Compactness conjecture for ADHM$_{1,k}$ monopoles

Let $M$ be the underlying manifold, and let $(\varepsilon_i,\Psi_i,{\bm\xi}_i,A_i)$ be a sequence of solutions of the blown-up ADHM$_{1,k}$ Seiberg--Witten equation satisfying

$$\slashed D_{A_i}\Psi_i=0,\qquad \slashed D_{A_i,C}{\bm\xi}_i=0,\qquad \varepsilon_i^2\varpi F_{A_i}=\mu(\Psi_i)+\mu({\bm\xi}_i),\qquad \lVert(\Psi_i,{\bm\xi}_i)\rVert_{L^2}=1,$$

with $\varepsilon_i\to0$. Compactness conjecture for ADHM$_{1,k}$ monopoles. After passing to a subsequence, there are a closed subset $Z\subset M$ of Hausdorff dimension at most one and a limiting triple $(0,{\bm\xi}_0^\infty,A_0^\infty)$ such that, outside $Z$ and up to gauge transformations, $(\Psi_i,{\bm\xi}_i,A_i)$ converges to $(0,{\bm\xi}_0^\infty,A_0^\infty)$ and $\varepsilon_i^{-1}(\Psi_i,{\bm\xi}_i-{\bm\xi}_0^\infty)$ converges to $(\Psi_1^\infty,{\bm\xi}_1^\infty)$. The limiting triple solves the limiting ADHM$_{1,k}$ Seiberg--Witten equation. Moreover, the section induced by ${\bm\xi}_0^\infty$ extends continuously across $Z$, and on the corresponding unbranched covers the components $(\tilde\Psi_{1,j},\tilde{\bm\xi}_{1,j},A_{0,j})$ solve the ADHM$_{1,k_j}$ Seiberg--Witten equations. This conjectural compactness statement would provide the analytic control needed for the ADHM monopole construction and its use in associative counts.