Right inverse of the Seiberg-Witten functional

Question

For closed 4 manifold X, we consider the derivative of the Seiberg-Witten functional, i.e. $$\Omega^1_2(X;\sqrt{-1}\mathbb R)\oplus\Gamma_2(S^+)\overset{D}{\to}\Omega^2_{+,1}(X;\sqrt{-1}\mathbb R)\Gamma_1(S^-),$$ where

1. $\Omega^i_j(X;\sqrt{-1}\mathbb R)$ means $\sqrt{-1}\mathbb R$-valued $i$-forms space with Sobolev $L^{2,j}$ norm;

2. $\Omega^2_\pm$ means space of the self/anti-self dual parts of 2-forms ;

3. $\Gamma_i(S^\pm)$ means the sections of $S^\pm$-spinor space with Sobolev $L^{2,i}$ norm;

4. $$D_{A,\psi}=\left(\begin{array}{cc}d^+&-Dq_\psi\\ \cdot\frac12\psi&D_A\\ \end{array}\right)$$ here

   • $Dq_\psi(\eta)=\psi\otimes\eta^*+\eta\otimes\psi^*-\frac{<\eta,\psi>+\overline{<\eta,\psi>}}2Id$,

   • $D_A$ means the the Dirac operator w.r.t. the connection $A$,

   • $\cdot$ means the Clifford action,

   • $d^+$ means the differential $d$ and orthogonal project to the self-dual part.

Q: Does $D$ admits a right inverse $R$, $DR=Id$?

Answer 1

This is found in any reference on SW-theory. I am implicitly assuming we have perturbed the SW-equations, with generic perturbation. The linearization (which includes the gauge-action) at a given SW-solution forms an elliptic complex. This operator is surjective, hence admits a right inverse.

I should clarify: the point of the perturbation is for all SW-solutions to be regular, which by definition means that the linearized operator is surjective at all SW-solutions.