Index
Define a polar space as follows:
Polar Spaces
1. Any line has at least two points.
2. Two points are on at most one line.
3. For every point p not on a line l, the number of points of l joined to p by a line is either 1 or v(l) (the total number of points on l )
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It follows that any linear space will be a polar space (this also means affine and projective spaces are polar spaces) and that any polar space is a near-linear space.
Let's look at a few other definitions. We say that a polar space S is degenerate if some point in S is collinear with all points of S. Note that the empty set is degenerate. It is easy to see that all linear spaces are degenerate polar spaces because it is an axiom that every point must be connected to every other via a line.
Another interesting idea when dealing with polar spaces is that of direct sums. Using direct sums it is possible to create new polar spaces from ones that already exist. Define {Si} as a family of pairwise disjoint polar spaces.
Definintion: A direct sum of the Si is the space S whose points are the points of the Si's and lines are the lines of the Si's. In addition, all pairs of points {p,q} are lines in S where p is in Si and q is in Sj with i and j distinct.
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We are combining polar spaces by joining each point of one polar space to every point of another. Let's look at an example.
The original polar spaces are the green and purples figures while the sum is represented by the entire figure. The red lines represent the new lines.
Now we define a polarity of a geometry.
Definintion: A polarity is a function f that maps the points of a geometry to the lines of the geometry and the lines to the points. Additionally, if the function is applied twice then you are left with the identity transformation. Also, incident point - line pairs are mapped to incident line - point pairs. (That is, suppose that point p lies on line l. Then point f(l) is on line f(p). In addition, if points p and q are connected by a line L, then under f, the lines f(p) and f(q) meet at the point f(L)).
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As an example let's define a polarity of the Fano Plane. Below are the generator only models (all three of which combine to create the actual polarity). Notice that for each of the three figures the black point maps to the line and the line maps to the black point.
If we combine these, we get the following figure (where like colors correspond to images under f ).
We can check that this is indeed a polarity. Let's check that our point - line incidence is preserved in the mapping. Above, we see that 1 lies on {0, 1, 3}. Then it should follow that f ({0, 1, 3} lies on f (1). Now f ({0, 1, 3}) = 0 and f (1) = {0, 5, 6), so 0 certainly lies on {0, 5, 6}. In addition, points 1, 4 and 6 lie on the line {1, 4, 6}. Note that f (1) = {0, 5, 6}, f (4) = {3, 4, 5}, f (6) = {1, 2, 5} and f ({1, 4, 6}) = 5. So we see that f (1), f (4) and f (6) meet at point 5 or f ({1, 4, 6}). You can continue checking the point - line incidence in this manner. This is indeed a legitimate polarity.
Definintion: An absolute point is a point that is contained in its image under a polarity (i.e., a point is mapped to a line with which it is incident). An absolute line is a line that is contained in its image under the polarity.
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In our diagram above, there are 3 absolute lines and 3 absolute points. In a projective plane, every absolute line of the polarity contains exactly one absolute point ( the dual statement is also true ). It is fairly trivial to see why this is true.
Suppose that l is absolute. Then f (l) lies on l and f (l) is absolute. If l contains more than one absolute point then there exist points p and q, both on l, such that f (l) must lie on both f (p) and f (q). We also know that p must be on f (p) and q is on f (q). So f (l) and p are both on f (p) but recall that they are also both on l . Additionally, f (l) and q are both on f (q) as well as l. These three points can not all lie on more than one line, so it must follow that f (p) = f (q) = l. But we know that f is 1-1 so p = q. But we assumed that p and q were distinct. From this contradiction, it follows that every absolute line of a polarity will contain exactly one absolute point.
It often proves very interesting to look at the sets of absolute points of a given polarity on a projective plane of order n. Let's look at the set of absolute points for the polarity of the Fano plane given above:
The absolute points are those in black and the absolute lines are the bold ones. One can see that the line on which the absolute points lie ( {0, 2, 4} ) is not an absolute line.
It is a true statement that if the order of the plane is even (in the case of the Fano plane, the order is 2), and a set contains n + 1 absolute points, then the set is a nonabsolute line.
If the order is odd and the set contains n + 1 elements, then the set is a nodegenerate conic section (i.e., one of the classical ovals in the plane)
Let's quickly give a definition of ovals before we show an example of an odd ordered plane:
Ovals in a projective plane
1. No three points of the set are collinear.
2. Every point of the set is contained in a unique tangent line, meaning a line that intersects the set in only one point.
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Let's look at a polarity on the projective plane of order 3.
Check for yourself that this is indeed a polarity. Now let's look at the set of absolute points:
One can see that the first axiom of an oval is easily satisfied. Now for the second. There must be a unique tangent line through each point of the set such that the line intersects the set in only one place. This is also easy to see.
If the order of the plane is a square, then then the set of absolute points contains n3/2+1 points. Let's look at the polarity for the projective plane of order 4.
Again, the colors correspond to the images under the mapping.
This set of abolute points, plus the set of nonabsolute lines form a unital.
Definition: A unital is a 2 - (n3+1, n +1, 1) design where n >1. Recall that this means the geometry contains n3 + 1 - many points, that every line contains n + 1 points and given 2 points there is exactly 1 line containing them.
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So in the case of the projective plane of order 4, we see that there are 42/3+1 = 9 absolute points. One can also see that for the unital we let n = 2. Thus we have a 2 - (9, 3, 1) design. Let's look at the set of absolute points:
It turns out that this unital is actually an affine plane of order 3. Here it is: (each line is colored differently to make it easier to differentiate between lines).
This looks a bit different from those we have already seen, but it really is isomorphic to any affine plane of order 3. Check for yourself.
As you can see, there are many interesting properties that apply to the set of absolute points for a given geometry.
Let's look at the set of polarities for the generalized quadrangle of order (2,2). Recall that we referred to this as the "doily."
Because we must preserve point-line incidence, we know that if p and q are on a line L, then line P and line Q intersect at point l.
In the case of the doily, every line contains 3 points and each point is incident with 3 lines. Thus, if points p, q and r are on a line L, then lines, P, Q and R intersect at point l.
Here is one polarity:
The figure on the right above shows the absolute points and lines of the polarity. The set of absolute points forms an ovoid: a set of points that has exactly one point in common with every line. In addtion, the absolute lines of the polarity create a spread: a partitioning of the point set. Here is one more polarity on the doily.
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PROBLEM 34: (a) Recall the example of the projective plane over the unit sphere from problem 15. Our point set of this projective plane consists of antipodal points and the line set is the set of great circles passing through these points. Show that the mapping of our points to the great circles, such that the image of a point corresponds to the circle that is the equator between the two antipodal points, is a polarity. Are there any absolute points?
(b) There are a total of six different polarities of the doily. Two are shown above. Find the other four.
(Hint: The ovoids of the other four polarities will look like the ovoid of either of the two polarities above. The spreads are different. Also remember that no two absolute lines can intersect at a point. )
Solutions
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Reading Project: Read chapters 6 and 7 from Polster.
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References: Batten, Polster
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