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## ContinuityWe consider a pair of
projectively-identical, skew rulers, one red, and one blue, made equivalent
by projectivity via an intermediate, green ruler.Consider just one, black, skew line of the set of skews
that directly link end-points of intervals on the red ruler to end-points of intervals on
the blue ruler, and how –
– to move that skew. When we know how to do that for one, we can generalise
to the others.
strictly projectivelyfigure 1 A red line in red point M joins a red point, say R, in the red ruler, to a point,
say Q, in the green ruler. A blue line in blue point
N joins that point Q
to a blue point, say B, in the blue ruler. The join of the red point
R in
the red ruler to the blue point B in the blue ruler is our single, black, skew
line.A ll these joins, and the points they join, are coplanar, in,
say, plane γ. If we wish to move the skew line, we might “move” the green point Q in its green line by swinging plane γ around the line joining M to N,
or, instead, “move” Q to swing γ. (Since lines have only
extension, and hence do not have location or position [points have that], we
here refer to "stations", rather than positions, of a line.) The question now is whether or not we can
move point
projectivelyQ, or rotate plane γ,
smoothly and continuously.If we simply assume that smooth, continuous
motion of an element is available (as we well might when seeing that an element,
such as Q, above, apparently may be “dragged” continuously on-screen), we should
remind ourselves that this
is a peculiar Euclideo-Newtonian assumption, of a kind rejected from
the start by non-Euclidean, Projective Geometry.
The on-screen dragging is in any case not smooth, but occurs in discrete,
digital steps. The smoothness and continuity are illusory.Now, we have a perspectivity in
Q,
meaning that every skew interval, RB, bridging the rulers, is equivalent to the
interpunctual interval MN, and that, therefore, each is equivalent to every other.Also, two or more such intervals formed by the , are equivalent.
Bear in mind that each skew is just that - skew - so can never touch any other -
this means that two, successive "stations" of the same skew, moved, moved skew same.cannot be
incident with each other in any way at all.It must follow that the passage from one skew "station" to the next cannot be continuous, but must be discrete |
## The “Fundamental” TheoremThe theorem shows that
arbitrary, contiguous, interpunctual intervals
per line, on distinct lines, can be made equivalent by projectivity.
It does a maximum of two, say that intervals on the not line are equivalent to each other.
sameYou may control this animation by left-clicking the buttons below it.
animation
figure 2 Notice that the two lines, red and blue, here called m and n, can
be, but need not be, coplanar. So in every
respect of geometric principle, except for the fact that these lines do not necessarily bear identical rulers, and that their colours
happen to have been swapped, figure 2 matches figure 1 in the leftmost column.
- The red point
**1**of fig. 2 is point**H**of fig. 1: - the intermediate line of fig. 2 is the green line of fig. 1:
- point
**G**is the incidence of this intermediate line with the blue line,**m**.
If m and n are skew, then (here undrawn)
bridging lines red 1 to blue 1, red 2 to blue 2 and red 3 to blue 3 are skew to each other,
and can neither envelope a plane curve, nor lie in a surface.If m and n are coplanar,
all the bridging lines (the black lines here) are too, and so each must meet every
other in a point, one per pair of lines. They then envelope a
curve, as tangents, but these meeting-points can not lie in that curve. If their several meeting-points happen to merge/degenerate into a single point,
there will be no curve of any kind.mightIt is Important to realise that
there is
from either figure 1 or figure 2 whether no way
to tellm and n are in
objective fact
coplanar or skew. However, it is always possible to find the
of skews on a plane – and these shadows must,
perforce, be
coplanar!“shadows” |