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Doing mathematics
means above all proving what one claims.
We have seen
that the stereographic projection
sends circles on the sphere
not going through the pole
to circles in the plane,
and now, we are going to prove it.
Even though this has been known for many centuries,
it is I, Bernhard Riemann
who will present this proof to you.
I am frequently honored since
one speaks today of the Riemann sphere.
Proving is much more than showing.
It is not enough to see in a movie
that some curve looks like a circle
to be sure that it is indeed a circle.
A mathematical proof must use reasoning
to be convincing and has to explain
why it is indeed a circle.
The great Euclid,
during the third century before Christ,
formulated the rules of the mathematical game,
in his book "The Elements".
A proof has to rely on facts
that, themselves, have to be proved.
But, one has to start with something
so that some statements have to be accepted without proof:
these are the axioms.
Therefore, mathematics appears as
a gigantic construction
whose foundations consists of the axioms
and such that each brick rests on the previous one.
In order to prove the theorem about the stereographic projection of circles,
we should in principle start with the axioms!
Of course, we have no time for that now...
We will assume that we already know the theorems of geometry
which are studied, say, in secondary school,
and we will prove this theorem.
Start with something simple,
the intersection of a sphere and a plane.
We see that if a plane cuts a sphere,
and if it is not tangent to the sphere,
then the intersection is a circle.
We can see it
but why is it true ?
How do we prove it?
So, let's consider an arbitrary plane, colored in blue.
We can draw the perpendicular
from the center C of the sphere to the plane.
Let's call P the foot of this perpendicular.
Consider two points A and B
on the intersection of the sphere and the plane
and let's look at the two triangles CPA and CPB.
They share a common side : CP.
Both have a right angle
since the angle at P is of course a right angle.
since the plane is perpendicular to CP.
But note that the hypotenuses AC and BC have the same length
because A and B are on the sphere
and are hence at the same distance from the center C.
But recall Pythagoras’ theorem!
Since our two right triangles
have two sides of the same length,
their three sides have the same length!
Hence, we proved that
PA and PB have the same length
that is, that A and B are
on the same circle with center P,
in the blue plane.
Therefore, we proved that
all points which are
both on the sphere and the plane
belong to some circle.
Does that imply that
all points on this circle
are on the sphere and on the plane?
A priori no ! We still have
to prove it !
Let A be a point which is in the sphere and the plane.
Consider the circle in the blue plane
with center P and that goes through A.
We will prove that this circle
is contained in the sphere
Let B be some point on that circle.
Look at the two triangles CPA and CPB.
They share a side : CP.
Both are right angle triangles.
since the angle at P is a right angle.
but the lengths of PA and PB are equal
since A and B are on the same circle with center P.
Again using Pythagoras’ theorem,
we conclude that the hypotenuses
have the same lengths.
CA equals CB.
This means that the point B
also lies on the sphere since
it is at the same distance from C as A.
That's it ! We proved that
when a plane cuts a sphere
the cross-section is a circle.
Let's look now at a diameter APB of our circle
and let's place it in the plane of the screen.
The blue plane appears as a straight line on the screen
and the sphere appears as a circle.
Let's draw the tangents to the circle at A and B.
They intersect in a point S.
Of course, the line CS is again
a symmetry axis for our figure.
Why ?
Well, because the triangles CAS and CBS are equal!
Why ? eh... because
they are both right-angles triangles
having a common hypotenuse
and the sides CA and CB have the same length!
Why?
Well, because these are two radii, of course !
You see,
if we had to go to the end of all the arguments,
this movie would be the longest in the history of the cinema.
Look!
We proved that any circle drawn on a sphere
can always be thought of
as the contact locus between a cone of revolution
and a tangent sphere.
If you like, the sphere is like ice cream
in a cone
Well, we should not
forget what our aim is!
To show that the stereographic projection
carries circles on circles!
Let's first prove
what mathematicians
call a lemma:
Here is the tangent plane to the sphere
at some point A, seen from the side.
Now, here is the tangent plane
at some other point B, also seen from the side.
These two planes intersect on a line d,
but presently we only see one point
since this line is perpendicular to the screen.
The figure that you are looking at
is symmetric with respect to
the bisecting line of the two lines that we see.
This 3 dimensional picture is symmetric
with respect to the bisecting plane of the two tangent planes.
Choose some plane containing the segment AB
It intersects the line d in a point M
unless it is parallel to d, of course.
The symmetry of the figure with respect to the bisecting plane
shows that AM and BM have the same length.
The triangle ABM is isosceles!
Here it is! That was our lemma!
Well, now we can prove our theorem,
using what we just learnt...
Consider a circle on the sphere which does not go through the north pole.
We want to show that its projection is a circle.
Look ! If instead of projecting onto the tangent plane to the south pole
we projected onto some other parallel plane
the famous theorem of Thales
would imply
that all the projections are similar.
Hence, in order to prove our theorem,
we may choose the projection plane
as we wish
(of course as long as it is parallel to the tangent plane to the south pole).
Well! Let's place our yellow circle in a cone!
Remember?
the ball of ice cream
in a cone with vertex S.
Well, we are going to project
onto the horizontal plane through S.
The point B projects on to a point D.
But... look at the figure!
The triangles AMB and DSB are similar!
Why?
Well, again Thales' theorem! Do you agree?
Now, remember our lemma!
The triangle ABM is isosceles.
Hence the same is true for the triangle BDS
so that BS
has the same length as DS.
When B moves along the yellow circle
the segment BS keeps tangent to the sphere.
Its length is therefore constant.
Since BS and DS have the same length,
the moving segment DS
also retains a constant length.
Let's see,
saying that DS has a constant length
means precisely that D
describes a circle with center S
so that the projection of our yellow circle
on the horizontal plane through S
is contained in a circle.
We have seen that
this implies by Thales' theorem
that the projection on the tangent plane to the south pole
is also contained in a circle!
And... this is what we wanted to prove!!!