On the complements of 3-dimensional convex polyhedra as polynomial images of \(\mathbb{R}^{3}\) (Q2878772)

The paper under review continues the study of polynomial and regular images of Euclidean spaces initiated in [\textit{J. F. Fernando} and \textit{J. M. Gamboa}, J. Pure Appl. Algebra 179, No. 3, 241--254 (2003; Zbl 1042.14035)] and continued in a long series of articles: [\textit{J. F. Fernando} and \textit{J. M. Gamboa}, Isr. J. Math. 153, 61--92 (2006; Zbl 1213.14109)]; [\textit{C. Ueno}, Rev. Mat. Iberoam. 24, No. 3, 981--988 (2008; Zbl 1156.14330)]; [\textit{J. F. Fernando} et al., Proc. Lond. Math. Soc. (3) 103, No. 5, 847--878 (2011; Zbl 1282.14101)]; [\textit{C. Ueno}, J. Pure Appl. Algebra 216, No. 11, 2436--2448 (2012; Zbl 1283.14024)] and [\textit{J. F. Fernando}, J. Pure Appl. Algebra 218, No. 9, 1745--1753 (2014; Zbl 1295.14052)].NEWLINENEWLINEA map \(f:\mathbb R^n\to\mathbb R^m\) is said to be polynomial if there exist \(f_1,\dots,f_m\in\mathbb R[{\mathtt x}_1,\dots,{\mathtt x}_n]\) such that \(f(x)=(f_1(x),\dots,f_m(x))\) for every point \(x\in\mathbb R^n\). It is a challenge problem to characterize geometrically the semialgebraic subsets \({\mathcal S}\subset\mathbb R^m\) that are polynomial images of \(\mathbb R^n\) for some positive integer \(n\). Indeed, just the case of one dimensional \({\mathcal S}\) is completely understood; see the last quoted paper by J. F. Fernando.NEWLINENEWLINEMotivation for the study of these sets comes from the remark that important problems in real algebraic geometry concerning semialgebraic sets \({\mathcal S}\) such as optimization or computation of trajectories can be reduced to the case \({\mathcal S}=\mathbb R^n\) if \({\mathcal S}\) is a polynomial image of an Euclidean space.NEWLINENEWLINEMost known polynomial images of Euclidean spaces have piecewise linear boundary, and this is also the case in the article under review. Let \({\mathcal K}\subset\mathbb R^n\) be a closed convex and possibly unbounded \(n\)-dimensional polyhedron and denote \({\mathcal S}:=\mathbb R^n\setminus{\mathcal K}\). Let \({\mathcal {\overline S}}\) be the closure of \({\mathcal S}\) in \(\mathbb R^n\) with respect to the euclidean topology. The authors prove that for \(n=3\) the semialgebraic sets \({\mathcal S}\) and \({\mathcal{\overline S}}\) are polynomial images of \(\mathbb R^n\). It is natural to ask if the result holds true for arbitrary \(n\). The key point in dimension \(3\) is the following property, that is exclusive for convex \(2\)-dimensional polyhedra.NEWLINENEWLINE{ Property. } For any convex polygon \({\mathcal K}\subset\mathbb R^2\) there exist a vectorial line \(L\) and a hyperplane \(H\) such that the projection \(\pi:\mathbb R^2\to H\) with direction \(L\) satisfies the equality \(\pi({\mathcal K})=\pi({\mathcal K}\cap H)\).NEWLINENEWLINEThe authors prove in an Appendix at the end of the article that the property above is not true for higher dimensions. Hence, to extend the main result to arbitrary dimension requires to develop another techniques.NEWLINENEWLINEA different approach is to enlarge the family of admissible maps. Recall that a map \(f:\mathbb R^n\to\mathbb R^m\) is said to be regular if there exist polynomials \(f_1,\dots,f_m,g\in\mathbb R[{\mathtt x}_1,\dots,{\mathtt x}_n]\) such that \(g^{-1}(0)=\varnothing\) and for every point \(x\in\mathbb R^n\), NEWLINE\[NEWLINE g(x)f(x)=(f_1(x),\dots,f_m(x)). NEWLINE\]NEWLINE Indeed, \textit{J. F. Fernando} and \textit{C. Ueno} have proved [Int. Math. Res. Not. 2014, No. 18, 5084--5123 (2014; Zbl 1328.14089)] that the complement \(\mathbb R^n\setminus{\mathcal K}\) of a convex polyhedron \({\mathcal K}\subset\mathbb R^n\) and the complement \(\mathbb R^n\setminus\mathrm{Int}({\mathcal K})\) of its interior are regular images of \(\mathbb R^n\). If \({\mathcal K}\) is moreover bounded, both \(\mathbb R^n\setminus{\mathcal K}\) and \(\mathbb R^n\setminus\mathrm{Int}({\mathcal K})\) are in fact polynomial images of \(\mathbb R^n\).NEWLINENEWLINEIt is worthwhile mentioning the nonexistence of a well developed theory to attack this kind of problems; whenever one proves that a set \({\mathcal S}\subset\mathbb R^m\) is a polynomial or regular image of some \(\mathbb R^n\) it is necessary (up to some inductive process) to explicitly construct a suitable map \(f:\mathbb R^n\to\mathbb R^m\) such that \(f(\mathbb R^n)={\mathcal S}\). This is also the case in this article, where the authors employ in a very clear and clever way a rather technical trick: to place the polyhedron in what they call first or second trimming position.NEWLINENEWLINEThis interesting article is very well structured and contains enlightening pictures.