To each mxn matrix Z such that
Z*Z < I
wehave associated a
"symmetry" operator J
defined in block matrix form as
The
pseudo-unitary group U(m,n) acts on the set of symmetries by
U:
J
↦ UJU*
Which
translates into linear fractional action on Z, written in a block
matrix form
U:
Z ↦ Z'
But
what if Z*Z = I? It looks as then we encounter a catastrophe. The
expression for J has then 0 in the denominator! J explodes. J is in such a case impossible! What to
do?
What
to do in general when we are facing a catastrophe? Fasten sit belts
and keep cold blood. Get smart! That is what we will do now. We
adjust and continue as if nothing has happened, except that we avoid
explosions. Explosions never do any good. Their effects are,
unfortunately long lasting. We have to learn how to control
explosions. And that is what we will do now.
We take the bull by the horns. We skip the intermediate steps and dive into the center of the cyclone. Do we really need J? Yes, it is useful, but do we need it NOW? How
did we get the formula for transformations of Z? We were considering
vectors of the form
When
Z*Z < I, we have (z,z) ≥ 0, and = 0 if and only if v=0. Thus,
if Z is fixed and v runs over Cn,
vectors z generate an n-dimensional positive subspace of X.
What
if Z*Z=I? Then ||Zv||2
= ||v||2
and vectors z generate a two-dimensional subspace V of X consisting of isotropic vectors: (z,z)=0. If n≤ m – it is a maximal
isotropic subspace.
From
now on we we will assume that n≤ m, otherwise we would have to make
some adjustements. Anyway, for applications in physics we usually take
m=n.
Now,
if V is maximal isotropic, and if U is in U(n,m), then V'=UV is also
maximal isotropic, since U is an isometry
Cn
to Cm. U is represented by an (m+n)x(m+n) A,B,C,D block matrix, while U is nxm matrix. Context IS important! Let us find a general form of a maximal isotropic subspace, say V. V is necessarily nxn dimensional (recall that we assume n≤ m). Let z be a non-zero vector of V. Written as a column vector {w,v}, it is evident that v is a non-zero vector, otherwise we would have (z,z)<0. If z,z' are two vectors in V, then if v=v' we must have w=w', otherwise z-z' would be negative and it should be isotropic. Thus, for z in V w is uniquely determined by v. By linearity is is easy to see that w must depend linearly on v, moreover we must have ||w||=||v||. Therefore z is of the form z={Uv,v}, where U is an isometry, i.e U*U=I. Conversely any such U detrmines a maximal isotropic subspace consisting of vectors {Uv,v}.
Now, let U be an element of U(n,m). Then Uv consists of vectors {U'v',v'}, for some other isometry U'. Writing U in a block matrix form, we get
U'v'=(AU+B)v
v'=(CU+D)v
exactly the same way as we did it before for Z.
The maximal isotropic subspaces of X form what it is called the Shilov boundary of the domain of maximal positive subspaces considered before.
In the coming posts we will discuss its relation to our four-dimensional spacetime for m=n=2.
P.S.1. In physics we start with m=n=2. The elements of X are called twistors. They are bi-spinors for six-dimensional extended spacetime with metric of signature (4,2) - the space of conformal relativity. Our division into blocks is nothing but a representation of a bi-spinor as a pair of spinors. Our U, elements of U(2) group, will correspond to the events of our four-dimensional space time. What is perceived as a simple point in our space time is, as we will see, a Plato projection of a null geodesic (light ray) in six dimensions. Transition from 6 to 4 (or from 4 to 6) is like a phase transition. Then we can go from 6 to 8, from 8 to 10, and from 10 to 12 (Burkhard Heim's world and seventh density) by adding each time two dimensions, always of signature (1,1). BTW: Distance between 4 and 12 is 8, so there is a place for octonions ....
P.S.2. And another by the way:
P.S.3. From an email I have sent to my Friend this morning:
P.S.5 Pretty soon we will have to ask a help from the PC, as there will be a need to do some tiresome calculations. Much of these calculations can be done with Mathematica as it can crunch both symbols and numbers, and also produce graphics. Unfortunately it is quite expensive - unless you are some kind of a student - then it gets cheaper. For a review of Mathematica see here. Symbolic noncommutative calculations are often simpler to deal wisth using FREE Reduce computer algebra software.
P.S.6. Good news! Physicists are now rediscovering EEQT!. Except that they rename it into a commercial name: "Hybrid Quantum-Classical Master Equations" (2014), and pretend they have discovered it all by themselves, like Lajos Diosi, who knows all my papers but will avoid quoting me.
Some of them, surprisingly, still remember that before "Hybrid Systems" there was EEQT. Example: "The constraints of post-quantum classical gravity" , by Jonathan Oppenheim and Zachary Weller-Davies 2022.
But their understanding of the subject is still rather superficial. Hopefully, with time, they will take all the goodies from my papers, re-own it, and sell to the wide public.
I fully understand why physicists in my Alma Mater town of Wroclaw will never quote me - according to them I have made "wrong choices", I am a "black sheep" - they are all "white wolves", following main stream and a "proper" religion and a "proper" politics. They think that one day I will feel sorry for my choices, perhaps after my death, so they hope.
P.S.7. It so happened that I am forced to visit the forest of homotopy groups. I have zero knowledge of this forest. Beasts are hiding behind every tree. I am completely dumb. Right now I will be looking at Trautman's "Double covers of pseudo-orthogonal groups". It's all new for me! Except, perhaps, of Clifford algebras, where my knowledge is slightly more than just simple zero. Adventures, adventures. Life is certainly not boring.
P.S.7. By "chance" (of course, as always) I have just received a message from Igor Bayak on Linkedin.com, announcing his paper "Chaotic dynamics of an electron". And what do we see in the Abstract?
Abstract
First, we construct the image of the torus on the two-layer shell of the sphere and note that the isometries of the image of the torus on the sphere generate the unitary group U(2), and then we establish that, as a result of the action of the modular group on the sphere, it is factorized in such a way that the minimal (one-element) equivalence classes are given by the set of primes.
The group U(2)! When I am just looking for the action of SU(2,2) (in fact O(4,2)) on its double covering space!
