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How many lines are there in the projective plane of $ \mathbb{F}_n $ ?

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  • $\begingroup$ I think this may be a subset of this question. I probably shouldn't vote to close this as a dup, because I answered that one. $\endgroup$ – Jyrki Lahtonen Mar 18 '17 at 11:52
  • $\begingroup$ I will tell you how many lines there are if you tell me how many points I am giving you on your homework. Or else you can convince me you've made some progress and that you'd benefit from my help. Then I'd help you too. $\endgroup$ – rschwieb Mar 20 '17 at 3:56
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    $\begingroup$ @rschwieb There are $n^2+n+1$ points because you can see the projective plane as $\Bbb F_n^2$ and the line to infinity ($n+1$ points). Isn't it? But I'm not able to count lines $\endgroup$ – B.nia Mar 20 '17 at 21:35
  • $\begingroup$ OK, thanks for letting me see where you currently stand. I've added a solution. $\endgroup$ – rschwieb Mar 21 '17 at 14:25
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The question is actually simpler than you think, but I'm going to start out from the context you supplied in the comments (which was helpful for me to gauge the solution.) I'll present several ways to see the same conclusion.

Method 1

Here is a way to count the lines if you can count the points (as you have demonstrated in the comments.)

In the projective plane, each line has $n+1$ points. Of course, any choice of two distinct points specifies a line. There are $\binom{n^2+n+1}{2}$ ways to do that. Of course, that is not the number of lines: we are duplicating the same lines over and over. The question is: how many times are we duplicating the same line?

Well, there are $\binom{n+1}{2}$ different ways to choose two points from a single line, and every time we do so, we are duplicating previous work. We should divide out by this factor, and when we do we get this

$$ \binom{n^2+n+1}{2}/\binom{n+1}{2}\\=\frac{(n^2+n+1)(n^2+n)}{2}/\frac{(n+1)n}{2}\\=\frac{(n^2+n+1)(n^2+n)}{2}\frac{2}{n^2+n}\\= n^2+n+1 $$

Gee, that's odd... this says there are as many lines as points. Is there a reason that should be the case? This leads us to backtrack and think about

Method 2

One important fact about projective planes over fields (or even division rings) is that they are self-dual. This duality states that every statement about points, lines and incidence in the plane can be changed in this way:

  • Replace the word "line" with "point"
  • Replace the word "point" with "line"
  • Replace the phrase "lies on" (as in "point lies on a line") with "contains"
  • Replace the phrase "contains" (as in "line contains a point") with "lies on".

In our case the statement is only about points: "There are $n^2+n+1$ points." By duality the transformed statement is also true: "There are $n^2+n+1$ lines."

Method 3

In the comments, you mentioned you had counted points using the model of the projective plane as a completion of the affine plane $F_n^2$. Another important model for the projective plane over $F$ is this: the points of the projective plane over $F$ are $1$-dimensional subspaces of the vector space $F^3$, and the lines of the projective plane are planes in $F^3$.

Here it is easy to count $1$-dimensional subspaces. $F_n^3$ has $n^3-1$ nonzero vectors, each of which generate a $1$-dimensinoal subspace. However, nonzero scalar multiples of a vector generates the same subspace, repeating work. We are repeating ourselves $n-1$ times, then. So there are $(n^3-1)/(n-1)=n^2+n+1$ $1$-dimensional subspaces of $F_n^3$ (aka points of the projective plane over $F_n$.) Everything mentioned in Method 1 can be translated into a statement about counting planes in $F_n^3$, and so the same number is reached. This is not really that different from Method 1, but I thought you might find the viewpoint useful.

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    $\begingroup$ Good to have the duality explained on our site as well! $\endgroup$ – Jyrki Lahtonen Mar 21 '17 at 14:44
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    $\begingroup$ Thank you very much. I' not yet familiar with the duality and you help me to understand this important principle. Really usefull! $\endgroup$ – B.nia Mar 21 '17 at 17:48

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