Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Question about a paper on Cosmic Topology

+ 3 like - 0 dislike
218 views

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.

asked Oct 1, 2016 in Mathematics by juancho (860 points) [ revision history ]

4 Answers

+ 4 like - 0 dislike

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 \(f\)

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

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

leading to :

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

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 igael (150 points) [ no revision ]

@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 :)

+ 2 like - 0 dislike

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 10, 2016 by juancho (860 points) [ no revision ]

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

+ 1 like - 0 dislike

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 juancho (860 points) [ no revision ]

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

+ 1 like - 0 dislike

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 juancho (860 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOver$\varnothing$low
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...