<tsao.doris@gmail.com> wrote in message news:1139303656.700012.130780@z14g2000cwz.googlegroups.com...

Hi,

I've finally started reading the Road to Reality. It's delightful, but
I find I'm having difficulties with many of the problems, of course w/
the hard ones, but also with many of the easy and medium ones. Is
there anywhere on the web where solutions are/can be posted by
interested readers. For a start, I scratched my head for hours the
other night trying to figure out why the projective map of a circle on
a sphere necessarily yields a circle on a disk, and why this mapping
preserves angles. I'm sure I could solve it w/ plug and chug using
coordinates, but Penrose's hint suggests a simpler, more elegant
solution. Anyone know the answer?

thanks,
doris

[added newsgroup sci.math]

It's a good thing that Penrose hasn't published the solutions yet.
That makes us work :-)
I have chosen to go analytic.

1) Preservation of circles:

Since a circle on the sphere must lie in some plane, an arbitrary
point (x,y,z) on the circle must satisfy the equations:
x^2 + y^2 + z^2 = 1
a x + b y + c z = d
for some suitable numbers a, b, c, d

Supposing the projected point has coordinates (X,Y,Z), we can
express the fact that (X,Y,Z) must lie on the line through
(x,y,z) and (0,0,-1) and on the equatorial plane Z = 0, by
stating that there is a number k such that
(X,Y,0) = (0,0,-1) + k [ (x,y,z) - (0,0,-1) ]
which gives, when solved for (x,y,z):
( x, y, z ) = ( X/k, Y/k, 1/k-1 ) .

So we know that the coordinates (X,Y) must satisfy the equations
X^2/k^2 + Y^2/k^2 + (1/k-1)^2 = 1
a X/k + b Y/k + c (1/k-1) = d .

Solving the first equation for k gives
k = 1/2 (X^2 + Y^2 + 1)
which gives when inserted into the second and rearranged
(c+d) X^2 + (c+d) Y^2 - 2 a X - 2 b Y - c + d = 0
which describes a circle in the X-Y plane.

2) To prove the preservation of angles, note that angles between lines
on the sphere are defined by great circles, and therefore by planes
through the origin (0,0,0).
So we can imagine two planes
a x + b y + c z = 0
a' x + b' y + c' z = 0
where we have normalized coefficients
a^2 + b^2 + c^2 = 1
a'^2 + b'^2 + c'^2 = 1
The dot-product of the normal unit vectors to the planes is then
given by
D1 = a a' + b b' + c c'

As we have seen before, these circles are projected to circles in
the X-Y plane:
c X^2 + c Y^2 - 2 a X - 2 b Y - c = 0
c' X^2 + c' Y^2 - 2 a' X - 2 b' Y - c' = 0
which, provided c # 0 and c' # 0, can be rearranged to
(X-a/c)^2 + (Y-b/c)^2 = 1/c^2
(X-a'/c')^2 + (Y-b'/c')^2 = 1/c'^2
which are circles with resp. centres (a/c,b/c) and (a'/c',b'/c') and radii
1/c and 1/c'.
The dot-product of the tangent unit vectors to the circles in a common
point is given by the dot-product of the normal (radial) unit vectors in
that common point. Supposing that (p,q) is a common point, we have
the normalized direction vectors
( (p-a/c)/(1/c), (q-b/c)/(1/c) )
( (p-a'/c')/(1/c'), (q-b'/c')/(1/c') )
or, shorter:
( c p - a, c q - b )
( c' p - a', c' q - b' ).
This result is also valid for the cases c=0 or c'=0 in which case a
circle degenerates to a straight line.
The dot-product of these vectors is
D2 = ( c p - a ) ( c' p - a' ) + ( c q - b ) ( c' q - b' )
which gives
D2 = c c' p^2 -(a c' + a' c) p + a a' + c c' q^2 - (b c' + b' c) q + b b' .

Since the point (p,q) is on both circles, we also have
(c p-a)^2 + (c q-b)^2 = 1
(c' p-a')^2 + (c' q-b')^2 = 1 ,
which can be written as
c p^2 - 2 a p + c q^2 - 2 b q - c = 0
c' p^2 - 2 a' p + c' q^2 - 2 b' q - c' = 0
which give after multiplying with c' resp c, adding and dividing by 2:
c c' p^2 -(a c' + a' c) p + c c' q^2 - (b c' + b' c) q = c c'.

Inserting the LHS of this equation into the expression for D2 gives
D2 = a a' + b b' + c c'
which proves that the dot-product and therefore the angle is preserved:
D1 = D2

I'm sure that the geometric proofs will be more elegant :-)

Dirk Vdm