i saw the figure 5 in gold


yet another duality diagram.

until now, this post from 2010
had my favorite such: “duality in
the fano plane”; seven “lines” through
seven “points”.

this one is a much-more-symmetric
picture of the situation represented
in this early version: ten “lines”
through ten points.

it recently occurred to me that
the “desargues configuration”
(aka “the situation represented here”)
could, in some sense, *best* be
visualized as the (set of) *opposite
faces* of an icosahedron.

but before i’d even made a decent
picture of that (hard-to-draw because
3-dimensional), along came this idea.
how it escaped me till now, i’ll never know.

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  1. Please explain this. It’s pretty, and I’d like to understand what it represents.

  2. so. draw a big pentagram (“star”).
    label the five lines (in any
    order you like): 1, 2, 3, 4, 5.

    (there are 120 [=5!] ways to
    label the lines; every fifth-grader
    should understand this but alas
    most college freshmen don’t
    and a great many never will.)

    the intersections… the *vertices* of the
    big pentagram… are now conveniently
    labeled with *two* “co-ordinates”:
    namely, the numbers associated to
    the lines intersecting there.
    {
    15, 14, 13, 12,
    25, 24, 23,
    35, 34,
    45
    }, say.

    there are 10 = \choose(5, 2)
    ( = 5 / [2! 3!] ) such co-ordinate
    pairs; one easily checks that
    there are indeed 10 vertices
    for the Big Star (the 5 “points”
    of the star and the 5 forming
    the inner pentagon).

    each point is then associated
    in the drawing with a certain
    “triangle”: three *other* points
    of the Star.

    for instance… *regardless of how
    the labels 1, 2, 3, 4, 5 are chosen*…
    the point labeled 12
    is associated with the triangle
    whose points are {34, 35, 45}.

    this phenomenon arises “because”
    the complement of a two-set
    chosen from {1,2,3,4,5}
    is a three-set: the triangle associated
    with a given vertex (a two-set)
    will be all the vertices whose
    co-ordinates do *not* include
    either co-ordinate of the given vertex.

    (and it’s this “complementation” computation
    that allows me to assert that the labelings
    from {1,2,3,4,5} are irrelevant in making the
    pairing between vertices and triangles;
    the complement of {1,2} is {3,4,5} *regardless*
    of where these objects occur on some diagram.)

    i hope this is easily understood thus far.
    now comes the “duality”… the pet concept
    of this whole blog for a couple years.

    consider {34, 35, 45} again
    (the triangle “dual to” to point 12).

    look at the duals of its vertices:
    34… {12, 15, 25}
    35… {12, 14, 24}
    45… {12, 13, 23}.

    ah-*ha*!
    the *intersection* of these is (gee!)
    12 again… the “starting point”.

    and so for any starting point:
    “my dual is the triangle whose
    points have duals that include
    me as one corner”.

    the Vertices and Triangles of the diagram
    at hand become Points and Lines from
    another point of view (where “the same”
    duality is maintained); for instance here:
    https://vlorbik.wordpress.com/2011/07/06/executive-summary/
    .
    someday soon i’ve gotta get all this
    organizized.




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