TITLE: Why no higher-genus polyhedra that are nearly regular?
QUESTION [9 upvotes]: It seems to be a fact that there are only five bounded connected non-selfintersecting polyhedra with identical regular-polygon faces and congruent vertices (i.e., you can pick a neighborhood of every vertex so that all the neighborhoods are congruent, this is sometimes called "locally vertex-transitive"), namely the platonic solids.
(Drop any one word in the above and you get more: unbounded allows an infinite linear chain of octahedra glued face to face; disconnected allows the union of two disjoint Platonic solids; self-intersecting allows two of the Kepler-Poinsot solids; non-identical allows Archimedean solids; irregular faces allows e.g. the noble disphenoid, and of course dropping congruent vertices allows a myriad of possibilities.) Note I have not included convex among the hypotheses.
However, I am hard-pressed to put together a proof of this. If you add the hypothesis that the polyhedron is genus 0, then Euler's formula shows you there must be an overall angle defect, so the same angle defect at each vertex since the vertices are congruent, and now you can do the usual three times the vertex angle must be less than a full circle, etc. (Even in the case of genus 1, Euler's formula just tells you that the angle sum must be 0 at each vertex, and then there are infinite solutions for both six triangles and four squares at each vertex -- basically infinitely long prisms, which are projectively speaking tori -- so you will really need to use the boundedness...)
Nevertheless, the genus-zero hypothesis doesn't seem to be necessary; there seem to be no higher-genus connected bounded non-selfintersecting polyhedra with identical regular-polygon faces and congruent vertices.  Can anyone outline or point me to a proof of this fact?

REPLY [2 votes]: Here is one reference where the author does not assume genus 0: He gives careful definitions and proofs:
Cromwell, Peter R., Polyhedra., Cambridge: Cambridge University Press. xiii, 451 p. (1999). ZBL0926.52014.
In Chapter 10 he gives a complete classification of embedded polyhedral surfaces $S$ in $E^3$ which are closed (compact, without boundary) and connected as topological surfaces and satisfy:

Each face is a regular convex polygon and all faces are congruent. 
Links of any two vertices are congruent. (In other words, for any two vertices $v, w$ of $S$ there are neighborhoods of these vertices in $S$ which are congruent in $E^3$.)  
All dihedral angles are equal. (Not sure if this is an independent condition.) 

All regular polyhedra listed by Cromwell have genus zero.
Edit. I do not have an answer to your precise question: The assumption that Cromwell makes about equality of all dihedral angles forces $S$ to be locally convex, hence, convex. However, with a bit stronger assumption than the one you make, your question does have positive answer: 
Theorem. Suppose that $S$ is a uniform 
(embedded) connected closed polyhedral surface in $E^3$, i.e. such that:
a. All faces of $S$ are regular polygons. 
b. The symmetry group of $S$ acts transitively on the set of vertices of $S$. 
Then:

$S$ is simply-connected. 
If, in addition, all faces of $S$ are congruent to each other, then $S$ is the boundary of one of the Platonic solids.  

Remark. A symmetry of $S$ is an isometry of $E^3$ which preserves $S$. 
A classification of all uniform polyhedral surfaces (with references) can be found in this Wikipedia article (which also allows for non-embedded surfaces). The conclusion of the theorem stated above comes from examination of the list. The references for completeness of the list are:

S. P. Sopov,   "A proof of the completeness on the list of elementary homogeneous polyhedra", Ukrainskiĭ Geometricheskiĭ Sbornik (8) (1970) 139–156.
J. Skilling, "The complete set of uniform polyhedra", Philosophical Transactions of the Royal Society of London, 278 (1975)  111–135

However, in your question instead of vertex-transitivity of the group of symmetries you only assume local transitivity:
For any two vertices $v, w$ of $S$ there exist neighborhoods $U_v, U_w$ of $v, w$ in $S$ and an isometry $g\in Isom(E^3)$ which carries $U_v$ to $U_w$. 
This is a weaker assumption than vertex-transitivity and I am not sure it suffices for the desired conclusion (even assuming, in addition, that $S$ has genus 0). 
Definition. A (closed, connected, embedded) polyhedral surface in $E^3$ is called pseudo-uniform if its faces are regular polygons and $S$ is locally vertex-transitive but not vertex-transitive. 
They are discussed in this Wikipedia article: 

There are two known pseudo-uniform polyhedra: the pseudorhombicuboctahedron and the pseudo-great rhombicuboctahedron. It is not known if there are any others; Branko Grünbaum conjectured that there are not, but thought that a proof would be "probably quite complicated".

The first of these two surfaces is convex but its faces are squares and triangles, hence, they are not all congruent. The second surface is not embedded (and, again, has non-congruent faces). As far as I can tell, Grunbaum's conjecture is still open. If true, it will imply the positive answer to your question. However, clearly, his conjecture is harder since he does not assume congruence of all faces. 
In 
Gillispie, Steven B.; Grünbaum, Branko, The ({4,5}) isogonal sponges on the cubic lattice, Electron. J. Comb. 16, No. 1, Research Paper R22, 28 p. (2009). ZBL1160.52011.
the authors define Platonic polyhedra as those which are locally vertex-transitive and have regular and congruent faces. They make no comments about the classification about such polyhedra (apart from the classical case of convex polyhedra). Using their terminology, your question can be formulated as:

Suppose that $S$ is a closed connected polyhedral surface in $E^3$ which is a Platonic polyhedron. Is it true that $S$ is convex?

Considering surfaces of genus $1$, it is easy to see that the only examples can occur when faces are triangles and all vertices are 6-valent; it is a polyhedron of type $P(3,6)$ in the standard terminology. For each vertex, up to congruence, there is a 3-dimensional family of deformations in $E^3$ of the star of the vertex...