The rigid-flexible value for symplectic embeddings of four-dimensional ellipsoids into polydiscs

We consider the embedding function cb(a)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$c_b(a)$$\end{document} describing the problem of symplectically embedding an ellipsoid E(1, a) into the smallest scaling of the polydisc P(1, b). Previous work suggests that determining the entirety of cb(a)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$c_b(a)$$\end{document} for all b is difficult, as infinite staircases can appear for many sequences of irrational b. In contrast, we show that for every polydisc P(1, b) with b>2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b>2$$\end{document}, there is an explicit formula for the minimum a such that the embedding problem is determined only by volume. That is, when the ellipsoid is sufficiently stretched, there is a symplectic embedding of E(1, a) fully filling an appropriately scaled polydisc P(λ,λb)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P(\lambda ,\lambda b)$$\end{document}. Denoted RF(b), this rigid-flexible (RF) value is piecewise smooth with a discrete set of discontinuities for b>2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b>2$$\end{document}. At the same time, by exhibiting a sequence of obstructive classes for bn=n+1n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b_n = \frac{n+1}{n}$$\end{document} at a=8\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a=8$$\end{document}, we show that RF is also discontinuous at b=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b=1$$\end{document}.