Yes, this is true for all polyhedral graphs. See the following reference:
My translation:
Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties:
(a) $\mathfrak{Q}$ is isomorphic to vertex-edge skeleton $\mathfrak{P}^1$ of $P$
(b) Every automorphism of $\mathfrak{P}^1$ is induced by a symmetry of the polyhedron $P$.
See this mathscinet link for an english summary of the result.
Yes, this is true for all polyhedral graphs. See the following reference:
My translation:
Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties:
(a) $\mathfrak{Q}$ is isomorphic to vertex-edge skeleton $\mathfrak{P}^1$ of $P$
(b) Every automorphism of $\mathfrak{P}^1$ is induced by a symmetry of the polyhedron $P$.
See this mathscinet link for an english summary of the result.
Yes, this is true for all polyhedral graphs. See the following reference:
My translation:
Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties:
(a) $\mathfrak{Q}$ is isomorphic to vertex-edge skeleton $\mathfrak{P}^1$ of $P$
(b) Every automorphism of $\mathfrak{P}^1$ is induced by a symmetry of the polyhedron $P$.
See this mathscinet link for an english summary of the result.
Yes, this is true for all polyhedral graphs. See the following reference:
My translation:
Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties:
(a) $\mathfrak{Q}$ is isomorphic to vertex-edge skeleton $\mathfrak{P}^1$ of $P$
(b) Every automorphism of $\mathfrak{P}^1$ is induced by a symmetry of the polyhedron $P$.
See this mathscinet link for an english summary of the result.
Yes, this is true for all polyhedral graphs. See the following reference:
Yes, this is true for all polyhedral graphs. See the following reference:
My translation:
Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties:
(a) $\mathfrak{Q}$ is isomorphic to vertex-edge skeleton $\mathfrak{P}^1$ of $P$
(b) Every automorphism of $\mathfrak{P}^1$ is induced by a symmetry of the polyhedron $P$.
See this mathscinet link for an english summary of the result.
Yes, this is true for all polyhedral graphs. See the following reference:
