If the polygonal presentation contains only one word, the following algorithm works:
- Add vertex symbols. Then, for each pair of edges with the same edge symbol, build one set containing the two initial vertices and one set containing the two terminal vertices.
- Build the union of all vertex sets that have at least one vertex in common until all resulting sets are disjoint.
- The number of vertices is the number of disjoint sets obtained in step 2. The number of edges is the number of different edge symbols and the number of faces is one.
Example:
Cube: $S = \langle a, b, c, d, e, f \mid aa^{-1}e^{-1}dd^{-1}gcc^{-1}g^{-1}e f bb^{-1}f^{-1} \rangle$ (c.f. A.P.'s answer)
Adding vertex symbols: $AaBa^{-1}Ce^{-1}DdEd^{-1}FgGcHc^{-1}Ig^{-1}KeLfM bNb^{-1}Of^{-1}A$
Building vertex sets:
Edge $a$: $\lbrace A, C \rbrace$, $\lbrace B, B \rbrace$;
Edge $b$: $\lbrace M, O \rbrace$, $\lbrace N, N \rbrace$;
Edge $c$: $\lbrace G, I \rbrace$, $\lbrace H, H \rbrace$;
Edge $d$: $\lbrace D, F \rbrace$, $\lbrace E, E \rbrace$;
Edge $e$: $\lbrace D, K \rbrace$, $\lbrace C, L \rbrace$;
Edge $f$: $\lbrace L, A \rbrace$, $\lbrace M, O \rbrace$;
Edge $g$: $\lbrace F, K \rbrace$, $\lbrace G, I \rbrace$
Building unions:
$\lbrace A, C \rbrace \cup \lbrace C, L \rbrace \cup \lbrace L, A \rbrace =
\lbrace A, C, L \rbrace$,
$\lbrace D, F \rbrace \cup \lbrace D, K \rbrace \cup \lbrace F, K \rbrace =
\lbrace D, F, K \rbrace$
Resulting disjoint vertex sets:
$\lbrace A, C, L \rbrace$, $\lbrace D, F, K \rbrace$, $\lbrace M, O \rbrace$,
$\lbrace G, I \rbrace$, $\lbrace B \rbrace$, $\lbrace E \rbrace$,
$\lbrace H \rbrace$, $\lbrace N \rbrace$
The number of disjoint vertex sets is 8. Thus, the number of vertices is 8.
When you join up the edges of the flat square to make a projective plane, what happens to its corners? Each corner gets identified with the opposite corner.
So at two points in the projective plane you have two corner bits of squares
identified. They may be flat, but round the corner, you have only $\pi$
worth of angle instead of $2\pi$. This means you can't extend the flat metric
to these corners, so the quotient isn't naturally a Riemannian manifold.
An easier illustration: consider the surface of a standard cube in $\Bbb R^3$.
Each face is flat. What about the edges? They aren't trouble, one can think
of each pair of adjacent faces as a $2\times1$ rectangle folded over, and
that has a flat metric. So if you remove the vertices, the surface of the
cube has a flat Riemannian metric all over it.
But those pesky vertices! If you make a little circuit about one of these,
in effect you're going through an angle of $3\pi/2$. If you parallel-transport a vector round the path, it come back turned through a right angle. This means
there's no way of extending this flat Riemannian metric to the vertex.
This works for every compact surface; you can express it as a simplicial complex,
and if you delete the vertices then you can put a flat metric on it. Alas
one can rarely extend it to the whole surface (certainly if it has nonzero
Euler characteristic).
Best Answer
You can see four squares in the figure. The two diagonally opposite squares form a face, as you can see from the gluing. That is why it has two faces in total.