Course:Harris, Fall 08: Diary Week 6

Mon:

• We looked at problems in identifying Wallpaper Symmetry Groups:
  • First look for rotations.
  • Then for reflections.
  • Then for glide-reflections; these can be difficult to spot.
• We reviewed what it means to have a subgroup of a group of symmetries.
  • In the Frieze Groups, the cyclic subgroups are all of the form {$E$,$X$}, where
    • $E$ is the identity transformation.
    • $X$ is a reflection or a rotation; in either case, $X^{2}$ = $E$, closing off the subgroup.
• The question was asked, why do I specify some symmetries with superscripts, some with subscripts?
  • Superscripts indicate multiplication: $TT=T^{2}$, i.e., $T^{2}$ indicates doing $T$ twice.
  • Subscripts are just labels: $R_{1}$ is a different rotation from $R_{2}$, and there is not necessarily any special significance to the choice of label.
• We spent the remainder of the period taking our first look at multiplying transformations in a complex manner, taking our example from D4.
  • We looked at two transformations within D4:
    • $R$ (rotation by 90 degrees)
    • $M_{1}$ (reflection across one of the reflecting lines)
      • We chose to look at D4 as exemplified in a square; $M_{1}$ was reflection across the NE-SW diagonal line.
  • We first looked at $RM_{1}$,i.e., if we do first $R$ and then $M_{1}$, what is the resulting transformation on the plane? It's got to be one of the other symmetries of the square, i.e., some other element of D4.
    • We found $RM_{1}=M_{4}$, reflection across the E-W line.
  • We then looked at $M_{1}R$, i.e., doing first $M_{1}$ and then $R$.
    • We found $M_{1}R$ = $M{}_{2}$, reflection across the N-S line.
  • Note: $RM_{1}$ and $M_{1}R$ are not equal! This is non-commutative "multiplication".

Wed:

• Groups worked on the D4 symmetry group (question 1 from Regular Triangle Symmetry Group Exploration), taking most of the period.
  • Hints for working on these kinds of problems:
    • $E$ times anything (and anything times $E$) is very easy.
    • Multiplication of $R^{n}$$R^{m}$ is always easy.
    • Any column (and any row) must contain all the elements of the group.
    • Orientation:
      • This is
        • preserved by any kind of rotation,
        • reversed by any kind of reflection.
      • Thus (with $R^{i}$ being any rotation and $M_{j}$ any reflection),
        • $R^{i}$$M_{j}$ must be an $M_{k}$ (because orientation is preserved, then reversed: net, reversed)
        • $M_{i}$$R^{j}$ must be an $M_{k}$ (because orientation is reversed, then preserved: net reversed)
        • $M_{i}$$M_{j}$ must be an $R^{k}$ (because orientation is reversed, then reversed: net, preserved)
  • Any progress towards the D3 group (question 8) well be added in as extra credit.
• Exam 1 will probably be next Wednesday, Oct. 8. Included:
  • Identification of symmetry groups as per quizzes and the Cathedral project.
  • Multiplying group elements, as per today's Exploration.
  • Ideas from the next Exploration, on Why There Are Only Three Regular Tessellations.

Fri:

• Exam 1 is moved to Friday of next week, Oct. 10 (I'm planning on getting a substitute instructor for that day, as I'll be heading to the airport for an afternoon flight).
  • You can use printed notes for Frieze and Wallpaper Groups.
  • Be able to identify a symmetry group from a pattern
  • Be able to build a pattern using a given motif and having a given symmetry group.
  • Be able to do identify the elements of a symmetry group (such as from looking at a pattern it describes).
  • Be able to do multiplication of elements of a rosette symmetry group.
  • Be able to answer questions on angles of polygons or why there are only three regular tessellations, as in Tessellations: Why There Are Only Three Regular Tessellations.
• We looked at the angle sum for any n-gon (n-sided polygon):
  • Any polygon can be subdivided into triangles, with
    • all triangle edges going between vertices of the polygon and
    • no triangles overlapping.
  • For an n-gon, it takes n-2 triangles to subdivide it:
    • Surely this is true for n = 3 (i.e., a triangle has 1 triangle "subdividing" it).
    • Suppose this formula were true for all polygons up to size N.
      • Then, what about for n = N+1? Given an (N+1)-gon, we can slice off two adjacent edges, say, A-B-C, replacing them with A-C, thus giving us an N-gon (sides AB and BC replaced by AC, so 1 fewer edge).
      • The N-gon can be subdivided into N-2 triangles (because we're supposing the formula true for n = N).
      • Then adding back in triangle ABC, we have N -2+1 = N-1 triangles subdividing the original (N+1)-gon.
      • That's the number we wanted, since N+1-2 = N-1.
    • Thus, if the formula holds for n = N, it also holds for n = N+1.
    • Thus, by the Principle of Induction, the formula holds for all n: Any n-gon can be subdivided into n-2 triangles.
  • The sum of all the angles of all the n-2 subdividing triangles, adds up to the angle-sum for the n-gon (we need the triangles to be non-overlapping for this, and also that the triangle edges go between vertices of the n-gon).
  • Thus, the angle-sum for the n-gon is n-2 times the angle-sum of a triangle, i.e., (n-2)180 degrees.
• We looked, then at the angle-formula for a regular n-gon:
  • Since an n-gon has n vertices, there are n angles.
  • We know the angle-sum (the sum of all those n angles) is (n-2)180.
  • We know all n angles, in a regular n-gon, are the same.
  • Thus, each of those angles must be (n-2)180/n.
• We changed the Exercises for Monday to questions 7-10 in Tessellations: Why There Are Only Three Regular Tessellations.