Let $P\to \Sigma$ be a $\mathrm{SU}(2)$-principal bundle over a smooth orientable closed genus $g$ real surface $\Sigma$. Let $\mathcal{A}$ be the space of connexions over $P$ and let $\mathcal{G}$ be the gauge group. Let $\Omega^k:=\Omega^k(\Sigma;\mathrm{Ad}P)$ be the space of $\mathrm{Ad}P$-valued differential $k$-forms on $\Sigma$ (here $\mathrm{Ad}P=P\times_\mathrm{Ad}\mathfrak{su}(2)$ is the adjoint bundle). On $\mathcal{A}$ lives not only a pretty canonical symplectic form but also Atiyah-Bott's moment map $$ F : \mathcal{A}\to \mathrm{Lie}(\mathcal{G})^* \; ; \quad A\mapsto F_A $$ whose hamiltonian action is the $\mathcal{G}$ action on $\mathcal{A}$ acting by pull-backs. Here $F_A\in \Omega^2<\mathrm{Lie}(\mathcal{G})^*$ is $A$'s curvature form. The Marsden-Weinstein reduction $$ \mathcal{M}^{\mathrm{fl}} := \mathcal{A}^{\mathrm{fl}}/\mathcal{G} = F^{-1}(0) \; /\!/ \; \mathcal{G} $$ is the so-called moduli space of flat connections over $\Sigma$. It is a symplectic orbifold whose irreducible part is smooth. If $\Sigma$ has genus $g\geq 2$, this irreducible part has dimension $6g-6$.
Consider now $\mathcal{O}\ne\{0\}$ be some coadjoint orbit inside $\mathrm{Lie}(\mathcal{G})^*$. Again, we can consider a Marsden-Weinstein reduction : $$ \mathcal{M}^{\mathcal{O}} := F^{-1}(\mathcal{O}) \; /\!/ \; \mathcal{G} $$
Question : What is the dimension of (the irreducible part of) $\mathcal{M}^{\mathcal{O}}$ ? Is it finite ?
Remark 1 : If my calculations are right, for $A$ irreducible I think we have the isomorphism $$ T_{[A]}\mathcal{M}^\mathcal{O}\cong\frac{\{\tau\in \Omega^1 | \mathrm{d}_A \tau\in \mathrm{im}(\mathrm{d}_A^2:\Omega^0\to\Omega^2)\}}{\mathrm{im}(\mathrm{d}_A:\Omega^0\to\Omega^1)} $$ (if $A$ is flat we recover the usual $T_{[A]}\mathcal{M}^{\mathrm{fl}}$). So the dimension I'm looking for should correspond to the dimension of that space. I just don't know how to compute it.
Remark 2 : According to the answer I got to this related question, the "Remark 1" can be reformulated as : $$ T_{[A]}\mathcal{M}^\mathcal{O}\cong \frac{\ker(\pi_A\circ\mathrm{d}_A|_{\Omega^1})}{\mathrm{im}(\mathrm{d}_A|_{\Omega^0})} $$ where $\pi_A$ is this quotient projection : $$ \pi_A : \Omega^2 \to \frac{\Omega^2}{\mathrm{im}(\mathrm{d}_A^2 : \Omega^0\to\Omega^2)} $$ Remark 3 : By giving $\Sigma$ a Riemannian metric and considering $\delta_A:\Omega^k\to\Omega^{k-1}$ the $L^2$-adjoint operator of $\mathrm{d}_A$ (i.e. the covariant coderivative) and the Laplacian $\Delta_A:=\mathrm{d}_A\delta_A+\delta_A \mathrm{d}_A:\Omega^k\to\Omega^k$, I also get that : $$ \ker(\Delta_A|_{\Omega^1}) < T_{[A]}\mathcal{M}^\mathcal{O} $$ But I'm not sure how "bigger" is $T_{[A]}\mathcal{M}^\mathcal{O}$ compared to the space $\ker(\Delta_A|_{\Omega^1})$ of harmonic $\mathrm{Ad}P$-valued differential 1-forms.
Remark 4 : the projection $\pi_A$ can be reformulated as $$ \pi_A : \Omega^2 \to \ker(\delta_A^2:\Omega^2\to \Omega^0) $$
2018-02-16 update : meanwhile I found that $T_{[A]}\mathcal{M}^{\mathcal{O}}_\Sigma \cong \ker(\Delta_{A}|_{\Omega^1})$. Now, my question becomes : for $A$ irreducible, does the dimension of $\ker(\Delta_{A}|_{\Omega^1})$ change if one takes a non-flat connection $A$ instead of a flat one ?