# Question about a paper on Cosmic Topology

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I am studying the paper  " Exact Polynomial Eigenmodes for Homogeneous Spherical 3-Manifolds"  (https://arxiv.org/pdf/math/0502566v3.pdf)

I am trying to obtain the polynomials given by the equation (27) on page 14 but I am unable to do it.  Please, can you tell what are the P vectors for the computations?

Many thanks.

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Consider this answer as a hint to find the path to your complete answer which needs a little bit hand computation.

It is detailed by Felix Klein in "Theorie des Ikosaeders in engerem Sinne" where he stated on the corresponding invariant polynomials in §13 in the well named chapter "Der Formenkreis des Ikosaeders".
Even if it is in german, it is possible to follow and get hints for the long check.

Then, the regular icosahedron has 12 vertices, 20 face centers and 30 edge midpoints.

For the 12 vertices, P contains these 12 points :
$0$
$\infty$
$\epsilon^{\nu}(\epsilon+\epsilon^4)$
$\epsilon^{\nu}(\epsilon^2+\epsilon^3)$ with $\nu \in \{ 0,1,2,3,4\}$

Taking them as roots of $f$

$f = \alpha \beta \prod_{\nu}(\alpha -\epsilon^{\nu}(\epsilon+\epsilon^4) \times \beta ) \prod_{\nu}(\alpha -\epsilon^{\nu}(\epsilon^2+\epsilon^3) \times \beta )$

$= \alpha \beta (\alpha^5-(\epsilon+\epsilon^4)^5\times \beta^5) (\alpha^5-(\epsilon^2+\epsilon^3)^5\times \beta^5)$

$I_{12}= \alpha \beta ( \alpha^{10}+11 \alpha^5 \beta^5 - \beta^{10})$

Then using $I_{12}$ to compute the 2 determinants :

$I_{20} = \frac{1}{121} \left| \begin{array}{ccc} \frac{\delta^2 f}{\delta z_1^2} & \frac{\delta^2 f}{\delta z_1 \delta z_2} \\ \frac{\delta^2 f}{\delta z_1 \delta z_2} & \frac{\delta^2 f}{\delta z_2^2} \end{array} \right|$

and

$I_{30} = \frac{1}{20} \left| \begin{array}{ccc} \frac{\delta f}{\delta z_1} & \frac{\delta f}{\delta z_2} \\ \frac{\delta H}{\delta z_1} & \frac{\delta H}{\delta z_2} \end{array} \right|$

you get, after some hand computation, the (27) of your reference, which are finally 3 very linked equations.

answered Oct 8, 2016 by (360 points)

@igael,  you are great.  Many, many thanks for your help. Now I am  obtaining the equations (27).  The Klein paper is here https://math.dartmouth.edu/~doyle/docs/ikos/scan/ikos.pdf

The computations were performed with  $$\epsilon={{\rm e}^{2/5\,i\pi }}$$

exactly ! great book ... happy if it helps :)

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One direct way to derive the first equation (27) is a follows.  We use Mathematica with the code:

<< PolyhedronOperations

PolyhedronData["Icosahedron", "VertexCoordinates"]

Then we obtain a list of the coordinates of the vertices of the icosahedron:

{{0, 0, -(5/Sqrt[50 - 10 Sqrt[5]])}, {0, 0, 5/Sqrt[
50 - 10 Sqrt[5]]}, {-Sqrt[(2/(5 - Sqrt[5]))],
0, -(1/Sqrt[10 - 2 Sqrt[5]])}, {Sqrt[2/(5 - Sqrt[5])], 0, 1/Sqrt[
10 - 2 Sqrt[5]]}, {(1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2), -(1/Sqrt[10 - 2 Sqrt[5]])}, {(
1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]), 1/
2, -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2), 1/Sqrt[
10 - 2 Sqrt[5]]}, {-((1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]])), 1/2,
1/Sqrt[10 - 2 Sqrt[5]]}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2) Sqrt[(5 + Sqrt[5])/(
5 - Sqrt[5])], -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], -(1/Sqrt[
10 - 2 Sqrt[5]])}, {(-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2) Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])],
1/Sqrt[10 - 2 Sqrt[5]]}, {(-1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], 1/Sqrt[10 - 2 Sqrt[5]]}}

Now we normalize such list:

[[0, 0, -1], [0, 0, 1], [-2/5*5^(1/2), 0, -1/5*5^(1/2)], [2/5*5^(1/2), 0, 1/5*5^(1/2)], [1/10*5^(1/2)+1/2, -1/10*(50-10*5^(1/2))^(1/2), -1/5*5^(1/2)], [1/10*5^(1/2)+1/2, 1/10*(50-10*5^(1/2))^(1/2), -1/5*5^(1/2)], [-1/10*5^(1/2)-1/2, -1/10*(50-10*5^(1/2))^(1/2), 1/5*5^(1/2)], [-1/10*5^(1/2)-1/2, 1/10*(50-10*5^(1/2))^(1/2), 1/5*5^(1/2)], [1/10*5^(1/2)-1/2, -1/20*(50-10*5^(1/2))^(1/2)-1/20*(50-10*5^(1/2))^(1/2)*5^(1/2), -1/5*5^(1/2)], [1/10*5^(1/2)-1/2, 1/20*(50-10*5^(1/2))^(1/2)+1/20*(50-10*5^(1/2))^(1/2)*5^(1/2), -1/5*5^(1/2)], [-1/10*5^(1/2)+1/2, -1/20*(50-10*5^(1/2))^(1/2)-1/20*(50-10*5^(1/2))^(1/2)*5^(1/2), 1/5*5^(1/2)], [-1/10*5^(1/2)+1/2, 1/20*(50-10*5^(1/2))^(1/2)+1/20*(50-10*5^(1/2))^(1/2)*5^(1/2), 1/5*5^(1/2)]]

From the last list we obtain the corresponding P vector given by

P := [[0, 1], [-1, a], [1, -2*5^(1/2)/(5+5^(1/2))], [1, -2*5^(1/2)/(-5+5^(1/2))], [1, 1/2*(5^(1/2)+5-I*(50-10*5^(1/2))^(1/2))/(5+5^(1/2))], [1, 1/2*(5^(1/2)+5+(50-10*5^(1/2))^(1/2)*I)/(5+5^(1/2))], [1, 1/2*(5^(1/2)+5+(50-10*5^(1/2))^(1/2)*I)/(-5+5^(1/2))], [1, -1/2*(-5^(1/2)-5+(50-10*5^(1/2))^(1/2)*I)/(-5+5^(1/2))], [1, -1/4*(-2*5^(1/2)+10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)/(5+5^(1/2))], [1, 1/4*(2*5^(1/2)-10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)/(5+5^(1/2))], [1, 1/4*(2*5^(1/2)-10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)/(-5+5^(1/2))], [1, -1/4*(-2*5^(1/2)+10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)/(-5+5^(1/2))]]

From this P and using the standard procedure we obtain the following polynomial

alpha/beta*(alpha/beta+1/a)*(alpha/beta+1/10*5^(1/2)*(5+5^(1/2)))*(alpha/beta+1/10*5^(1/2)*(-5+5^(1/2)))*(alpha/beta-2/(5^(1/2)+5-I*(50-10*5^(1/2))^(1/2))*(5+5^(1/2)))*(alpha/beta-2/(5^(1/2)+5+(50-10*5^(1/2))^(1/2)*I)*(5+5^(1/2)))*(alpha/beta-2/(5^(1/2)+5+(50-10*5^(1/2))^(1/2)*I)*(-5+5^(1/2)))*(alpha/beta+2/(-5^(1/2)-5+(50-10*5^(1/2))^(1/2)*I)*(-5+5^(1/2)))*(alpha/beta+4/(-2*5^(1/2)+10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)*(5+5^(1/2)))*(alpha/beta-4/(2*5^(1/2)-10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)*(5+5^(1/2)))*(alpha/beta-4/(2*5^(1/2)-10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)*(-5+5^(1/2)))*(alpha/beta+4/(-2*5^(1/2)+10+(50-10*5^(1/2))^(1/2)*I+(50-10*5^(1/2))^(1/2)*5^(1/2)*I)*(-5+5^(1/2)))

Simplifying such polynomial and with $$a = 0$$ we obtain

$${\alpha}^{11}\beta-\alpha\,{\beta}^{11}+11\,{\alpha}^{6}{\beta}^{6}$$

which is the first polynomial in the equation (27).  The computations were performed using Mathematica and Maple.

answered Oct 9, 2016 by (1,130 points)

well, this may show that the relations are true (if you trust Mathematica), but not why...

 Courageous man ! you did the right thing ... ( Do you mean that you take 1/a=0 ? ) Now, it remains to do the same with I_20 and I_30. PS : I have a tool to convert expressions in latex, I'll suggest an edit later. @ArnoldNeumaier : please, could you say what is missing which can be added or evoked ? TY

@igael, many thanks for your comment. I am representing  $\infty =1/a$ with $a =0$. Now, I have a Mathematica code for the computation of  $I_{30}$.  I am trying to do the same for $I_{20}$. I will try to edit with latex.  All the best.

@igael: If you trust Mathematica, nothing is missing except for the insight into the truth of the results. If you don't trust it you'd have to check all the computations yourself, which is a mess and error-prone - no human would ever do such computations.

Note that symbolic computations may be faulty; for example, they conclude from c*x=c that x=1, even when you later specialize to c=0.

@ArnoldNeumaier : indeed, there is always a risk of error ... TY

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One direct way to derive the second  equation (27) is a follows.  We use Mathematica with the code:

<< PolyhedronOperations

PolyhedronData["Icosahedron", "Edges"]

Then we obtain a list of the edges of the icosahedron as line primitives:

GraphicsComplex[{{0, 0, -(5/Sqrt[50 - 10 Sqrt[5]])}, {0, 0, 5/Sqrt[
50 - 10 Sqrt[5]]}, {-Sqrt[(2/(5 - Sqrt[5]))],
0, -(1/Sqrt[10 - 2 Sqrt[5]])}, {Sqrt[2/(5 - Sqrt[5])], 0, 1/Sqrt[
10 - 2 Sqrt[5]]}, {(1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2), -(1/Sqrt[10 - 2 Sqrt[5]])}, {(
1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]), 1/
2, -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2), 1/Sqrt[
10 - 2 Sqrt[5]]}, {-((1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]])), 1/2,
1/Sqrt[10 - 2 Sqrt[5]]}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2) Sqrt[(5 + Sqrt[5])/(
5 - Sqrt[5])], -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], -(1/Sqrt[
10 - 2 Sqrt[5]])}, {(-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2) Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])],
1/Sqrt[10 - 2 Sqrt[5]]}, {(-1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], 1/Sqrt[10 - 2 Sqrt[5]]}},
Line[{{1, 3}, {1, 5}, {1, 6}, {1, 9}, {1, 10}, {2, 4}, {2, 7}, {2,
8}, {2, 11}, {2, 12}, {3, 7}, {3, 8}, {3, 9}, {3, 10}, {4, 5}, {4,
6}, {4, 11}, {4, 12}, {5, 6}, {5, 9}, {5, 11}, {6, 10}, {6,
12}, {7, 8}, {7, 9}, {7, 11}, {8, 10}, {8, 12}, {9, 11}, {10,
12}}]]

From such list we obtain the coordinates (normalized) of the edge midpoints of the icosahedron:

[[-1/10*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2), 0, -1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2)], [1/8*(10-2*5^(1/2))^(1/2)+1/40*5^(1/2)*(10-2*5^(1/2))^(1/2), 1/4-1/4*5^(1/2), -1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2)], [1/8*(10-2*5^(1/2))^(1/2)+1/40*5^(1/2)*(10-2*5^(1/2))^(1/2), -1/4+1/4*5^(1/2), -1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2)], [1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/8*(10-2*5^(1/2))^(1/2), -1/2, -1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2)], [1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/8*(10-2*5^(1/2))^(1/2), 1/2, -1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2)], [1/10*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2), 0, 1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2)], [-1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/8*(10-2*5^(1/2))^(1/2), 1/4-1/4*5^(1/2), 1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2)], [-1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/8*(10-2*5^(1/2))^(1/2), -1/4+1/4*5^(1/2), 1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2)], [-1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/8*(10-2*5^(1/2))^(1/2), -1/2, 1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2)], [-1/40*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/8*(10-2*5^(1/2))^(1/2), 1/2, 1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2)], [-1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2), 1/4-1/4*5^(1/2), 0], [-1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2), -1/4+1/4*5^(1/2), 0], [-3/40*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2), -1/2, -1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [-3/40*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2), 1/2, -1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2), 1/4-1/4*5^(1/2), 0], [1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2), -1/4+1/4*5^(1/2), 0], [3/40*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2), -1/2, 1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [3/40*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2), 1/2, 1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)+1/4*(10-2*5^(1/2))^(1/2), 0, -1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [1/20*5^(1/2)*(10-2*5^(1/2))^(1/2), -1/4-1/4*5^(1/2), -1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [1/4*(10-2*5^(1/2))^(1/2), -1/4-1/4*5^(1/2), 0], [1/20*5^(1/2)*(10-2*5^(1/2))^(1/2), 1/4+1/4*5^(1/2), -1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [1/4*(10-2*5^(1/2))^(1/2), 1/4+1/4*5^(1/2), 0], [-1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)-1/4*(10-2*5^(1/2))^(1/2), 0, 1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [-1/4*(10-2*5^(1/2))^(1/2), -1/4-1/4*5^(1/2), 0], [-1/20*5^(1/2)*(10-2*5^(1/2))^(1/2), -1/4-1/4*5^(1/2), 1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [-1/4*(10-2*5^(1/2))^(1/2), 1/4+1/4*5^(1/2), 0], [-1/20*5^(1/2)*(10-2*5^(1/2))^(1/2), 1/4+1/4*5^(1/2), 1/10*5^(1/2)*(10-2*5^(1/2))^(1/2)], [0, -1, 0], [0, 1, 0]]

From the last list we obtain the corresponding P vector given by

P := [[1, -2*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)/(20+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2))], [1, -1/2*(-5*(10-2*5^(1/2))^(1/2)-5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))/(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, 1/2*(5*(10-2*5^(1/2))^(1/2)+5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))/(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, -1/2*(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)+20*I)/(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, 1/2*(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)+20*I)/(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, -2*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)/(-20+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2))], [1, 1/2*(5*(10-2*5^(1/2))^(1/2)+5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))/(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, -1/2*(-5*(10-2*5^(1/2))^(1/2)-5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))/(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, 1/2*(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)+20*I)/(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, 1/2*(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)-20*I)/(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2))], [1, -1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2)+1/4*I-1/4*I*5^(1/2)], [1, -1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2)-1/4*I+1/4*I*5^(1/2)], [1, -1/4*(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)/(10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2))], [1, -1/4*(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)-20*I)/(10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2))], [1, 1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2)+1/4*I-1/4*I*5^(1/2)], [1, 1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2)-1/4*I+1/4*I*5^(1/2)], [1, 1/4*(-3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)/(-10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2))], [1, -1/4*(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)/(-10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2))], [1, 1/2*(10-2*5^(1/2))^(1/2)*(5+5^(1/2))/(10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, -1/2*(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))/(10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, 1/4*(10-2*5^(1/2))^(1/2)-1/4*I-1/4*I*5^(1/2)], [1, 1/2*(5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))/(10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, 1/4*(10-2*5^(1/2))^(1/2)+1/4*I+1/4*I*5^(1/2)], [1, 1/2*(10-2*5^(1/2))^(1/2)*(5+5^(1/2))/(-10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, -1/4*(10-2*5^(1/2))^(1/2)-1/4*I-1/4*I*5^(1/2)], [1, 1/2*(5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))/(-10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, -1/4*(10-2*5^(1/2))^(1/2)+1/4*I+1/4*I*5^(1/2)], [1, -1/2*(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))/(-10+5^(1/2)*(10-2*5^(1/2))^(1/2))], [1, -I], [1, I]]

From this P and using the standard procedure we obtain the following polynomial

(alpha/beta+1/20/(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)*(20+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)))*(alpha/beta+2/(-5*(10-2*5^(1/2))^(1/2)-5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))*(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta-2/(5*(10-2*5^(1/2))^(1/2)+5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))*(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta+2/(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)+20*I)*(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta-2/(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)+20*I)*(20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta+1/20/(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)*(-20+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)))*(alpha/beta-2/(5*(10-2*5^(1/2))^(1/2)+5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))*(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta+2/(-5*(10-2*5^(1/2))^(1/2)-5^(1/2)*(10-2*5^(1/2))^(1/2)-10*I+10*I*5^(1/2))*(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta-2/(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)+20*I)*(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta-2/(5^(1/2)*(10-2*5^(1/2))^(1/2)-5*(10-2*5^(1/2))^(1/2)-20*I)*(-20+5^(1/2)*(10-2*5^(1/2))^(1/2)+5*(10-2*5^(1/2))^(1/2)))*(alpha/beta-1/(-1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2)+1/4*I-1/4*I*5^(1/2)))*(alpha/beta-1/(-1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-1/8*(5-5^(1/2))^(1/2)*2^(1/2)-1/4*I+1/4*I*5^(1/2)))*(alpha/beta+4/(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)*(10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)))*(alpha/beta+4/(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)-20*I)*(10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)))*(alpha/beta-1/(1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2)+1/4*I-1/4*I*5^(1/2)))*(alpha/beta-1/(1/8*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+1/8*(5-5^(1/2))^(1/2)*2^(1/2)-1/4*I+1/4*I*5^(1/2)))*(alpha/beta-4/(-3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)-5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)*(-10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)))*(alpha/beta+4/(3*(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)+5*(5-5^(1/2))^(1/2)*2^(1/2)+20*I)*(-10+(5-5^(1/2))^(1/2)*5^(1/2)*2^(1/2)))*(alpha/beta-2/(10-2*5^(1/2))^(1/2)/(5+5^(1/2))*(10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta+2/(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))*(10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta-1/(1/4*(10-2*5^(1/2))^(1/2)-1/4*I-1/4*I*5^(1/2)))*(alpha/beta-2/(5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))*(10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta-1/(1/4*(10-2*5^(1/2))^(1/2)+1/4*I+1/4*I*5^(1/2)))*(alpha/beta-2/(10-2*5^(1/2))^(1/2)/(5+5^(1/2))*(-10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta-1/(-1/4*(10-2*5^(1/2))^(1/2)-1/4*I-1/4*I*5^(1/2)))*(alpha/beta-2/(5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))*(-10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta-1/(-1/4*(10-2*5^(1/2))^(1/2)+1/4*I+1/4*I*5^(1/2)))*(alpha/beta+2/(-5^(1/2)*(10-2*5^(1/2))^(1/2)+5*I+5*I*5^(1/2))*(-10+5^(1/2)*(10-2*5^(1/2))^(1/2)))*(alpha/beta-I)*(alpha/beta+I)

Simplifying such expression we obtain  14411518807585587200000000000000 times the following polynomial

$$522\,{\alpha}^{25}{\beta}^{5}+{\beta}^{30}-10005\,{\beta}^{10}{\alpha} ^{20}-522\,{\beta}^{25}{\alpha}^{5}-10005\,{\beta}^{20}{\alpha}^{10}+{ \alpha}^{30}$$

which is precisely the second polynomial in the equation (27).

answered Oct 10, 2016 by (1,130 points)

tex : I hacked a hack : just replace the ":=" assignation which is not recognized by a "=" sign. I added automatic new lines suiting for the datas and the products above

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One direct way to derive the third polynomial in  equation (27) is a follows.  We use Mathematica with the following code:

<< PolyhedronOperations`

PolyhedronData["Icosahedron", "Faces"]

in order to obtain the list of the faces of the icosahedron as polygon primitives.  The result is

GraphicsComplex[{{0, 0, -(5/Sqrt[50 - 10 Sqrt[5]])}, {0, 0, 5/Sqrt[
50 - 10 Sqrt[5]]}, {-Sqrt[(2/(5 - Sqrt[5]))],
0, -(1/Sqrt[10 - 2 Sqrt[5]])}, {Sqrt[2/(5 - Sqrt[5])], 0, 1/Sqrt[
10 - 2 Sqrt[5]]}, {(1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2), -(1/Sqrt[10 - 2 Sqrt[5]])}, {(
1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]), 1/
2, -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2), 1/Sqrt[
10 - 2 Sqrt[5]]}, {-((1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]])), 1/2,
1/Sqrt[10 - 2 Sqrt[5]]}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])), -(1/2) Sqrt[(5 + Sqrt[5])/(
5 - Sqrt[5])], -(1/Sqrt[10 - 2 Sqrt[5]])}, {-((-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]])),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], -(1/Sqrt[
10 - 2 Sqrt[5]])}, {(-1 + Sqrt[5])/(
2 Sqrt[10 - 2 Sqrt[5]]), -(1/2) Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])],
1/Sqrt[10 - 2 Sqrt[5]]}, {(-1 + Sqrt[5])/(2 Sqrt[10 - 2 Sqrt[5]]),
1/2 Sqrt[(5 + Sqrt[5])/(5 - Sqrt[5])], 1/Sqrt[10 - 2 Sqrt[5]]}},
Polygon[{{2, 12, 8}, {2, 8, 7}, {2, 7, 11}, {2, 11, 4}, {2, 4,
12}, {5, 9, 1}, {6, 5, 1}, {10, 6, 1}, {3, 10, 1}, {9, 3, 1}, {12,
10, 8}, {8, 3, 7}, {7, 9, 11}, {11, 5, 4}, {4, 6, 12}, {5, 11,
9}, {6, 4, 5}, {10, 12, 6}, {3, 8, 10}, {9, 7, 3}}]]

Now this result is translated to Maple with the following code:

P0:=subs(H=sqrt((5 + sqrt(5))/(5 - sqrt(5)))/2,G=(-1 + sqrt(5))/2,F=(1/2)*(1 + sqrt(5)),C=sqrt(2/(5 - sqrt(5))),A= (1/sqrt(10 - 2*sqrt(5))),B=5/sqrt(50 - 10*sqrt(5)),[[0, 0, -B], [0, 0, B], [-C, 0, -A], [C, 0, A], [A*F, -(1/2), -A], [A*F, 1/2, -A], [-A*F, -(1/2), A], [-A*F, 1/2, A], [-A*G, -H, -A], [-A*G, H, -A], [A*G, -H, A], [A*G, H, A]]);

Faces:=[[2, 12, 8], [2, 8, 7], [2, 7, 11], [2, 11, 4], [2, 4, 12], [5, 9, 1], [6, 5, 1], [10, 6, 1], [3, 10, 1], [9, 3, 1], [12, 10, 8], [8, 3, 7], [7, 9, 11], [11, 5, 4], [4, 6, 12], [5, 11, 9], [6, 4, 5], [10, 12, 6], [3, 8, 10], [9, 7, 3]];

With such data we compute the coordinates (normalized) of the centers of the faces of the icosahedron using the following Maple code

with(geom3d):

for j from 1 to 20 do ps:=[point(A,seq(P0[Faces[j][1]][i],i=1..3)),point(B,seq(P0[Faces[j][2]][i],i=1..3)),point(C,seq(P0[Faces[j][3]][i],i=1..3))]:centroid(G,ps):V[j]:=coordinates(G): end do:

P1:=factor(expand(((-5/3*(5+5^(1/2))/(-5+5^(1/2))^3)^(-1/2))*[seq(V[j],j=1..20)])):

Now we determine the corresponding stereographic P vector using the command

P:=simplify([seq([1,(P1[i][1]+I*P1[i][2])/(1-P1[i][3])],i=1..20)]):

Finally, we compute the corresponding invariant polynomial using the following Maple code

eq1:=[seq(alpha/beta-P[i][1]/P[i][2],i=1..nops(P))]:

eq2:=product(eq1[i],i=1..nops(P)):

eq3:=expand(numer(simplify(eq2)));

eq3A:=eq3/coeff(eq3,beta,20);

Then, we obtain 3518437208883200000000000000 times the following polynomial

$$228\,{\beta}^{15}{\alpha}^{5}+494\,{\beta}^{10}{\alpha}^{10}+{\beta}^{ 20}+{\alpha}^{20}-228\,{\alpha}^{15}{\beta}^{5}$$

which is precisely the third polynomial in the equation (27).

answered Oct 11, 2016 by (1,130 points)

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