This post is thought to be an answer to a comment by Bjab from the previous post.
I do not know who has discovered the
stereographic projection and when? I don't know, But it was certainly
a great discovery. If we are to believe Wikipedia:
“The stereographic projection was
known to Hipparchus, Ptolemy and probably earlier to the Egyptians.
It was originally known as the planisphere projection.[2]
Planisphaerium by Ptolemy is the oldest surviving document that
describes it. One of its most important uses was the representation
of celestial charts.[2] The term planisphere is still used to refer
to such charts.”
Here is the picture
Here we embed the real axis with the
coordinate x into the unit circle. The point P' on the x-axis is
projected, here from the north pole N, onto the point P on the circle. If we denote by X the
coordinate of P' and by x,y the two coordinates of P, we have the
following transformation formulae (taken from Wikipedia):
We are going to make one change here:
we do not like the north pole! It is on the ocean. We prefer to stay
on the land, therefore we will choose the south pole as the
projection origin. The formulae will be almost the same, only the
sign of the z-axis with change into the opposite. Thus we will use
(X,Y) = (x/(1+z), y/(1+z))
(x,y,z) = ( 2X/(1+X2+Y2),
2Y/(1+X2+Y2), (1-X2-Y2)/(1+X2+Y2)
)
Notice that th image of the whole plane
is the sphere minus the south pole. The south pole, z=-1, escapes to
“infinity” on the plane. Thus the sphere is the plane plus
one-point – the infinity. It is called a one-point
compactification of the Cartesian plane.
Now nothing prevents us from
considering a one point compactification of the three-space with
coordinates X,Y,Z. We simply need to add on dimension. Thus we
consider for-dimensional space with coordinates x,y,z,v and use a
straightforward extension of the above formulae:
(X,Y,Z) = (x/(1+v), y/(1+v), z/(1+v))
(x,y,z,v) = ( 2X/(1+X2+Y2+Z2),
2Y/(1+X2+Y2+Z2), 2Z/(1+X2+Y2+Z2),
(1-X2-Y2-Z2)/(1+X2+Y2+Z2))
We may be having problems with imagining the three-sphere in four dimensions, but the algebra is as simple as before. Algebra rules!
We
are able now to construct a one-point compactification of Rn
for any n. It will be a unit sphere in Rn+1. We
simply add one coordinate and use a straightforward generalization of
the formulae above.
But:
it is not yet the end of the story. For our story to have a happy
end, as any good story should, we have to go through yet another adventure.
Let
us go back to only one coordinate x. The line becomes a circle. But
there, spontaenously, comes the observation that a circle comes from
an intersection of the cone with the plane, as on the graphics below:
Do
not pay attention to any detail on the picture (I have borrowed it
from a random paper
on Researchgate), except that the circle is created by the
intersection of the light cone with a constant t=1 plane.
Guided
by this new brave idea we add an extra variable playing the role of
"time". But since it is just an extra variable, not a real
"time", we call it w, and we will set w=-1. Why "-1"
instead of 1? Well both are good, but to be in agreement with a
certain number of conventions used in the literature, lest us agree
for -1. Ok?
We
also simplify our notation to easily cover a general case. So we will
write X for the vector with coordinates X1,..., Xn,
and we write X.X or X2 for the sum of squares of the
coordinates X1,..., Xn . Similarly we write x
for the vector with coordinates x1,..., xn . Our
formula for the embedding reads now
(x,v,w)
= ( 2X/(1+X2), (1-X2)/(1+X2), -1 )
The
sum of squares of the first n+1 coordinates, x2+v2
is now automatically equal 1 (it can be verified independently
though), so the point (x,v,w) is on the surface of the cone
x2+v2-w2
= 0.
The
coordinate w=1 intersects this cone, the intersection is the sphere
Sn.
We
are not quite happy yet. In the formula above we have denominators
(1+X2), and we do not like them, even if here they are
doing no harm. Our cone is made of generator lines. These are
straight lines from the origin, along the cone. It is these lines
that are important, not the particular intersection. Thus we replace
the formula above by multiplying all coordinates on the right by the
1/2 of the common denominator for n+1 first coordinates. We end up
with a new embedding formula:
(x,v,w)
= ( X, ½ (1-X2), -½ (1+X2) )
We
still have a point on the cone x2+v2-w2
= 0. The mapping above is one to-one. We will look at it as a mapping
from Rn to generator lines of the cone in Rn+2.
Soon we will specify n that is of interest for us to be n=4, thus the total space Rn+2 will be (4+2=6) six-dimensional.
To
feel more "at home" with our formulas let us do a little
exercise. Namely, let us take a generator line on the cone x2+v2-w2
= 0 and find the point X to which it corresponds. In the future
instead of "generator line" we will simply use the term
"line". Thus we should have
(x,v,w)
= a ( X, ½ (1-X2), -½ (1+X2) ),
where
a is a proportionality factor, telling us that (x,v,w) and a ( X, ½
(1-X2), -½ (1+X2) ) are on the same line. So,
we should have
x =
aX,
v =
a(1-X2)/2,
w = -a(1+X2)/2.
Here
x,v,w are given, and we want to calculate a,X. Subtracting the two
last equations we find
v-w =
a.
Thus,
as long as v≠ w we have a≠ 0 and from the first equation
X=x/(v-w).
When v=w, and we are on the cone x2+v2-w2
= 0, x must be 0. This is one generator line (0,v,v), v∈R
.So
far being brave did not lead us to any trouble. So now we dare to be
even more brave. First of all we specify n=4. By X we mean a vector
with coordinates X1,X2,X3,X4.
But now we will think of X4 as "time coordinate".
Therefore by X2 we will now mean
X2
= (X1)2 + (X2)2 + (X3)2
- (X4)2.
Notice
that we have changed the coordinate indices from upper to lower to
avoid the confusion about what X2 means.
So,
we will keep the formula:
(x,v,w)
= ( X, ½ (1-X2), -½ (1+X2) )
but
with the above understanding what X and X2 mean.
Altogether
we thus have the signature (+++-) for spacetime, + for v, and – for
w. The total signature of the six-dimensional space is (4,2), more
specifically (+++-+-).
We will check now carefully if we got into some trouble this way?
Let
us analyze the map (*) X ↦ (x,v,w) from M to V. We notice that it
is one-to-one, we have an injection. Indeed if X≠X' then the images
are also different, simply because, as it is evident form (*), x=X,
and it is sufficient for just one coordinate of two points to be
different for these two points to be different. Moreover, (x,v,w) and
(x',v',w') are certainly not on the same line.
But
it is not surjective. Let us find the set on which the inverse map is
not defined. In the purely Euclidean case it was just one point, and
now what are we going to get? To answer this question we follow the
previous method and try to express X in terms of (x,v,w). Thus we
write again
x =
aX,
v =
a(1-X2)/2,
w = -a(1+X2)/2.
And
try to solve for a and X in terms of x,v,w, assuming x2+v2-w2
=0. As before v-w=a, therefore as long as v≠ w, we have a≠ 0 and
X=x/(v-w).
The
"infinity" now is defined as before by v=w. Before it was
just one point (one generator line). But now? If v=w, from x2+v2-w2
=0 we deduce x2 =
0. We now need to find algebraic equations describing our (projective) manifold.
But
now x2 =
(x1)2
+ (x2)2
+ (x3)2
- (x4)2,
and we cannot deduce that x=0 in the Euclidean case. We have a
3-dimensional surface. Let us analyze this surface remembering that
we are dealing with lines. We are on the quadric in described in
coordinates x,v,w by the formula
x2+v2-w2
= 0
In
projective geometry these are called homogeneous coordinates. Writing
explicitly:
(x1)2
+ (x2)2
+ (x3)2
- (x4)2
+ v2
- w2
= 0
or
(x1)2
+ (x2)2
+ (x3)2
+ v2
= (x4)2
+ w2
= a
The
common value a above cannot vanish, since if it vanishes, all x,v,w
would be zero, and the origin is excluded, as it does not define any
line. Thus a>0. We can therefore replace x,v,w by (x,v,w)/√ a ,
and get a pair of equations
(x1)2
+ (x2)2
+ (x3)2
+ v2
= 1
(x4)2
+ w2
= 1
Now
we set w=v (our "infinity"):
(x1)2
+ (x2)2
+ (x3)2
+ v2
= 1
(x4)2
+ v2
= 1
We
have two equations for five variables. Thus our "conformal infinity" now is not a
point, it is a three-dimensional surface in a five-dimensional space
of coordinates (x,v). We will explain why this term "conformal" in the coming posts. The point is that the conformal transformations which are singular in space time, are nicely realized by linear transformations from the group SO(4,2) acting in our 6D space of variables x,u,v.
Skipping one space dimension, say x3,
we get a two-dimensional surface in a four-dimensional space
(x1)2
+ (x2)2
+ v2
= 1
(x4)2
+ v2
= 1
Removing
also the variable x2
we have the
intersection of two cylindrical surfaces in 3D:
(x1)2
+ v2
= 1
(x4)2
+ v2
= 1
And
these were our equations form the previous two posts, though the
names of the coordinates were different.
P.S.1. If we were to take care about the physical dimensions, we would have to replace "1" in all the above equations by some constant R (or R squared, as the case may be) - where R is the "radius of the universe". I. Segal et co used this finite number R to avoid the need for renormalization of quantum field theory at low energies.
P.S.2. Why couldn't we take v=w and x^2=0 as the solution? The problem is that if v=w is different from 0, then x=0 belongs to the solution. While if v=w=0, we have to exclude x=0. This is not a simple algebraic equation. We have to proceed differently.
P.S.3. There is still one problem with our solution. If (x,v,w) satisfies our equations, then (-x,-v,-w) also satisfies them. But these two points are on the same line, so they should be identified. In other words: we have to take the quotient of our solution set by an equivalence relation. We will discuss it later when we will be talking about topology and differential structure on the compactified Minkowski space.
