Homework 8

Lori Thomson
Math 5337

  1. The prices of the items are as follows:
    Name Symbol Cost
    Handle o 2
    Cross Cap x 1
    Mirror * 1
    Gyration Points 2 1/2
    3 2/3
    4 3/4
    ... ...
    n (n-1)/n
    Kaleidoscopic Points 2 1/4
    3 2/6
    4 3/8
    ... ...
    n (n-1)/2n
    Notice as n gets larger, the costs of gyration points and kaleidoscopic points also get larger. Also, as n approaches infinity, the cost of a gyration point approaches $1 and the cost of a kaleidoscopic point approaches $.50.

    If I buy a handle, I can buy only one because it uses my entire $2. If I buy a handle, I can't buy anything else.

    
    	1.  o
    

    If I don't buy a handle, I can buy a cross cap for $1, which leaves me with $1 that I can spend on another cross cap or a mirror...

    
    	2.  xx
    	3.  x*
    

    but I can't buy any kaleidoscopic points because the mirror that is needed for kaleidoscopic points used up my last dollar. However, I can still buy gyration points. I will certainly need more than one gyration point to use up my whole dollar because gyration points always cost less than $1 [ (n-1)/n < 1 for n >= 2]. The least amount of money I can spend on more than one gyrations is what I spend when buying two order 2 gyration points, which is exactly $1.

    
    	4.  22x
    

    Since that is both the least I can spend and the most I can spend, it is the only way I can buy both cross caps and gyration points. Now I have exhausted all possible ways to buy cross caps and handles.

    Moving on to mirrors... I can buy two mirrors at $1 each to exactly use my $2.

    
    	5.  **
    

    But if I buy just one mirror, then I have $1 left over to buy gyration points and kaleidoscopic points. Starting with the most expensive... gyration points:

    I have the same restriction as I did when buying gyration points with the cross cap. If I buy only gyration points with my mirror, then my only choice is to buy two 2-fold gyration points:

    
    	6.  22*
    

    If I buy a mixture of gyration points and kaleidoscopic points, then I can trade in one of the 50-cent 2-fold gyration points for two 25-cent 2-fold kaleidoscopic points.

    
    	7.  2*22
    

    Similar to previous arguments, If I keep the 2-fold gyration point and the mirror, and I spend the remaining $.50 on kaleidoscopic points, my only choice is to buy two 2-fold kaleidoscopic points: I can't use the whole 50 cents on just one kaleidoscopic point because each point costs less than $.50 [ (n-1)/2n < .5 for n >=2 ]. Furthermore, the least I can spend on two or more kaleidoscopic points is $.50, which is when I buy two 2-fold kaleidoscopic points.

    Now if I choose a 3-fold gyration point instead of 2-fold, the 3 and the mirror cost me 1+2/3 leaving me with 1/3 of a dollar. I canít buy any more gyration points with that since the cheapest gyration point is $.50. If I buy a 2-fold kaleidoscopic point for a quarter, I will have 1/3 - 1/4 = 1/12 left, but I canít buy anything for 1/12 of a dollar, so I canít buy the 2-fold kaleidoscopic point. If I buy a 3-fold kaleidoscopic point for 1/3 of a dollar, Iíve used up my money exactly.

    
    	8.  3*3
    

    All other kaleidoscopic points cost more that 1/3 of a dollar.

    So now I must see what happens if I buy a 4-fold gyration point and a mirror. Those cost me 1+3/4, leaving me with $.25. The only thing I can buy with $.25 is one 2-fold kaleidoscopic point; everything else costs more than $.25.

    
    	9.  4*2
    

    If I buy a 5-fold gyration point and a mirror, it costs me 1+4/5 dollars leaving me with 1/5 of a dollar which I canít buy anything with. In fact, for all n>=5, 2-[1+(n-1)/n] <= 1/5. So if I buy a mirror, I canít buy any gyration points of order greater or equal to 5.

    If I buy a mirror and do not buy gyration points, then I have $1 left to spend on kaleidoscopic points. Let d = 1 = the number of dollars I have left to spend on kaleidoscopic points, which cost d(n-1)/2n dollars each. (The reason for introducing d will make sense later.)

    If I buy a 2-fold kaleidoscopic point for d(1/4) dollars, I have d-d/4 = 3d/4 dollars left.
    If I buy a second 2-fold kaleidoscopic point, I would have 3d/4 - d/4 = d/2 dollars left. As shown before, the only way to spend the remaining d/2 dollars ($.50) only on kaleidoscopic points is to buy two 2-fold kaleidoscopic points:

    
    	10.  *2222
    

    If instead of buying a second 2-fold kaleidoscopic point, I buy a 3-fold kaleidoscopic point at d(2/6) dollars, I have 3d/4 - d(2/6) = 5d/12 dollars left. I could still buy another kaleidoscopic point of order 3, 4, 5 or 6, for d(2/6), d(3/8), d(4/10) or d(5/12) dollars, respectively. One order 6 kaleidoscopic point would use the remaining money exactly:

    
    	11.  *632
    

    If I buy any of the other three orders (3, 4 or 5) I must buy them in combination. Their costs with common denominators are d(40/120), d(45/120), and d(48/120) dollars, respectively. Any two of these would add up to more than the d(50/120) dollars that I have left to spend, so Iíve exhausted all the possibilities of having a mirror boundary with both 2-fold and 3-fold kaleidoscopic points.

    If instead of buying that 3-fold kaleidoscopic point, I buy a 4-fold kaleidoscopic point, at d(3/8) dollars, I have 3d/4 - d(3/8) = 3d/8 dollars left. With this I can buy only another 4-fold kaleidoscopic point. (Smaller order kaleidoscopic points have been exhausted previously, and larger order kaleidoscopic points are too expensive.)

    
    	12.  *442
    

    If instead of buying the 4-fold kaleidoscopic point I buy an order 5 or greater kaleidoscopic point at a price greater than the 4-fold, I will be left with less money than the previous example which was just enough to buy an order 4 kaleidoscopic point. Now I canít afford anything else. There are no other possibilities that include a mirror and an order 2 kaleidoscopic point.

    I will now look for possibilities that include a mirror and an order 3 kaleidoscopic point. After buying those two pieces I am left with d-d(2/6) = 2d/3 dollars to spend on order 3 or greater kaleidoscopic points.

    I can start by buying a second order 3 kaleidoscopic point. Now I have 2d/3 - d(2/6) = d/3 dollars left. With this I can buy exactly one more order 3 kaleidoscopic point. Higher order kaleidoscopic cost too much.

    
    	13.  *333
    

    Instead of buying the second order 3 kaleidoscopic point, I can buy an order 4 kaleidoscopic point for d(3/8) dollars, leaving me with 2d/3 - d(3/8) = 7d/24 dollars, which is not enough to buy any order 4 or greater kaleidoscopic point. Kaleidoscopic points with higher order than 4 will just cost more leaving me with less. Therefore there are no other possibilities that include a mirror and an order 3 kaleidoscopic point.

    I will now look for possibilities that include a mirror and an order 4 kaleidoscopic point. After buying these two pieces, I am left with d-d(3/8) = 5d/8 dollars to spend on order 4 or greater kaleidoscopic points.

    I will start by buying a second order 4 kaleidoscopic point. Now I have 5d/8 - d(3/8) = 2d/8, which is not enough to buy any kaleidoscopic point of order 4 or greater. If instead of the second order 4 kaleidoscopic point I bought a higher order kaleidoscopic point, I would just have even less money remaining and still could not buy anything with it. Therefore, there are no possibilities that include only a mirror and kaleidoscopic points of order 4 or higher.

    All possibilities including only a mirror and kaleidoscopic points have been covered. The only type left are those that include only gyration points. For these I have the entire $2 to spend on gyration points. This time let d = 2 = the number of dollars I have to spend on gyration points, which cost d(n-1)/2n dollars each. (Since d = 2, the cost is (n-1)/n each.) Using the same arguments as were given for d = 1 and kaleidoscopic points, four analogous groups arise when d = 2 and I am buying gyration points:

    
    	14.  2222
    	15.  442
    	16.  632
    	17.  333
    

    I have enumerated all of the possibilities for spending exactly $2 in the orbifold shop. There are exactly seventeen $2 orbifolds. The costs of the pieces of an orbifold have been chosen so that even money exchanges won't change the Euler characteristic of the group and thus won't change the topology of the group. If any one $2 orbifold (say "o") gives a pattern with the same Euler characteristic as the Euclidean plane, then they all do. So all $2 orbifolds correspond to real plane symmetry groups. Since every wallpaper pattern can be represented by an orbifold, there are at most seventeen different types of symmetry found in wallpaper patterns.


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