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.

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  1. {\Bbb F}_4 = \{0, 1, a, b\}
    as a set. its addition and multiplication
    tables are

    +01ab
    001ab
    110ba
    aab01
    bba10

    *01ab
    00000
    101ab
    a0ab1
    b0b1a
    .

    consider F_4 \times F_4
    (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-ordinates
    for 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.




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