Index
(a) Let's let our point set be {1, 2, 3, ... ,15} and so v = b = 15. We know that this is of order (2,2) and thus each line is incident with 3 points and each point is incident with 3 lines. So, without loss of generality, let us pick a line that contains 3 points:
Both points 1 and 3 must lie on other lines which can not meet (every point is on 3 lines and 2 points are contained in a unique line) . Let's let these lines be {3,4,5} and {1,9,10}.
There must be some line on point 9 that meets {3, 4, 5}, but in addition there must be another line on point 9 which is distinct from each of the lines already created. Let's call this line {7, 8, 9}. There must be a line that joins point 5 to {7, 8, 9} so lets call this line {5, 6, 7}.
Now, because there does not yet exist a line through point 1 that meets {5, 6, 7}, there must be a line between 1 and 6 (1 already lies on a line that meets a line incident with every other point in the figure thus far). So, we will draw a line from 1 to 6. By the same reasoning, there must be a line between 3 and 8, a line between 5 and 10, a line between 7 and 2 and a line between 9 and 4. There also must be a third point on each of these lines, so we shall create one:
Let's look at point 2. It is only on 2 lines thus far, therefore we know that it must be on the on a line that meets lines {4, 9} and {5, 10} because there is not yet a line through 2 that meets these two lines. But because point 2 must lie on exactly 3 lines, and it already lies on 2, the line through 2 that meets {4, 9} and {5, 10} must be the same line. We also know that lines {4, 9} and {5, 10} must each contain a third point. Thus, the line through 2 that meets {4, 9} and {5, 10} will also contain the third points of these lines.
By similar resoning, we can find the third lines through points 4, 6, 8 and 10. This will ultimately complete the generalized quadrangle and we will have 15 lines and 15 points.
(b). Here is one example of an isomorphism between the two geometries. Use the labeling of the doily above and let f be our isomorphism.
1 a (5, 3)
2 a (1, 2)
3 a (4, 6)
4 a (1, 3)
5 a (2, 5)
6 a (1, 4)
7 a (3, 6)
8 a (1, 5)
9 a (2, 4)
10 a (1, 6)
11 a (5, 6)
12 a (4, 3)
13 a (2, 6)
14 a (4, 5)
15 a (2, 3)
From this, it is trivial to see where the lines map. Let's look at a labeling of the doily such that each point is labeled with a duad:
This helps us to see that the lines of the doily do indeed map to legitimate lines of the syntheme-duad geometry.
(c). Here are the 15 different synthemes:
It is certainly possible to model these on the generalized quadrangle that we are used to - the doily. Here is such a modeling:
It is also possible to model this on a three dimensional figure. This figure is isomorphic to the other figures we have seen that represent the generalized quadrangle. The first figure is the view from above. The other two are the hidden points that appear "underneath" the figure.
|
|
