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I have a question about constructing projective plane over $\mathbb{F}_3$. We first establish seven equivalence classes $P= \{ [0,0,1], [0,1,0], [1,0,0], [0,1,1], [1,1,0], [1,0,1], [1,1,1] \}$.

Given a triple $(a_0, a_1, a_2) \in \mathbb{F}^3_3 \setminus (0, 0, 0)$ we define the line $L(a_0, a_1, a_2)$ as follows: $L(a_0, a_1, a_2) := \{ [x_0; x_1; x_2] \in P : a_0x_0 + a_1x_1 + a_2x_2 = 0 \}$.

It is quite easy for $L(0,0,1), L(0,1,0), L(1,0,0)$, because here we take the points which have zero on the first, second and third coordinate.

It gets more difficult for me for $L(0,1,1)$, because here we need to have $x_1+x_2=0$. Do we treat the coordinates of the points as elements of $\mathbb{F}_2$ or $\mathbb{F}_3$?

There are $26$ nonzero triples in $\mathbb{F}^3_3 $.

Do we check all 26 $L(x_0, x_1, x_2) $ sets?

Please help, because I really want to understand it.

Thank you.

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2 Answers 2

up vote 6 down vote accepted

You are describing a method for getting the Fano plane, which by definition is the projective plane over $\mathbb F_2$.

For getting the projective plane over $\mathbb F_3$ in a similar way, do the following:

Select a set of projective representatives of the nonzero elements of $\mathbb F_3^3$. One way to do this is to select only the vectors whose first nonzero entry is a $1$. Since $\mathbb F_3$ has $2$ units, you end up with $(3^3 - 1) / 2 = 13$ vectors. Those vectors are called coordinate vectors and give you the $13$ points of the projective plane.

You can also use the coordinate vectors for the description of the lines: Each coordinate vector $v$ corresponds to the line containing all the points $w$ such that the scalar product of $v$ and $w$ equals $0$ (so $\langle v,w\rangle = 0$). In this way, there are $13$ lines containing $4$ points each.

Typically, the projective plane over a field $K$ is defined in this way:

The subspaces of $K^3$ of dimension $1$ are the points, and the subspaces of $K^3$ of dimension $2$ are the lines. It may be worth to convince yourself that the above construction is compatible with this description.

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What does first nonzero entry mean? There are 9 vectors whose first entry is 1, so it must mean something else. –  Sandy May 26 '13 at 9:07
The first nonzero entry of $(0,1,0)$ is $1$, too! –  azimut May 26 '13 at 9:08
You should understand that finite projective planes are primarily combinatorial objects. Their main feature is: For each pair of distinct points, there is exactly one line connecting those two points. Of course it is nice to visualize those planes, but it gets harder as he field size increases. The plane over $\mathbb F_3$ can still be visualized in a reasonable way, see geometrie.tuwien.ac.at/havlicek/img/pg23_240x240.gif for a common way to do it. –  azimut May 26 '13 at 9:31
Yes, exactly. The projective plane over $\mathbb{F}_4$ hast 21 points and 21 lines. –  azimut May 26 '13 at 12:23
In general, in the projective plane over $\mathbb F_q$, each line contains $q+1$ points, and through each point there are $q+1$ lines. The number of points and lines is $(q^3 - 1)/(q-1) = q^2 + q + 1$. Google brought up this picture of the plane over $\mathbb F_4$: maa.org/editorial/mathgames/21lines.gif You see, this is already quite involved. –  azimut May 26 '13 at 13:15

The points and lines are not triples in $\mathbb {F}^3_3$, but they are elements of $\mathbb{F}^3_2$. So the coordinates of points and lines are in $\mathbb{F}_2$.

For example, your line $L(0,1,1)$ contains the points $[x_0,x_1,x_2]$ with $x_1+x_2=0$. Thus $x_1=x_2=1$ or $x_1=x_2=0$. So the possibilities are: $[0,1,1]$ , $[1,1,1]$ , $[1,0,0]$.

Edit: First the question was about the Fano plane, I provided my answer in that case.

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