Zero-energy scattering and the real Bers image on the line

Let $\mathrm{Diff}_{\mathcal S}(\mathbb R)$ be the group of orientation-preserving diffeomorphisms $φ$ of the line with $φ'-1$ Schwartz and $φ(x)-x\to0$ as $x\to-\infty$, and let $β(φ)=\frac12S(φ)$ be half the Schwarzian derivative. We determine the image of $β$. The coordinate $w_φ=\frac12(\logφ')'$ identifies $\mathrm{Diff}_{\mathcal S}(\mathbb R)$ with the zero-mean hyperplane $\mathcal S_0(\mathbb R;\mathbb R)$ and turns $β$ into $w\mapsto w'-w^2$, so the question is which real Schwartz $q$ equal $w'-w^2$ for some $w$ of zero mean. Nonnegativity of the Schrodinger operator $H_q=-\partial_x^2-q$ decides which $q$ admit such a $w$ at all, and says nothing about the mean of $w$. The mean is a scattering invariant. Let $T_q$ and $R_q$ be the transmission and reflection coefficients of $H_q$. For every real $w\in\mathcal S(\mathbb R)$, with $q=w'-w^2$, we prove $T_q(0)=\mathrm{sech}(\int_{\mathbb R} w\,dx)$ and $R_q(0)=-\tanh(\int_{\mathbb R} w\,dx)$, so $\int_\mathbb R w\,dx$ is read off the scattering matrix at zero energy. Thus $q\inβ(\mathrm{Diff}_{\mathcal S}(\mathbb R))$ if and only if $H_q$ has no negative eigenvalues and $R_q(0)=0$, equivalently $T_q(0)=1$, and $φ\mapsto R_{β(φ)}$ is a bijection onto an explicit set of Schwartz reflection coefficients. We also compute the differential of $β$, whose range depends on the ambient topology. On the Schwartz space, that range is closed and split of codimension two, with normal functionals the first variations of the Wronskian of the two zero-energy Jost solutions and of $R_q(0)$. In every $W^{k,1}$ realization the second functional is unbounded. The range is then a dense proper subspace of the kernel of the first, hence neither closed nor split, and the operator has no bounded inverse on its range.