Instructor:Frieze Exercises Solutions

Frieze Exercises

1. • There are vertical and horizontal mirror lines. Two-fold rotational symmetry: centers at both the center of the rectangle as well as at the midpoint of the vertical side (It's a pmm2)
   • Vertical mirror lines. Glide reflections. Two-fold rotation at the center of the triangle and at the midpoint of the (diagonal) sides. (It's a pma2)
   • Vertical mirror lines. (It's a pm11)
2. p1m1, pm11, p1a1, p112
3. Supposed to draw patterns using this motif for all seven frieze groups.
4. From left to right:
   • pm11
   • p112 : The under and over crossings prevent it from being pmm2
   • p112 : The crossings and some of the small triangles break the symmetry.
   • pmm2
5. There's glide reflections, and if we ignore the wavy lines (water?) it is p1a1
6. Supposed to draw four patterns with group pma2.
7. The composition of two glide reflections is a translation.
8. If you were to rotate the strip through any other angle, then the resulting strip would be at an angle with the original. The only way it could ever match up with the original is by doing a half-turn (ie 180 degrees).
9. • A, M, T, U, V, W, Y - pm11
   • B, C, D, E - p1m1
   • F, G, J, L, P, Q, R - p111
   • H, I, O, X - pmm2
   • N, S, Z - p112
   When you compare with rosettes, the C1 rosettes become p111, the D1 rosettes split into pm11 and p1m1, the D2 rosettes become pmm2 and the C2 rosettes become p112.
10. p111 = $F_{1}$; p1a1 = $F_{1}^{3}$; pm11 = $F_{1}^{2}$; p112 = $F_{2}$; pma2 = $F_{2}^{2}$; p1m1 = $F^{1}$; pmm2 = $F_{2}^{1}$. The subscript 2 means the group has order 2 rotations.
11. It is pma2