[This thread continues a marginally related thread; see this URL: -- djr]
	99/analytic_var
==============================================================================
From: Dave Rusin <rusin@math.niu.edu>
Subject: Re:  Rotation group? (it's me again)
Date: Mon, 6 Sep 1999 15:48:21 -0500 (CDT)
Newsgroups: [missing]
To: dario@fis.unico.it
Keywords: determining symmetry groups of a family of functions.

Hi, sorry for the delay in communications.

Let me preface my remarks by the warning that I'm not 100% sure what i
is being asked, so I may be giving incorrect information; but I'll try
to explain what I'm thinking, so you can decide whether or not this is
applicable to your situation.

To summarize my remarks: I think the largest group of symmetries of your 
family of functions is the orthogonal group  O(3)  of three variables (applied
simultaneously to  (x1,x2,x3) and to (x4,x5,x6) -- the "diagonal action").

================

>I believe all these belong to the rotation group O(6), is it correct?
I don't know what "belong to" means, but if you mean "invariant under",
then the answer is no; it seems as though you recognized this later
since you later asked about subgroups. But we can decide what is the
largest group under which all these functions are invariant. That is, we ask 
for all possible functions  h : R^6 -> R^6  which make  F(h(v)) = F(v)  
for all vectors  v  in R^6  and all functions  F  in your family of functions.
(These functions are the "symmetries" of your family.)


We begin with the functions  F(x1, x2, x3, x4, x5, x6)=
>f(r1)*f(r2)
So we're looking for functions  h : R^6 -> R^6  which make
   f(sqrt(h1^2+h2^2+h3^2))*f(sqrt(h4^2+h5^2+h6^2)) = f(r1)*f(r2)
for all vectors in  R^6  and for all functions  f: R->R   -- is that right?
Taking  f(x)=x  and  f(x) = exp(x)  we conclude  h  must preserve both
r1*r2  and  r1+r2, and so must preserve the set {r1, r2}. In other words,
either (h1,h2,h3) has the same length as (x1,x2,x3) and (h4,h5,h6) has
the same length as  (x4,x5,x6), or vice versa. In either case, the length
of (h1,h2,h3,h4,h5,h6) will be the same as the length of (x1,x2,x3,x4,x5,x6)
(which is  sqrt(r1^2+r2^2) ).

So yes, the only functions we want are functions  h  which preserve lengths
in  R^6. Assuming you are only interested in linear functions  h, this
means  h  lies in  O(6). But not every element of  O(6)  preserves
the set of lengths of the two vectors in  R^3. For example, the function
h(x1,x2,x3,x4,x5,x6)=(x1,x3,x5,x2,x4,x6) certainly lies in  O(6), but it
is not in general true that  x1^2+x3^2+x5^2  will equal either of
x1^2+x2^2+x3^2 or x4^2+x5^2+x6^2.

If you think about vectors of the form  (x1,x2,x3,0,0,0), you see
h4^2+h5^2+h6^2 only equals x4^2+x5^2+x6^2 if h4=h5=h6=0, that is, if
h carries all these vectors to the subset  R^3 x {0}  of  R^6. Extending this
idea a bit, we conclude that the linear functions  h  which preserve all
the functions  F = "f(r1)*f(r2)"  are precisely those of the form
    h( v1, v2 ) = ( A(v1), B(v2) )  or  ( A(v2), B(v1) )
with  A, B  being arbitrary orthogonal transformations  R^3 -> R^3.
Here I've written the vector  v  of  R^6  as split into two sets of three
coordinates  v=(v1,v2).

If you prefer a matrix formulation, you see the only symmetries allowed
so far are those of the form
   [ A  0 ]        [ 0 A ]
   [ 0  B ]    or  [ B 0 ]
for a pair of 3x3 orthogonal matrices  A  and  B.

The group of matrices of the first form is the direct product  O(3) x O(3);
I don't know if there's a name for the larger group of all these functions,
including both  O(3)xO(3)  and the other coset.


By the way, this seems to be what you intended in 
your reference to the symmetric group: a set of  N  numbers  r_i  is
uniquely specified as the set of solutions to a polynomial equation
X^N - (r1+r2+...+rN) X^(N-1) + ... +- (r1*r2*...*rN), 
so if  h  is a function on  R^6  which must make  F(h(v))=F(v)  for each
of the elementary symmetric functions 
 F(a1, a2, ..., aN) = a1+a2+...+aN
 F(a1, a2, ..., aN) = a1*a2 + a1*a3 + ... + a(N-1)*aN
 etc.
then  h  can only be a permutation of the  N  coordinates.


Now let's look for functions which also preserve functions of the form F=
>(x1*x2 + y1*y2 + z1*z2) * f(r1)*f(r2)
Now since we know  h  is supposed to preserve  f(r1)*f(r2), it is clear
that it will preserve this function if and only if it preserves the first
factor too. This is just an inner product. You may prefer to view these
geometrically: recall that the inner product satisfies
	       v . w = ||v|| * ||w|| * cos(theta)
where  theta is the angle between  v  and  w. Since we have already
noted that  h  will preserve the set of lengths  r1=||v||  and  r2=||w||,
it follows that  h  preserves this dot product if and only if it
preserves the angle between  v  and  w.

This greatly reduces the possible symmetries  h. In the notation I used before,
    h( v1, v2 ) = ( A(v1), B(v2) )  or  ( A(v2), B(v1) )
we must have  A = B ! For when we specialize this equation to the case
v2=v1, we clearly have  v1  and  v2  separated by an angle of  0; if
angles are to be preserved, then  A(v1)  and  B(v1)  will have to be the
same vector; this can happen for all vectors  v1  iff  A = B.

With this limitation, we see the entire symmetry group is contained in
O(3) (the orthogonal group of rotations and reflections in  R^3) or
this 2-fold cover of it including matrices of the form  
     [ A 0 ]
     [ 0 A ]
with orthogonal matrices  A.


Your next class of functions,
>1+(x1*x2 + y1*y2 + z1*z2) * f(r1)*f(r2)
is just the previous ones added to a constant, and so these are
automatically invariant under all the symmetries of the previous set.


Finally, your functions of the form
>(x1*x2 + y1*y2 + z1*z2) * (f(r1)g(r2) - f(r2)g(r1))
may be divided by the functions of type 2; the symmetries of both 
families of functions are then the symmetries of the previous types 
which also preserve the quotient
     ( g(r2)/f(r2) - g(r1)/f(r1) )
Now, if this is to be invariant for all functions  f  and  g, then
taking in particular  f(r) = 1  and  g(r) = r  shows the symmetries
must preserve  r2 - r1.  That will be true of symmetries which preserve
r1 and r2 separately, but we have noticed before that there is also
the possibility that  h(r2) = r1  and  h(r1) = r2; now we see that
in this case,  h(r2-r1) won't equal r2-r1 so these symmetries can be
excluded.


My conclusion, then, is that the only linear functions  h : R^6 -> R^6
under which all functions  F  of the types you have described
remain invariant are the rotations  h(v1,v2) = (A(v1), A(v2)).
In other words, the symmetry group is isomorphic to the rotation group O(3).

================

>The second step would be to equate all those functions (and much more) to
>zero and understand the symmetry of the resulting "nodal surfaces" (we
>call them this way since they are the "nodes" of solutions of the
>schrodinger equation)

If you want to study the zero-set (or "variety")  Z  determined by a set
of functions  F, and if it's true that  F(h(v)) = F(v)  for all  F
and all v, then of course  F(h(v))=0  iff  F(v) = 0,  i.e.  h(v)  lies in
Z  iff  v  does; therefore, the set  Z  is invariant under the action
of all such symmetries  h.  In your case, this only means your set  Z
is invariant under the action of performing any rotation or reflection
simultaneously in the first 3 coordinates and in the last 3 coordinates.

Of course, there may be other symmetries of the zero-set  Z. As I
remarked above, the simple symmetry  h(v1,v2) = (v2,v1)  does NOT preserve
your functions of the last type. Nonetheless, this symmetry changes
such functions according to  F(h(v)) = - F(v), and so in particular
we still have  F(h(v))=0  iff  F(v)=0 ! So the group of symmetries of
the _variety_ is already larger by a factor of  2, including the
symmetries of the _set of equations_ as well as the swap  (v1,v2) -> (v2,v1).


It's correct for you to think of  Z  as part of  R^6, of course, but
you can also think of the elements of  Z  as pairs of points in  R^3.
Our observations above simply show that from one pair of points in  Z
you can get others by performing arbitrary rotations of the configuration
in  R^3  --  and from the previous paragraph, we also see that a pair
of points in  R^3  is either in  Z  or not --  it doesn't depend on 
whether we take a given pair on one order ot the other.

================

I don't know if I have a really good answer to the question, 
>How do I proceed to classify the functions and their zeros? 
There is a body of knowledge called "Invariant Theory", now stretching
back over a century, which studies the relationships between actions
of groups and the things invariant under those actions. Perhaps a
specialist in this area could help.

To give you an indication of the difficulty let me remark that to
every polynomial in one variable having integer coefficients we may
asociate a finite group -- the Galois group -- which is also a
measure of symmetry. Roughly speaking the only way to _compute_ the
Galois group of a polynomial is to know in advance what all the
possible answers are, and then to develop a checklist of tests to use
to distinguish one answer from another, until only one (right) answer remains.

I fear it might be true in your case too that the general question of
"what is the group of symmetries of this ideal?" may only be answered by
knowing all the possible answers, and then excluding the wrong ones.
This is likely to be problematic, since we cannot know all the subgroups
of an infinite group.

The only _general_ approach I can think of right away would be to
assume a linear map  h: R^3N -> R^3N  preserves each of the polynomials
in a set of polynomials; set  F(h(v))=F(v)  for each  F; and compare
coefficients of all monomials in the coordinates of  v. This gives a
number of equations which together describe conditions on the entries
of the matrix  h. This method will, eventually, describe your subgroup 
as an algebraic group -- that is, the group itself is the subgroup of
the 3N-by-3N matrix group  GL(3N) determined by the vanishing of a
large number of equations in  (3N)^2 variables ! This looks very
complicated in general, and I don't know if the answer will be
satisfying.

I would hope that your more general situation has more structure to it
so that one could analyze it as I did in the beginning of this letter,
instead of using the cumbersome approach of the previous paragraph
(which could be used even in a completely unstructured setting).

================

I hope this is of some use to you. Perhaps you could send me a paper
(a URL would be fine) in which I could see the kind of application you
have in mind for this. Then I might be able to help you better.
(Maybe I should confess that I only studied one semester of physical
chemistry and one semester of quantum theory; when we depart Mathematics
for Science, I get easily lost.)

dave
==============================================================================

From: Dario Bressanini <dario@fis.unico.it>
Subject: Rotation group? (it's me again)
Date: Mon, 6 Sep 1999 10:30:13 +0200 (MET DST)
Newsgroups: [missing]
To: Dave Rusin <rusin@math.niu.edu>

On Fri, 20 Aug 1999, Dave Rusin wrote:

> Let me know if I can be of more help.
> 
> dave

Hi, Thanks again for your help, I followed you example in some other
simple cases and I was able to understand a bit more the surfaced i need
to study.

I also realized that the question I posed initially to you was too much
general, and I need to take advantage of all the information I have on my
surfaces. So before embarking on the task of approximating the various
surfaces i need to know to which group they belong.
The question now is "how do I do it"
Let's take a concrete example,

Let's work in R^6 with vectors (x1,y1,z1,x2,y2,z2) 
and use r1=Sqrt[x1^2+y1^2+z1^2]  and r2=Sqrt[x2^2+y2^2+z2^2]

Now consider the functions (f() and g() are generic functions)

f(r1)*f(r2)
(x1*x2 + y1*y2 + z1*z2) * f(r1)*f(r2)
1+(x1*x2 + y1*y2 + z1*z2) * f(r1)*f(r2)
(x1*x2 + y1*y2 + z1*z2) * (f(r1)g(r2) - f(r2)g(r1))

I need to classify them according to symmetry.
I believe all these belong to the rotation group O(6), is it correct?

We chemists are familiar with discrete point symmetry group
(C2v, D3h etc....) but not with continuous N-dimensional groups.
In the Character tables of point groups I usually find the classification
of simple functions of x,y and z, according to which Irrep they transform,
Is there some similar table for O(n) groups?

It is clear to me that the first and second function have a different
symmetry, since, for example, if I change (x1,y1,z1) -> (-x1,-y1,-z1)
the first function is unchanged but the second changes sign.
But to what subgroup of O(6) tdo they belong?
(and also there is the permutational symmetry involved: x1<->x2 etc...)

The second step would be to equate all those functions (and much more) to
zero and understand the symmetry of the resulting "nodal surfaces" (we
call them this way since they are the "nodes" of solutions of the
schrodinger equation)


So, to summarize my question: How do I proceed to classify the functions
and their zeros? (The functions, in R^3N ALWAYS involve r1,r2 etc.... and
some combination of the coordinates, and furthermore always belong to Sn,
the symmetric permutation group)

Thanks again.

Dario