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 $R$, $DR=Id$?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.