sqrt{2}\epsilon\\\
0&0\end{array}\right)$
$\displaystyle\rho(\bar{\epsilon})(\alpha,\beta)=(0,\sqrt{2}\bar{\epsilon}\alpha)$
$\displaystyle\iff\left(\begin{array}[]{cc}0&0\\\
\sqrt{2}\bar{\epsilon}&0\end{array}\right)$

Following the discussion in [26, Section 3.2], we can write a connection $B$
on $S^{1}\times\varSigma$ in the form

$B=C(\theta)+c(\theta)d\theta$

where $C(\theta)$ is a one parameter periodic family of $U(2)$ connections on
$\varSigma$ and $c(\theta)\in\varOmega^{0}(\varSigma,\mathfrak{su}(2))$ is
also another periodic family. The curvature becomes

(30) $F_{B}=F_{C(\theta)}+d\theta\wedge\left(\frac{\partial
C(\theta)}{\partial\theta}+d_{C(\theta)}c(\theta)\right)$

Moreover, the Dirac operator can be written as [compare with [26, Lemma 3.5],
[67, Eq 5.12], [62, Section 3.1]]

$D_{B}=\left(\begin{array}[]{cc}i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)&\sqrt{2}\bar{\partial}_{C(\theta)}^{*}\\\
\sqrt{2}\bar{\partial}_{C(\theta)}&-i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)\end{array}\right)$

Therefore, the Dirac equation for $(B,(\alpha,\beta))$ reads

$\begin{cases}i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)\alpha+\sqrt{2}\bar{\partial}_{C(\theta)}^{*}\beta=0\\\
\sqrt{2}\bar{\partial}_{C(\theta)}\alpha-i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)\beta=0\end{cases}$

where $\alpha\in\varOmega^{0}(E)$ and $\beta\in\varOmega^{0,1}(E)$.

In order to understand the curvature equation, decompose the curvature in
terms of a co-frame as

$F_{B}^{0}=F_{\theta,e^{1}}d\theta\wedge e^{1}+F_{\theta,e^{2}}d\theta\wedge
e^{2}+F_{e^{1},e^{2}}e^{1}\wedge e^{2}$
$*F_{B}^{0}=F_{\theta,e^{1}}e^{2}-F_{\theta,e^{2}}e^{1}+F_{e^{1},e^{2}}d\theta$

Since $\frac{\epsilon+\bar{\epsilon}}{\sqrt{2}}=e^{1}$,
$\frac{i(\bar{\epsilon}-\epsilon)}{\sqrt{2}}=e^{2}$ then
$\rho(e^{1})=\left(\begin{array}[]{cc}0&-\epsilon\\\
\bar{\epsilon}&0\end{array}\right)$,
$\rho(e^{2})=\left(\begin{array}[]{cc}0&i\epsilon\\\
i\bar{\epsilon}&0\end{array}\right)$

$\rho(*F_{B}^{0})=\left(\begin{array}[]{cc}iF_{e^{1},e^{2}}&\left(iF_{\theta,e^{1}}+F_{\theta,e^{2}}\right)\epsilon\\\
\left(iF_{\theta,e^{1}}-F_{\theta,e^{2}}\right)\bar{\epsilon}&-iF_{e^{1},e^{2}}\end{array}\right)$

Thus the curvature equation reads (compare with [31, p.10])

$\begin{cases}iF_{e^{1},e^{2}}+(\alpha\otimes\alpha^{*})_{0}-(\varLambda\beta\otimes\beta^{*})_{0}=0\\\
\left(iF_{\theta,e^{1}}-F_{\theta,e^{2}}\right)\bar{\epsilon}+2(\beta\otimes\alpha^{*})_{0}=0\end{cases}$

The expression $(\varLambda\beta\otimes\beta^{*})_{0}$ denotes that a
contraction with the symplectic form on $\varSigma$ is implicit.

######  Theorem 44.

Suppose $(B,(\alpha,\beta))$ satisfies the $SO(3)$ vortex equations on
$S^{1}\times\varSigma$, for the spin-u structure $V_{can}=(\mathbb{C}\oplus
K_{\varSigma}^{-1})\otimes E$. Here $\alpha\in\varOmega^{0}(E)$ and
$\beta\in\varOmega^{0,1}(E)$. Write $B=C(\theta)+c(\theta)d\theta$. Then

(31) $\displaystyle
i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)\alpha+\sqrt{2}\bar{\partial}_{C(\theta)}^{*}\beta=0$
$\displaystyle\sqrt{2}\bar{\partial}_{C(\theta)}\alpha-i\left(\frac{\partial}{\partial\theta}+c(\theta)\right)\beta=0$
$\displaystyle
iF_{e^{1},e^{2}}+(\alpha\otimes\alpha^{*})_{0}-(\varLambda\beta\otimes\beta^{*})_{0}=0$
$\displaystyle\left(iF_{\theta,e^{1}}-F_{\theta,e^{2}}\right)\bar{\epsilon}+2(\beta\otimes\alpha^{*})_{0}=0$

In fact, all solutions are of the form $(B,(\alpha,0))$ or $(B,(0,\beta))$.

In the first case, we get a solution of the form

$\displaystyle\bar{\partial}_{C}\alpha=0$ $\displaystyle
F_{C}-i(\alpha\otimes\alpha^{*})_{0}=0$

which can be identified with an $SO(3)$ vortex on the bundle $E$.

In the second case, we get a solution of the form

$\displaystyle\bar{\partial}_{C}^{*}\beta=0$ $\displaystyle
F_{C}+i(\varLambda\beta\otimes\beta^{*})_{0}=0$

which can be identified via Serre duality with an $SO(3)$ vortex on the bundle
$K_{\varSigma}\otimes E^{-1}$.

In particular, if we assume that $\deg E>2\deg K_{\varSigma}$, then this
second type of moduli space of $SO(3)$ vortices is empty.

######  Remark 45.

The previous theorem states that on $S^{1}\times\varSigma$ the solutions to
the $SO(3)$ monopole equations decompose into the union of two moduli spaces
of $SO(3)$ vortices on the Riemann surface. This is the analogue of what
happens for the solutions to the $SO(3)$ monopole equations on a Kahler
surface, where the moduli space also decomposes into the union of two moduli
spaces of stable pairs on the Kahler surface (see [63, Theorem 6.3.10] and
[31, Section 1.2]).

###### Proof.

We follow the proof in [68, Proposition 3.1], [26, Proposition 3.6], which
applies with cosmetic changes to our situation.

Suppose that $(B,(\alpha,\beta))$ is irreducible in that $B$ is irreducible
and $(\alpha,\beta)$ non-vanishing. This forces the the stabilizer of this
$SO(3)$ monopole to be Id under the determinant one gauge group. Moreover, one
can use a gauge transformation to assume the connection is in “temporal”
gauge, which means that $B=C(\theta)$ instead of the more general case
$B=C(\theta)+c(\theta)d\theta$. So we can assume that $c(\theta)=0$ in the
equations (31)

Since $i\bar{\partial}_{C(\theta)}^{*}=\varLambda\partial_{C(\theta)}$ on
$(0,1)$ forms, the first equation (31) can be written as

$-\frac{\partial}{\partial\theta}\alpha+\sqrt{2}\varLambda\partial_{C(\theta)}\beta=0$

Differentiate this with respect to $\theta$ to find

$0=-\frac{\partial^{2}}{\partial\theta^{2}}\alpha+\sqrt{2}(\frac{\partial}{\partial\theta}\varLambda\partial_{C(\theta)})\beta+\sqrt{2}\varLambda\partial_{C(\theta)}\frac{\partial\beta}{\partial\theta}$

Because of the second equation in (31),
$\frac{\partial}{\partial\theta}\beta=-\sqrt{2}i\bar{\partial}_{C(\theta)}\alpha$,
so we obtain

$0=-\frac{\partial^{2}}{\partial\theta^{2}}\alpha+\sqrt{2}(\frac{\partial}{\partial\theta}\varLambda\partial_{C(\theta)})\beta-2i\varLambda\partial_{C(\theta)}\underbrace{\bar{\partial}_{C(\theta)}\alpha}_{(0,1)\text{
form}}$

Using the Kahler identity
$i\bar{\partial}_{C(\theta)}^{*}=\varLambda\partial_{C(\theta)}$ one more time
we end up with

$0=-\frac{\partial^{2}}{\partial\theta^{2}}\alpha+\sqrt{2}(\frac{\partial}{\partial\theta}\varLambda\partial_{C(\theta)})\beta+2\bar{\partial}_{C(\theta)}^{*}\bar{\partial}_{C(\theta)}\alpha$

Take inner product with $\alpha$, and use integration by parts and the third
equation in (31) to conclude

$0=-\int_{\check{\varSigma}}\left\langle\frac{\partial^{2}\alpha}{\partial\theta^{2}},\alpha\right\rangle+\sqrt{2}\int_{\check{\varSigma}}\left\langle(\alpha\otimes\beta^{*})_{0}\beta,\alpha\right\rangle+2\int_{\check{\varSigma}}|\bar{\partial}_{C(\theta)}\alpha|^{2}$

The integrand of the term in the middle is

$\left\langle|\beta|^{2}\alpha-\frac{1}{2}\left\langle\beta,\alpha\right\rangle_{\tilde{E}}\beta,\alpha\right\rangle=|\beta|^{2}|\alpha|^{2}-\frac{1}{2}|<\beta,\alpha>_{\tilde{E}}|^{2}$

Integrate from $0$ to $1$ to obtain (recall that we are thinking of
$S^{1}\times\varSigma$ as $[0,1]\times\varSigma$ with the ends identified)

$|\frac{\partial}{\partial\theta}\alpha|^{2}+\sqrt{2}\left(|\beta|^{2}|\alpha|^{2}-\frac{1}{2}|<\beta,\alpha>_{\tilde{E}}|^{2}\right)+2|\bar{\partial}_{C(\theta)}\alpha|^{2}=0$

So $\alpha$ is $\theta$ independent and by Cauchy Schwarz since
$|<\beta,\alpha>_{\tilde{E}}|^{2}\leq|\beta|^{2}|\alpha|^{2}$ we conclude that
either $\alpha$ or $\beta$ vanishes identically. Because of the curvature
decomposition (30) we also find that $\frac{\partial C}{\partial\theta}=0$
thus all the data pulls back from the surface $\varSigma$, and the claim of
the theorem is now straightforward.

For the last claim, suppose that $\deg E>2K_{\varSigma}$. Then
$\deg(K_{\varSigma}\otimes E^{-1})<0$ so if $(C,\beta)$ denotes an $SO(3)$
vortex on the bundle $K_{\varSigma}\otimes E^{-1}$ , we have
$c_{1}(L)<\frac{1}{2}\deg(K_{\varSigma}\otimes E^{-1})<0$ for any holomorphic
sub-line bundle of $E$ for which $\beta\in H^{0}(L)$ and this is clearly
impossible. Thus the moduli space of $SO(3)$ vortices must be empty. ∎

Now we specialize to the case that $\varSigma=T^{2}$. The canonical bundle of
an elliptic curve is trivial, and choose the $U(2)$ bundle over $T^{2}$ with
$\deg E=1$. Now we discuss the solutions to the $SO(3)$ monopole equations
$(B,\varPsi)$ on $S^{1}\times T^{2}$.

$\bullet$ For the irreducible $SO(3)$ monopoles, where $B$ is an irreducible
$SO(3)$ connection, and $\varPsi$ non-vanishing, by the previous theorem (44),
$\varPsi=\alpha\oplus 0$ or $\varPsi=0\oplus\beta$, and $B$ pulls back from a
connection $C$ on $T^{2}$. Moreover, $(C,\alpha)$ is an $SO(3)$ vortex on $E$
and there are no $SO(3)$ vortices of the second kind since $\deg E=1>0=2\deg
K_{T^{2}}$. As described in Example (37), the dimension of the moduli space of
$SO(3)$ vortices on $E$ is two, which becomes one dimensional after taking the
quotient by the $S^{1}$ action.

$\bullet$ The moduli space of flat connections on $T^{3}$. As described in
[54, Section 4.1], there are two flat connections on $T^{3}$. One corresponds
to the flat connection (representation) on $T^{2}$, while the other is
obtained from this one by applying a twist to the representation on $T^{2}$.

$\bullet$ The Seiberg-Witten solutions which appear associated to our choice
of spin-u structure must satisfy the perturbed Seiberg Witten equations (27):

$\begin{cases}*F_{B_{L}}+\rho^{-1}(\psi\psi^{*})_{0}-\frac{1}{2}*F_{B^{\det}}=0\\\
D_{B}\psi=0\end{cases}$

Here $\psi$ corresponds to a section of
$\underbrace{K_{\varSigma}^{1/2}\otimes L}_{\simeq
L}\subset\underbrace{K_{\varSigma}^{1/2}\otimes E}_{\simeq E}$ for a splitting
of $E=L\oplus(L^{*}\otimes\det E)$. Thus $\psi$ satisfies the abelian vortex
equations

$\begin{cases}*_{T^{2}}F_{C}-\frac{i}{2}|\psi|^{2}=\frac{1}{2}*_{T^{2}}F_{C^{\det
E}}\\\ \bar{\partial}_{C}\psi=0\end{cases}$

The usual Chern-Weil argument says that

$\deg L+\frac{1}{4\pi}\int_{T^{2}}|\psi|^{2}=\frac{1}{2}\deg E=\frac{1}{2}$

Since $\deg L\geq 0$, this forces $\deg L=0$ and thus $\psi$ is a holomorphic
_function_ on $T^{2}$. Thus it is constant and up to gauge there is a unique
solution. In this way we find that there is a unique solution to these
perturbed Seiberg-Witten equations on $T^{3}$.

In particular, for this choice of $U(2)$ bundle on $T^{3}$, the moduli space
of $SO(3)$ monopoles on $T^{3}$ consists of a one-dimensional moduli space (an
interval) with one endpoint corresponding to the Seiberg-Witten monopole, the
other endpoint corresponding to the flat connection which pulls back from the
flat connection on $T^{2}$, and an additional flat connection not connected to
this one dimensional moduli space. Based on this picture, we define:

######  Definition 46.

Suppose that $Y$ is a closed oriented 3-manifold. The _framed monopole Floer_
__ groups of $Y$, denoted $HM^{\\#}(Y)$, is defined as the monopole Floer
homology $HM(Y\\#T^{3},\omega,\varGamma)$. Here $\omega$ is a non-balanced
perturbation which is a harmonic representative of $-\frac{\pi
i}{2}c_{1}(E)\in H^{2}(T^{2};i\mathbb{R})\subset H^{2}(T^{3};i\mathbb{R})$ for
the $U(2)$ bundle $E$ over $T^{2}$ of degree one. Since $\omega$ is a non-
balanced perturbation, the definition of the monopole Floer homology requires
the use of a local coefficient system $\varGamma$.

######  Remark 47.

Some choices of local systems are described in [55, Chapter VIII] as well as
in the author’s paper [32].

As explained in the previous section, the fact that we are using a non-
balanced perturbation means that there are no reducible solutions to the
Seiberg-Witten equations on the 3 manifold $Y\\#T^{3}$. In fact, there are
only finitely many irreducible Seiberg-Witten monopoles which make up the
chain complex $CM(Y\\#T^{3},\omega)$ which computes
$HM(Y\\#T^{3},\omega,\varGamma)$. Notice that

$HM(Y\\#T^{3},\omega,\varGamma)=\bigoplus_{\mathfrak{s}\\#\mathfrak{s}^{\prime}}HM(Y\\#T^{3},\omega,\varGamma,\mathfrak{s}\\#\mathfrak{s}^{\prime})$

where $\mathfrak{s}\\#\mathfrak{s}^{\prime}$ denotes a spin-c structure on
$Y\\#T^{3}$. Observe that our notation for the framed monopole Floer groups
$HM^{\\#}(Y)$ suggest that they depend on $Y$, not on $Y\\#T^{3}$. In order to
make this more precise we need to use the K nneth formula from [57, Section
6].

In the notation of Kutluhan, Lee, Taubes, $M_{1},M_{2}$ are two three
manifolds and $M_{\\#}$ denotes its connected sum. The Floer complex for
$M_{1}$ is assumed to come from a _nonbalanced perturbation._ In particular,
the complex for $M_{\\#}$ uses a non-balanced perturbation as well. The
connected sum theorem relates $CM_{*}(M_{\\#})$ with
$S_{U_{\sqcup}}(\hat{C}_{*}(M_{\sqcup}))$. The latter refers to
$\hat{C}_{\sqcup}\otimes\mathbb{Z}[y]$, where $y$ acts as a degree 1 chain map
[57, Section 4.1]. If we write this as $\hat{C}_{\sqcup}\oplus
y\hat{C}_{\sqcup}$ , then the differential takes the block form

$D_{\sqcup}=\left(\begin{array}[]{cc}\hat{\partial}_{\sqcup}&0\\\
\hat{U}_{\sqcup}&-\hat{\partial}_{\sqcup}\end{array}\right)$

Proposition 6.7 in [57, Section 6] then states that there is a chain homotopy
equivalence between the complex
$C(M_{\\#},\mathfrak{s}_{\\#},\omega_{\\#},\varGamma_{\\#})$ and the complex
$S_{U_{\sqcup}}(\hat{C}_{*}(M_{\sqcup},\mathfrak{s}_{\sqcup},\omega_{\sqcup};\varGamma_{\sqcup})$.

In our case, $M_{1}=T^{3}$, $M_{2}=Y$. The differential
$\hat{\partial}_{\sqcup}=1\otimes\partial_{\natural}^{\\#}(M_{2})+\partial_{o}^{o}(M_{1})\otimes
1)$ reduces to $1\otimes\partial_{\natural}^{\\#}(Y)$. Likewise,
$\hat{U}_{\sqcup}=1\otimes\hat{U}_{p_{2}}-\hat{U}_{p_{1}}\otimes 1$ reduces to
$1\otimes\hat{U}_{p_{2}}$. Thus, it is not difficult to see that in our case
the complex
$S_{U_{\sqcup}}(\hat{C}_{*}(M_{\sqcup},\mathfrak{s}_{\sqcup},\omega_{\sqcup};\varGamma_{\sqcup})$
is isomorphic to $S_{U_{Y}}(\hat{C}_{*}(Y,\mathfrak{s};\varGamma))$, but this
is precisely the complex which computes the monopole Floer group
$\widetilde{HM}(Y,\varGamma)$ [15, Section 8], the only difference is that we
now end up with a twisted version of this group. In other words, we have

######  Theorem 48.

The framed monopole Floer group $HM^{\\#}(Y)$ is isomorphic to the group
$\widetilde{HM}(Y,\varGamma)$. Moreover, the Euler characteristic of
$HM^{\\#}(Y)$ (hence $\widetilde{HM}(Y,\varGamma)$) is the same as the one of
$HI^{\\#}(Y)$.

######  Remark 49.

The equality of the Euler characteristics was explained near the end of the
introduction to this work.

##  8\. $SO(3)$ Monopoles on Seifert Manifolds

Now we proceed to analyze the $SO(3)$ monopole equations on a Seifert manifold
$Y$. We will follow [74] and [67]. Unfortunately, our notation and some of our
conventions are not quite isomorphic to those used in either reference, so we
have to redo some of their computations.

For the identies we require we need to have very good control on all the
constants appearing in our formulas, so at the cost of redudance we will
reprove some of the results in each paper.

First of all, for an $SO(3)$ connection ${}^{\circ}\nabla$ on the cotangent
bundle $T^{*}Y$ which is not necessarily the Levi-Civita connection, we say
that a connection

$\nabla^{S}:\varGamma(S)\rightarrow\varGamma(T^{*}Y\otimes S)$

is a spinorial connection with respect to ${}^{\circ}\nabla$ if all vector
fields $v$ on $Y$, sections $\varPhi$ of $S$ and one forms $\theta$ we have

(32)
$\nabla_{v}^{S}(\rho(\theta)\varPhi)=\rho(^{\circ}\nabla_{v}\theta)\varPhi+\rho(\theta)(\nabla_{v}^{S}\varPhi)$

If $(e^{i})$ is a coframe and the connection matrix of ${}^{\circ}\nabla$ is
given by 333Our convention is chosen in such a way that if
${}^{\circ}\nabla^{TY}$ is the induced connection on the tangent bundle then
${}^{\circ}\nabla^{TY}e_{i}=\sum_{j=1}^{3}\omega_{ji}e_{j}$ where $e_{i}$ an
orthonormal frame

${}^{\circ}\nabla e^{i}=\omega e^{i}=\sum_{j=1}^{3}\omega_{ji}\otimes e^{j}$

where $\omega\in\varOmega^{1}(Y)\otimes\mathfrak{so}(3)$ then then the
connection matrix of $\nabla^{S}$ is

(33) $-\frac{1}{4}\sum_{i,j=1}^{3}\omega_{ij}\otimes\rho(e^{i}\wedge
e^{j})+bId_{S}$

where $b$ is any imaginary valued one form.

With this baggage out of the way we can start defining some special geometric
structures on $Y=S(L)$. Recall that $S(L)$ is a principal $S^{1}$ bundle. As
such, we can find a connection $i\eta$ for $S(L)$ and we assume that it is of
constant curvature. We give $Y$ the metric

$g_{Y}=\eta\otimes\eta+\pi^{*}(g_{\check{\varSigma}})$

This is a locally homogeneous Riemann metric on $Y=S(L)$. The global 1 form
induces a reduction in the structure group of $TY$ to $SO(2)$; the kernel of
$\eta$ is naturally identified with the pull back of the orbifold tangent
bundle of $\varSigma$, so that we have orthogonal splittings

$\begin{cases}TY\simeq\mathbb{R}\frac{\partial}{\partial\varphi}\oplus\pi^{*}(T\check{\varSigma})\\\
T^{*}Y\simeq\mathbb{R}\eta\oplus\pi^{*}(T^{*}\check{\varSigma})\end{cases}$

Here $\frac{\partial}{\partial\varphi}$ is the vector field dual to $\eta$
with respect to $g_{Y}$. Notice that $\frac{\partial}{\partial\varphi}$ has
unit length with respect to this metric and in fact it is a Killing vector
field for the metric $g_{Y}$, that is, the Lie derivative of the metric
$g_{Y}$ with respect to $\frac{\partial}{\partial\varphi}$ vanishes.

Moreover, for any vector field $v_{\check{\varSigma}}$ dual to the pull back
of a 1 form from $\check{\varSigma}$, we have that
$\left[v_{\check{\varSigma}},\frac{\partial}{\partial\varphi}\right]=0$. If
$\nabla_{\check{\varSigma}}^{LC}$ is the Levi-Civita connection on
$T^{*}\check{\varSigma}$, we give $T^{*}Y$ the so called adiabatic connection

(34) $\nabla^{\infty}=d\oplus\pi^{*}(\nabla_{\check{\varSigma}}^{LC})$

In particular,

$\begin{cases}\nabla^{\infty}\eta=0\\\
\nabla^{\infty}\pi^{*}(\check{\theta})=\pi^{*}(\nabla_{\check{\varSigma}}^{LC}\check{\theta})\end{cases}$

where $\check{\theta}$ is any 1 form on $\check{\varSigma}$. Since $d\eta$ is
constant, let $c_{\eta}$ be the constant determined by

(35) $d\eta=2c_{\eta}\pi^{*}(\mu_{\check{\varSigma}})$

Using Chern-Weil theory one deduces that

(36)
$c_{\eta}=-\frac{\pi\int_{\check{\varSigma}}c_{1}(L)}{\text{vol}(\check{\varSigma})}=-\frac{\pi\deg(\check{L})}{\text{vol}(\check{\varSigma})}$

The following estimate will allows us to compare two Dirac operators
associated to the adiabatic and Levi-Civita connections.

######  Lemma 50.

(Comparing Dirac Operators, [67, Lemma 5.5]) Let $(S,\rho)$ be a spin-c
structure over $Y=S(L)$ and $\nabla^{S},\nabla_{LC}^{S}$ a pair of connections
which are spinorial with respect to $\nabla^{\infty}$ and $\nabla^{LC}$
respectively, which induce the same connection on the determinant line bundle
of $S$. Then, for any $\theta\in\varOmega^{1}(Y,\mathbb{R})$ , we have

$\nabla_{LC,\theta^{\\#}}^{S}=\nabla_{\theta^{\\#}}^{S}+\frac{c_{\eta}}{2}\left(\rho(\theta)-2\left\langle\theta,\eta\right\rangle\rho(\eta)\right)$

where $\theta^{\\#}$ denotes the vector field which is $g_{Y}$ dual to
$\theta$. If $D_{LC}$ and $D^{\infty}$ are the corresponding Dirac operators
then

$D_{LC}=D^{\infty}-\frac{1}{2}c_{\eta}$

In particular, this implies that the Dirac operator $D^{\infty}$ is self
adjoint, since $D_{LC}$ and multiplication by a real scalar valued function
are self adjoint.

###### Proof.

Recall the Cartan’s structure equations [22, Section 4.3]. Let
$\\{e^{0},e^{1},e^{2}\\}$ be a (local) orthonormal frame of $T^{*}Y$. Define
the connection matrix for the Levi Civita connection $\nabla^{LC}$ by

$\nabla^{LC}e^{i}=\sum_{j=0}^{2}\omega_{ji}^{LC}\otimes e_{j}$

then

$de^{i}=\sum_{j=0}^{2}e^{j}\wedge\omega_{ij}^{LC}=-\sum_{j=0}^{2}\omega_{ij}^{LC}\wedge
e^{j}$

These can be conveniently rewritten as

$d\left[\begin{array}[]{c}e^{0}\\\ e^{1}\\\
e^{2}\end{array}\right]=-\left[\begin{array}[]{ccc}0&\omega_{01}^{LC}&\omega_{02}^{LC}\\\
\omega_{10}^{LC}&0&\omega_{12}^{LC}\\\
\omega_{20}^{LC}&\omega_{21}^{LC}&0\end{array}\right]\wedge\left[\begin{array}[]{c}e^{0}\\\
e^{1}\\\ e^{2}\end{array}\right]$

Choose $e^{0}=\eta$, and $e^{1},e^{2}$ the pull back from an orthonormal frame
on $\check{\varSigma}$. We obtain the system of equations

(40) $\displaystyle\begin{array}[]{ccccc}d\eta&=&&-\omega_{01}^{LC}\wedge
e^{1}&-\omega_{02}^{LC}\wedge e^{2}\\\ de^{1}&=&-\omega_{10}^{LC}\wedge
e^{0}&&-\omega_{12}^{LC}\wedge e^{2}\\\ de^{2}&=&-\omega_{20}^{LC}\wedge
e^{0}&-\omega_{21}^{LC}\wedge e^{1}\end{array}$

Given that

$d\eta=2c_{\eta}\mu_{\check{\varSigma}}=2c_{\eta}e^{1}\wedge e^{2}$

our first equation in 40becomes

$2c_{\eta}e^{1}\wedge e^{2}=e^{1}\wedge\omega_{01}^{LC}-\omega_{02}^{LC}\wedge
e^{2}$

Therefore we have

$\displaystyle\omega_{01}^{LC}=c_{\eta}e^{2}$
$\displaystyle\omega_{02}^{LC}=-c_{\eta}e^{1}$

Also, if we let $\omega_{\check{\varSigma}}^{LC}$ be the connection form for
the Levi-Civita connection, since pull back commutes with exterior
differentiation then the structure equations for
$\omega_{\check{\varSigma}}^{LC}$ are

(41) $d\left[\begin{array}[]{c}e^{1}\\\
e^{2}\end{array}\right]=-\left[\begin{array}[]{cc}0&\omega_{\check{\varSigma},12}^{LC}\\\
\omega_{\check{\varSigma},21}^{LC}&0\end{array}\right]\wedge\left[\begin{array}[]{c}e^{1}\\\
e^{2}\end{array}\right]$

Comparing 41 with 40 and using the antisymmetry of the $\omega_{ij}$ we obtain
the equations

$\displaystyle c_{\eta}e^{2}\wedge e^{0}-\omega_{12}^{LC}\wedge
e^{2}=-\omega_{\check{\varSigma},12}^{LC}\wedge
e^{2}\implies(-c_{\eta}e^{0}-\omega_{12}^{LC}+\omega_{\check{\varSigma},12}^{LC})\wedge
e^{2}=0$ $\displaystyle-c_{\eta}e^{1}\wedge e^{0}-\omega_{21}^{LC}\wedge
e^{1}=-\omega_{\check{\varSigma},21}^{LC}\wedge e^{1}\implies
e^{1}\wedge(-c_{\eta}e^{0}+\omega_{21}^{LC}-\omega_{\check{\varSigma},21}^{LC})=0$

Clearly we obtain

$\omega_{21}^{LC}=c_{\eta}e^{0}+\omega_{\check{\varSigma},21}^{LC}$

Therefore the connection matrix for the Levi-Civita connection $\omega^{LC}$
becomes 444As mentioned in [67], we are writing the connection matrix for $TY$
which differs by a sign from the connection matrix for $T^{*}Y$ , hence
explaining the difference sign with their paper

(48)
$\displaystyle\left(\begin{array}[]{ccc}0&\omega_{01}^{LC}&\omega_{02}^{LC}\\\
\omega_{10}^{LC}&0&\omega_{12}^{LC}\\\
\omega_{20}^{LC}&\omega_{21}^{LC}&0\end{array}\right)=\left(\begin{array}[]{ccc}0&c_{\eta}e^{2}&-c_{\eta}e^{1}\\\
-c_{\eta}e^{2}&0&-c_{\eta}e^{0}-\omega_{\check{\varSigma},21}^{LC}\\\
c_{\eta}e^{1}&c_{\eta}e^{0}+\omega_{\check{\varSigma},21}^{LC}&0\end{array}\right)$

The connection matrix for $\nabla^{\infty}$ is easier to find since it is a
reducible connection and trivial in the first summand; it is simply

(49) $\left(\begin{array}[]{ccc}0&0&0\\\
0&0&-\omega_{\check{\varSigma},21}^{LC}\\\
0&\omega_{\check{\varSigma},21}^{LC}&0\end{array}\right)$

Therefore the difference between connection matrices is

$\nabla^{LC}-\nabla^{\infty}\iff\left(\begin{array}[]{ccc}0&c_{\eta}e^{2}&-c_{\eta}e^{1}\\\
-c_{\eta}e^{2}&0&-c_{\eta}e^{0}\\\
c_{\eta}e^{1}&c_{\eta}e^{0}&0\end{array}\right)=(\tilde{\omega}_{ij})$

Using formula 33 the corresponding spin connections differ by

$\displaystyle\nabla_{LC}^{S}-\nabla^{S}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\sum_{i,j=0}^{2}\tilde{\omega}_{ij}\otimes\rho(e^{i})\rho(e^{j})$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{i<j}\tilde{\omega}_{ij}\otimes\rho(e^{i})\rho(e^{j})$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\left(c_{\eta}e^{2}\otimes\rho(e^{0})\rho(e^{1})-c_{\eta}e^{1}\otimes\rho(e^{0})\rho(e^{2})-c_{\eta}e^{0}\otimes\rho(e^{1})\rho(e^{2})\right)$
$\displaystyle=$
$\displaystyle\frac{c_{\eta}}{2}\left(e^{0}\otimes\rho(e^{1})\rho(e^{2})+e^{1}\otimes\rho(e^{0})\rho(e^{2})-e^{2}\otimes\rho(e^{0})\rho(e^{1})\right)$
$\displaystyle=$
$\displaystyle\frac{c_{\eta}}{2}\left(-e^{0}\otimes\rho(e^{0})+e^{1}\otimes\rho(e^{1})+e^{2}\otimes\rho(e^{2})\right)\rho(\mu_{Y})$
$\displaystyle=$
$\displaystyle\frac{c_{\eta}}{2}\left(-e^{0}\otimes\rho(e^{0})+e^{1}\otimes\rho(e^{1})+e^{2}\otimes\rho(e^{2})\right)$

where we used $1_{S}=\rho(\mu_{Y})=\rho(e_{0})\rho(e_{1})\rho(e_{2})$. To
check the first formula in the lemma
$\nabla_{LC,\theta^{\\#}}^{S}=\nabla_{\theta^{\\#}}^{S}+\frac{c_{\eta}}{2}\left(\rho(\theta)-2\left\langle\theta,\eta\right\rangle\rho(\eta)\right)$
notice that it is linear in $\theta^{\\#}$ and so we just need to verify it on
the coframe for each covector $e^{0},e^{1},e^{2}$.

We will just check it for $\theta=e^{0}$ and $\theta=e^{1}$ since
$\theta=e^{2}$ is entirely analogous to the second case. In the case
$\theta^{\\#}=e_{0}$ because of the previous calculation

$\displaystyle\nabla_{LC,e_{0}}^{S}-\nabla_{e_{0}}^{S}$
$\displaystyle=\frac{c_{\eta}}{2}\left(-e^{0}\otimes\rho(e^{0})+e^{1}\otimes\rho(e^{1})+e^{2}\otimes\rho(e^{2})\right)e_{0}$
$\displaystyle=-\frac{c_{\eta}}{2}\rho(e^{0})$

On the other hand

$\displaystyle\frac{c_{\eta}}{2}\left(\rho(e^{0})-2\left\langle
e^{0},\eta\right\rangle\rho(\eta)\right)$
$\displaystyle=\frac{c_{\eta}}{2}\left(\rho(e^{0})-2\rho(\eta)\right)$
$\displaystyle=-\frac{c_{\eta}}{2}\rho(e^{0})$

and so both formulas agree. When $\theta=e^{1}$ our long calculation implies
that

$\displaystyle\nabla_{LC,e_{1}}^{S}-\nabla_{e_{1}}^{S}$
$\displaystyle=\frac{c_{\eta}}{2}\left(-e^{0}\otimes\rho(e^{0})+e^{1}\otimes\rho(e^{1})+e^{2}\otimes\rho(e^{2})\right)e_{1}$
$\displaystyle=\frac{c_{\eta}}{2}\rho(e^{1})$

On the other hand

$\begin{array}[]{ccc}\frac{c_{\eta}}{2}\left(\rho(e^{1})-2\left\langle
e^{1},\eta\right\rangle\rho(\eta)\right)&=&\frac{c_{\eta}}{2}\rho(e^{1})\end{array}$

Which verifies the formula. The statement about the Dirac operators
$D_{LC}=D^{\infty}-\frac{1}{2}c_{\eta}$ also follows because

$\displaystyle D_{LC}-D^{\infty}$
$\displaystyle=\frac{c_{\eta}}{2}\left(-\rho(e^{0})\rho(e^{0})+\rho(e^{1})\rho(e^{1})+\rho(e^{2})\rho(e^{2})\right)$
$\displaystyle=\frac{c_{\eta}}{2}\left(1-1-1\right)$
$\displaystyle=-\frac{c_{\eta}}{2}$

∎

As a corollary, we also have

######  Corollary 51.

([67, Corollary 5.7]) The anticommutator of the Dirac operator with Clifford
multiplication by a one form $\theta$ is given by the formula

(50)
$\\{D^{\infty},\rho(\theta)\\}=-\rho((*d+d*)\theta)-2\nabla_{\theta^{\\#}}^{S}+2c_{\eta}<\theta,\eta>\rho(\eta)$

###### Proof.

This follows from the analogous formula for the Levi-Civita operator [10,
Proposition 3.45]

$\\{D_{LC},\rho(\theta)\\}=\rho((*d+d*)\theta)-2\nabla_{LC,\theta^{\\#}}^{S}$

Since $D_{LC}=D^{\infty}-\frac{c_{\eta}}{2}$ and
$\nabla_{LC,\theta^{\\#}}^{S}=\nabla_{\theta^{\\#}}^{S}+\frac{c_{\eta}}{2}\left(\rho(\theta)-2\left\langle\theta,\eta\right\rangle\rho(\eta)\right)$
we find that

$\\{D^{\infty},\rho(\theta)\\}-c_{\eta}\rho(\theta)=\rho((*d+d*)\theta)-2\nabla_{\theta^{\\#}}^{S}-c_{\eta}(\rho(\theta)-2\left\langle\theta,\eta\right\rangle\rho(\eta))$

from which the result is now obvious. ∎

For our purposes we are interested in the case where we twist our spinor
bundle with a rank two hermitian bundle $E$, endowed with a $U(2)$ connection
$B$. Then we have a twisted connection

$\nabla^{S,B}:\varGamma(S\otimes E)\rightarrow\varGamma(T^{*}Y\otimes S\otimes
E)$

defined in the usual way

$\nabla^{S,B}(s\otimes s_{E})=(\nabla^{S}s)\otimes
s_{E}+s\otimes(\nabla^{B}s_{E})$

Here $s$ is a section of $S$, while $s_{E}$ a section of $E$. The twisted
Dirac operator is then

$D_{S,B}=\sum_{i=0}^{2}\rho(e^{i})\nabla_{e_{i}}^{S,B}=\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B}+\rho(e^{1})\nabla_{e_{1}}^{S,B}+\rho(e^{2})\nabla_{e_{2}}^{S,B}$

Observe first of all that since $D_{S}$ [called $D^{\infty}$ before] is self-
adjoint thanks to Lemma (50), we know that $D_{S,B}$ is a self-adjoint
operator as well. Hence

$D_{S,B}^{*}D_{S,B}=D_{S,B}^{2}$

and our objective is to find an useful decomposition for the square of this
operator, in other words, a Weitzenbock-type formula. Tautologically we define

$D_{2,S,B}\equiv
D_{S,B}-\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B}$

Then

$D_{S,B}^{2}=\left(\rho(\eta)\cdot\nabla_{\frac{\partial}{\partial\varphi}}^{S,B}\right)^{2}+D_{2,S,B}^{2}+\left\\{\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B},D_{2,S,B}\right\\}$

decomposes $D_{S,B}^{2}$ as a sum of three operators, the first two which are
evidently non-negative. The third term is an anti-commutator, which we need to
analyze explicitly.

######  Lemma 52.

The term
$\left\\{\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B},D_{2,S,B}\right\\}$
is equal to

$\left\\{\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B},D_{2,S,B}\right\\}=\rho(\eta)\rho(e^{1})F_{S,B}(\partial_{\varphi},e_{1})+\rho(\eta)\rho(e^{2})F_{S,B}(\partial_{\varphi},e_{2})$

######  Remark.

This computation should be compared with the formula given in [67, Lemma 5.8].
There $E$ would be a rank complex line bundle, and the formula would read in
terms of an orthonormal frame

$\left\\{\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B},D_{2,S,B}\right\\}=\rho(\eta)\left(\rho(e^{1})(db)_{\eta,e_{1}}+\rho(e^{2})(db)_{\eta,e_{2}}\right)1_{S}$

where $b$ is the imaginary valued one form which appears in 33 and we
decomposed it as

$db=(db)_{\eta,e_{1}}\eta\wedge e^{1}+(db)_{\eta,e_{2}}\eta\wedge
e^{2}+(db)_{e_{1},e_{2}}e^{1}\wedge e^{2}$

###### Proof.

Clearly,

$\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S},D_{2,SB}\right\\}=\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S},\rho(e^{1})\nabla_{e_{1}}^{S,B}\right\\}+\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S},\rho(e^{2})\nabla_{e_{2}}^{S,B}\right\\}$

where $\partial_{\varphi}=\frac{\partial}{\partial\varphi}$ , and it suffices
to find the first term since the other one is analogous.

For a spinor $\varPsi$ of $S\otimes E$ we have

$\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B},\rho(e^{1})\nabla_{e_{1}}^{S,B}\right\\}\varPsi=\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B}\left(\rho(e^{1})\nabla_{e_{1}}^{S,B}\varPsi\right)+\rho(e^{1})\nabla_{e_{1}}^{S,B}\left(\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B}\varPsi\right)$

Now we use the fact that the connection is spinorial, i.e,
$\nabla_{v}^{S,B}(\rho(\theta)\varPsi)=\rho(\nabla_{v}^{\infty}\theta)\varPsi+\rho(\theta)(\nabla_{v}^{S,B}\varPsi)$.
In our case the anticommutator becomes

$\displaystyle\rho(\eta)\rho(\nabla_{\partial_{\varphi}}^{\infty}e^{1})\left(\nabla_{e_{1}}^{S,B}\varPsi\right)+\rho(\eta)\rho(e^{1})\nabla_{\partial_{\varphi}}^{S,B}\left(\nabla_{e_{1}}^{S,B}\varPsi\right)$
$\displaystyle+$
$\displaystyle\rho(e^{1})\rho\left(\nabla_{e_{1}}^{\infty}\rho(\eta)\right)\left(\nabla_{\partial_{\varphi}}^{S,B}\varPsi\right)+\rho(e^{1})\rho(\eta)\nabla_{e_{1}}^{S,B}\left(\nabla_{\partial_{\varphi}}^{S,B}\varPsi\right)$

Now, for our adiabatic connection we have that $\nabla^{\infty}\eta=0$ and
$\nabla^{\infty}\pi^{*}(\check{\theta})=\pi^{*}(\nabla_{\varSigma}^{LC}\check{\theta})$
so the first term in each row of the previous expression disappears. Also,
using the Clifford relations $\rho(e^{1})\rho(\eta)=-\rho(\eta)\rho(e^{1})$
the computation for the anticommutator becomes

$\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B},\rho(e^{1})\nabla_{e_{1}}^{S,B}\right\\}\varPsi=\rho(\eta)\rho(e^{1})[\nabla_{\partial_{\varphi}}^{S,B},\nabla_{e_{1}}^{S_{c}}]\varPsi$

To compute the commutator recall that because $\nabla^{S,B}$ is a connection
on a vector bundle it has a curvature $F_{S,B}$ given by [76, Section 3.3.3]

$[\nabla_{v}^{S,B},\nabla_{w}^{S,B}]-\nabla_{[v,w]}^{S,B}=F_{S,B}(v,w)$

Since $\partial_{\varphi}$ commutes with any vector field which is dual to a
form that pulls back from $\varSigma$ we have that
$[\partial_{\varphi},e_{1}]=0$ and so

$\begin{array}[]{c}[\nabla_{\partial_{\varphi}}^{S,B},\nabla_{e_{1}}^{S,B}]=F_{S,B}(\partial_{\varphi},e_{1})\end{array}$

Since a similar result holds for $\partial_{\varphi}$ and $e_{2}$ we have just
found that

$\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B},D_{2,SB}\right\\}=\rho(\eta)\rho(e^{1})F_{S,B}(\partial_{\varphi},e_{1})+\rho(\eta)\rho(e^{2})F_{S,B}(\partial_{\varphi},e_{2})$

∎

######  Remark 53.

Notice that as [67] point out before Lemma 5.8, the term
$\left\\{\rho(\eta)\nabla_{\partial_{\varphi}}^{S,B},D_{2,SB}\right\\}$ ends
up being a zeroth-order operator, not a first order operator.

Now we write the $SO(3)$ monopole equations on a Seifert manifold. Locally, we
can decompose the curvature in the same way as in the previous section, namely

$\displaystyle F_{B}^{0}=F_{\eta,e^{1}}\eta\wedge
e^{1}+F_{\eta,e^{2}}\eta\wedge e^{2}+F_{e^{1},e^{2}}e^{1}\wedge e^{2}$
$\displaystyle*F_{B}^{0}=F_{\eta,e^{1}}e^{2}-F_{\eta,e^{2}}e^{1}+F_{e^{1},e^{2}}d\eta$

and we obtain:

######  Theorem 54.

Suppose $(B,\varPsi)$ satisfies the $SO(3)$ vortex equations on a Seifert
manifold $Y$, for the spin-u structure
$V_{can}=(\mathbb{C}\oplus\pi^{*}(K_{\check{\varSigma}}^{-1}))\otimes\pi^{*}(\check{E})$.
Write $\varPsi=\alpha\oplus\beta$ with
$\alpha\in\varOmega^{0}(\pi^{*}(\check{E}))$ and
$\beta\in\varOmega^{0,1}(\pi^{*}(K_{\check{\varSigma}}^{-1}\otimes\check{E}))$.

Then with respect to the $SO(3)$ monopole equations

$\displaystyle
iF_{e^{1},e^{2}}+(\alpha\otimes\alpha^{*})_{0}-(\varLambda\beta\otimes\beta^{*})_{0}=0$
$\displaystyle\left(iF_{\eta,e^{1}}-F_{\eta,e^{2}}\right)\bar{\epsilon}+2(\beta\otimes\alpha^{*})_{0}=0$
(51)
$\displaystyle\rho(\eta)\nabla_{\frac{\partial}{\partial\varphi}}^{S,B}\varPsi+D_{2,S}\varPsi=0$

all solutions are of the form $(B,(\alpha,0))$ or $(B,(0,\beta))$.

In the first case, we get a solution of the form

$\displaystyle\bar{\partial}_{C}\alpha=0$ $\displaystyle
F_{C}-i(\alpha\otimes\alpha^{*})_{0}=0$

which can be identified with a $SO(3)$ vortex on a bundle $\check{E}^{\prime}$
which satisfies that $\det\check{E}^{\prime}\simeq\det\check{E}$.

In the second case, we get a solution of the form

$\displaystyle\bar{\partial}_{C}^{*}\beta=0$ $\displaystyle
F_{C}+i(\varLambda\beta\otimes\beta^{*})_{0}=0$

which can be identified via Serre duality with an $SO(3)$ vortex on the bundle
$K_{\check{\varSigma}}\otimes\check{E}^{\prime-1}$, where again $\tilde{E}$ is
a $U(2)$ bundle satisfying $\det\check{E}^{\prime}\simeq\det E$.

In particular, if we assume that
$c_{1}(\check{E})>2c_{1}(K_{\check{\varSigma}})$, then this second type of
moduli space of stable pairs is empty.

######  Remark 55.

a) Notice that implicitly we chose a connection $C^{\det}$ on $\det\check{E}$,
and we are considering all $U(2)$ connections on $\pi^{*}(\check{E})$ which
induce the predetermined connection $\pi^{*}(C^{\det})$ on
$\pi^{*}(\det\check{E})$. Using [67, Propositon 5.3], there is a bijection
between bundles such connections on orbifold line bundles over
$\check{\varSigma}$, and the usual connections on line bundles over $Y$, whose
curvature pulls up from $\check{\varSigma}$ and whose fiberwise holonomy is
trivial.

Thus, this is why the bundle $\pi^{*}(\check{E})$ “remembers” the choice of
connection we made downstairs (i.e, on the orbifold $\check{\varSigma}$).
Since there are several $U(2)$ orbifold bundles with the same $\det\check{E}$
(up to isomorphism), we need to consider all the distinct choices, which
differ in their isotropy data. But at least there are only finitely many
isomorphism classes to consider.

b) One could also be interested in understanding how the abelian vortices that
appear in these moduli spaces are related to those appearing in MOY [67]. As
Theorem 1 of MOY states, the Seiberg-Witten solutions that appear there can be
identified with two copies of the space of effective orbifold divisors over
$\varSigma$ with orbifold degree less than
$-\frac{\chi(\varSigma)}{2}=\frac{c_{1}(K_{\varSigma})}{2}$. For simplicity
suppose that the $a_{i}$ are coprime and that $\check{\varSigma}$ admits orbi-
spin bundles [67, Definition 5.13].

This means that there is a square-root $K_{\check{\varSigma}}^{1/2}$ of
$K_{\check{\varSigma}}$ in the sense that
$2c_{1}(K_{\check{\varSigma}}^{1/2})=c_{1}(K_{\check{\varSigma}})$. In our
setup this would require all the $a_{i}$ to be odd integers and moreover
$K_{\check{\varSigma}}^{1/2}$ is unique (see the proof of Corollary 5.17 in
MOY). Thus, if we take for example $\det\check{E}\simeq
K_{\check{\varSigma}}\otimes\check{L}_{0}$ this is a valid choice in the sense
that $K_{\check{\varSigma}}$ is an even power of $\check{L}_{0}$ thus
$\det\check{E}$ is an odd power of $\check{L}_{0}$. However, this choice for
$\det\check{E}$ is not within the vanishing condition stated in our theorem
thus we obtain:

1) Moduli spaces of $SO(3)$ vortices on $\check{E}^{\prime}$: now abelian
vortices can arise provided
$c_{1}(\check{L})<\frac{1}{2}c_{1}(K_{\check{\varSigma}}\otimes\check{L}_{0})=\frac{1}{2}c_{1}(K_{\check{\varSigma}})+\frac{1}{2a_{1}\cdots
a_{n}}$ .

2) Moduli spaces of $SO(3)$ vortices on
$K_{\check{\varSigma}}\otimes\check{E}^{\prime-1}$: now abelian vortices can
arise provided
$c_{1}(\check{L})<\frac{1}{2}c_{1}(K_{\check{\varSigma}})-\frac{1}{2a_{1}\cdots
a_{n}}$ .

This seems the closest one can get to the constraint that appears on MOY.
However, because of this choice for $\det\check{E}$ it is not clear that the
moduli space of $SO(3)$ vortices is smooth at the reducibles so it seems
better to assume $c_{1}(\check{E})>2c_{1}(K_{\check{\varSigma}})$ instead.

###### Proof.

We follow the strategy of [67, Theorem 4]. If we start with the equation
$D_{S,B}\varPsi=0$, and apply $D_{S,B}$ to it, we obtain

$0=\left(\rho(\eta)\cdot\nabla_{\frac{\partial}{\partial\varphi}}^{S,B}\right)^{2}\varPsi+D_{2,S,B}^{2}\varPsi+\rho(\eta)\rho(e^{1})F_{S,B}(\partial_{\varphi},e_{1})\varPsi+\rho(\eta)\rho(e^{2})F_{S,B}(\partial_{\varphi},e_{2})\varPsi$

Now take inner product with $\varPsi$ and integrate over $Y$ in order to
obtain

(52)
$0=\left|\nabla_{\partial_{\varphi}}^{S}\varPhi\right|^{2}+|D_{2}\varPhi|^{2}+\frac{1}{2}\left\langle\rho(\eta)\rho(e^{1})F_{\eta,e^{1}}\varPsi+\rho(\eta)\rho(e^{2})F_{\eta,e^{2}}\varPsi,\varPsi\right\rangle$

where $F_{\eta,e^{1}}=F_{S,B}(\partial_{\varphi},e_{1})$ and
$F_{\eta,e^{2}}=F_{S,B}(\partial_{\varphi},e_{2})$.

Under a slight abuse of notation, recall from last section that

$\displaystyle\rho(\eta)=\left(\begin{array}[]{cc}i&0\\\
0&-i\end{array}\right)$ $\displaystyle\rho(e^{1})=$
$\displaystyle\left(\begin{array}[]{cc}0&-\epsilon\\\
\bar{\epsilon}&0\end{array}\right)$
$\displaystyle\rho(e^{2})=\left(\begin{array}[]{cc}0&i\epsilon\\\
i\bar{\epsilon}&0\end{array}\right)$

Thus

$\displaystyle\rho(\eta)\rho(e^{1})=\left(\begin{array}[]{cc}0&-i\epsilon\\\
-i\bar{\epsilon}&0\end{array}\right)$
$\displaystyle\rho(\eta)\rho(e^{2})=\left(\begin{array}[]{cc}0&-\epsilon\\\
\bar{\epsilon}&0\end{array}\right)$

Recall that if we were to write $\varPsi$ locally as
$\varPsi=\sum_{i,j=1}^{2}c_{ij}\varPsi_{i}^{S}\otimes\varPsi_{j}^{E}$, where
$\varPsi_{1}^{S},\varPsi_{2}^{S}$ is an orthonormal basis for $S$ and
$\varPsi_{1}^{E},\varPsi_{2}^{E}$ an orthonormal basis for $E$, then for any
one form
$\rho(\theta)\varPsi=\sum_{i,j=1}^{2}c_{ij}\left(\rho(\theta)\varPsi_{i}^{S}\right)\otimes\varPsi_{j}^{E}$
, while for any section $\xi$ of $\mathfrak{su}(E)$,
$\xi\varPsi=\sum_{i,j=1}^{2}c_{ij}\varPsi_{i}^{S}\otimes(\xi\varPsi_{j}^{E})$.
In other words, the Clifford multiplication action on spinors $\varPsi$ is
uncoupled from the action of $\mathfrak{su}(E)$ on spinors.

Therefore, we can write

$\displaystyle\rho(\eta)\rho(e^{1})F_{\eta,e^{1}}\varPsi$
$\displaystyle\rho(\eta)\rho(e^{2})F_{\eta,e^{2}}\varPsi$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&-i\epsilon\\\
-i\bar{\epsilon}&0\end{array}\right)F_{\eta,e^{1}}\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&-\epsilon\\\
\bar{\epsilon}&0\end{array}\right)F_{\eta,e^{2}}\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&-i\epsilon\\\
-i\bar{\epsilon}&0\end{array}\right)\left(\begin{array}[]{c}F_{\eta,e^{1}}\alpha\\\
F_{\eta,e^{1}}\beta\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&-\epsilon\\\
\bar{\epsilon}&0\end{array}\right)\left(\begin{array}[]{c}F_{\eta,e^{2}}\alpha\\\
F_{\eta,e^{2}}\beta\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}-i\epsilon F_{\eta,e^{1}}\beta\\\
-i\bar{\epsilon}F_{\eta,e^{1}}\alpha\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}-\epsilon F_{\eta,e^{2}}\beta\\\
\bar{\epsilon}F_{\eta,e^{2}}\alpha\end{array}\right)$

And so we must understand

$\displaystyle\left\langle\left(\begin{array}[]{c}-\epsilon(iF_{\eta,.e^{1}}+F_{\eta,e^{2}})\beta\\\
-\bar{\epsilon}(iF_{\eta,e^{1}}-F_{\eta,e^{2}})\alpha\end{array}\right),\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)\right\rangle$ $\displaystyle=$ $\displaystyle
2\left\langle\left(\begin{array}[]{c}(\alpha\otimes\beta^{*})_{0}\beta\\\
(\beta\otimes\alpha^{*})_{0}\alpha\end{array}\right),\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)\right\rangle$ $\displaystyle=$ $\displaystyle
2\left\langle\left(\begin{array}[]{c}|\beta|^{2}\alpha-\frac{1}{2}<\beta,\alpha>_{E}\beta\\\
|\alpha|^{2}\beta-\frac{1}{2}<\alpha,\beta>_{E}\alpha\end{array}\right),\left(\begin{array}[]{c}\alpha\\\
\beta\end{array}\right)\right\rangle$ $\displaystyle=$ $\displaystyle
2\left(|\beta|^{2}|\alpha|^{2}-\frac{1}{2}|<\beta,\alpha>_{E}|^{2}+|\beta|^{2}|\alpha|^{2}-\frac{1}{2}|<\alpha,\beta>_{E}|^{2}\right)$

Just as in the case of $S^{1}\times\varSigma$, by Cauchy-Schwarz this is non-
negative so going back to the equality (52) we conclude that

$\begin{cases}\nabla_{\partial_{\varphi}}^{S}\alpha\equiv 0\\\
\nabla_{\partial_{\varphi}}^{S}\beta\equiv 0\\\ D_{2,S,B}\varPsi\equiv 0\\\
\alpha\equiv 0\text{ or }\beta\equiv 0\end{cases}$

Once we know this, the argument is identical to the one given by [67, Theorem
4].

Namely, the first two equations say that $\alpha,\beta$ actually pullback from
the orbifold Riemann surface. Likewise, the connection $B$ will pullback from
an orbifold bundle $\check{E}^{\prime}$ over $\check{\varSigma}$ which
satisfies $\det\check{E}^{\prime}\simeq\det\check{E}$ . Finally, the operator
$D_{2}$ can then be identified with a twisted
$\bar{\partial}\oplus\bar{\partial}^{*}$ operator [67, p. 709], so this is why
$D_{2,S,B}\varPsi\equiv 0$ together with the equation for $F_{e^{1},e^{2}}$
gives a solution means that we have a solution to the $SO(3)$ vortex
equations.

The vanishing result uses and the identification with the stable pairs on
$\check{\varSigma}$ is exactly the same argument as the one we used for
$S^{1}\times\varSigma$. ∎

######  Example 56.

The Poincar Homology Sphere

As is well know the Poincar homology sphere can be identified with the
Brieskorn homology sphere $\varSigma(2,3,5)$, and as a Seifert manifold, the
corresponding orbifold is $S^{2}(2,3,5)$.

In this case the generator for the topological Picard group is an orbifold
line bundle with $c_{1}(\check{L}_{0})=\frac{1}{30}$. The isotropy invariants
consists of three integers $b_{1},b_{2},b_{3}$ satisfying

$\displaystyle 1\leq b_{1}<2$ $\displaystyle 1\leq b_{2}<3$ $\displaystyle
1\leq b_{3}<5$

These numbers are found by the requirement that

$\frac{1}{\prod\alpha_{i}}-\sum_{i=1}^{n}\frac{\beta_{i}}{\alpha_{i}}=\frac{1}{30}-\frac{15b_{1}+10b_{2}+6b_{3}}{30}$

be an integer (the background degree of $\check{L}_{0}$) . It is easy to check
that the only choice which works is

$b_{1}=b_{2}=b_{3}=1$

Thus we will write $\check{L}_{0}(1,1,1)$ to emphasize the isotropy data.

Recall that the Seifert invariants for the canonical bundle were given in
equation (12). Thus, our non-vanishing condition reads in this case

$c_{1}(\check{E})>2c_{1}(K_{\check{\varSigma}})=2\left(0-2+3-\frac{1}{2}-\frac{1}{3}-\frac{1}{5}\right)=2\left(\frac{30-15-10-6}{30}\right)=-\frac{1}{15}$

In particular, it suffices to take $\det\check{E}\simeq\check{L}_{0}$. Now
suppose that $\check{E}^{\prime}$ is a $U(2)$ bundle with
$\det\check{E}^{\prime}\simeq\check{L}_{0}$. Write the isotropy invariants of
$\check{E}^{\prime}$ as
$((b_{1}^{-},b_{1}^{+}),(b_{2}^{-},b_{2}^{+}),(b_{3}^{-},b_{3}^{+}))$. The
conditions these integers must satisfy are

$0\leq b_{i}^{-}\leq b_{i}^{+}<a_{i}$

and also

$b_{i}^{-}+b_{i}^{+}\equiv 1\mod a_{i}$

since they must give the isotropy data of $\check{L}_{0}$.

We find by trial and error that the only options which work are

$\displaystyle((0,1),(0,1),(0,1))$ $\displaystyle((0,1),(0,1),(2,4))$
$\displaystyle((0,1),(0,1),(3,3))$ $\displaystyle((0,1),(2,2),(0,1))$
$\displaystyle((0,1),(2,2),(2,4))$ $\displaystyle((0,1),(2,4),(3,3))$

In order to analyze these bundles in a systematic way we recall the Facts
(12), as well as some other important things to keep in mind when analyzing
the different moduli spaces:

  1. (1)

Because $g=0$, $n=3$, if $n_{0}=\\#\\{i\mid b_{i}^{-}=b_{i}^{+}\\}\geq 1$,
then $\check{E}^{\prime}$ admits no projectively flat connections. Moreover,
the space of projectively flat connections is connected and for three marked
points of expected dimension zero (this can be seen from the proof of our
Lemma (21)).

  2. (2)

If the moduli space $\mathcal{M}(\check{\varSigma},\check{E}^{\prime})$ admits
no moduli space of abelian vortices, then there are no irreducible vortices
inside this moduli space.

  3. (3)

The abelian vortices which can arise must satisfy
$c_{1}(\check{L})<\frac{1}{2}c_{1}(\check{L}_{0})=\frac{1}{2}\det(\check{E})$.
Since $\check{L}=\check{L}_{0}^{l}$ for some integer $l$ and $l\geq 0$ because
otherwise $H^{0}(\check{L})$ vanishes, we must have $\check{L}$ is the trivial
line bundle (isotropy $(0,0,0)$ and $c_{1}(\check{L})=0$). Thus in this case
any isotropy data which does not contain a $0$ in each of the three pairs
$(b_{i}^{-},b_{i}^{+})$ will have an empty moduli space of irreducible $SO(3)$
vortices.

  4. (4)

Moreover, the dimension of the moduli space of irreducible $SO(3)$ vortices
can be read from (18)

$2\left((g-1)+c_{1}(\det\check{E})+(n-n_{0})-\sum_{i=1}^{n}\frac{b_{i}^{-}+b_{i}^{+}}{a_{i}}\right)=2\left(-1+\frac{1}{30}+3-n_{0}-\sum_{i=1}^{n}\frac{b_{i}^{-}+b_{i}^{+}}{a_{i}}\right)$

  5. (5)

The index for an abelian monopole will be according to equation (25)

$\displaystyle
2\left(g-1+c_{1}(\det\check{E})-2c_{1}(\check{L})+\sum_{i\mid\epsilon_{i}=1}\frac{b_{i}^{+}-b_{i}^{-}}{a_{i}}+n_{-}+\sum_{i\mid\epsilon_{i}=-1}\frac{b_{i}^{-}-b_{i}^{+}}{a_{i}}\right)$
$\displaystyle=$ $\displaystyle
2\left[-1+\frac{1}{30}+\sum_{i\mid\epsilon_{i}=1}\frac{b_{i}^{+}-b_{i}^{-}}{a_{i}}+n_{-}+\sum_{i\mid\epsilon_{i}=-1}\frac{b_{i}^{-}-b_{i}^{+}}{a_{i}}\right]$

With this information, we can make the following table [the dimensions refer
to those before we take the quotient by the circle action]:

Isotropy | $\mathcal{M}^{*}(\check{\varSigma},\check{E}^{\prime})$ | # Proj. Flat | # Abelian Vortices
---|---|---|---
$((0,1),(0,1),(0,1))$ | non empty, of dim $2$ | one | one vortex of index $2$
$((0,1),(0,1),(2,4))$ | empty, exp. dim$=$$0$ | one (isolated) | no vortices (wrong isotropy)
$((0,1),(0,1),(3,3))$ | empty, exp. dim$=$$-2$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,2),(0,1))$ | empty, exp. dim$=$$-2$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,2),(2,4))$ | empty, exp. dim$=$$-4$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,4),(3,3))$ | empty, exp. dim$=$$-4$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)

Therefore, we conclude that the solutions to the $SO(3)$ monopole equations on
the Poincar homology sphere consist of: a) the trivial $SU(2)$ connection, b)
two irreducible flat $SU(2)$ connections, c) one Seiberg-Witten monopole, d) a
one dimensional moduli space of $SO(3)$ monopoles which serves as a
“cobordism” between the Seiberg-Witten monopole and one of the irreducible
flat connections.

A more realistic example to consider is the following:

######  Example 57.

The Brieskorn Homology Sphere $\varSigma(2,3,7)$

In this case we are looking at $S^{2}(2,3,7)$. The generator $\check{L}_{0}$
now satisfies $c_{1}(\check{L}_{0})=\frac{1}{42}$ and its isotropy must
satisfy that

$\frac{1}{42}-\frac{21b_{1}+14b_{2}+6b_{3}}{42}$

is an integer. By trial and error we find that the only choice is

$b_{1}=1,b_{2}=2,b_{3}=6$

Write $\check{L}_{0}(1,2,6)$. Now the vanishing condition reads

$c_{1}(\check{E})>2c_{1}(K_{\check{\varSigma}})=2\left(0-2+3-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}\right)=\frac{1}{21}$

So we need to look at the moduli spaces of $SO(3)$ vortices with
$\det\check{E}\simeq\check{L}_{0}^{3}$. Notice that the isotropy is now
$\check{L}_{0}^{3}(1,0,4)$. In this case the expected dimension would need to
be

$2\left(2+\frac{1}{14}-\frac{b_{1}^{-}+b_{1}^{+}}{2}-\frac{b_{2}^{-}+b_{2}^{+}}{3}-\frac{b_{3}^{-}+b_{3}^{+}}{7}\right)$

Notice that in this case $b_{1}^{-}+b_{1}^{+}=1$ and $b_{2}^{-}+b_{2}^{+}=3$
so this forces the expected dimension to be

$2\left(\frac{1}{2}+\frac{1}{14}-\frac{b_{3}^{-}+b_{3}^{+}}{7}\right)=1+\frac{1-2b_{3}^{-}-2b_{3}^{+}}{7}$

and since one of the $b_{3}^{\pm}$ must be non-trivial, this forces the
dimension after taking the quotient by the $S^{1}$ action to be negative, thus
we end up with empty moduli space of $SO(3)$ vortices, which is not
interesting.

Thus, we need to increase $c_{1}(\det\check{E})$ in order to make the moduli
spaces of $SO(3)$ vortices to have positive expected dimension. The next
choice is thus $\det\check{E}\simeq\check{L}_{0}^{5}$. Notice that the
isotropy is now $\check{L}_{0}^{5}(1,1,2)$. In this case the formula for the
expected dimension reads

$2\left(2+\frac{5}{42}-\frac{b_{1}^{-}+b_{1}^{+}}{2}-\frac{b_{2}^{-}+b_{2}^{+}}{3}-\frac{b_{3}^{-}+b_{3}^{+}}{7}\right)$

Again, $b_{1}^{-}+b_{1}^{+}=1$, but now we can have $b_{2}^{-}+b_{2}^{+}=1$
and $b_{3}^{-}+b_{3}^{+}=2$ in which case the dimension would be

$2\left(2+\frac{5}{42}-\frac{1}{2}-\frac{1}{3}-\frac{2}{7}\right)=2$

so that means we will find positive dimensional moduli spaces of $SO(3)$
vortices!

The isotropy
$((b_{1}^{-},b_{1}^{+}),(b_{2}^{-},b_{2}^{+}),(b_{3}^{-},b_{3}^{+}))$ of our
$U(2)$ bundles must satisfy

$\begin{cases}0\leq b_{1}^{-}\leq b_{1}^{+}\leq 1&b_{1}^{-}+b_{1}^{+}-1\equiv
0\mod 2\\\ 0\leq b_{2}^{-}\leq b_{2}^{+}\leq 1&b_{2}^{-}+b_{2}^{+}-1\equiv
0\mod 3\\\ 0\leq b_{3}^{-}\leq b_{3}^{+}\leq 2&b_{3}^{-}+b_{3}^{+}-2\equiv
0\mod 7\end{cases}$

Again after trial and error one finds the isotropies

$\displaystyle((0,1),(0,1),(0,2))$ $\displaystyle((0,1),(0,1),(1,1))$
$\displaystyle((0,1),(0,1),(3,6))$ $\displaystyle((0,1),(0,1),(4,5))$
$\displaystyle((0,1),(2,2),(0,2))$ $\displaystyle((0,1),(2,2),(1,1))$
$\displaystyle((0,1),(2,2),(3,6))$ $\displaystyle((0,1),(2,2),(4,5))$

In addition to the facts used for the case of the Poincar homology sphere,
notice that:

  1. (1)

The moduli spaces of abelian vortices which can arise are associated to
$\check{L}_{0}(1,2,6)$, $\check{L}_{0}^{2}(0,1,5)$ and the trivial one
$\check{L}_{triv}$. The background degrees are

$\begin{cases}\deg_{B}\check{L}_{triv}=0\\\
\deg_{B}\check{L}_{0}(1,2,6)=\frac{1}{42}-\frac{1}{2}-\frac{2}{3}-\frac{6}{7}=-2\\\
\deg_{B}\check{L}_{0}^{2}(0,1,5)=\frac{2}{42}-\frac{0}{2}-\frac{1}{3}-\frac{5}{7}=-1\end{cases}$

so the space of abelian vortices to consider is really just
$\check{L}_{triv}$, since recall from MOY that these moduli spaces of abelian
vortices are isomorphic to $\text{Sym}^{\deg_{B}\check{L}}(\varSigma)$, thus
they are empty if the background degree is negative.

  2. (2)

From Facts (12), the space of irreducible projectively flat connections is
empty if and only if there exists a vector $(\epsilon_{i})$, with
$\epsilon_{i}=\pm 1$, such that $n_{+}+\deg_{B}(\det\check{E})\equiv 1\mod 2$
and $n_{+}-\sum_{i=1}^{n}\frac{\epsilon_{i}(b_{i}^{+}-b_{i}^{-})}{a_{i}}<1-g$.
Here $n_{\pm}=\\#\\{i\mid\epsilon_{i}=\pm 1\\}$.

  3. (3)

Moreover, the dimension of the moduli space of irreducible $SO(3)$ vortices
can be read from (18)

$2\left((g-1)+c_{1}(\det\check{E})+(n-n_{0})-\sum_{i=1}^{n}\frac{b_{i}^{-}+b_{i}^{+}}{a_{i}}\right)=2\left(-1+\frac{5}{42}+3-n_{0}-\sum_{i=1}^{n}\frac{b_{i}^{-}+b_{i}^{+}}{a_{i}}\right)$

  4. (4)

The index for an abelian monopole will be according to equation (25)

$\displaystyle
2\left(g-1+c_{1}(\det\check{E})-2c_{1}(\check{L})+\sum_{i\mid\epsilon_{i}=1}\frac{b_{i}^{+}-b_{i}^{-}}{a_{i}}+n_{-}+\sum_{i\mid\epsilon_{i}=-1}\frac{b_{i}^{-}-b_{i}^{+}}{a_{i}}\right)$
$\displaystyle=$ $\displaystyle
2\left[-1+\frac{1}{42}+\sum_{i\mid\epsilon_{i}=1}\frac{b_{i}^{+}-b_{i}^{-}}{a_{i}}+n_{-}+\sum_{i\mid\epsilon_{i}=-1}\frac{b_{i}^{-}-b_{i}^{+}}{a_{i}}\right]$

With this information, we can make the following table [the dimensions refer
to those before we take the quotient by the circle action]:

Isotropy | $\mathcal{M}^{*}(\check{\varSigma},\check{E}^{\prime})$ | # Proj. Flat | # Abelian Vortices
---|---|---|---
$((0,1),(0,1),(0,2))$ | non empty, of dim $2$ | one | one vortex of index $2$
$((0,1),(0,1),(1,1))$ | empty, exp. dim$=0$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(0,1),(3,6))$ | empty, exp. dim$=0$ | one | no vortices (wrong isotropy)
$((0,1),(0,1),(4,5))$ | empty, exp. dim$=$$0$ | empty (see below) | no vortices (wrong isotropy)
$((0,1),(2,2),(0,2))$ | empty, exp. dim$=$$-2$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,2),(1,1))$ | empty, exp. dim$=$$-4$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,2),(3,6))$ | empty, exp. dim$=$$-4$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)
$((0,1),(2,2),(4,5))$ | empty, exp. dim$=$$-4$ | empty since $n_{0}\geq 1$ | no vortices (wrong isotropy)

The reason why the bundle with isotropy $((0,1),(0,1),(4,5))$ is empty is that
we can take the isotropy vector $\epsilon=(1,-1,-1)$, so $n_{+}=1$ and
$\frac{1}{2}-\frac{1}{3}-\frac{1}{7}>0$ which verifies the criterion mentioned
before for the emptiness of the space of irreducible projectively flat
connections.

Thus one more time our picture is similar to the case of the Poincar homology
sphere. On $\varSigma(2,3,7)$ we have the trivial connection, two irreducible
flat connections, one Seiberg-Witten monopole, and a “cobordism” of $SO(3)$
monopoles between this Seiberg-Witten monopole and one of the irreducible flat
connections.

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