	Classifying quadratic polynomials over familiar fields
			(rusin, 1/22/95)

(In what follows we will not distinguish between the polynomials 
F and cF  when  c  is a non-zero constant.)

Let  F  be a quadratic equation, H=(1/2) ( partial F/ partial xi xj )
the (Hessian) matrix of its quadratic form. This may be done formally,
that is, we don't need the metric of the reals and complexes.
Note that (either by the use of Taylor series or by direct substitution)
we can select any point  (x0, y0) and write
F(x,y) = F(x0,y0) + a(x-x0) + b(y-y0) + (x-x0,y-y0).H(x-x0,y-y0)

1) If  H=0, then  F  is really linear, say F = ax+by+c. If, say,
b<>0,  the substitution  y'= ax+by+c, x'=x   is invertible, and in 
the new coordinates,  F = y. Thus  the solution set is the line.

(If both  a=0  and  b=0, the equation is F=c, whose solution set is
either empty or all of C^2 depending on whether or not  c=0. I guess
this ought to be case zero.)

Note that this provides a classification of linear curves up to 
affine equivalence: all are equivalent to F=0 with F=0, F=1, or F=y.

2) If  H  is not zero but is singular, let  v1  be a vector in the kernel
of  H.  If v2  is linearly independent of  v1, then  v2.Hv2 isn't zero;
if it were, then for any   r  and  s, we'd have (rv1+sv2).H(rv1+sv2)=
=(rv1+sv2).H(sv2) [v1 is in the kernel]
=(sv2).H(rv1+sv2) [H is symmetric]
=(sv2).H(sv2)     [v1 is in the kernel]
=(s^2).v2.Hv2  =0  by assumption.
Yet  v.Hv isn't identically zero, for then  F  would be linear, and so H=0,
which we've excluded.
	So  A=v2.Hv2 isn't zero. Choose  s  to be 1/sqrt(A). (Note that
I'm assuming  A  is a square in the field!). Then use a linear substitution
making  v1  and  v2  the basis vectors (1,0) and (0,1). In this basis,
H must be the matrix
         [  0   0  ]
         [  0   1  ]
and so in the present coordinates,  F = c + ax+by + y^2  (using the
Taylor expansion as above with  (x0,y0)=(0,0). )  By completing the
square we may assume  b = 0.  If  a<>0, we can replace  x  by  ax+c to
get  F = y^2+x.  If a=0, we get  F = y^2  if  c=0  and  F = y^2-1  if
c<>0 (using the susbtitution  y'=y/sqrt(-c). Observe again that  -c  has
to be a square.)

Thus the equivalence classes are F = y^2+x, y^2, y^2-1.

These are the equivalence classes in which the quadratic part of  F  is
the square of a non-zero linear term. They have a whole line of critical
points in the middle case, and no critical points otherwise.

(3) If  H  is non-singular, we invoke the classification of quadratic forms
to show that up to a (strictly) linear change of variables, we may assume
H is the identity matrix, i.e.,  F = c + ax+by+ (x^2+y^2). Then we may
use a translation to complete the square and assume  a=b=0. It is then
easiest to use the substitution  x'=x+iy, y'=x-iy  to write  F = c + xy
If  c<>0 we may substitute y'=y/c and get the equation  F=xy+1. Thus the
equivalence classes are  F= xy, xy+1.

Observe that these are the quadratic polynomials with a unique critical 
point. The affine maps we use move the critical point to the origin,
so that  c, above, may be interpreted as the value of  F  at the critical 
point.

Note that if  F  is given its coefficients randomly, then  H  becomes a
random symmetric polynomial, which is non-singular with probability 1.
In that case it will have a unique critical point; the probability that
the value of  F  there is precisely zero, is zero. Thus it is almost
surely the case that  F  is equivalent to  F=xy+1  over the complexes.

Affine equivalence of sets is the narrowest equivalence relation containing
both similarity and dilation along a single axis. In particular, 
equivalent sets are homeomorphic (even diffeomorphic) although not
necessarily congruent or even similar. The homeomorphism type of the
solution sets to these equivalence classes of equations is as follows:
	F=0      all of  C^2
	F=1      empty set
	F=y      C
	F=y^2    C (same solution set as F=y)
	F=y^2-1  C union C (two parallel lines)
	F=y^2+x  C (homeomorphic but not similar to the case  F=y)
	F=xy     two intersecting lines
	F=xy+1   C-{0}

Note that some curves can have the same solution set, but this requires
the two equations to be powers of the same irreducible equation.
(Hilbert Nullstellensatz).

The same analysis works when the coefficients lie in  R.
(1) The classes are those of the equations F=0, F=1, and F=y
(2) The classes are those of F=y^2, F=y^2+x, and F=y^2+1 and F=y^2-1.
	(Here I am using the fact that we are considering the equations
	F=y^2 and F=-y^2 equivalent, although the equivalence comes from
	multiplying the equation by -1 rather than from an affine substitution.
	Otherwise, this, and most other classes, splits into two eq. classes)
(3) A non-singular quadratic form over the reals is equivalent to either
	x^2+y^2 or -(x^2+y^2) OR x^2-y^2, and these are not equivalent to
	each other. It is in the latter case that we get the equivalence
	classes F=xy, F=xy+1; in the former case, we can substitute 
	x=x'sqrt(|c|) and y'=y'sqrt(|c|) to get x^2+y^2=+1 or -1 if c<>0,
	but have x^2+y^2=0 otherwise. 
Here are the divisions of the complex classes and a description of their
real geometries:
	F=0      	all of  R^2
	F=1      	empty set
	F=y      	R
	F=y^2    	R (same solution set as F=y)
      /	F=y^2-1  	R union R (two parallel lines)
      \	F=y^2+1  	empty set
	F=y^2+x  	R (parabola)
      /	F=xy     	two intersecting lines
      \ F=x^2+y^2	singleton
      /	F=xy+1   	R-{0} (hyperbola)
      |	F=x^2+y^2+1	empty set
      \	F=x^2+y^2-1	S^1 (circle)

Over the rationals, a similar division of equivalence classes takes place.
The first split class breaks up into classes  F=y^2-c  where  c  ranges over
equivalence classes of  (Q^x)/(Q^x)^2, the second into the classes of
F=x^2-cy^2, and the last into F=x^2-cy^2 -d where  c  and  d  range 
independently over  (Q^x)/(Q^x)^2.

Under birational equivalence, some of these equivalence classes
coalesce.  For example, the rational functions G(x,y)=(y, x+y^2) and
H(x,y)=(y-x^2, x) are inverses and transform F=y to F=y^2+x.  The
functions G(x,y) = (x, y-1/x) and H(x,y)=(x,y+1/x) are inverses and
transform the equations F=xy and F=xy+1. In addition, G(x,y)=((y+1)/x,(y-1)/x)
and H(x,y)=(2/(x-y), (x+y)/(x-y)) are inverses and show F=y^2-1 is
equivalent to these two. The map G(x,y) = (x/y, 1/y) is its own inverse
and shows F=xy+1 and F=y^2+x to be equivalent. Thus, up to birational
equivalence, the only quadratics over  C  are F=0, F=1, and F=y !
Here are the solution sets in CP^2:

	F=0      all of  CP^2
	F=1      point at infinity only
	F=y^2    (same solution set as F=y)
      /	F=y      Riemann sphere
      \	F=y^2+x  ditto
     /	F=xy     Two touching spheres
     |	F=y^2-1  ditto
     \	F=xy+1   Riemann sphere

The real versions use the same maps when available. For the new polynomials,
F=x^2+y^2+1, F=y^2+1, and F=x^2+y^2, the solutions are so sparse as to
not be a problem. I'll leave F=x^2+y^2-1 as an exercise.
Thus this time we have the equivalence classes of  F=0, F=1, F=y, and the
empty set, F=x^2+y^2+1.

	F=0      	all of  RP^2
      	F=x^2+y^2+1	empty set
      /	F=1      	point at infinity only
      |	F=y^2+1  	ditto
      \ F=x^2+y^2	ditto
	F=y^2    	(same solution set as F=y)
     /	F=y      	circle
     \	F=y^2+x  	ditto
      \	F=x^2+y^2-1	ditto
      /	F=xy     	two touching circles
      |	F=y^2-1  	ditto
      \	F=xy+1   	S^1 (circle)
