The Asymptotic Shape Conjecture for Kostant-positive permutations

Let $\lambda=(\lambda_1,\ldots,\lambda_r)\in\Lambda_m$ be a Young diagram with $\lambda_1$ boxes in its first row, and let $\lambda^{(n)}$ be obtained by adding a first row of $n$ boxes, where $n\geq\lambda_1$. Let $\mathbf{k}_{m+n}^{+}(\lambda^{(n)})$ and $\mathbf{k}_{m+n}^{-}(\lambda^{(n)})$ count the Kostant-positive and Kostant-negative permutations in $\mathrm S_{m+n}$ of shape $\lambda^{(n)}$, respectively, and let $\mathtt{SYT}_{m+n}(\lambda^{(n)})$ denote the standard Young tableaux of that shape. Asymptotic Shape Conjecture. For every $m\geq0$ and $\lambda\in\Lambda_m$, the proportion of permutations of shape $\lambda^{(n)}$ that are Kostant positive satisfies

$$\frac{\mathbf{k}_{m+n}^{+}(\lambda^{(n)})}{|\mathtt{SYT}_{m+n}(\lambda^{(n)})|^2}=\frac{\mathbf{k}_{m+n}^{+}(\lambda^{(n)})}{\mathbf{k}_{m+n}^{+}(\lambda^{(n)})+\mathbf{k}_{m+n}^{-}(\lambda^{(n)})}\longrightarrow1$$

as $n\geq\lambda_1$ tends to infinity. This was known for fully commutative permutations, equivalently one-row shapes, and the paper proves it for shapes $(1,1)$ and $(2,1)$; the general assertion remains open.