For convex polyhedra you have Steinitz's theorem characterizing them as the 3-connected planar graphs. My question is not about spheric tilings, but about periodic tilings of the euclidean plane. Is it here also the case that 3-connectivity corresponds with convexity?
It is easy to construct an example of a 2-connected periodic tiling that is not convex. My guess is that the symmetry group prohibits some tilings from being convex and thus there are also 3-connected periodic tilings that are not convex, but I can't seem to be able to construct a counter example to support this guess.
Is there a known counterexample for this? Or is it actually true? Any reference or hint would help.
edit: I was looking for a periodic tiling which has no equivariantly equivalent (i.e. topologically equivalent and the same symmetry group) tiling that is convex and that is 3-connected when you look at the segments as graph edges. My feeling is that in case of the euclidean plane there might be a tiling in which you can't convexify one face, without needing to make another one nonconvex to keep the same symmetry group.