Linear fractional action of U(n,m)

 First we will review of definitions and results from previous posts.

We have discussed indefinite metric complex vector space X, endowed with a scalar product (z,z') of signature (n,m), where m,n ≥ 1. Let ei (i=1,2,...,n) be a basis in X. Then each z in X can be decomposed into the basis vectors

z = z1 e1+...+zm+n em+n

We call the basis orthonormal if the scalar product, when written in this basis, takes the form

(z,z') = - z1*z'1 -...- zm*z'm -.+ zm+1*z'm+1 +...+ zn*z'n ,         (*)

where zi* denotes the complex conjugate to zi.

Selecting an orthonormal basis in X, we identify X with Cm+n, = Cm⊕Cn. The vectors in X we then write as columns z={w,v}, with w ∈ Cm and v ∈ Cn. The scalar product (z,z') takes then the form

(z,z') = -w*w' + v*v'

where * applied to vectors in Cm and Cn denotes the hermitian conjugate.

We denote by J the set of all maximal positive subspaces of X. A subspace V of X is called positive if the scalar product (z,z') restricted to V is positive definite. By Sylvester's Law of Inertia 





each such V is necessarily n-dimensional. We are interested in the set J of all such subspaces. In a convenient parametrization by mxn complex matrices Z with Z*Z<I (that we will introduce below), the pseudo-unitary group U(n,m) will act on J by (generalized) linear fractional transformations. In mathematics J is an example of a bounded symmetric domain of type I(one).

Note: There are also infinite-dimensional generalizations - see "Bounded Symmetric Domains in Banach spaces" by Cho-Ho Chu. We will restrict ourselves to the finite-dimensional case.

Strictly speaking we are interested in the group U of all isometries of X endowed with the scalar product (z,z'). But once we have selected an orthonormal basis of X, then U becomes identified with U(n,m) - the group of all complex (m+n)x(m+n) matrices preserving the scalar product (*). Introducing the diagonal  block matrix  J₀

 J₀= diag(-I_m, I_n) ,

we have written the condition on matrices U from U(n,m) as

U^†J₀ U = J₀

_{where the dagger} † denotes the hermitian conjugate. We notice that J₀ itself  is in U(n,m).

Note: We are using bold letters to denote (m+n)x(m+n)  matrices.

We have defined  J is as the set of all maximal positive subspaces of X. Equivalently we could have defined J as the set of all linear operators J acting on X satisfying the three conditions:

1) J=J*, 

2) J²=I, 

3) the sesquilinear form (z,z')_J defined by  

(z,z')_J =(z,Jz')

 is positive definite. 

Note: Notice that J₀ ∈ J.

If V is a maximal positive subspace of X, and if W is its orthogonal complement, then J corresponding to V is defined as the unique linear operator defined as the identity on V and as minus identity on V. Conversely, if J satisfies the conditions 1),2),3), then its eigenspace belonging to the eigenvalue +1 is a maximal positive subspace of V.  This follows by an elementary linear algebra.

We have shown that every J ∈ J is of the form :

Where Z is an mxn matrix satisfying Z*Z<I (which is equivalent to ZZ*<I) uniquely determined by J. Moreover the maximal positive subspace determined by J (that is the eigensubspace belonging to the eigenvalue +1) consists of all vectors z of the form

Let now U be an isometry of X (equipped with the indefinite scalar product (z,z')). It is elementary to show that if J is in J, i.e. J satisfies the conditions 1)-3), the J'=UJU* also satisfies these conditions. It is also elementary to show that if V is the eigensubspace of J belonging to the eigenvalue +1, and if V' is the eigensubspace of J' belonging to the eigenvalue +1, then

V' = UV.

Therefore vectors of V' are again necessarily of the form

Here Z' is another mxn matrix satisfying Z'*Z"<I, determined uniquely by Z and by U. We will now find an explicit form of Z'.

To this end we write U,z and z'  in a block  form and calculate the result:

Now, if z is nonzero, then z' must be also nonzero, since U is invertible. It follows then that the nxn matrix CZ+D must be invertible. Indeed, if z is nonzero, then also v is nonzero. If there existed nonzero v such that (CZ+D)=0, then we would have (z',z')≤0, while we should have (z',z')>0. Therefore, setting v'=(CZ+D)v , we get

Since this should hold now for any v, comparing with the prvious expression for z' we get

And this is our final formula - a generalized linear fractional transformation. It automatically follows that if Z*Z<I, then Z'*Z'<I.

In the next post we will discuss what happens to this formula when we leave the safe ground and  try to do something "forbidden", namely extend the above transformation formula to Z such that Z*Z=I. For m=n=2 such Z parametrize points of the "Shilov boundary" - the compactified Minkowski spacetime of events equipped with the flat conformal causal (light-cone) structure.

P.S.1. Everything presented in this note requires only elementary linear algebra. In particular I did not use any computer algebra software, like for instance Mathematica, or Reduce, which I love to use when it helps. 

Well, I used one line of code (which I am not particularly proud about) to get the formula (8) from The Sound of Silence:

Reduce[2 x + x y + 2 x Sqrt[1 + y] == y, y]

The function graph above this formula comes with the code.

P.S.2 (31-12-22) Started reading Christopher Langan's "Introduction to Quantum Metamechanics". Observations from the first page: Langan rightly complains about the state of quantum mechanics. Mentions the need for "post-quantum mechanics" (post-QM). (I think he borrowed this term from Jack Sarfatti?) But then Langan writes about it: "Because this theory is necessarily a metatheory (or theoretical metalanguage) of QM, it is called Quantum Metamechanics or QMM)."

I do not see any necessity for post-QM to be a metatheory. What I see is the necessity of having a better theory than the standard QM.

P.S.3 31-12-22 13:00 Encouraged by Irina Eganova I have started reading "World as Space and Time" by <a href="https://en.wikipedia.org/wiki/Friedmann_equations">A.A. Friedman</a>. Beautifully written! The book (in Russian) is accessible for reading online <a href="https://reallib.org/reader?file=583994&pg=7">here</a>

P.S.4 Searching the net for Friedman and "space-time boundary" I have stumbled upon "Category Theory in Physics, Mathematics and Philosphy", Ed. Marek Kuś and Bartłomiej Skowron, Springer 2019, and there the paper by Michael Heller and Jerzy Król "Beyond the Space-Time Boundary". Interesting reading though only superficially related to my own projects. The authors claim that " The standard geometric tools on M do not allow one “to cross the boundary”. Well it all depends on what they call "standard".

Happy new Year! May your dreams come true! But while chasing your dreams pay close attention to reality left and  right!

Domino effect

 

Do I agree with the above? Mostly - yes. But not always. For instance at the end of the last post I told my Reader what I intend to do in the next post. But I did not show it FIRST. I am showing it only now.

But in general it seems to be true that we often tend to disperse our energy on telling the world what we intend to do: 

"World, you will see how great things I will do! I will do THIS and THAT. The day will come when you see and realize how great I am!"  

And then we have lie to ourselves when we realize that, in fact, our true destiny is in doing something else. 

Anyway yesterday became today and today I will show how to get nice expressions for C and D of the previous post. This is its continuation.

Let Z be mxn  (m,n >0) complex matrix. So Z is, in general, rectangular, not necessarily square, matrix. Let Z* be its hermitian conjugate (complex conjugate transpose). Then Z* is nxm. Moreover ZZ* is mxm, while Z*Z is nxn. We will denote by I the unit matrix, whether it is mxm or nxn,  will depend on the context.

We first notice that 

ZI = IZ

Now we will use the associativity of matrix multiplication:

Z(Z*Z) = (ZZ*)Z

Z (Z*Z)(Z*Z)=ZZ*ZZ*Z=(ZZ*)(ZZ*)Z

and in general, for any n >= 0

Z(Z*Z)ⁿ =(ZZ*)ⁿZ

This is our "domino effect" from the title of this post: we push from the left with Z and Z falls down on the right. In between powers of Z*Z get replaced by the same powers of ZZ*.

Thus for any analytic (representable as a convergent power series) function f of a complex variable we have

Z f(Z*Z) = f(ZZ*) Z 

Similarly

Z* f(ZZ*) = f(Z*Z) Z*= 

Nice? Nice!

Notice that we are taking powers of square matrices. Taking powers of a non-square matrix would not make sense!

In the previous post we have obtained the following formula for B:

B = 2(I-ZZ*)^-1 Z

We also know that C=-B*. We can now use the domino effect formula to obtain

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

So

C = -2(I-Z*Z)^-1 Z*

It remains to calculate D. In the previous post we have obtained:

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

and 

B = 2(I-ZZ*)^-1 Z

Thus 

B*B = 4Z*(I-ZZ*)^-2 Z

Using the domino effect formula we can rewrite it as

B*B = 4(I-Z*Z)^-2 Z*Z

Now I+B*B can be easily calculated to give

I+B*B=(I-ZZ*)^-2 (I-ZZ*)²  + 4(I-Z*Z)^-2 Z*Z = (I-ZZ*)^-2 (I+ZZ*)² 

and therefore 

D = (I-ZZ*) (I+ZZ*) 

 P.S.1. Here is a continuation of P.S.1 from the post Eine Klein Al Gebra . My comments on the book "Mistakes we made: But not by me" by Carol Tavris and Elliott Aronson. In a chapter "Cognitive Dissonance: The Engine of Self-Justification" the authors give us a serious warning: how easily we fall into self-justification! For instance a smoker will try to find all possible (often irrational, neglecting completely rational ones) arguments to convince themselves and others that smoking is good for you. Or take this argument from the book:

>Some scientific evidence for the power of irrevocability comes from a clever study of the mental maneuverings of gamblers at a racetrack. The racetrack is an ideal place to study irrevocability because once you’ve placed your bet, you can’t go back and tell the nice man behind the window you’ve changed your mind. In this study, the researchers simply intercepted people who were standing in line to place two-dollar bets and other people who had just left the window. The investigators asked them how certain they were that their horses would win. The bettors who had placed their bets were far more certain about their choice than the folks waiting in line. Yet nothing had changed except the finality of placing the bet. People become more certain they are right about something they just did if they can’t undo it.

While the above is certainly true enough, it is dangerously only partially true. Here is why: a person that decides to quit smoking must often show a very strong will power to convince himself constantly that the decision was right. Writing down arguments supporting this decision and rereading them again and again may be of help.

>Another example from ""Narcissistic Personality Disorder How to Spot the Subtle Signs of a Narcissist and Continue to Thrive After an Encounter" by Tony Sayers:

George had just about had enough of his father’s behavior. He didn’t like being belittled, torn down, and compared to his dad, and he hated having his hard work underappreciated simply because he was ‘just his father’s son.’ So, after their last heated argument, he decided to just walk away and leave their relationship at that. Whether they’d ever be on good terms again, he was uncertain. But he was happy to finally be free from his dad’s abuse.

In the weeks following his falling out with his dad, George started to feel a strong sense of isolation and guilt. He felt as though he had wronged his father, and struggled to resist the urge to reconcile, knowing full well that it would only give his dad the fuel he needed to make George feel bad about protecting and defending himself.

Here we have again: a difficult but necessary decision. Also in such cases self-justification is not a bad thing. It is a ncessity. Similar examples can be found, for instance in "Energy Vampires. How to protect yourself from toxic people with narcissistic tendencies" by the same author and in "The Borderline Personality Disorder. Survival Guide" by Alexander Chapman and Kim L. Gratz. In similar cases taking a difficult decision and keeping to it is the only solution to otherwise never ending problems with constant ups and downs.

P.S.2. As I am recently (encouraged by Laura)  studying Christopher Langan's  "philosophy", here is something related: 

The Results & Features of a Person with a High IQ | Jordan Peterson

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 

Eine Kleine Al Gebra - Romanze: Andante (C major)

 Xmas 2022

Romanze: Andante

The 2nd movement is also a short one when compared to other contemporary pieces of the “Romantic” era. It has a gentle and slower tempo, with a recurring A section and a B section similar to the 1st movement. ...

Therefore We will go slowly and gently, in a romantic mood,  with the parametrization of the set of symmetries (or maximal positive subspaces of X) as defined in the previous post. But first of all, dear Reader:

Please do backups of your hard drives. Today. An email that I received today from a Friend contained this part:

"...everything I wrote, all the articles I had collected over the years, everything for myself, which is thousands of files, all of it crashed...

As if someone decided that it would not be enough for me to have Carcinoma, it is necessary that the Disk, on which all my Life is, is irreparably peeled off. Well, it has peeled off..."  

Backup!

Make your backup. Save your files. Save your life. Today! 

Lest us start with the star. 

Recall from the previous post:

Given an orthonormal basis X can be identified with C^m+n.

Let † denote the standard hermitian conjugate (conjugate transpose). Then, (z,z') can be written as

(z,z') = z^†J₀z'

where J₀ is the diagonal (m+n)x(m+n) matrix J₀= diag(-I_m, I_n) and I_m, I_n are the mxm and nxn unit matrices respectively. The group U(n,m) is the set of all matrices U satisfying

U^†J₀ U = J₀

Let A be an (m+n)x(m+n) matrix written in a block form as A ={{A,B},{C,D}}. We denote by A* its conjugate with respect to the indefinite scalar product (z,z'). Then A* is given by 

A* = J₀A^†J₀  

or, in block form 

A* ={{A*,-C*},{-B*,D*}}

where * on the right, applied to the blocks, denotes the ordinary hermitian conjugate. It's confusing, isn't it? Anyway...

Allelujah!

Another version here!

P.S.1 Today from Irina Eganova in Novosibirsk I received a copy of the paper by M. M. Lavrent'ev "Ad disputandum". At the end the author recommends the ideas of the late Polish nuclear physicist M. Gryzinski. Grzyzinski has his own, mostly classical,  ideas about explaining many quantum phenomena. Some time ago I have looked into Gryzinski's ideas and criticised them. I do like Lavrent'ev. He is certainly sincere. But is he right about Gryzinski?Now I will have to look at Gryzinkski's ideas again.

P.S.2 From my working desk. Here is a piece  from Lavrent'ev paper that concerns us, as we are interested in merging the two very different theories, General Relativity and Quantum Mechanics, into one new simple and clear conceptual, theoretical and mathematical framework. Lavrent'ev is quoting and discussing Ginzburg. (slightly corrected automatic translation from Russian to English below may not represent adequately all the subtleties of the text)

1. Monopolism in science: loss of logic

"The fact that I do not at all abuse the charge of pseudoscience, when it comes to ideas and constructions that I do not share, is clear, I think, from the discussion with Academician A.A. Logunov. I have a negative attitude to his criticism of general relativity and to his own relativistic theory of gravitation, I wrote about it."

Having read these weighty sounding lines of Academician A.L. Ginzburg any reader can draw a conclusion that the position of Academician A.A. Logunov is erroneous. A. A. Logunov's position is erroneous, because if it is so clearly treated negatively it means that it has been unqualified fallacy. It cannot be otherwise! So publicly to express one's negative attitude (exactly negative!) to a theory is admissible only in one case: if it is shown to be erroneous. It is admissible only in the one and only case that it is shown to be erroneous.

But then we go on to read: "At the same time, in the article [2] I specifically emphasize that since Logunov's views have not been rigorously refuted, to declare them "pseudoscience" would be unacceptable and, of course, I did not and I do not do it."

But where is the logic?! If "Logunov's views are not strictly refuted" (and there is no such thing as a non-strict refutation!), then what is the reason for being negative to them! In this case it is not only inadmissible to declare these views as pseudoscience, it is also inadmissible to propagate (which, by the way, lasts more than one year) one's own personal rejection of them. The great Plato condemned such "discussions" more than two thousand years ago. He called them "eristics” (from the Gr. eris - dispute, discord) and defined them as a purely sophistical dispute for the sake of argument, self-assertion. It is astonishing that such a gaffe was overlooked by such a representative editorial board.The editorial board of Vestnik RAN, or perhaps such a "logica" has already become the order of the day.

Translated with www.DeepL.com/Translator (free version)

 

Eine Kleine Al Gebra

 From Wikipedia:

Algebra (from Arabic ‏الجبر‎ (al-jabr) 'reunion of broken parts

And it is time for starting such a reunion. 

It is said:

Matthew 9:17 King James Bible. "Neither do men put new wine into old bottles; else the bottles break, and the wine runeth out, and the bottles perish; but they put new wine into new bottles, and both are preserved."

I will be putting old wine into old bottles Algebra is old and conformal group that includes similarity transformations (as above so below) is probably even older.

Quod est superius est sicut quod inferius, et quod inferius est sicut quod est superius.

That which is above is like to that which is below, and that which is below is like to that which is above.

So here it comes: Eine Kleine Al Gebra. But first the old music-wine in old bottles

Allegro

Let m,n ≧ 1 be two integers. Let X be a complex vector space of complex dimension m+n. 

Let X be equipped with a fixed sesquilinear form (z,z'), antilinear in the first argument, linear in the second one.

We assume that X admits a basis e₁,...,e_m,e_m+1,...,e_{m+n  such that, with respect to this basis, the scalar product (z,z') takes the form}

(z,z') = -z₁* z'₁ - ... -z_m* z'_m + z_m+1* z'_m+1 + ... + z_n* z'_n

where the star * stands for the complex conjugation. Such a basis will be called orthonormal. We say that (z,z') is a hermitian scalar product of signature (n,m). Notice the order: (n,m). n plus signs, m minus signs.

Any two orthonormal bases are related by a transformation from the group U(n,m). We will write matrices of U(m,n) in a block form

U =

A  B

C D

where the matrices A,B,C,D are repectively mxm,mxn,nxm,nxn.

Given an orthonormal basis X can be identified with C^m+n.

Let † denote the standard hermitian conjugate (conjugate transpose). Then, (z,z') can be written as

(z,z') = z^†J₀z'

where J₀ is the diagonal (m+n)x(m+n) matrix J₀= diag(-I_m, I_n) and I_m, I_n are the mxm and nxn unit matrices respectively. The group U(n,m) is the set of all matrices U satisfying

U^†J₀ U = J₀

Writing U in a in block matrix form as above, the condition for U to be in U(n,m) translates into:

A*A - C*C = I_m, D*D - B*B = I_n

A*B - C*D = 0, B*A - D*C = 0

In the following, for the ease of notation,  we will use * to denote the ordinary hermitian conjugate for mxm,mxn,nxm, and nxn matrices.

We notice that A*A = I_m + C*C and C*C is non-negative. Therefore A*A ≧ I_m, and, in particular, A is invertible. D is invertible for a similar reason.

From Wikipedia:

In mathematics, the Grassmannian Gr(k, V) is a space that parameterizes all k-dimensional linear subspaces of the n-dimensional vector space V

We will be interested in in a subset of Gr(n,X), Namely we will be interested in the set J of all  n-dimensional subspaces of X on which the scalar product (z,z') is positive definite. Choosing an orthonormal bases one such subspace is evident: it consist of column vectors u,v, u from C^m. v from Cⁿ. for which u=0.

It is more or less evident that U(n,m) transforms J into itself and that the action of U(n,m) on J is transitive. The stability subgroup is U(m)xU(n), therefore J can be identified with the manifold of equivalence classes (cosets) U(n,m)/(U(m)xU(n)).

We will need an explicit parametrization of J.

Let us do some Christmas Eve talk. Say V is an n-dimensional subspace of X on which our scalar product is positive definite. It is then necessarily a maximal subspace with this property, and that because our scalar product is of signature (n,m). 

Let W be the orthogonal complement of V. Then on W the scalar product is necessarily negative definite. Let E be the orthogonal projection on V. Then E*=E and EE=E. Here, and in the following, * applied to linear operators acting on X will denote the hermitian conjugate with respect to the scalar product (z,z') on X. The eigenvalues of E are 1 and 0. 1 on V and 0 on W.

Then F=I-E is the orthogonal projection on W - the orthogonal complement of V. We also have F*=F and FF=F. The eigenvalues of F are 0 and 1. 0 on V and 1 on W. Subspaces V and W are mutually orthogonal.

Let us define 

J=E-F.

Then J=J* and JJ=I. The eigenvalues of J are -1 and 1. 1 on V and -1 onW.

Let us define a sesquilinear form (z,z')_J = (z,Jz'). We claim that (z,z')_J is positive definite. Indeed, let z be any nonzero vector in X. Then z decomposes into z=v+w, where v is in V and w is in W. Thus Jv=v, Jw=-w. But then (z,z)_J=(z,Jz)=(v+w,J(v+w)=(v+w,v-w)=(v,v)-(w,w). But if z is not zero, then v or w (or both) must be non-zero. For a non-zero v, (v,v) is positive and for a non-zero w, -(w,w) is positive. Therefore (z,z)_J is positive. 

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. 

Let us see that the converse is also true. Namely, Let J has the above properties. Define E=(I+J)/2, F=(I-J)/2. Then, because of 1) and 2), we have that E*=E, EE=E, F*=F, FF=F, EF=FE=0, E+F=I. E and F are two complementary orthogonal projection operators. Let V be the subspace belonging to eigenvalue 1 of J, W be the subspace belonging to the eigenvalue -1 of J. E projects on V, F projects on W. Moreover, if a non-zero v is in V, then Jv=v and, because of 3),  (v,v)=(v,Jv)>0. Similarly, if a nonzero w is in W then Jw=-w and, because of 3),  -(w,w)=(w,-w)=(w,Jw)>0. Thus V is maximal positive and W is maximal negative subspace of X. QED.

Our next step is parametrization of such operators J. We will do that in the next post. Here we just note that these J are very special elemenets of the group U(n,m) (Why so?).  We will cal them "symmetries".

Merry Christmas!

P.S.1 I am slowly (mainly during breakfasts) reading the book "Mistakes we made: But not by me" by Carol Tavris and Elliott Aronson. 

As I have promised, here are my first initial comments on the book "Mistakes we made: But not by me" by Carol Tavris and Elliott Aronson. Today I will deal only with the Itroduction and with quite general comments.

It starts like that:

INTRODUCTION

Knaves, Fools, Villains, and Hypocrites:

How Do They Live with Themselves?

Mistakes were quite possibly made by the administrations in

which I served.

—Henry Kissinger, responding to charges that he

committed war crimes in his role in the United

States’ actions in Vietnam, Cambodia, and South

America in the 1970s

And a liitle further on we find the following biting remark:

When Henry Kissinger said that the administration in which he’d served may have made mistakes, he was sidestepping the fact that as national security adviser and secretary of state (simultaneously), he essentially was the administration. This self-justification allowed him to accept the Nobel Peace Prize with a straight face and a clear conscience.

So far so good. There are many more illustrative examples of self-justification in the introduction. But then there is a pragraph that have rised a red frag in my mind. Here it is:

By understanding the inner workings of self-justification, we can answer these questions and make sense of dozens of other things people do that otherwise seem unfathomable or crazy. We can answer the question so many people ask when they look at ruthless dictators, greedy corporate CEOs, religious zealots who murder in the name of God, priests who molest children, or family members who cheat their relatives out of inheritances: How in the world can they live with themselves? The answer is: exactly the way the rest of

us do.

It is evident to me that the authors neglect here important discoveries about workings of human minds: not all brains are wired the same way. If we read carefully „Inside the Criminal Mind” by Stanton E. Samenow, we learn that

„There are people who would be “criminals” no matter where they live. These are individuals for whom to be someone in life is to do the forbidden, whatever the forbidden might be.

…

I submit that there is a major difference between the person who tells an occasional lie and an individual who lies as a way of life. The criminal lies to cover his tracks (he has a lot to conceal) and to get out of a jam that he has created for himself.

However, he also lies about the most minuscule matters even when there is no ostensible reason. He’ll say that he went to one store when he really went to another. Some people in the mental health field will conclude that this is pathological or compulsive. This is not the case. The lie that makes no sense does make sense when you understand the mentality of the liar.”

Not all men are wired the same way. Some are wired differently. And not all mistakes are equal. Results do count. A politician has more responsibilities than a receptionist in an inn. There is a difference between a mistake that costs human lifes and a mistake that has no serious  consequencesat all. It is therefore important to know ourslves, self-observe, consciously compensate for all defects in our wiring. And take responsibility for all results of our actions, and for our negligance of actions, when such are needed.