From: " " Subject: Polynomials and Transforms Date: Thu, 25 Mar 1999 07:15:20 -0800 To: rusin@vesuvius.math.niu.edu Keywords: relations among hypergeometric functions [all reformatted -- djr] Dear Dave: First of all thank you for responding to my rather undetailed problem. As you were the only one I figure its best to e-mail you directly. I kept the nature of the transforms vague since I was trying to keep it simple and hoped for a general approach. However, as you indicated, I need to specify the actual transforms. So if I may, I will specify them here and further hope that there is an avenue to persue. The problem is a bit more complicated since I don't even have isomorphism. Take the four Appell functions. They are hypergeometric functions of two variables. If we consider only the variables (x,y) and transform the polynomials x+y=0 and x-y=0 as in the initial posting then the following are the transforms between these functions and themselves: 1 F1 (x,y) <---> F1 ( x/(x-1), y/(y-1) ) 2 F1 (x,y) <---> F1 ( x/(x-1) (y-x)/(1-x) ) 3 F1 (x,y) <---> F1 ( x, (x-y)/(1-y) ) 4 F1 (x,y) <---> F2 ( x, 1-x/y ) 5 F1 (x,y) <---> F2 ( x 1-y ) 6 F1 (x,y) <---> F2 ( .5 (1-( (x-1) (y-1))^(1/2)+ (x y)^(1/2)), .5 (1-( (x-1) (y-1))^(1/2)-(x y)^(1/2)) ) 7 F1 (x,y) <---> F3 ( x, y/(y-1) ) 8 F1 (x,y) <---> F4 ( x^2/((1-x) (y-x)), y/((1-x) (y-x)) ) 9 F2 (x,y) <---> F2 ( x/(x-1), y/(1-x) ) 10 F2 (x,y) <---> F2 ( x/(x+y-1), y/(x+y-1) ) 11 F2 (x,y) <---> F4 ( x (1-y), y (1-x) ) 12 F2 (x,y) <---> F4 ( 4 x/(1+(1-y)^(1/2))^2, ( (1-(1-y)^(1/2))/(1+(1-y)^(1/2)) )^2 ) 13 F2 (x,y) <---> F4 ( x^2/(2-x-y)^2, y^2/(2-x-y)^2 ) 14 F3 (x,y) <---> F2 ( 1/x, 1/y ) 15 F4 (x,y) <---> F2 ( (1/((1-x+y) (1+x-y)))^(1/2), (4 x y/((1-x+y) (1+x-y)))^(1/2) ) 16 F4 (x,y) <---> F4 ( 1/x,1/y) ) I have looked at the invariants for polynomials under the bilinear transform and, as you mentioned genus would be one. However, the quadratic transforms change this. In my original posting I mentioned some observed partial invariants, but I would much prefer tighter and more rigourous invariant(s). Is there an approach or is this like the 3 n+1 problem that is currently unsolvable? I understand a bit of group theory, but I see that the above transforms form a category (a small one I believe) and hence a much reduced set of known properties. If you have gotten this far, I thank you for you attention, interest and any advice you are willing to offer. -----== Sent via Deja News, The Discussion Network ==----- http://www.dejanews.com/ Easy access to 50,000+ discussion forums ============================================================================== From: Dave Rusin Subject: Re: Polynomials and Transforms Date: Thu, 25 Mar 1999 12:37:12 -0600 (CST) Newsgroups: [missing] To: josegarn@my-dejanews.com I still don't know if I understand your question. You have these four functions (with which I am not familiar) of two variables. You list 16 correspondences; I assume I am to read "F1(x, y) <--> F2( blah, blah) " as saying that "there is an explicit relationship between the value of F1 at a point (x,y) and the value of F2 at (blah, blah)". Maybe you are even saying they are equal. If this much is correct, then before we go further I'd like to restate the situation a little so that we can discuss just a single function, say F1. From #5 it appears you can (with fudge factors?) replace any F2(x,y) by F1(x, 1-y), so I would like to do so throughout the list. That way I never have to refer to any F2's. Likewise I use #14 to replace F3(x,y) with F2(1/x, 1/y) and thus with F1(1/x, 1- 1/y); there's no need to mention F3. I'd like to get rid of one of F1 or F4, too. Your relation #11 lets me get rid of the former in terms of the latter, although you have more F1's in your formulas than F4's, but the problem with doing that is that you'd need to introduce square roots. I have a problem with this, since square roots are not always defined, almost never unique, and extend poorly to complex numbers (is your function F4 defined for almost all complex pairs (x,y) ? ). But then, I see you've used square roots in some of your other formulas, too. Is this the way the formulas were given, or were you playing around with some other formulas to get these? You'd have to explain to me what this means, for example > 6 F1 (x,y) <---> F2 ( .5 (1-( (x-1) (y-1))^(1/2)+ (x y)^(1/2)), > .5 (1-( (x-1) (y-1))^(1/2)-(x y)^(1/2)) ) since by taking all combinations of signs, the right side of this correspondence seems to refer to 16 different points! Are all of them related to F1(x,y) ? Assuming you can explain what your formulas mean, it then seems to me that you intend to look at some points in the domain of F4 as somehow "equivalent": declare (x,y) to be equivalent to (x',y') if there is a chain of correspondences in your list which together let you calculate F(x',y') from F(x,y). This is indeed an equivalence relation, allowing us to consider the set D / ~ of equivalence classes (where D is the domain of F -- the plane? C x C ? [0,1]x[0,1]?) It's not unreasonable to hope that there is a nice open set in D which contains one and only one representative from each equivalence class. This is what occurs when, for example, you have a function f defined on the upper half plane in the complex plane, and you know f(z) = f(z+1) and f(z) = f(1/z); the open set you want is a fundamental domain for the group Gamma of fractional linear transformations generated by a(z)=z+1 and b(z)=1/z. That's a famous example, and you can then tile the halfplane with images of this set under combinations of a's and b's. Then for any function h you have f(z) = f(h(z)) for all z iff h is in this group. This sounds like what your original question was: how to recognize whether a transformation h is among the set of transformations which preserve a function. But another famous example should give you pause. Define a transformation a(z) = z^2 + c for some constant c. If this c is outside the Mandlebrot set (you know, that fractal thing you see on T-shirts) then this map a is chaotic. You have really no chance of deciding whether or not two points z are equivalent under iterates of this map, so if you had a function f with f(z) = f(z^2+c), it would be difficult to decide whether or not the value of f at two points was equal or not. I suspect your example is more like the modular group than the fractal example, but I don't know that and can't really decide unless I know more about what the domain of F4 is, what the correspondences mean, and what the role of the square roots is. It seems like this information would be available in whatever source you were using to learn of these functions F1-F4 in the first place. dave ============================================================================== From: " " Subject: Re: Polynomials and Transforms Date: Thu, 25 Mar 1999 11:58:48 -0800 Newsgroups: [missing] To: "Dave Rusin" Thanks for getting back so fast. I will try and explain a bit better the problem. The functions F1,F2,F3 and F4 are Appell double hypergeometric functions and can be expressed by double variable series with 4 or 5 parameters. They are generalizations of the Gaussian Hypergeometric function 2F1[a,b,c|z]. They were defined in 1880 by Appell. They can be analytically continued by their Barnes Integral representation into the entire complex plane. The transforms of one function into the other are equalities. For example: F1[a,b,b',c|x,y]=(y/x)^(-b') F2[b+b',a,b',c,b+b'|x,1-x/y] all the others are similar and apply pretty much in the full complex plane even if the series representation is not convergent. You have correctly identified that any F1 or F3 can be fully represented by F2. The F4, as far as anyone knows can not be done fully. I would like to know which set of polynomials I get buy the 16 transforms, and then later perhaps that would also give me an understand of how the parameters transform too. To avoid complications, it is safest to kept the values under the square root positive. There are of course many, many references to these functions, the most classic is the Bateman manuscript project or Appell and Kampe de Feriet's orginal work of 1926. I think you have help me narror down a bit the possiblities and I hope that it is not fractal! However, even that would say something. I hope that above helps and that I'm not becoming a pain! Jose -- On Thu, 25 Mar 1999 12:37:12 Dave Rusin wrote: [above -- djr] ============================================================================== From: " " Subject: Re: Polynomials and Transforms Date: Thu, 25 Mar 1999 22:41:18 -0800 Newsgroups: [missing] To: "Dave Rusin" You seem to understand the problem fairly well now. However, the F2 is a single, double variable of 5 paramters. Your understand of my correspondence notation is correct. I only used it to simplify the e-mail and it seems to have complicated matters! The square roots if taken only on one Riemann surface as is natural for simplification does not allow the continuation to the whole complex plane, however, as I mentioned before, there exists Barnes integral representation of it. Your comment on the <---> is the correct interpretation. I is an important point to know if the 16 transforms is complete. There is no known proof of it but it seems likely in that most have been known for quite a while. This list is a compilation of mine from several sources and from more general considerations seems to be complete. So the problem indeed, reduces to whether a given polynomial, or variety is in the 'solution set' or not. I guess the quadratic transforms rule out the chance of having a 'classic' invariant such as genus. I was hoping that there was something about these transforms that has an invariant property in common with the bilinear functions. My observations work as originally stated but I'm mystified by how well they work and have to wonder if they are trying to tell me there is something better. Thanks for all the consideration! Jose -- On Thu, 25 Mar 1999 15:58:35 Dave Rusin wrote: [above -- djr] -----== Sent via Deja News, The Discussion Network ==----- http://www.dejanews.com/ Easy access to 50,000+ discussion forums ============================================================================== From: Dave Rusin Subject: Re: Polynomials and Transforms Date: Mon, 5 Apr 1999 04:10:00 -0500 (CDT) Newsgroups: [missing] To: josegarn@my-dejanews.com I thought some more about your question, and I have some comments which you might find useful. This is a very long message, so let me just summarize the situation: All the information from the transformations you originally described amount to one invertible mapping, one quadratic mapping, one quadratic relation, and the action of a moderately large finite group. I don't know if this will be very useful or not. The finite group action can be used to reduce you to translates of a single fundamental domain as I suggested in a previous letter; this is good. The quadratic transformations are likely to display chaotic behaviour, particularly when combined by composition; this is bad. In your original question you mentioned rational functions; this suggests an algebraic aspect which I have not really considered. ------------------ First of all, I think I now know the right way to describe the problem. Let P be the domain of these functions F_i; you can visualize P as the x-y plane, or take it to be the complex 'plane' C x C or an open subset of that -- whatever your choice, it doesn't seem to matter much. I actually want to take four copies of P which I will distinguish as P[1], P[2], ...; then the points on them can be described as (a,b)[1], (a,b)[2], etc. -- again, a and b can be real, complex, etc. I have in mind that P[i] will be the domain of your functions F_i. You might want to visualize these P[i] as being a set of horizontal planes stacked above each other in space. Now I will interpret each of your "correspondences" as a pairing of points in one P[i] with another (possibly the same space). That is, we are defining an equivalence relation among the points in the union of the spaces P[1], ..., P[4]; it will be the equivalence relation generated by your 16 families of correspondences: 1 (x,y)[1] ~ ( x/(x-1), y/(y-1) )[1] for any (x,y) in P; ... 4 (x,y)[1] ~ ( x, 1 - x/y )[2] for any (x,y) in P; ... Some of your formulas refer to square roots; we will have to be a little careful about just which points are defined to be equivalent to which others. This seems to be the topic under discussion for you; you seem to have formulas which let you calculate F_i(x,y) from F_j(x',y') if (x,y)[i] ~ (x',y')[j] under this equivalence relation. (Of course, in order actually to be able to USE this, you will have to be able to trace just which of the 16 correspondences is used in which order to _demonstrate_ that (x,y)[i] ~ (x',y')[j] . I won't follow that too much.) Your original question seems to be, given some particular rational functions h(x,y) and k(x,y), can we tell whether or not (x,y)[4] ~ (h(x,y),k(x,y))[4] (say) is true for all (x,y) in P. So I think this equivalence relation captures the main points of the discussion. Let's try to see which points are equivalent to which others. ------------------ First of all, we can remove two of the planes, P[2] and P[3]. Your correspondence 5 F1(x,y) <---> F2( x, 1-y ) simply states that (x,y)[1] ~ (x, 1-y)[2] for all (x,y). Roughly speaking, if you turn the plane P[2] upside down, around the line y=1/2, you're just glueing planes P[1] and P[2] together point by point. So all other references to points in P[2] can be replaced by a reference to a point in P[1]. For example, you use a correspondence 4 F1(x,y) <---> F2( x, 1-x/y ) which I interpret as saying that (x,y)[1] ~ (x, 1 - x/y)[2] for all (x,y) in P. Since the latter point is now equivalent to (x, x/y)[1] under this glueing, we have (using the transitivity of an equivalence relation) the information (x,y)[1] ~ (x, x/y)[1] . Likewise, your correspondence #14 allows us to identify the points in P[3] with those in P[2] (and hence with those in P[1]) as (x,y)[3] ~ ( 1/x, 1/y )[2] ~ ( 1/x, 1 - 1/y)[1]. With this in mind I can present the information you supplied as a set of equivalences among points in P[1] and P[4]. Here I am re-ordering the correspondences and dropping obviously redundant ones: 1. (x,y)[1] ~ ( 1/x, 1/y )[1] 2. (x,y)[1] ~ ( x/(x-1), y/(y-1) )[1] 3. (x,y)[1] ~ ( x, (x-y)/(1-y) ) [1] 4. (x,y)[1] ~ ( x/(x-1), (y-x)/(1-x) )[1] 5. (x,y)[1] ~ ( x, x/y )[1] 6. (x,y)[1] ~ ( x/(x-y), (1-x)/(y-x) )[1] 7. (x,y)[1] ~ ( x*y, (1-y)*(1-x) )[4] 8. (x,y)[1] ~ ( (1/2)*(1-((x-1)*(y-1))^(1/2)+(x*y)^(1/2)), 1 - (1/2)*(1-((x-1)*(y-1))^(1/2)-(x*y)^(1/2)) )[1] 9. (x,y)[1] ~ ( x^2/((1-x)*(y-x)), y/((1-x)*(y-x)) )[4] 10. (x,y)[1] ~ ( x^2/(2-x-(1-y))^2, (1-y)^2/(2-x-(1-y))^2 )[4] 11. (x,y)[1] ~ ( 4* x/(1+y^(1/2))^2,((1-y^(1/2))/(1+y^(1/2)) )^2 )[4] 12. (x,y)[4] ~ ((1/((1-x+y)*(1+x-y)))^(1/2), 1 - (4*x*y/((1-x+y)*(1+x-y)))^(1/2) )[1] 13. (x,y)[4] ~ ( 1/x,1/y)[4] Again, we want to know which points on P[1] union P[4] are equivalent to which others. ------------------ Now, I have separated the first six of these statements for a reason. Each of them can be described by introducing a certain function f_i : P[1] -> P[1] and stating that p ~ f_i(p) for all points p in P; for example, f1(x,y)=(1/x,1/y) for relation #1. These six functions f_i happen to be invertible, and so by taking composites we get a _group_ G of functions f : P[1] -> P[1]; the first six correspondences then show that all the orbits of this group are in the same equivalence class. This is a very interesting group. It turns out that it has order precisely 120; that is, all composites of these six functions form only 120 different functions. (I'm pretty sure that it's isomorphic to the symmetric group Sym(5), although I haven't yet found a set of five things being permuted by all these f's ). Stated another way, the information from these six lines is that each point in P[1] is equivalent to precisely 120 others (including itself). I proved this simply by taking a generic point p=(x,y) in P, defining a set S = { p }, and then repeatedly replacing S by S union f_i(S) for each of the six functions f_i; eventually I got a set S with 120 elements with S = f(S), so that S is simply the orbit of p under S. We find for example that G includes maps like f(x,y) = (y,x), (namely f = f2 o f5 o f6 o f5 ), f(x,y) = (1-x, 1-y), (namely f=f1 o f2 o f1) etc. -- some not very obvious from the original six correspondences. Now we can divide up P into 120 regions which (except along the boundaries) are disjoint, tile the plane P, and are permuted by G. (At least we can in principle. About 30 elements of G leave a curve invariant, so these curves are in the boundaries of the translates of the fundamental domain. It looks like the region bounded by y=x, y=1-x, and y=2-1/x is a fundamental domain, although this set may have to be further split. I haven't tried to find a fundamental domain for the action of G on the _complex_ "plane" P.) To recap: I have now reduced the problem to just two planes, with sets of 120 points equivalent in the first plane, and various points in the two planes equivalent to each other through relations #7-#13. Our next goal will be to reduce to a single plane, P[4]. ------------------ I turn to relation #7, which identifies points on P[1] with some on P[4]. This pairing is not one-to-one, but it's close: the only point in P[1] which is glued to a given point in P[4] besides (x,y) itself is (y,x), and as it turns out this is already equivalent to (x,y) under the action of G. So the first six relations merely link sets of 60 points at a time in P[4]. We'll use relation #7 to move the other information from the other relations so that ALL references are to P[4] only. Two of the relations are easy to fix. The last, #13, already binds only P[4]: 13. (x,y)[4] ~ ( 1/x,1/y)[4] and the one before it, #12, may be restated by a direct application of #7: 12. (x,y)[4] ~ ((1/((1-x+y)*(1+x-y)))^(1/2), 1 - (4*x*y/((1-x+y)*(1+x-y)))^(1/2) )[1] implies 12'. (x,y)[4] ~ ((x^2-2*x*y-1+y^2-2*(w*x*y)^(1/2))/w^(3/2), 2*(x*y)^(1/2)*(-w^(1/2)+x^2-2*x*y-1+y^2)/w^(3/2) )[4] where w=1-(x-y)^2. The other correspondences can also be transferred to P[4], but we have to be a bit more creative when doing so. For example, equation #8 relates two points on P[1], but each corresponds to a point on P[4] via #7, so really #8 simply tells us that ( x*y, (1-y)*(1-x) )[4] ~ ( 1+(x*y)^(1/2), (x+y)/4 + (x*y)^(1/2) / 2 )[4] If we call the point on the left (X,Y)[4], then this relation states 8'. (X, Y)[4] ~ ( 1 + X^(1/2), (1+X-Y)/4 + X^(1/2) / 2 )[4] for all (X,Y)[4] which are paired with some (x,y)[1]. (In the complex case, this means, "... for all (X,Y) in P[4]".) Now I wish to do the same thing for the other relations 9-11. It's just a little tricky to accomplish this: in their present forms, the remaining relations can't easily be expressed with reference to points (X,Y) in P[4], since in order to go back to P[1] I would, apparently, have to extract some square roots to determine (x,y)[1]. But I can avoid that problem by restating the correpondences 9-11 above in a more symmetric way. In order to do this, let me recall that the whole point of our exercise is to study the equivalence class _generated by_ these relations, and so in particular there's no problem replacing any given point in P[1] with any of the 120 points it's conjugate to under the action of G. Consider, for example, relation #9, which asserts that point (x,y) in plane P[1] is to be equivalent to the point ( x^2/((1-x)*(y-x)), y/((1-x)*(y-x)) ) in plane P[4]. In the presence of the first 6 relations, we know that then a set of 120 points in P[1] is equivalent to this point in P[4]. Among them is (x',y'):=(x/(x-1), x(x-y)/(y(x-1)) ). Expressing the coordinates of the point in P[4] in terms of these x' and y' shows that #9 simply says (x',y')[1] ~ ( x'*y', x'*y' + 1 - (x'+y') )[4]. Since this last expression is now symmetric in x' and y' we may express in terms of the coordinates of the point (X,Y) on P[4] to which (x',y')[1] has already been attached. (In this particular case we see relation #9 has added _nothing_ beyond what is known from the first seven relations.) Likewise we can proceed with the other relations: from 10. (x,y)[1] ~ ( x^2/(2-x-(1-y))^2, (1-y)^2/(2-x-(1-y))^2 )[4] and (x',y') := (x, x/y) ~ (x,y) on P[1] we get (x',y')[1] ~ (x'^2*y'^2/(-y'+x'*y'-x')^2, (-y'+x')^2/(-y'+x'*y'-x')^2)[4] Now the point on the left is equivalent #7 to (X,Y) := ( x'*y', 1+x'*y'-x'-y') in P[4], AND since we have achieved a symmetric expression in x' and y' we obtain an equivalence of two points in P[4]: 10". (X,Y)[4] ~ ( X^2/(Y-1)^2, ((1+X-Y)^2-4*X)/(Y-1)^2 )[4] The same preparation turns #11 into: (x',y')[1] ~ (x,y)[1] ~ (4*x'*y'/(y'+2*x'^(1/2)*y'^(1/2)+x'), (y'-2*x'^(1/2)*y'^(1/2)+x')/(y'+2*x'^(1/2)*y'^(1/2)+x'))[4] But again (x',y')[1] is by #7 equivalent to (X,Y)[4]=(x'*y', 1+x'*y'-(x'+y'))[4], and we see we may express #11 in terms of X and Y only: 11'.(X,Y)[4] ~ (4*X/(2*X^(1/2)+1+X-Y), (-2*X^(1/2)+1+X-Y)/(2*X^(1/2)+1+X-Y))[4] To recap again: we have now reduced the problem to studying a single plane P=P[4] and an equivalence relation on it determined by two things: a partition into sets of 60 via the action of the group G somewhere else, and a set of five relations we will assume hold for all (x,y) in P: 13. (X,Y) ~ ( 1/X,1/Y ) 10". (X,Y) ~ ( X^2/(Y-1)^2, ((1+X-Y)^2-4*X)/(Y-1)^2 ) 8'. (X,Y) ~ ( 1 + X^(1/2), (1+X-Y)/4 + X^(1/2) / 2 ) 11'. (X,Y) ~ ( 4*X/(2*X^(1/2)+1+X-Y), (-2*X^(1/2)+1+X-Y)/(2*X^(1/2)+1+X-Y) ) 12'. (X,Y) ~ ( (X^2-2*X*Y-1+Y^2-2*(w*X*Y)^(1/2))/w^(3/2), 2*(X*Y)^(1/2)*(-w^(1/2)+X^2-2*X*Y-1+Y^2)/w^(3/2) ) where w=1-(X-Y)^2. The next step is to consider what is probably meant by the square roots. ------------------ Consider equation 8'. I assume the two references to X^(1/2) are to refer to the same number both times. I could call this x, say, and then replace X by x^2. For simplicity I'd rather use a slightly different replacement: {X = (x-1)^2, Y = x^2-4*y }; then 8' becomes ( (x-1)^2, x^2-4*y ) ~ ( x, y ) ; reading backwards and renaming the variables we have a replacement 8". (X,Y) ~ ( (X-1)^2, X^2-4*Y ) In exactly the same way we replace 11' by 11". (X,Y) ~ (X^2/(Y-1)^2, (-2*X-2*X*Y+Y^2-2*Y+1+X^2)/(Y-1)^2) but this turns out to be precisely the same relation as #10". Finally we have 12'. (X,Y) ~ ((X^2-2*X*Y-1+Y^2-2*(w*X*Y)^(1/2))/w^(3/2), 2*(X*Y)^(1/2)*(-w^(1/2)+X^2-2*X*Y-1+Y^2)/w^(3/2) ) where w=1-(X-Y)^2. which I have been unable to simplify into something more meaningful. So now the situation is that we wish to identify the equivalence classes in the plane P which result from the sets of 60 and the relations 13. (X,Y) ~ ( 1/X,1/Y ) 8". (X,Y) ~ ( (X-1)^2, X^2-4*Y ) 10". (X,Y) ~ ( X^2/(Y-1)^2, ((1+X-Y)^2-4*X)/(Y-1)^2 ) 12'. (X,Y) ~ ( (X^2-2*X*Y-1+Y^2-2*(w*X*Y)^(1/2))/w^(3/2), 2*(X*Y)^(1/2)*(-w^(1/2)+X^2-2*X*Y-1+Y^2)/w^(3/2) ) where w=1-(X-Y)^2. All the except the last are now square-root-free. We may write relation #13 as asserting that p ~ f(p) for all p in P, where f is the function f(X,Y)=(1/X,1/Y). Note that f is of order 2 (f o f = identity map) so the effect of relation #13 is simply to enlarge the sets of equivalent points to sets of 120 elements each (again). We compute that if x = (X+Y-1)/(Y-1), y = X*Y/(Y-1)^2, then the right side of 10" may be written ( (x-1)^2, x^2-4*y ), and so in the presence of relation 8" this tells us (X,Y) ~ (x,y), so we replace equation 10" by the statement that p ~ f(p) for all p in P, where f is the function f(X,Y) = ( (X+Y-1)/(Y-1), X*Y/(Y-1)^2 ). This is similar to the other relations discussed before except that this f has _infinite order_ : each point p is now equivalent to a doubly-infinite sequence of points ..., f^(-1)(p), p, f(p), f(f(p)), ... You can probably find a fundamental domain for the action of f on P (a subset of P which contains one representative from each orbit). Of course the equivalence classes of a point will be even larger than this simple sequence, since before each application of f or f^(-1) we can pass from a point to one of the 120 points equivalent to it from the other relations. There's probably a way to sort out the two patterns of fundamental domains, but it's likely to get quite complicated unless some iterate of f can be shown to permute the sets of 120 or something. Similarly, relation 8" itself is of the form p ~ Q(p) for a certain function Q. This time Q is _quadratic_ (it's two-to-one), not invertible. Thus even ignoring the action of the group of order 120 or the map f above, we are still going to get complicated -- possibly even chaotic behaviour. The points equivalent to p using just this information are the points in the sets (Q^i)^(-1) (Q^j(p)) for various i and j. If you think about the simple maps Q(z) = z^2+c on the complex plane, you know that quadratic maps are already quite capable of producing chaotic results. Finally, I try to simplify 12' a bit. First I write it in the form 12". (X,Y) ~ ( -(w1+2*w2)/w1^2, -2*w2*(1+w1)/w1^2 ) where w1^2=1-(X-Y)^2, w2^2=X*Y Now, we can combine the information in #10" with the equivalences of sets of 60 points in P; among those 60 are (X,Y) ~ (1/Y, X/Y) and (X,Y) ~ (Y/X, 1/X). Combining with #10" and changing variables, we conclude that (x,y) ~ Q(x,y) = (1-2*(x+y)+(x-y)^2, (x-y)^2) for all (x,y). Then we may rewrite 12", replacing the (X,Y) on the left by Q(X,Y). This then gives a relation which is symmetric in X and Y, on both sides of the relation. That is, since we know (1-2*(X+Y)+(X-Y)^2, (X-Y)^2) ~ ( -(w1+2*w2)/w1^2, -2*w2*(1+w1)/w1^2 ) (where w1^2=1-(X-Y)^2, w2^2=X*Y) for any X and Y, we may write this as (1-2*S+S^2-4*P, S^2-4*P) ~ ( -(w1+2*w2)/w1^2, -2*w2*(1+w1)/w1^2 ) (where w1^2=1-S^2+4*P, w2^2=P) for any S and P, i.e. (2-2*S-w1^2, 1-w1^2) ~ ( -(w1+2*w2)/w1^2, -2*w2*(1+w1)/w1^2 ) for any w1 and w2, where S^2 = 1 - w1^2 + 4 w2^2. It is possible to parameterize the set of these combinations: simply take w1 = -2*m/(1+m^2-n^2), w2 = -n/(1+m^2-n^2), S = (-1+m^2-n^2)/(1+m^2-n^2) for any m and n. Thus we may write this last relation as (4*(1-n^2)/(1+m^2-n^2)^2, (1-2*m^2-2*n^2+m^4-2*m^2*n^2+n^4)/(1+m^2-n^2)^2) ~ (1/2*(m+n+m^3+m^2*n-m*n^2-n^3)/m^2, 1/2*(1+m^2-n^2-2*m)*n/m^2) for any m and n. So to recap a third time: we have now reduced the problem to studying a single plane P and an equivalence relation on it determined by three things: (1) a partition into sets of 120 via inversion and the action of the group G somewhere, (2) an invertible mapping f(X,Y) = ( (X+Y-1)/(Y-1), X*Y/(Y-1)^2 ) and a quadratic polynomial mapping Q(X,Y) = ( 1-2*(X+Y)+(X-Y)^2, (X-Y)^2 ) with which we impose the equivalences p ~ f(p) and p ~ Q(p) for all p in P, and (3) the relation between certain pairs of points in P, (4*(1-n^2)/(1+m^2-n^2)^2, (1-2*m^2-2*n^2+m^4-2*m^2*n^2+n^4)/(1+m^2-n^2)^2) ~ (1/2*(m+n+m^3+m^2*n-m*n^2-n^3)/m^2, 1/2*(1+m^2-n^2-2*m)*n/m^2) assumed to hold for all m, n. I would be willing to believe these pieces of information are redundant, but I certainly see no relations among them yet. ------------------ I'm not really sure what all this gets you. It's now easier to generate a list of the points equivalent to a given point, in a way which is likely to give fewer redundancies than would have resulted from your original 16 relations. Given a point p we can compute Q^j(p) for some small j and then apply (Q^i)^(-1) for some small values of i>=0 to get an initial list. To each of these we can apply f and f^(-1) a few times to get more equivalent points. Then every point can be replaced by a set of 120 points derived by retreating to plane P[1], applying the operations f_1, f_2, and f_5 enough times, and then returning to P[4] (and adding (1/X, 1/Y) to the list for each (X,Y) already there). Repeat ad infinitum. Toss in transformations 12' whenever possible. Not very interesting. I'm not really sure where to take the matter from here. Perhaps if you have a specific example or further question I can try to work out some more details. I can't imagine it's not already known at least that there's an action of the symmetric group Sym(5) on this domain in an interesting way. You did say this topic is over one hundred years old, right? dave ============================================================================== [Here are some notes I kept during this analysis --djr] The transformation (x,y) -> (y/x,(y-1)/(x-1)) has order 5 Original set of relations, retyped for uniformity: 1 F1(x,y) <---> F1( x/(x-1), y/(y-1) ) 2 F1(x,y) <---> F1( x/(x-1), (y-x)/(1-x) ) 3 F1(x,y) <---> F1( x,(x-y)/(1-y) ) 4 F1(x,y) <---> F2( x, 1-x/y ) 5 F1(x,y) <---> F2( x, 1-y ) 6 F1(x,y) <---> F2( (1/2)*(1-((x-1)*(y-1))^(1/2)+(x*y)^(1/2)), (1/2)*(1-((x-1)*(y-1))^(1/2)-(x*y)^(1/2)) ) 7 F1(x,y) <---> F3( x, y/(y-1) ) 8 F1(x,y) <---> F4( x^2/((1-x)*(y-x)), y/((1-x)*(y-x)) ) 9 F2(x,y) <---> F2( x/(x-1), y/(1-x) ) 10 F2(x,y) <---> F2( x/(x+y-1), y/(x+y-1) ) 11 F2(x,y) <---> F4( x*(1-y), y*(1-x) ) 12 F2(x,y) <---> F4( 4 x/(1+(1-y)^(1/2))^2,((1-(1-y)^(1/2))/(1+(1-y)^(1/2)) )^2 ) 13 F2(x,y) <---> F4( x^2/(2-x-y)^2, y^2/(2-x-y)^2 ) 14 F3(x,y) <---> F2( 1/x, 1/y ) 15 F4(x,y) <---> F2((1/((1-x+y)*(1+x-y)))^(1/2),(4*x*y/((1-x+y)*(1+x-y)))^(1/2) ) 16 F4(x,y) <---> F4( 1/x,1/y ) Rewritten so as not to mention F2, F3: (Use #5: F2(x,y) <---> F1(x,1-y); and #14: F3(x,y) <---> F2( 1/x, 1/y ) <---> F1( 1/x, 1 - 1/y) ) Then reorder equations, change variables to 'F1(x,y)=' whenever possible. Also drop eq. 9 (now says the same as eq #2). What remains: 7 F1(x,y) <---> F1( 1/x, 1/y ) 1 F1(x,y) <---> F1( x/(x-1), y/(y-1) ) 3 F1(x,y) <---> F1( x, (x-y)/(1-y) ) 2 F1(x,y) <---> F1( x/(x-1), (y-x)/(1-x) ) 4 F1(x,y) <---> F1( x, x/y ) 10 F1(x,y) <---> F1( x/(x-y), (1-x)/(y-x) ) 6 F1(x,y) <---> F1( (1/2)*(1-((x-1)*(y-1))^(1/2)+(x*y)^(1/2)), 1 - (1/2)*(1-((x-1)*(y-1))^(1/2)-(x*y)^(1/2)) ) 11 F1(x,y) <---> F4( x*y, (1-y)*(1-x) ) 8 F1(x,y) <---> F4( x^2/((1-x)*(y-x)), y/((1-x)*(y-x)) ) 13 F1(x,y) <---> F4( x^2/(2-x-(1-y))^2, (1-y)^2/(2-x-(1-y))^2 ) 12 F1(x,y) <---> F4( 4 x/(1+y^(1/2))^2,((1-y^(1/2))/(1+y^(1/2)) )^2 ) 15 F4(x,y) <---> F1((1/((1-x+y)*(1+x-y)))^(1/2),1 - (4*x*y/((1-x+y)*(1+x-y)))^(1/2) ) 16 F4(x,y) <---> F4( 1/x,1/y)