1. Use the costs presented in this chapter to prove that the seventeen plane group symbols listed in our table are all the orbifold symbols that cost exactly $2.
To begin, the chart below shows the cost of the various items for an orbifold.
Name || Handle | Gyration Points --------------------------------------------------- Symbol || o | 2 3 4 5 6 ... n --------------------------------------------------- Cost || 2 | 1/2 2/3 3/4 4/5 5/6 ... (n-1)/n Name || Cross Cap / Mirror | Kaleidoscopic/Corner Points ----------------------------------------------------------------- Symbol || x or * | 2 3 4 5 6 ... n ----------------------------------------------------------------- Cost || 1 | 1/4 2/6 3/8 4/10 5/12 ... (n-1)/2n
To start with, let's buy a Handle. Because it costs exactly $2 to buy a handle, that is all we would be able to afford with only $2 to spend. Thus, we can have a handle and nothing else. this notation is o and represents a wallpaper pattern that has only translational symmetry and nothing else.
If we purchase a cross cap, it costs $1 and we have $1 left. We could purchase another crosscap for $1 and then the cost would total exactly $2. The notation for this would be xx.
In addition, we could purchase one cross cap and one mirror which, because they cost $1 each would total exactly $2. This notation would be x*.
Alternatively, we could purchase one cross cap and then spend the remaining $1 on 2 gyration points. This would give us a notation of 22x.
These are the only possibilities with a crosscap because we can't have kaleidoscopic points unless we have a mirror and when we purchase both a cross cap and a mirror we are out of money. Also, the only gyration points that will total $1 are 2 period 2 gyration points. The rest will either put us over or leave a remainder.
If we purchase a mirror, we still have $1 left over, like we do with the cross cap. However, we can spend this money in slightly different ways.
We could purchase another mirror for another $1 and we would then be out of money. this notation would be **.
Instead of purchasing a second mirror, we could look at possible gyration point combinations. The most obvious one is two period 2 gyration points. That would total $2 (which is similar to the crosscap and two gyration points). The notation would be 22*.
If we purchase one period two gyration point we would still have 50 cents left over and we could purchase two kaleidoscopic points. The notation would be2*22.
Those are the only combinations we could have with the period two gyration point.
If we have a period three gyration point, we could also purchase a 3-mirror kaleidoscopic point. The notation would be 3*3. This is the only period 3 gyration point combination possible - nothing else is 33 cents.
We could purchase a period 4 gyratoin point, which would leave us with a quarter. The only thing that costs a quarter is a 2-mirror kaleidoscopic point. The notation would be 4*2.
Purchasing anything bigger than period 4 gyroscopic points leaves us with values less than a quarter, therefore, there are no other possible combinations of gyroscopic points. Now let's look at combinations of kaleidoscopic points.
We are still purchasing the mirror to begin with, so we are still looking for mirror points that total $1.
One possible combination would be to buy 4 2-way kaleidoscopic points. Since they are $.25 each, that would total $1. The notation would be *2222. If I bought only three 2-ways, I would need a quarter to make up but the only thing that costs a quarter is another 2-way, so the notation would be the same. This is also true if I purchase 2, 2-way kaleidoscopic points- I would need another half, but nothing else will total $.50. We will leave the purchase of 1, 2-way kaleidoscopic point for later.
We could also purchase 3 3-way kaleidoscopic points for $1 total. The notation would be *333. If I purchase less than one, I need another third or another two-thirds to total $1 and there is no other way to make this combination.
If we look at 4-way kaleidoscopic points, we can purchase a maximum of two and then we need to purchase something else to make up the difference. The only thing we could purchase would be another 2-way kaleidoscopic point. The notation for this would be *442. If I purchase only one 4-way kaleidoscopic point, I would need 5/8 more to make up the difference, however, I cannot get any kaleidoscopic points to add up to this so there are no more 4-way kaleidoscopic combinations.
We can't buy a 5-way kaleidoscopic point because we would need 3/5 or 6/10 more. There are no combinations of lower kaleidoscopic points that sum to 3/5.
If we purchase a 6-way kaleidoscopic point, it would cost 5/12 which means we have 7/12 left. Purchasing a 3/12 (2-way) would leave us with 4/12 left which is a 3-way kaleidoscopic point. This combination would have the notation *632.
If we purchase a 5-way mirror, we would need 11/60 more, but this just is not possible. Also, if we were to purchase 2 6-way kaleidoscopic points, we would need 2/12 or 1/6 more, but again, we have no number smaller than 1/4, therefore, this is not possible.This applies to all larger values of kaleidoscopic points as well.
These are all the possible combinations of patterns that have mirror symmetry because if we did NOT purchase a mirror, then we could NOT purchase kaleidoscopic points.
Now we will look at purchasing only gyration points. This is the only category left.
We could purchase four period 2 gyration points which would total $2. The notation would be 2222.
If we purchased three period 2 points, we would need another 1/2, but all the other gyration points total more than that. If we purchased two period 2 points, we would need another $1, but again, everything else is more than one half (but less than$1) and therfore we cannot total $1 without going over.
If we purchase only 1 period 2 gyration point, we would need another $1.50. Because two period 4 gyration points would total $1.50, we could make this combination. The notation would be442. WE could also get $1.50 by purchasing one period 3 gyration point (2/3) and one period 6 gyration point(5/6). The notation for this pattern would be 632.
We could purchase 3 period 3 gyration points for $2 total. The notation would be 333.
If I purchase two period 3 gyration points, I need another 2/3 which would be the same as the last one. If I purchase only 1 period three gyration point, I would need another 4/3, but none of the other gyration points will total that.
I cannot purchase more than two period 4 points because they would cost too much. We have already discussed purchasing just two. If I purchased only one, I owuld need another 5/4, but I cannot get this combination because all of the higher period points cost too much.
Both period 5 and period 6 gyration points (as well as all the rest) would require one or two purchased. If you purchased two, the amount left over would be less than 1/2, therefore, two is not a possibility. If we look at one purchased plus something else, we know we cannot purchase something bigger, for the same reason listed above, and if we think about purchasing something less, we have already discussed all those options when we started with the smaller ones. Therefore, there are no gyration only patterns with anything bigger than period 6.
Explain in your own words why the fact that there are exactly seventeen $2 orbifolds implies that there are only seventeen different symmetries a wallpaper pattern can have.
Because we know that the amount of money we have left from $2 when we purchase various symmetries will, when two is divided by the amount remaining, give us the order of the orbifold. In this case, because the amount remaining from our $2 when we purchase various symmetries is 0, we divide two by 0 to determine the order of our orbifold. In this case, 2/0 is undefined or infinite. What this means is we actually have a plane because it's the only surface we can actually have an infinite number of orbifolds..
4. Part I: An example of a symmetrical pattern that has a square as its orbifold would be a 2*2 pattern such as the one shown below. 2*22 means that there is one order 2 gyration point and two 2-mirror kaleidoscopic points.
The pattern shown below also has a square at its orbifold, but it has two kaleidoscopic points and one gyration point.
4. Part II: A pattern whose orbifold has one ninety degree kaleidoscopic corner and an order four gyration point is shown below.
I know this pattern has this orbifold because I can start off by drawing many lines of reflection and trying to find the smallest polygonal region (which will be the one that has no more mirror symmetries within it). Once I've found the smallest region, I identify a mirror string and begin to "walk" around it as if I was an ant. What I look for is to see how many different views there are. For example, since the smallest polygonal region of this pattern is a square, I start at one corner and look at the pattern. I then walk to the next corner and realize that this is exactly the same thing I saw from the last corner, which means there is only one kaleidoscopic point.