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"Foundations of Projective Geometry" by Hartshorne says the following:

The completion of the affine plane of four points is a projective plane with 7 points.

The affine plane of $4$ points is essentially a paralellogram $ABCD$. The completion will contain $A,B,C,D,[AB],[AD],[AC],[BD]$. Here $[AC]$ is the point of intersection of all lines parallel to $AC$ with the line at infinity (in other words it is an ideal point).

Hence I am getting $8$ points instead of $7$. Where am I going wrong?

  • I think you only have 6 points here: $[AB]$ and $[CD]$ are the same point, because the lines $AB$ and $CD$ are parallel, and similarly $[AC]$ and $[BD]$ are the same. On the other hand, you don't have $[AD]$, so that gets you back up to 7. –  Jul 02 '14 at 12:10
  • @AsalBeagDubh- Sorry I meant something else. I have edited the question. ABCD taken counter-clockwise. –  Jul 02 '14 at 12:23
  • OK, now you have all the points. As I mentioned in my previous comment, two of the items in your list are the same: $[AC]=[BD]$. Do you see why? –  Jul 02 '14 at 12:25
  • @AsalBeagDubh- Which two? $[AC]$ and $[BD]$ are not the same! They are not co-incident or parallel. –  Jul 02 '14 at 12:27
  • They are parallel. Don't be deceived by a picture. –  Jul 02 '14 at 12:27
  • @AsalBeagDubh- According to my diagram, they are the two diagonals of the parallelogram, which intersect at a point. Am I missing something? –  Jul 02 '14 at 12:28
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    The basic point is this: there is no reason the picture you draw (a parallelogram in the plane) should accurately reflect the true properties of the affine plane (which is an abstract system). Perhaps the following question might be suggestive: you say these two lines intersect in a point. Which of the four points of the plane is it? –  Jul 02 '14 at 12:33
  • @AsalBeagDubh- I think I have some idea where you're going through this. If $AC$ and $BD$ were not parallel, then there would be a fifth point on the plane, which is a contradiction. Is it sometimes impossible to draw geometric representations of the Affine or Projective planes? –  Jul 02 '14 at 13:05
  • Yes, that's what I was getting at: the "obvious" picture makes it look like these lines intersect, but at a point that is different from all of $A$,$B$,$C$, and $D$. But there is no such point! (And in fact you can check rigorously, using the axioms, that they do not intersect at the points $A$, $B$, $C$ and $D$ either.) As for pictures: pictures in the plane can still be useful, perhaps, as long as you understand what they mean --- e.g. in this case, that the "apparent" intersection of $AB$ and $CD$ isn't "really there" in your affine plane. In fact, there is a very famous picture of... –  Jul 02 '14 at 13:12
  • the "completion" you ask about in your question. This is usually called the Fano plane: http://en.wikipedia.org/wiki/Fano_plane. –  Jul 02 '14 at 13:14

1 Answers1

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To paraphrase what has been said so that the question is answered:

There are only three distinct pencils of parallel lines: $\{\overleftrightarrow{AB}, \overleftrightarrow{CD}\}$, $\{\overleftrightarrow{AC}, \overleftrightarrow{BD}\}$ and $\{\overleftrightarrow{AD}, \overleftrightarrow{BC}\}$. These give rise to three ideal points in the projective completion.

The mistake made was thinking $\overleftrightarrow{AC}$ was not parallel to $\overleftrightarrow{BD}$ apparently because of a visual illusion of intersecting diagonals while modeling the geometry as the corners of square.

We can also count everything in another way. Since each line has two points, the underlying field is $F_2$. The projective plane then corresponds to one dimensional subspaces of $\mathbb F_2^3$, of which there are $8-1$ (there are seven distinct generators of $1$ dimensional subspaces, since $0$ will not work.)

rschwieb
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