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)