Does the automorphism group of a cone determine the cone? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:40:03Z http://mathoverflow.net/feeds/question/76191 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76191/does-the-automorphism-group-of-a-cone-determine-the-cone Does the automorphism group of a cone determine the cone? Colin Tan 2011-09-23T08:52:21Z 2011-09-23T09:56:16Z <p>A cone is a $R_+$-module. That is, a cone is an abelian monoid that is closed under nonnegative real scalar multiplication. An automorphism of a cone is a bijective $R_+$-linear map. That is a map $f:C\to C$ such that $f(\alpha x +\beta y)=\alpha f(x) + \beta f(y)$ for all $\alpha,\beta \ge 0$ and $x,y\in C$.</p> <p>Suppose I have an inclusion of cones $C\subset D$. If the automorphism group $Aut(C)$ and $Aut(D)$ are isomorphic, does that imply that $C=D$? You may assume that these cones are finite-dimensional, in the sense that there is an $R_+$-linear map from each of these cones to some euclidean space $R^d$.</p> http://mathoverflow.net/questions/76191/does-the-automorphism-group-of-a-cone-determine-the-cone/76195#76195 Answer by doetoe for Does the automorphism group of a cone determine the cone? doetoe 2011-09-23T09:55:29Z 2011-09-23T09:55:29Z <p>Unless there is something I misunderstand, a cone can be isomorphic to a strict subcone (hence no). Take e.g. for D a cone in $\mathbb{R}^2$ generated by two linearly independent vectors, and for C the cone generated by two linearly independent vectors in D that don't generate D (i.e. that are not on the boundary).</p> http://mathoverflow.net/questions/76191/does-the-automorphism-group-of-a-cone-determine-the-cone/76196#76196 Answer by Sergei Ivanov for Does the automorphism group of a cone determine the cone? Sergei Ivanov 2011-09-23T09:56:16Z 2011-09-23T09:56:16Z <p>No. Consider solid angles in $\mathbb R^2$. They all (except the half-plane) are isomorphic, yet one may be strictly included in another.</p> <p>Furthermore, a generic cone in $\mathbb R^d$, $d\ge 3$, has a trivial automorphism group, so you may have $Aut(C)=Aut(D)$ in the strongest possible sense but $C\ne D$.</p>