From: lrudolph@panix.com (Lee Rudolph) Subject: Re: find a continuous real function Date: 4 Jul 2000 08:55:08 -0400 Newsgroups: sci.math Summary: [missing] ags@seaman.cc.purdue.edu (Dave Seaman) writes: >In article <8jq1qi$ejb$1@nnrp1.deja.com>, wrote: >>Let R be the field of real numbers. > >>I need a continuous bijection f:RxR -> RxR such that f^-1 is continuous >>and f(a,0)=(a^2, (a-1)^2 ), for all real number a in R. > > f(x,y) = (x^2+y,(x-1)^2+y) To be more general: Suppose K is an algebraically closed field of characteristic 0 (for instance, C), and (p,q): K -> K^2 is a "polynomial embedding of the line in the plane" (for instance, (p,q)(t)=(t^2,(t-1)^2)), that is, a pair of polynomials over K such that (1) for all s, t in K, if (p,q)(t)=(p,q)(s), then t=s (so (p,q) is an injection of sets), and (2) for all t in K, (p'(t),q'(t)) is not (0,0) (so (p,q) is a "smooth immersion"). Then by a theorem of Abhyankar & Moh and Suzuki, there are "polynomial automorphisms of the line and the plane" A:K -> K and B:K^2 -> K^2 (for instance, A(t)=t and B(x,y)=(x^2+y,(x-1)^2+y)), that is, polynomial bijections whose inverses are also polynomials, such that B(t,0)=(p(A(t)),q(A(t)). In other words, up to automorphism there is only one embedding of the line in the plane. (It's also true, and a theorem of Zaidenberg & Lin, that up to automorphism every polynomial *injection* of K in K^2, which is *not* an embedding, is of the form (p,q)(t)=(t^m,t^n) with m,n>1, GCD(m,n)=1.) To apply the A&M-S theorem over a non-algebraically-closed field (like R), you simply have to check that the given polynomial embedding stays an embedding when the scalars are extended to the algebraic closure. As to *finding* B, well, that can be automated. A theorem of Jung (which is a corollary of A&M's proof of the theorem; maybe even simply of the truth of the theorem, it's been some time since I've checked) tells you that the polynomial automorphisms of K^2 are generated by the linear automorphisms GL(2,K) and the "polynomial shears" B(x,y)=(x,y+r(x)), where r is an arbitrary polyomial. So if you want to find B for (p,q)(t)=(t^2,t^2-2t+1), first find a linear map B_1 which makes the degree of one component strictly smaller than that of the other (say, B_1(x,y)=(x,y-x)), then find a (conjugate of a) polyomial shear with r a suitable monomial to reduce the larger of the two degrees (say, B_2(x,y)=(x-(y^2/4),y)), and iterate as necessary. (The existence of r is entailed by the hypothesis that (p,q) is an embedding, via the A&M proof.) Here, B_1(t^2,(t-1)^2) = (t^2,-2t+1), B_2(t^2,-2t+1)=(t-(1/4),-2t+1), so we can let B_3(x,y) be any linear automorphism which takes (t-(1/4),-2t+1) to (t,1), and B_4 be the shear B_4(x,y)=(x,y-1). Then B = B_4 o B_3 o B_2 o B_1 works (and for a suitable choice of B_3, you will get Dave's f). Lee Rudolph