What surface is obtained from the square [0,1]^2 by identifying each edge with the opposite edge after translating it halfway (mod 1)?

[... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] [... spoiler space ...] Two handles. Each edge of the square is really two edges, so we have an octagon with sides abcdABCD where I hope the meaning of the letters is obvious. Now we can cut-and-paste in various ways; here's one. 1. Slice into the following bits along three parallel lines: cdx, bXAz, aZBy, YCD. 2. Splice the triangles along d/D to give cYCx, in other words a tube with x at one end and Y at the other. 3. Splice the quadrilaterals along b/B to give aZXAzy, then along a/A to give a tube with zy at one end and ZX at the other, then along z/Z to give a torus with a hole whose boundary is Xy. 4. Noting that in the tube the ends of x are identified, as are those of Y, do the same in the boundary of the hole in the torus so it now has a figure-eight hole whose boundary is Xy. 5. Now splice the tube on, obviously producing a double torus (= sphere with two handles). I bet there's a neater way to do it. Maybe some magic involving writing down the right group to quotient the hyperbolic plane by? -- g