From: teke@jocosus.matematik.su.se (Torsten Ekedahl) Newsgroups: sci.math.research Subject: Re: seeking rational functions as group operations Date: Wed, 10 Nov 1993 00:30:26 GMT In article <199311081959.AA06422@mp.cs.niu.edu> David Rusin writes: > I am looking for examples of rational functions which define an associative > binary operation on (most of) a vector space (over the rationals, say). > For example on a one-dimensional vector space (Q itself) we have the two > polynomials F(x,y)=x+y and F(x,y)=xy which define the group operations on > (Q,+) and (Q^*, \times). But it is a typical exercise to show that > F(x,y) = x+y+xy is also associative [using x-> x+1 we see that this is the > same group as Q^*]. Similarly, F(x,y)=(xy)/(x+y) works almost everywhere > [using x -> 1/x we recognize the group (Q,+)] although there is a difficulty > when x=-y which I am happy to ignore. > > Question 1. Are all example similar to these (e.g. of the form (S^-1) o F o S > where F(x,y)=x+y or F(x,y)=xy, and S is a rational function of one variable > with rational inverse)? > > Question 2. What is known about rational functions in more variables? For > example, ordinary matrix multiplication defines a polynomial function > F: Q^8 -> Q^4. Are higher-dimensional examples related (in the same sense > as in Question 1, perhaps) to something like Lie subgroups of GL_n(Q)? > > Question 3. Is there a non-abelian example "smaller" than GL_2 ? > > dave rusin@math.niu.edu > Ps - I am looking for something more concrete and less encompassing than > references to unirational subvarieties of algebraic groups or whatever. I am going to answer your questions only in the case when you also assume that there is an identity element and an inverse given by a rational function (given an identity element the inverse always exists as a transcendental function but may not be rational, though it actually may in the 1-variable case). Without those assumptions I don't think very much is known. The answer to question 1 is no. It is yes if if S is allowed to have complex coefficients (or any other algebraically closed field containing Q). It is also yes if you allow S to be a power series. Over Q you can start with a rational number q and then consider the ring of expressions a+b sqrt(q), i.e. Q^2 with a certain multiplication (this needs to said if q is a square). Then we define the norm of such an element as a^2+b^2q. The elements of norm 1 (a^2+b^2q=1) form a subgroup of the group of invertible elements. These elements can also be given a rational parametrisation: For each t look for solutions of the form a = 1+s, b = st. This gives 2 qt - 1 -2t a = ------- b = ------ 2 2 qt + 1 qt + 1 and there is also a rational inverse. Using this parametrisation to transfer the group structure given by multiplication on elements of norm 1 one gets a 1-dimensional rational group law. When q = -1 a law equivalent to (x+y)/(1-xy), which is mentioned in Rusin's follow up note, is obtained. The following is then true: a) Any 1-dimensional group law is equivalent to one if this form. The additive law, F=x+y, is obtained when q=0, the multiplicative law F=xy is obtained when q=1. b) Two such laws associated to q and q' are equivalent iff q' = r^2q, where r is a non-zero rational number. c) For q not 0 the law associated to q becomes rationally equivalent to the multiplicative group if coefficients involving the square root of q are allowed. To answer question 2 I will first replace Q by an algebraically closed field K. Then indeed the answerm is an emphatic yes. That is if you by Lie subgroup of GL_n(K) mean the subgroups defined as the zero sets of polynomial equations and which are also connected. With that understood then we have the following. a) Any rational group law comes from a Lie subgroup. b) Two rational group laws are equivalent iff the Lie groups are isomorphic. c) Any Lie group corresponds to a rational group law. a) and b) follows from a result of Weil which says that "any rational group law (in a slightly extended sense than that used here) has an everywhere regular model and any birational group isomorphism is regular" together with a small argument which excludes an abelian variety part. (I guess I shouldn't have said this as this is exactly the kind of statement that, according to the PS, was not asked for :-).) c), on the other hand follows from a structure theorem which shows that any Lie group in our sense has an open subset which is isomorphic to an open subset of some affine space. If we go back to the case when our base field is Q then a) and b) are still OK. c), however, no longer holds. The reason can be seen already in the 1-dimensional case where we had to make an extra argument to show that the elements of norm 1 had a rational parametrisation. In the higher dimensional case this isn't even true so that we will have Lie groups which are not given by rational group laws. The answer to question 3 is yes. By realising pairs (a,b) with a not 0 as affine transformations x |-> ax+b we get a 2-dimensional non-abelian group law F(a,b,a',b') = (aa',ab'+b). -- Torsten Ekedahl teke@matematik.su.se