From: "steve degroof" <steve.degroof@canrem.com>
Subject: mode & dimensions
Reply-To: "steve degroof" <steve.degroof@canrem.com>
Organization: Canada Remote Systems
Distribution: alt
Date: 17 Nov 92 22:28:44 EST

Pay attention. There will be a short quiz following this
lecture. :-)
> Just a short question on the need for a seven-dimensional
> universe for the anchor worlds to be different from one
> another.  I think you mean 3 dimen- sional anchor universes.
> If a given universe consisting of three spatial dimensions
> can be specified by a time co-ordinate and an n-dimensional
> universe vector, then for five anchors we need four
> dimensions to express the universe if it is to be
> equidistant from the other anchor universes (is this not the
> same number of dimensions needed for no three of the anchor
> universes to be collinear).  Thus, we have 3 spatial
> dimensions, three universe dimensions, and a time dimension,
> for a total of seven dimensions.
>
>> No, if one draws a pentagon(in two dimensions) then no three
>> points are collinear.  Four is needed to make them
>> equidistant.
Exactly. For an object to have 5 equidistant vertices, it
must be in a coordinate system with *at least* 4 dimensions.
There could be more, but there can't be fewer. Such an
object is bounded by 5 tetrahedra, each of which is bounded
by 4 triangles, each of which is bounded by 3 lines, each of
which is bounded by 2 points.

> The thing is though, when discussing things in more than two
> or three dimensions, what does pentagonal mean.  If you mean
> a literal pentagon, as drawn, then eliminating an anchor and
> replacing it would be difficult since four points of the
> pentagon would be drawn correctly and the fifth would have
> to be out of place unless we distort the shape of the
> universe. That's why I think that the shape of the virtual
> mode is that of a 5-cell (that's the name of the
> 4-dimensional shape made by 5 equidistant points). If one
> removes one of the corners of a 5-cell then one can still
> find a fifth point that is equidistant from the other 4.
> (Try thinking of removing one corner from an equilateral
> triangle.  you can still find a different point opposite the
> first corner that is equidistant from the other two points)
Right, assuming the triangle is in 2-space, the number of
possible vertices is exactly 2. This is always true when the
number of vertices in the object is 1 more than the number
of dimensions. If, however, the numer of vertices and the
number of dimensions are equal, then the number of alternate
vertices is infinite. Think of an equilateral triangle in
3-space. If you remove one vertex, you can choose from an
infinite number of alternate vertices which are also
equidistant to the two remaining points. Those points, by
the way, always descibe a circle. Same goes for a
line in 2-space and a tetrahedron in 4-space. So, the
minimum number of (space) dimensions required if you want an
infinite number of alternate anchor points is 5. There is no
maximum.

Now, about the anchor points being in different worlds:
Assuming the number of (space) dimensions which defines a
world is 3, then the number of anchor points which can be in
the same world is in the range 0 to 4. There's no way that
all five can be in the same world and still be equidistant.
Only three can be on the surface of the same planet; if a
fourth exists in the same world, it's either in the ground
or in the air (4 points make a tetrahedron).

Now, assume all the planets in all the worlds are the same
size and in the same 3-space coordinates. If you want all
the anchor points to be on the surface of a planet, you're
going to need at least 6 dimensions. With 5 dimensions, at
least one anchor point has to be off the surface.

OK, that's 6. Now, if you want any point on the lines
connecting the anchor points to be on the surface of the
planet, you must the ensure that each line runs
perpendicular to 3-space. Otherwise, the lines will cut
chords through the planets. That means a minimum of 7
(space) dimensions. With 7, each anchor point must be at the
same 3-space coordinates (the same geographical location).
According to the book, this isn't the case.

I don't know if 8 dimensions solves this last problem; my
brain blew a fuse at that point.

Then again, it doesn't really matter; it's only a story,
right? ;-)

SD
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