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Give a (convex) polytope $P\subset\Bbb R^d$ (the convex hull of finitely many points). Consider its linear and projective symmetry groups:

\begin{align} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\OO}{O} \Aut_{\GL}(P)&:=\{T\in\GL(\Bbb R^d)\mid TP=P\}\subseteq\GL(\Bbb R^d),\\ \Aut_{\PGL}(P)&:=\{T\in\PGL(\Bbb R^d)\mid TP=P\}\subseteq\PGL(\Bbb R^d). \end{align}

Clearly, $\Aut_{\GL}(P)\subseteq \Aut_{\PGL}(P)$, but we do not necessarily have equality.

Question: Is there always a transformation $T\in\PGL(\Bbb R^d)$ so that $\Aut_{\GL}(TP)=\Aut_{\PGL}(TP)$?

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As pointed out by Guillaume, I should exclude the case when $P$ is a simplex, since then $\smash{\Aut_{\PGL}(P)}$ is infinite. Alterantively, my intention is to consider two symmetries the same if they induce the same permutation on the vertices of the polyope.

Note that the answer is No if we consider finite sets of points rather than polytopes, because the projective symmetry group of any $d+2$ points in $\smash{\Bbb R^d}$ is isomorphic to the symmetric group $\mathrm{Sym}(d+2)$, which is not possible with just linear symmetries.

The question is motivated from the well-known fact that the linear symmetries of a polytopes $P$ are not more general than its orthogonal symmetries, in the sense that there always is a transformation $\smash{T\in\GL(\Bbb R^d)}$ with $\smash{\Aut_{\OO}(TP)=\Aut_{\GL}(TP)}$.

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  • $\begingroup$ This seems to fail for a triangle, since the affine symmetry group is finite but the projective symmetry group is infinite. $\endgroup$
    – Guillaume Aubrun
    Commented Mar 23, 2021 at 12:47
  • $\begingroup$ @GuillaumeAubrun Am I assuming a non-standard way to talk about projective symmetries? If two symmetries (affine or projective) permute the vertices in the same way then it is the same symmetry, isn't it? $\endgroup$
    – M. Winter
    Commented Mar 23, 2021 at 12:49
  • $\begingroup$ The issue I point is even more transparent in dimension $1$. The map $x \mapsto \frac{x}{2-x}$ belongs to the projective symmetry group of the polytope $[0,1]$ ; its action on vertices is the same as the identity map $x \mapsto x$. Do you want to consider these maps as the same? You may want to assume that your polytope is not a simplex. $\endgroup$
    – Guillaume Aubrun
    Commented Mar 23, 2021 at 12:59

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