The Sound of Silence

26-12-2022

Hello darkness, my old friend

I've come to talk with you again

Because a vision softly creeping

Left its seeds while I was sleeping

And the vision that was planted in my brain

Still remains

Within the sound of silence

So, here we will continue with my vision softly creeping - the vision of symmetric spaces (perhaps non-commutative). We will derive a general form of symmetries defined in the previous two posts. Not yet though the definite form. One step at a time.

We recall from Eine Kleine Al Gebra

This way with each n-dimensional subspace V of X on which the scalar product is positive definite we have associated a linear operator J on X such that 

1) J=J*, 

2) JJ=I, 

3) and (z,Jz') is positive definite. 

We will find a general form of such J. 

Writing J in a block form J={{A,B},{C,D}} we find from 1) that 

A=A*, D=D* and C=-B*.

The matrices A,B,C,D are respectively mxm,mxn,nxm,nxn, and * for these matrtices denotes the standard hermitian conjugation (i.e. complex conjugate transpose).

Then 2) leads to

A² =  I_m + BB*

D² =  I_n + B*B

AB + BD = 0

We will  now use the positivity condition 3). Let u in Cⁿ be a non-zero vector, and let z={u,0}. Then (z,Jz) = -u*Au should be positive, therefore A is a (hermitian) negative definite matrix. It follows that

(4) A= -(I_m+BB*)^1/2

Similartly, taking z={0,v} we deduce that D is positive definite, therefore

(5) D = (I_n+B*B)^1/2

The Reader is now encouraged to apply singular value decomposition to the matrix B in order to deduce that the condition AB+BD=0 is satisfied automatically. This way we have found a general form of J:

 The matrix B can be arbitrary, A and D must be given by the above expressions, C = -B*.

But this is not yet a form that is convenient to use. Th action of the group U(n,m) on matrices B that define J happens to be inconvenient.

Let us recall that we are working in an orthonormal basis. Such a basis allows us to identify X with C^m+n , and, in particular, detrmines a split of X into the direct sum of a positive subspace, spanned by n last vectors of the basis, and a negative subspace spanned by the first m vectors of the basis. Of course different orthonormal bases will determine the same split. If the first m vectors are rotated by a unitary matrix in U(m) and the last n vectors of the basis are rotated by a unitary matrix in U(n), the split will stay the same. What we need in the following is really a split, not a basis, but, nevertheless we will assume that we have selected a basis and that we are working with C^m+n . No harm will be done by such an assumption.

So, let J be as above, and let V be the subspace of X composed of eigenvectors of J belonging to the eigenvalue +1. The scalar product (z,z') is therefore positive definite on V Let z be a nonzero vector from V. Thus Jz=z. We write z and J in blockmatrix form. Thus z is a column vector z={w,v}, w from C^m, v from C^n.. J = {{A,B},{C,D}}, with B artbitrary mxn matrix, C=-B*, while A and D are completely determined by B,  and are given by the expressions above.  The eigenvalue equation  Jz=z translates then to:

Aw+Bv=w
Cw+Dv=v.

We rewrite the first equation as (here and below we will write simply I for mxm and nxn unit matrices)

(I-A)w = Bv

Now, from (4)  A is negative definite, therefore I-A is invertible. Therefore we may write

w = (I-A)^-1Bv

Let us define mxn matrix Z as

(6) Z =  (I-A)^-1B = ( I + (I + BB*)^1/2 )^-1B,

so that the eigenvalue +1 eigenspace V of J is spanned by vectors of the form z={Zw,w}.

We will now find the conditions on Z and solve the above equation expressing B through Z and Z*.

From (6) we get

(6a) Z* = B*( I + (I + BB*)^1/2 )^-1,

therefore

ZZ* = ( I + (I + BB*)^1/2 )^-1BB*( I + (I + BB*)^1/2 )^-1

or

 (7) ZZ* = ( I + (I + BB*)^1/2 )^-2 BB*

Let us set y=ZZ*, x=BB* and plot y as a function of x. We get

We see that the function is monotonous, and that it maps the interval [0,infinity) to [0,1). Thus the condition on Z is 

(7a) ZZ* < I 

We can easily find the expression of x in terms of y:

(8) BB* = 4ZZ*/(I - ZZ*)²

From that we instantly get

(I+BB*)^1/2 = (I+ZZ*)/(I-ZZ*)

and thus, from (4)

(9) A = - (I+ZZ*)/(I-ZZ*),

while from (6) we obtain

B = 2(I-ZZ*)^-1 Z

It remains to calculate C=-B* and D. We will do that in the next post. Then we will describe the action of U(n,m) on J in terms of action on Z. We will obtain very nice and manegable linear fractional transformations.

P.S.1 Good news. Have just received email from a Friend in Vienna. The email started with very kind words: "... Your brilliant paper "Random walk on quantum blobs" appeared in Open Systems & Information Dynamics.". And indeed, I have just checked and it appeared - just today:  Open Systems & Information Dynamics