Let $X$ be a compact spin 4 manifold with boundary $Y$, for any configuration $(A,\Psi)\in \mathcal C(X)$, we define the topological energy as $$\mathscr E^{top}(A,\Psi)=\frac14\int_XF_A\wedge F_A-\int_Y(\Psi|_Y,D_B\Psi|_Y)+\int_Y(H/2)|\Psi|^2,$$ where $B$ denotes the restriction of $A$ to $Y$ and $H$ denotes the mean curvature.
Q: Why $\mathscr E^{top}(A,\Psi)=-CSD(A|_Y,\Psi|_Y)+C$, here $C$ depends on the base connection $B_0$ of $Y$ and topology of spin bundle?
PS: The formula is written in the book Monopole and Three manifold Page 98.