### duality in the 2-dimensional projective space over the field with 4 elements

new drawing today.

you can check my work…

as of course i already *have*:

start with any “cross”, C, and

verify on five cross-diagrams

(by shading one “circle” on each…

namely the one whose position

corresponds to that of C

in the “big picture”) that

the “Point-at-C” belongs

to each of the “Lines-through-C”.

i’ll edit in some algebra; probably soon.

but the point now is that i *don’t* need

to refer to any but purely *visual*

calculations.

somewhat to my surprise, i’ve decided

in middle age to become much more

of a “visual learner”. in my case, this

amounts to “how can i represent results

from abstract algebra as pictures?”.

it still counts.

March 25, 2011 at 5:34 pm

the real title is .

March 25, 2011 at 10:55 pm

as a set. its addition and multiplication

tables are

+01ab

001ab

110ba

aab01

bba10

*01ab

00000

101ab

a0ab1

b0b1a

.

consider

(ordered pairs [x,y] with

x, y \in {0,1,a,b}…). tack on

a “z co-ordinate” of 1 to each point

to obtain the so-called

homogenous co-ordinatesfor the “finite points” of P_2(F_4):

(a,a,1) (a,b,1) | (b,a,1) (b,b,1)

(a,0,1) (a,1,1) | (b,0,1) (b,1,1)

————————————-

(0,a,1) (0,b,1) | (1,a,1) (1,b,1)

(0,0,1) (0,1,1) | (1,0,1) (1,1,1)

the other five points… forming

“the line at infinity” are then

(1,0,0) [the center in my drawing]

(0,1,0) [the bottom]

(a,1,0) [left-hand]

(b,1,0) [right-hand] and

(1,1,0) [top].

“dualizing” can now be expressed by

considering *lines* having co-ordinates

[X,Y,Z].

a point (p,q,r) is then “on the line” [X,Y,Z]

if and only if Xp + Yq + Zr = 0

(as an element of F_4 [of course!]).

a^2 = b

b^2 = a

a+b+1 =0

ab = ba = 1

and suchlike equations come into play

over and over in calculating out the drawing.

after a while F_4 gets pretty familiar.