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| I am kind of a Platonist. But I'm not the only one! |
We continue from Part 5. Let us first recall the definitions.
We consider a 6-dimensional real vector space V with a scalar product of signature (4,2). We select an orthonormal basis in this space, ei (i=0,...,5) , so that
(e0,e0)=(e1,e1,)=(e2,e2)=(e3,e3)=1, (e4,e4)=(e5,e5)= -1.
with vectors ei being orthogonal to each other:
(ei,ej) = 0 for i≠j.
We call this space, with a selected basis, R4,2. A general vector x of V is then written as
x = x0e0+...+x5e5.
We define Q as the set of all isotropic (i.e. (u,u)=0 for any nonzero vector u on the line) lines through the origin 0 in V. Alternatively Q can be written as (see Part 5):
Q = {[u]∈P(V): (u,u) = 0},
where P(V) is the real projective space of V: P(V)=(V∖{0})/∼, with x∼y if and only if there exists a real λ such that y=λx. We now want to prove the following proposition:
Proposition. The following formula defines an explicit isomorphism between the space of oriented spheres (discussed in Lie Sphere Geometry Part 4: oriented spheres) and Q:
(m,t) ⟼ x(m,t) = [m+sin(t)e4+cos(t)e5]. (1)
Proof. Let us start with analyzing the meaning of the formula (1). On the left hand side we have m and t. Together they define an oriented sphere as explained in Part 4. But in Part 4 we had coordinates x1,...,x4, while here we have coordinates x0,...,x5. How should we understand (1)? On the right hand side we see m+cos(t)e5+sin(t)e4. This suggests that m in (1) should be understood as m=m0e0+m1e1+m2e2+m3e3. So our S3 in Part 4 is now in the space of coordinates x0,...,x3 instead of x1,...,x4 as it was previously. Since m is assumed to be on the unit sphere S3, we have (m,m)=1. We also have (m,e4)=(m,e5)=0.
Then we find that
(x(m,t), x(m,t)) = (m,m)-sin2(t)-cos2(t)=1-1=0,
so that x(m,t) is an isotropic vector in V. It is certainly a non-zero vector (Why?), so it defines a point of Q.
Next, we remember from Part 4 that (m, t) and (-m, t+π mod 2π) define the same oriented sphere. Do they define the same point of Q? We easily see, from the properties of sin and cos functions, that x(-m,t+π mod 2π) = -x(m,t). Therefore indeed (1) defines a well defined map from the set of oriented spheres (S3⨉S1)/Z2 into Q.
It remains to be shown that the map is one-to one and onto. Let us first check that the map is one-to-one. Suppose [x(m,t)]=[x(m',t')]. That means x(m',t')=λx(m,t), or, explicitly
λ(m+cos(t)e5+sin(t)e4)=(m'+cos(t')e5+sin(t')e4) .
Since 1=cos2(t)+sin2(t) = cos2(t')+sin2(t')=1, it follows that λ=±1. If λ=1, then t'=t and m'=m. If λ=-1, then t'=t+π mod 2π, and m'=-m. Thus (m,t) and (m',t') define the same oriented sphere. So our map is injective.
We
will show now that it is also surjective. The simplest way of doing
this is by explicitly calculating the inverse map from Q to (S3⨉S1)/Z2. Let's do it. We take any point [u] in Q. Then (u,u)=0 and u≠0. That means
(u0)2+(u1)2+(u2)2+(u3)2-(u4)2-(u5)2 = 0,
or
(u0)2+(u1)2+(u2)2+(u3)2 = (u4)2+(u5)2 .
But u≠0, therefore (u4)2+(u5)2 > 0. Set u'=λu, where λ = ( (u4)2+(u5)2 )-½ . Then [u']=[u]. But now (u'4)2+(u'5)2= 1, therefore there exists a unique t in [0,2π) such that u'4=cos(t), u'5=sin(t). Set m=u'0e0+u'1e1+u'2e2.+u'3e3. Then m fulfills (m,m) =1, so for each point [u] in Q there is oriented sphere (m,t) for which x(m,t)=[u]. QED.
The mystery of zero-radius spheres
The space (S3⨉S1) looks like a homogeneous space - none of its points is better than any other. And (S3⨉S1)/Z2
or isomorphic to it Q doesn't look any worse. In fact, we will see
later on that both are homogeneous spaces under the action of SO(4,2).
And yet among the oriented spheres in S3
there are those with zero radius and no orientation. The same applies to
our experiences in the 3D world we live in. How can these two different
views be reconciled? That is a bugging question. Here is my attempt to
answer it, and details I hope to give in future posts.
What
we see, what we experience, is a projection, like a shadow of a
higher-dimensional structure on a lower-dimensional wall. A sphere
becomes a point, but the sphere itself is just a shadow. Change the
position of the lamp, and the sphere will become a point and the point
will become a sphere.
Now,
formally it looks like a reasonable answer, but after some thinking the
next question comes to mind: how can we do it in practice? Can we? Or
is it just mathematics that has nothing to do with the physical world?
I
think we can, and, I think, it has to do a lot with physics. Special
Relativity deals with inertial frames. But there are no truly inertial
frames in nature. Lorentz transformations are transformations between
inertial frames. It is an idealization valid as long as accelerations
can be neglected. This is our overwhelming experience. We do not have
much experience with accelerations. Well, fighter jet pilots have some,
but still their experiences are local and short, not more than second's
long 10 Gs. That's next to nothing on the Universe scale. What happens
inside elementary particles? We have no idea. Their internal structure
is still a pure guess. We have models that answer some questions, but
leave other questions open. I will approach this subject from the side
of mathematics. The physical interpretation and possible application of
the mathematical formulas is a hard and demanding work. And, anyway,
what we do here is just playing with one particular model, a model that,
perhaps, catches some of the properties of Reality, but leaves aside
other important properties.
So, in the next post we study group action on Q. And after that I will propose a way how to get rid of taking quotient by Z2,
which bothers me aesthetically. Why must we take this quotient? To
take another shadow for reality? As you see I am kind of a Platonist.
But I'm not the only one!
