When can a folded polygon be isometrically (locally) embedded into R^3? I am interested in 3-D representations of various things that naturally live in a non-simply-connected compact surface. There is the usual way of producing a compact surface of any orientable or non-orientable genus, by associating a boundary word to the edges of a polygon to indicate how they are to be glued. But there may or may not be an isometric embedding of this polygon quotient into R^3, not even a local one (isometric immersion). As a simple example : "a a^-1 b b^-1 c c^-1" generates a simply connected surface, but if you put this word around a regular hexagon and fold accordingly, the result is a squashed tetrahedron - not even an immersion. You can make it work by altering the angles of the hexagon slightly. 
Is there a known method to decide, given the angles at each vertex of the unfolded polygon, and a boundary word, whether an isometric embedding, or immersion, exists? (More ambitiously I would like it to decide whether it is ugly, but that sounds like it might be an AI-complete question.)
Addendum
The faces are to be pasted as Joseph O'Rourke indicated. Thanks for the clarification. 
So yes, there would be a finite number of points with angle other than 2 pi (they could actually be greater - multiple vertices of the original polygon can get identified.) Everywhere else it's flat. Self-intersection is unavoidable for non-orientable cases, but for example the regular hexagon case, the "embedding" is not injective even locally.
And never mind "ugly". A decision procedure for that would probably be a sentient algorithm (AI-complete question).
 A: I offer a clarification, one substantive point, and some references, but not a full answer.
The clarification is in response to X.M.Du's question.  I assume what Daniel means is this. Start with a geometric hexagon (or any polygon) embedded in the plane, with edge lengths and angles.  Label the edges with symbols.  Glue edges with the same symbol together, with the convention that $a^{-1}$ glues oppositely oriented to $a$.  Daniel is asking if, knowing the angles and gluing instructions, is there a way to decide if the resulting object is embedded/immersed.
The one point I'd like to make is that, if the angles and gluing instructions are such that each point
of the resulting manifold is surrounded by at most $2 \pi$, then Alexandrov's theorem answers the question:  The result is realized by a unique convex polyhedron. Alexandrov's amazing theorem is discussed in this MO question, in Lectures on Discrete and Polyhedral Geometry, in
Geometric Folding Algorithms: Linkages, Origami, Polyhedra, and in Alexandrov's 1950 book, recently translated into English: Convex Polyhedra, Springer, 2005.
I can't resist this tangent. One very neat application of this theorem is the following.  Take any convex shape (smooth or polygonal). Choose two
boundary points that partition the perimeter into equal halves.  Glue the two
halves together.  The result is a unique convex body.  For smooth convex shapes, this is called
a D-form.  This is discussed in Geometric Folding Algorithms above.
Without the nonnegative curvature condition needed for Alexandrov's theorem, I am not aware of any general method beyond
ad hoc techniques applied to small examples.
The work of Jügen Bokowski is the closest I know to what you seek.
He found the first realization of Dyck's regular map:
"On Heuristic Methods for Finding Realizations of Surfaces," 2008.
I would be interested to learn more from other responses
to Daniel's question.
