I doubt you can improve much on an algebraic parametrization unless you get really lucky. One thing I can mention is that you probably should parametrize the first three ellipses using different trig variables. You then will have two remaining equations in the three trig variables. You can solve for five of the six trig functions in terms of the sixth one. If you have some nice insight at this point (which I did not) then maybe you will find simplifications that can be used.

This will leave you with some fairly nasty algebraic functions of that last trig, and you will still be stuck with square roots which means you will not avoid branch cut problems.

I'm guessing this is more or less what you did originally, but with rational rather than trig parametrizations of the three ellipses.

Hi Daniel 

Thank you for your suggestion and answer ! I tried several different combinations  and also parameter s for b ellipse, but final result did not look much easier than this.

Thank you for your answer though, because what's really hunting me is that I might have missed some obvious trick and parametrization here. 

I am glad that you had the time to look into this and agree with me. To be able to parametrize the variety even this way was already good and as I said enough for me. Sometimes getting proper parametrizations for varieties is just impossible and an open problem I think.

Thank you again for your answer Daniel ! 

kind regards

Samuli Piipponen

Dear Samuli,

                        I've checked this in Magma, and your curve is a genus 1 curve - ie, it is an elliptic curve once you take the projective closure in 6-D projective space P^6 and desingularise (and choose a base point).

This means that you will NOT be able to parametrise by trigonometric functions (without taking roots or whatever of some of the trigonometric functions at the end or some other construction which will give branching functions as you have). If you COULD parametrise by trig functions, you would have a rational map from a complex cylinder to your curve C, which is the same analytically as a rational map from the punctured complex plane to C,. This would extend to a holomorphic map from the Riemann sphere (P^1) to the projective desingularisation of C. But no non-constant holomorphic map exists from P^1 to a curve of genus > 0!

Mike Harrison

Dear Samuli,

Your curve C is actually equivalent to the genus 1 curve E with defining equation y^2=x^3-102x^2+2592x over the real numbers (actually, only using the square roots of 2 and 3). I've explicitly computed (using Magma again: it's very handy :-)) a birational isomorphism from E to C and that gives a parametrisation of C using only rational polynomial functions and square-roots, if you are interested. The expressions are a bit involved. Here it is:

Write a = SquareRoot(2), b = SquareRoot(3) [say the positive square roots].

Let x be any real number (the parametrising variable).

y = Squareroot(x^3-102x^2+2592x)=Squareroot(x(x-48)(x-54))

y can be either square-root. If you want to stick to real values, you need 0 (0,1/3,1,1/b,-1/a,0)

Also, the one point in C that is not included in the image of the parametrisation is (0,1/3,-1/b,1/b,1/a,0), which is actually the image of 'the point at infinity' of E.

Besides the case y=0 (occuring for those specific 3 values of x),  the parametrisation map is not defined when Q1,Q2 or Q3=0.

Q2 or Q3 =0 for 4 values of x and Q1=0 when x^2 - 120*x + 2592=0 or x^2 - 104*x + 2592=0 and y is the specific root -D/(2*a). This together gives 12 values for (x,y) pairs. In fact, these 12 (x,y) pairs map to the 'points at infinity' of C, if you take the parametrisation map as a map between projective curves.

Hi Michael 

Thank's for you conformation ! I also came with rational parametrization where you have to consider special points at infinity, where you would have to change your parametrization to an other rational one. Unfortunately rational parametrizations are not very practical for me since the components a_i,b_i,c_i in my application represent Euler parameters in Euler rotation matrices. 

In this context dealing with trigonometric functions is little bit easier and natural for me. But thank you for confirming my fears that no obvious 'easy' parametrization seem to be available .

I can manage the required computations with parametrization I have although it requires little bit labor work. 

Thank's to both of you for your answers and time you dedicated to look into the question !

kind regards 

Samuli Piipponen

Let a0=cos(t); a1=sin(t)/sqrt(3);

b0=(2sqrt(3)+sqrt(3)cos(2t)+3sin(2t))/3sqrt(5+4cos(2t));

b1=-(2sqrt(3)+sqrt(3)cos(2t)-3sin(2t))/3sqrt(5+4cos(2t));

c1=(2cos(t)+sqrt(3)sin(t))/(sqrt(6)sqrt(5+4cos(2t)));

c2=-sqrt(6)cos(t)/sqrt(5+4cos(2t)).

This gives a parametrization of your curve in terms of t with the worst feature being a square root of a trig function. It is periodic with period 2 pi, so you have a closed curve.

I have checked this with Mathematica, and it appears to work.

Cheers,

David Singer

Slight mistake. In the formula for c1, it should start with 3cos(t).

Hi David 

Thank you so much for your answer and effort. This would indeed look better than

one of my trials, which looked similar although messier. Instead of changing branches of trig. functions it involved changing +sqrt to - sqrt at some point. I already constructed a code using my original parametrization, but I just might try this one also, so thank you so much !!! 

Yes from the origins of the problem it was fairly clear to me from the beginning that the curve should and would be a smooth Jordan curve in R6  and quite possibly here an union of 4 smooth Jordan curves (because of the +- choices for b:s and c:s in sqrt:s here) which all represent the same rotation matrices in SO(3)  due to reflection symmetry of Euler parameters. From your constructions this look's perhaps more visible. 

In C6  of course the curve/variety should consist only of one component (not 4 ) since the variety should itself be irreducible. Although I initially suspected that it still might be a union of exactly 4 irreducible varieties/components, because of the same reasoning I gave above (antipodal points in Euler's representation of SO(3) correspond to same rotation).

I just might still check this with program called Bertini (https://bertini.nd.edu/). If Bertini actually gives 4 components this would mean that I am symbolically working over to small extension field of Q. For symbolic computations for the ideal I am using the program Singular (http://www.singular.uni-kl.de/).  

But my application is very real and it will not make a difference which one I pick of the 4 real curves I pick.

Thank you again David for you answer and your effort !

kind regards

Samuli Piipponen 

Hi Samuli,

No problem.

David's 'biquadratic' parametrisation (with sin(t) and sqrt(5+4cos(2t)) as the square roots of  quadratic polys in cos(t)) is much nicer than mine and sounds like the kind of thing you were looking for.

Mike.

Hi Samuli,

I ran this using Bertini and found that the variety decomposes over the complex field into two curves of degree 8 (not four as you conjectured).  The attached text file, main_data, contains 8 points on each irreducible component, obtained from a random hyperplane section.  It also echos the input file I used so you can run it too.  (Remove the RandomSeed configuration line to get a new random hyperplane section.)

Hope that is useful info for you,

Charles

Hi Charles

Thank you so much for your anwer ! I came up with exactly the same result using Bertini. But I think now there is no conradiction here. What has happened I think is the following:

The variety represents 2 smooth curves in C6 . If you take t as a 'complex parameter' t\in C in D.Singers parametrization the function f sqrt(5+4cos(2t)), will definitely not remain bounded. since cos(2t) is not bounded.

However as you look the variety in R6it will have Euclidiean topologically 4 components.

So if we abobt Euclidean topology I think

1) in R6  the variety V(K) consists of 4 components, which are smooth Jordan 

curves.

2) In C6 the variety V(K) consists of 2 components--> In Zariski topology it should also consist of two components, so in fact K should still be an intersection of two prime ideals which means that I am initially working over too small extension field of Q.

The (+,+),(-.-),(+,-) & (-,+) choices for b:s and c:s in square roots. I think all

these were possible. Although I made rigorous test with only 1 of them which worked and was enough for me :-).

So in R6 the situation looks now fine to me and I am in fact glad I and Charles checked teh V(K) with Bertini since now K should be decomposed into two primes. Just need the correct minimal polynomial for the extension field. This is worth of little research although not mandatory for me. So you really do not have to search it for me. The results I have here are already enough for me. And I have already done infact all the necesssary computations.

In fact considering the structure of V(K) in  C6 the stucture of the varieties/curves is not probably what you would expect if you consider the application I am. The V(K) might not eaven be closed in C6 .

In this application the restriction of the variety V(K) to R6 should infact represent smooth Jordan curve(s) so in this sense D.Singer's parametrization(s) were probably most suitable ones. Well I only checked the one he gave here, but used my original one for furher computations.

If people here are interested you might want to search the field extension in which K brakes into two. Although if the required larger extension is transcendental don't bother :-). Although from the origins I would conjecture its not.

But thank you Charle's for confirming my Bertini computations also !!

kind regards

Samuli Piipponen

I went through some of the algebra of rational parametrizations (no trigs), to see if it gave anything useful. Jury is out on that last; this might be what you already had or it might be an obfuscated variant of what @Michael Harrison showed. Anyway, here goes.

We first parametrizethe pair  c1, c2 using r, b0, b1 using s, and a0,a1 using t. In Mathematica notation (less error prone if I just cut and paste):

{c1 -> (Sqrt[2/3]*r)/(1 + r^2) + (-1 + r^2)/(Sqrt[2]*(1 + r^2)),

c2 -> (Sqrt[2/3]*r)/(1 + r^2) - (-1 + r^2)/(Sqrt[2]*(1 + r^2)),

b0 -> (2*s)/(1 + s^2) + (-1 + s^2)/(Sqrt[3]*(1 + s^2)),

b1 -> (2*s)/(1 + s^2) - (-1 + s^2)/(Sqrt[3]*(1 + s^2)),

a0 -> (-1 + t^2)/(1 + t^2), a1 -> (2*t)/(Sqrt[3]*(1 + t^2))}

Plugging into the last two polynomials gives two equations in the variables (r,s,t).

{3 - 6 Sqrt[3] r - 3 r^2 + 2 Sqrt[3] t + 4 r t - 2 Sqrt[3] r^2 t -

3 t^2 + 6 Sqrt[3] r t^2 + 3 r^2 t^2,

-Sqrt[2] (-Sqrt[3] + 6 r +Sqrt[3] r^2 - 8 Sqrt[3] r s + Sqrt[3] s^2 -

6 r s^2 - Sqrt[3] r^2 s^2 + 4 Sqrt[3] s t - 8 r s t -

4 Sqrt[3] r^2 s t + Sqrt[3] t^2 - 6 r t^2 - Sqrt[3] r^2 t^2 +

8 Sqrt[3] r s t^2 - Sqrt[3] s^2 t^2 + 6 r s^2 t^2 +

Sqrt[3] r^2 s^2 t^2)}

A Groebner basis for this, over the rational function field C(t) and with r>s in lex order, is as below.

{2 t - 2 t^3 + s^2 (-2 t + 2 t^3) + s (-3 + 2 t^2 - 3 t^4),

3 + 2 Sqrt[3] t - 3 t^2 + r^2 (-3 - 2 Sqrt[3] t + 3 t^2) +

r (-6 Sqrt[3] + 4 t + 6 Sqrt[3] t^2)}

So this has four components and I guess that means that the curve has no more than four components; I defer to pretty much everyone else in this thread on that matter. As for a parametrization, these can be used to write r and s as algebraics in t. As others have noted, the worst you get are square roots involving the parametrization variable t. The four branches are similar except for +/- choices in front of the radicals. Here is one of them.

{r -> (3 Sqrt[3] - 2 t - 3 Sqrt[3] t^2 - Sqrt[

36 - 56 t^2 + 36 t^4])/(-3 - 2 Sqrt[3] t + 3 t^2),

s -> (-3 + 2 t^2 - 3 t^4 - (1 + t^2) Sqrt[9 - 14 t^2 + 9 t^4])/(

2 (2 t - 2 t^3))}

Not sure if this is any simpler than what you already have though.

Have a look at I. R. Shafarevich, Algebra I, paragraph 1, 2,3,4 at least, where there are many interesting examples of "coordinatisations". 

Hi All

Thank you for everyone for your efforts and dedication to this problem. I have already 

done all the computations I need and from my part you can consider this case closed :-). 

So if you your own research problems going on no need to waste time anymore with this one :-) . 

Of course if you want to continue to think about this I wont stop you, but personally I do not need any more answers ! 

So thank's again to all of you ! Also I emphasize that I am from small UN from small country so this network seems really promising to me. 

with best regards

Samuli Piipponen