From: Dave Rusin <rusin@math.niu.edu>
Date: Thu, 30 Apr 1998 15:36:07 -0500 (CDT)
To: [Permission pending]
Subject: Test for rationality of an algebraic surface?

I don't know if sci.math.num-analysis is one of the newsgroups you
scan, but there was a question there recently which you might be able to
resolve; one interpretation of the question is whether or not the surface
defined by an equation

	x^2 + y^2 + (z^2 + A)(z^2 + B)(z^2 + C) = 0

is rational, for generic values of  A,B,C (in  Q, say).  I can show that
it is, but only by using maps defined over an extension field, which in
particular will be complex. (There is a parameterization of the form
x=x(z), y=y(z), z=z,  where x(z) and y(z) are cubic polynomials in  z; there
is one degree of freedom in the selection of these polynomials).

Question: whether there is a reasonable way to determine whether there is
a rational parameterization defined over a real subfield.

The original poster's query was for a parameterization of the set of lines
simultaneously tangent to two given spheres. 

dave

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

Date: Thu, 30 Apr 1998 17:03:26 -0400
From: [Permission pending]
To: rusin@math.niu.edu
Subject: Re:  Test for rationality of an algebraic surface?

>I don't know if sci.math.num-analysis is one of the newsgroups
>you scan,

Sorry, it isn't.

> x^2 + y^2 + (z^2 + A)(z^2 + B)(z^2 + C) = 0

> Question: whether there is a reasonable way to determine whether
>there is a rational parameterization defined over a real subfield.

Clearly a necessary condition is that one of A,B,C be negative,
else the surface has no rational points over R.

On the other hand it would seem if at least two of A,B,C are negative
then the surface has two components, and so again cannot be birational
with P^2(R) since P^2(R) is connected.

You write:

> There is a parameterization of the form x=x(z), y=y(z), z=z,
>where x(z) and y(z) are cubic polynomials in  z; there
>is one degree of freedom in the selection of these polynomials

I don't think that this is quite enough -- you would also need
to show that each (x,y,z) in an open set on your surface comes
from a unique such parametrization, and that the "one degree
of freedom" is controlled by a curve of genus 0.  (Assuming the
latter, the surface is rationally dominated by P^2, so is rational
at least over C by Luroth, but perhaps not over R.)

Once these two things are verified it would presumably be easy
to check that the relevant genus-0 curves have points over R,
and thus are birational with P^1(R).

Questions of rationality over Q are in general much harder,
but presumably not of much interest to numerical analysts.

Sincerely,
[Permission pending]
==============================================================================
From: Dave Rusin <rusin@math.niu.edu>
Date: Thu, 30 Apr 1998 21:09:51 -0500 (CDT)
To: [Permission pending]
Subject: Re:  Test for rationality of an algebraic surface?

Thanks.

The degree of freedom is in fact precisely the symmetry group of the circle:
we are modelling the set of lines tangent to two spheres, which admits
an action of the circle rotating the lines around the axis joining
the centers. The curve y^2+(cubic in t^2)=0 picks out one point in each
orbit in a naural way. I thought at first that this made the surface into
a product of a circle and an hyperelliptic curve, but of course there's
no projection to the second factor; I have to remember not to use a
topologist's category in algebraic geometry!

There is a natural correspondence between the signs of the cubic and the
geometry of the situation, which I can't remember without my notes, but
once case corresponded to disjoint balls, one to nested balls, and one to
balls with intersecting boundaries.

I have no clue why this person was posting to num-analysis; I look at it
sometimes to learn something by osmosis, and end up answering the occasional
number-theory question from people who have no clue what the fields are.

Thanks again for your response.

dave
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[This file slightly edited from the original letters.-- djr