Let $S$ be a (basic) semialgebraic set, for which we know the constraints (i.e. we know the polynomial inequalities and equalities which define the set). Is it necessarily the case that the cone $C = \mathbb{R}^+ S$ is a semialgebraic set, and if so, can one determine the inequalities and equations which define $C$ (i.e. is there an algorithm to find the constraints)?


  • what if $S$ is a spectrahedron?
  • what if $S = \text{conv}(X)$ where $X$ is a real algebraic set?
  • 1
    $\begingroup$ Isn't it obvious? I may miss the whole point, but it seems to me that if $S$ lies in $\mathbb R^n=\{x=(x_1,\ldots,x_n)\}$, defined by $E_j(x)=0$ and $I_j(x)\geq0$, then the subset $C\subset \mathbb R^{n+1}=\{(x,t)\}$ is given by $t\geq 0$ and $E_j(x/t)=0$ and $I_j(x/t)\geq 0$, constraints which translate directly as polynomial constraints in $(x,t)$ because each $E_j,I_j$ is polynomial and $t\geq 0$. $\endgroup$ – Loïc Teyssier Dec 1 '17 at 11:45

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