A D-dimensional totally unsupported polytope, necessarily has at least D+1 convex vertices (i.e. which are on the convex hull). Say it has C, where C>=D+1. * If D=2, it must also have C concave vertices for a simple polygon otherwise a supported edge would occur. * If D=3 and C=4, then it must also have at least 2 nonconvex vertices since it it had 0 it'd be convex, and if it had 1 then by considering all possible combinators for the edge graph you can see there would have to be a support face. And if D=3, C>=5 at least 1 nonconvex vertex would be required. Then, IF the faces are triangles Euler formula allows you to see at least 8 faces would be needed. So I claim the above argument PROVES that the "alternating hexagon" and the "Asimov pringle" are both minimal examples, in the sense the former has the min # edges, and the latter the min # triangle faces (provided all faces triangles and interior is topologically a ball). I tend to doubt allowing nontriangle faces would help, but have not proven it (and even my proof above of what I did prove. feels a bit handwavy)... if so then Asimov pringle is minimal-faced for all polyhedra whose interior has the topology of a ball.