-1
$\begingroup$

Please explain this https://math.stackexchange.com/a/463369/672948 in a simpler way. I am not from higher mathematics background and these terms are quite hard to understand.

I am clear upto finding $\mathbb{P}^2(F)$ ie.. $P_1=(0,0,1)$,... $P_{13}=(1,0,0)$. What does the below line mean?

equate, for example, the point $(2,1,2)$ with $2(2,1,2)=(1,2,1)=P_6$ and the point $(1,2,0)$ with the point $2(1,2,0)=(2,1,0)=P_{12}$ ?

How did they find this for a line ? $$ L_U=\{P_6,P_3,P_{13},P_9\}.$$

$\endgroup$
  • $\begingroup$ Welcome to stackexchange. In order to help you I think you need to edit the question to provide more context. How much not necessarily higher mathematics do you know? Linear algebra? Finite fields? Equivalence relations? It would also help us to know why you want to know - for an application, or just curiosity, or some other reasons. You may also find useful information at this wikipedia page, where your $13$ point plane is discussed: en.wikipedia.org/wiki/Projective_plane $\endgroup$ – Ethan Bolker May 11 at 12:17
1
$\begingroup$

The concept you need to understand here is a finite field with $p^n$ elements where $p$ is a prime number. In this particular case the finite field has a prime number of elements where $p=3$. The three elements can be written as $\ \{0,1,2\}\ $ with the understanding that arithmetic with these elements are done using Modular arithmetic with modulus $p=3$. In particular, for multiplication you need to know that $\ 2\cdot 2=1\ $ besides the obvious $$ 1\cdot 1 = 1,\;\; 1\cdot 2 = 2\cdot 1 = 2,\;\; 0\cdot x=x\cdot 0 =0. \tag{1}$$ For addition you need to know that $\ 1+1=2\ $ and

$$ 2+2=1,\;\; 2+1 = 1+2 = 0,\;\; 0+x=x+0=x. \tag{2} $$

Given the point $\ P_6 = (1,2,1),\ $ for example, you multiply each component by $\ 2\ $ to get an equivalent point. That is, $$ 2P_6 = 2(1,2,1) = (2\cdot 1,2\cdot 2,2\cdot 1) = (2,1,2) \tag{3} $$ is an equivalent way to express the same point. Notice that this is just the same as replacing all the $2$s with $1$s and $1$s with $2$s and leaving $0$s the same.

Now, we can construct a line given one point and a direction where the direction point has last coordinate $0$. For example, given the point $\ P_3 = (0,2,1)\ $ and the direction point $\ P_{13} = (1,0,0)\ $ we add multiples of the direction point to the original point. Thus, $\ P_3+P_{13} = (0,2,1)+(1,0,0) = (1,2,1) = P_6,\ $ then $\ P_3+2P_{13} = (0,2,1)+(2,0,0) = (2,2,1) = P_9\ $ and then $\ P_2+3P_{13} = P_3\ $ which we started with since $3=0$. The direction point $\ P_{13}\ $ is also considered the point of the line "at infinity", and so now this line has four points $\ \{P_3,P_6,P_9,P_{13}\}.$ The other $12$ lines are constructed in a similar way, each with one of the four different directions.

Note: The line "at infinity" $\{P_{10},P_{11},P_{12},P_{13}\}$ contains all of the direction points "at infinity".

NOTE: The sum of equivalent points is not usually equivalent to the sum of the original points. For example, $\ P_1+P_2 = (0,1,2) = 2P_3\ $ and $\ P_1+2P_2 = 2P_{10}.\ $ Here $\ P_2\ $ is equivalent to $\ 2P_2\ $ but $\ 2P_3\ $ is not equivalent to $\ 2P_{10}.$ However, it does give us the four points $\ \{P_1,P_2,P_3,P_{10}\}\ $ of another line.

$\endgroup$
  • $\begingroup$ If I take a point $\ P_{1}\ $ and $\ P_{2}\ $ as a direction point . Then $\ P_{1}\ $ + $\ P_{2}\ $ = $(0,1,2)$ = $\ P_{3}\ $. But then $\ P_{3}\ $ + $\ P_{2}\ $ is $(0,3,2)$ which equal $2(0,3,2) = (0,0,1)=$$\ P_{1}\ $ .So I get only 3 points...? $\endgroup$ – user672948 May 12 at 5:55
  • $\begingroup$ So how do I know which point and direction I need to consider for all 13 lines? $\endgroup$ – user672948 May 12 at 6:07
0
$\begingroup$

Well, the projective plane $P^2(K)$ over a field $K$ (say rational or real numbers in the examples below) consists of three subsets:

  1. $A^2(K) = \{(a:b:1)\mid a,b\in K\}$. This set corresponds to the affine plane $K^2=\{(a,b)\mid a,b\in K\}$.

  2. $L^2(K) = \{(a:1:0)\mid a\in K\}$. This is a line at infinity.

  3. $\{(1:0:0)\}$. This is a point at infinity.

In the first case, the last coordinate is normalized to 1. For instance, the point $(2:1:2)$ corresponds to $(1:1/2:1)$ by multiplying the first point by $1/2$.

In the second case, the last coordinate is $0$ and the 2nd coordinate is normalized to 1. For instance, the point $(2:3:0)$ corresponds to $(2/3:1:0)$ by multiplying the first point by $1/3$.

In the third case, the 2nd and 3rd coordinates are $0$ and the first coordinate is normalized to 1. For instance, the point $(4:0:0)$ corresponds to $(1:0:0)$ by multiplying the first point by $1/4$.

Hope it helps!

$\endgroup$
  • $\begingroup$ Yes but how did they arrive with line $ L_U=\{P_6,P_3,P_{13},P_9\}.$?What if I want all the lines for n = 3, means(total 3^2+3+1 = 13) lines? $\endgroup$ – user672948 May 11 at 14:05
  • $\begingroup$ The combinatorial design called projective plane is constructed differently. The link gives a construction of one of 13 lines comprising the combinatorial design. $\endgroup$ – Wuestenfux May 11 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy