MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x,y\in C$.

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 $R^d$.

share|cite|improve this question
thank you for your answers. – user2529 Sep 24 '11 at 3:31
up vote 9 down vote accepted

No. Consider solid angles in $\mathbb R^2$. They all (except the half-plane) are isomorphic, yet one may be strictly included in another.

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 $C\ne D$.

share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.