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For real $q$, write $\bar{f}(k)=\overline{f(\bar{k})}$. The reduction symmetry and unit determinants give the scattering data parametrizationsTo see the symmetry, conjugate the coefficient matrices by $C=\left(\begin{array}{cc}0& 1\\ \lambda & 0\end{array}\right)$ after taking complex conjugates at $\bar{k}$; $C{\sigma }_3{C}^{-1}=-{\sigma }_3$ and the real-$q$ reduction restore the original Lax pair. DefineIf ${\mu }_j^{(l)}$ denotes column $l$, bounded columns produce the following normalized matrices:Each divisor is exactly the determinant of the unscaled column pair, so $\det M_j=1$. The eigenfunctions and divisors tend to $I$ and one, respectively, giving $M\to I$ at infinity. The contour for the sectorial half-line mKdV Riemann-Hilbert reconstruction isIt is six rays, not just the real axis or a four-quadrant contour. For the displayed meromorphic construction assume the divisors have no zeros on the contour; contour zeros require an indented or otherwise regularized formulation.

Here are explicit jump data without ambiguity about which side is used. DefineThe connection formulas give $M_j={\mu }_2{e}^{i\theta {\hat{\sigma }}_3}{C}_j$. Orient each ray outwards and let plus mean its left side. If that ray separates domains $D_i$ on the minus side and $D_j$ on the plus side, its jump isThis specifies every jump using only the spectral connection matrices; each jump has determinant one. Boundedness and consistency at the junction are inherited from the original eigenfunctions when the spectral data satisfy their global constraint.

If $a,\bar{a},d_{+}$ or $d_{-}$ has zeros in its corresponding domain, their simple zeros generate poles and require residue conditions. For example, at a simple zero $k_0$ of $a$ in $D_3$,At a simple zero of $d_{+}$ in $D_1$, the analogous coefficient for the first column is $\lambda \bar{b}{e}^{-2i\theta }/(A{d}_{+}^{\prime })$ times the second column; in $D_4$ it is $b{e}^{2i\theta }/(\bar{A}{d}_{-}^{\prime })$ for the second column. The $\bar{a}$ residue in $D_2$ is the conjugate first-column relation. These formulas assume the other displayed denominators are nonzero there; multiple or coincident zeros need their corresponding higher-order conditions. In a sufficiently small-data zero-free regime, the normalized Riemann-Hilbert problem has only the jumps. Finally the order-one term in ${\mu }_x-ik[{\sigma }_3,\mu ]=Q\mu$ givesThe column normalizations do not change this off-diagonal leading coefficient.