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The Fredholm Kuranishi reduction has the following form. Let $F$ be a smooth map between real Hilbert spaces with $F(0)=0$ whose derivative $L=DF(0)$ is a Fredholm operator. Split the source as $K\oplus V$, where $K=\ker L$, and the target as $R\oplus C$, where $R=\mathrm{im}L$ and $C$ represents the finite-dimensional cokernel. The restriction $L:V\to R$ is a bounded isomorphism. The implicit function theorem applied to $P_{R}F(k+v)=0$ solves it uniquely near zero as $v=v(k)$, with $v(0)=Dv(0)=0$. Thusis a smooth finite-dimensional obstruction map, and the local zero set of $F$ is the graph over ${\kappa }^{-1}(0)$. If a compact group acts preserving the problem, average inner products to choose invariant splittings; uniqueness in the implicit function theorem makes the construction equivariant.

Apply this to the vector-bundle curvature equation near $A$. Complete connections in $L_{\ell }^2$ and the unitary bundle gauge group in $L_{\ell +1}^2$ with integer $\ell \ge 3$, so the required multiplication and gauge operations are smooth. The Coulomb slice for unitary connections imposes $d_A^{\ast }a=0$. Its local existence follows from the gauge-fixing equation: the derivative in the gauge direction is $d_A^{\ast }{d}_A$, invertible on the orthogonal complement of its parallel kernel by the elliptic estimate. The local slice theorem then says that its remaining identifications are precisely by $\mathrm{Stab}(A)$.

On this slice the equation isIts kernel is the harmonic representative space $H_A^1$, and its cokernel is $H_A^2$. The hypothesis $H_A^2=0$ makes the obstruction target zero. Therefore the solutions near $A$ are an equivariant smooth graph over $H_A^1$, and the gauge-orbit neighborhood is a neighborhood of zero in $H_A^1/\mathrm{Stab}(A)$.

It remains to compute this representation, not just its dimension. A reducible SU2 connection with circle holonomy preserves $E=L\oplus L^{-1}$. In this splitting,The diagonal part is a trivial real connection. Its degree-one deformation space is the space of ordinary harmonic one-forms. To check this directly, if $d^{\ast }a=0$ and $d^{+}a=0$, then $da$ is an exact anti-self-dual two-form. Stokes theorem givesso $da=0$ and $a$ is harmonic. This neutral deformation space vanishes because $b_1(X)=0$. Hence all of $H_A^1$ is in the off-diagonal part, where the complex structure of $L^2$ commutes with both operators; it is a complex vector space.

Using part (a), part (b), and $H_A^2=0$,Parameterize the stabilizer by $u_{\lambda }=\mathrm{diag}({\lambda }^{-1},\lambda )$, with $\lambda \in S^1$. Under our convention $A^u=u^{-1}Au+u^{-1}du$, its action on an off-diagonal perturbation isConsequently the equivariant graph identifies a neighborhood of $[A]$ with a neighborhood of the origin inThis proves the exact scalar weight requested, with a stabilizer parametrization consistent with the chosen gauge convention. The ineffective kernel is $\{\pm 1\}$. The orbit space is also the cone on ${\mathbb{CP}}^{d-1}$, because the weight-two circle has the same nonzero orbits as the ordinary scalar circle. This is the local cone at an unobstructed reducible SU2 instanton.