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The Bring sextic of equilateral pentagons

Lyle Ramshaw, retired

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WOLFRAM MATERIALS for the ARTICLE:

Lyle Ramshaw, (2022).
The Bring sextic of equilateral pentagons.
arXiv:2205.08196 [math.MG]
https://doi.org/10.48550/arXiv.2205.08196
Full article in PDF

enter image description here

The Bring sextic of equilateral pentagons
by Lyle Ramshaw
May 14, 2022

This is the calculations of the paper “The Bring sextic of equilateral pentagons” in detail.

Saving this notebook from within Mathematica — even without having done any evaluations or edits — writes out, as the file “EquilateralPentagonMap.m”, a Mathematica package that exports the conformal mapping projection of the paper as a routine named PoincareCoordinates. That routine is both specified and implemented in section [A9] below. (The package "EquilateralPentagonMap.m" consists of the cells in section [A9] that are InitializationCells, with gray backgrounds.)

Both this notebook and that package assume that the data file “ApproximatingPolynomials.m” will be available in the directory from which the notebook or package is read.

[A1]: The singular points of
E
4

Figure 3 depicts the polygon space

of equilateral quadrilaterals. It consists of three loops of decomposables, each of which is a geodesic under the bag-of-edges metric

. But what does

say about the three intersection points?

At a singular point of a polygon space, all of the

-gon’s edges lie along a single line. Perturbing their phases thus leads to a failure-to-close vector that is always orthogonal to that line, to first order. With only one dimension of failure-to-close vectors, rather than two, the dimension of the tangent space jumps up, at a singular point, from

n-3

n-2

; for

, from a tangent line to a tangent plane. And the metric

gives that entire tangent plane a Euclidean structure.

Consider the bottom intersection

θ=(0,π,0,π)

, where the four edges of the quadrilateral go back and forth over the same segment twice each way. We can leave

by opening up that quadrilateral into an L that is traveled forward and backward, with the angle of the L at either the left or the right. There are tangent vectors to

that point either way — for example,

∂

left

=(0,0,1,1)

and

∂

right

=(0,-1,-1,0)

. Their sum

∂

left

∂

right

=(0,-1,0,1)

is also a tangent vector to

, lying in the tangent plane to

at the singular point

. Note, though, that we can’t move more than an infinitesimal distance in the direction

∂

left

∂

right

while remaining in

. Opening up on both sides simultaneously while keeping the edge lengths fixed produces quadrilaterals that close up only to first order, not to second order. To have them close up, looking like

, we’d have to lengthen their crossing edges or shorten their noncrossing edges. Such bowtie-shaped quadrilaterals are modeled in the Grassmannian



(2,4)

, since we can balance the lengthening and shortening so as to preserve the required fixed perimeter.

We can determine the angle between the vectors

∂

left

and

∂

right

by applying, to them and to their sum, our formula in (1) for the bag-of-edges metric

, where

re(4)

. Taking the six pairs

(i,j)

in (1) in lexicographic order, we have

(

∂

left

)(0+1+1+1+1+0)/161/4

and



∂

right

(1+1+0+0+1+1)/161/4

, while



∂

left

∂

right

(1+0+1+1+4+1)/161/2

. It follows that



∂

left

==

∂

right

1/2

, while



∂

left

∂

right

1

, which implies that

∂

left

and

∂

right

are orthogonal.

Another way to see that this must be so is to reason as follows: For the manifold-with-singularities

(

)

to have 48 isomorphisms, we must be able to map the intersection

to any of the three intersections, say

′

, and to map the tangent plane at

to the tangent plane at

′

in one of eight ways, leaving a binary choice about how to map the two halves of the top loop. In order to have eight different ways to map the tangent plane, all three intersections must be orthogonal.

[A2]: Figures 2 depicts
+
E
5

We here verify that Figure 2 depicts the polygon space

with the orientation

, the orientation suggested by the argument in Section 7.2. Since Figures 7 and 8 have the same orientation as Figure 2 where they overlap, they also depict

. Let’s consider the isosceles trapezoid along the horizontal axis in Figure 2, which has an orange top. If we rotate that trapezoid clockwise (CW) to make its double-length edge horizontal, it has phase vector

(2π/3,π,-2π/3,0,0)

, taking the edges in rainbow order from red to purple.

◼

The tangent vector

(0,1,0,0,1)

points along the ditri-pentagon loop that runs diagonally up and to the right. As we move in that direction, we rotate the orange and purple edges CCW in lockstep, passing through a child’s house and then leaving the decatract of convex-and-CCW pentagons at the purple-topped isosceles trapezoid.

◼

The tangent vector

(0,-1,0,-1,0)

points along the ditri-pentagon loop that runs diagonally up and to the left. As we move in that direction, we rotate the orange and blue edges CW in lockstep, passing through a child’s house and leaving the decatract at the blue-topped trapezoid. So the vectors

(0,1,0,0,1)

and

(0,-1,0,-1,0)

, in that order, are a positive basis for the tangent space to

, when that tangent space has the orientation of Figure 2.

◼

The vector

(1,1,1,1,1)

is a positive basis for the real line



of rigid rotations.

◼

The orange edge

is CCW from the red edge

. Hence, rotating

a bit CCW gives a failure-to-close vector that is CCW of the analogous vector that would result from rotating

a bit CCW. Thus, the vectors

(1,0,0,0,0)

and

(0,1,0,0,0)

, in that order, are a positive basis for the plane



of failures to close.

Figure 2 depicts

because this 5-by-5 determinant is positive:

In[]:=

Det[{{0,1,0,0,1},{0,-1,0,-1,0},{1,1,1,1,1},{1,0,0,0,0},{0,1,0,0,0}}]

Out[]=

[A3]: Relations among the vertex angles

We begin with the two equations in (3):

In[]:=

cosSum=1+Cos[

]+Cos[

]Cos[

]-Sin[

]Sin[

]+Cos[

]Cos[

]-Sin[

]Sin[

]

Out[]=

1+Cos[

]+Cos[

]Cos[

]+Cos[

]Cos[

]-Sin[

]Sin[

]-Sin[

]Sin[

]

In[]:=

sinSum=Sin[

]+Sin[

]Cos[

]+Cos[

]Sin[

]-Sin[

]Cos[

]-Cos[

]Sin[

]

Out[]=

-Cos[

]Sin[

]-Sin[

]-Cos[

]Sin[

]+Sin[

]+Cos[

]Sin[

]+Cos[

]Sin[

]

In[]:=

algSubs=JoinTableCos[

](1-

)(1+

),{k,4},TableSin[

]2

(1+

),{k,4}

Out[]=

Cos[

]

,Cos[

]

,Cos[

]

,Cos[

]

,Sin[

]

,Sin[

]

,Sin[

]

,Sin[

]



In[]:=

commonDenom=Product[1+

,{k,4}]

Out[]=

(1+

)(1+

)1+

(1+

)

In[]:=

cos

=commonDenomFactor[cosSum/.algSubs]

Out[]=

5+3

-4

-3

-4

In[]:=

sin

=commonDenomFactor[sinSum/.algSubs]

Out[]=

-2

-2



We first use a Gröbner basis to eliminate

. The second argument to GroebnerBasis is a list of variables in which the first is the most expensive, that is, the one that the basis should try the hardest to avoid.

In[]:=

gbOmit4=GroebnerBasis[{

cos

sin

},{

}][[1]]//Factor

Out[]=

(1+

)

(1+

)

1+



-15-3

+16

-3

+16

-8

-3



The first polynomial in the resulting basis is indeed free of

. We ignore the quadratic factors, which are positive:

In[]:=

relate123=-Last[gbOmit4]

Out[]=

15+3

-16

-8

-16

Eliminating

is similar and has a slightly simpler result:

In[]:=

gbOmit2=GroebnerBasis[{

cos

sin

},{

}][[1]]//Factor

Out[]=

(1+

)

1+



(1+

)

5+9

-3

-8

-3



In[]:=

relate134=Last[gbOmit2]

Out[]=

5+9

-3

-8

-3

[A4]: Compatible values for
t
2
and
t
3

We introduce the new names

and

for

and

. We then compute the two possible values for

in terms of

and

In[]:=

useXY={

x,

y}

Out[]=

{

x,

y}

In[]:=

t3vals=Solve[relate1340,

]/.useXY//Simplify

Out[]=





4(1+

)y-

15+22

-9

(-3+

)

(22+60

)

(-3+

)(1+

)

,



4(1+

)y+

15+22

-9

(-3+

)

(22+60

)

(-3+

)(1+

)



The radicand here is the discriminant of relate134 as a quadratic polynomial in

, except that Simplify has pulled out a factor of

Sqrt[4]2

In[]:=

radQ=Discriminant[relate134,

]/4/.useXY

Out[]=

15+22

-9

+22

+60

-9

The possible values for

are symmetric:

In[]:=

t2vals=t3vals/.{

,xy,yx}

Out[]=





4x(1+

15+22

-9

+(22+60

)

(-3+

)

(1+

)(-3+

)

,



4x(1+

15+22

-9

+(22+60

)

(-3+

)

(1+

)(-3+

)



To see which signs for the radicals in

and

are compatible, we substitute our two solutions for each into both

cos

and

sin

. It turns out that

cos

doesn’t care about the signs, but

sin

insists that the signs in the formulas for

and

be the same:

In[]:=

MatrixForm[Table[SameQ[0,Simplify[

cos

/.useXY/.t2vals[[i]]/.t3vals[[j]]]],{i,2},{j,2}]]

Out[]//MatrixForm=



True	True
True	True



In[]:=

MatrixForm[Table[SameQ[0,Simplify[

sin

/.useXY/.t2vals[[i]]/.t3vals[[j]]]],{i,2},{j,2}]]

Out[]//MatrixForm=

True	False
False	True

[A5]: The first fundamental form of
(
E
5
,
e
5
)

We denote the “bag-of-edges first fundamental form” as “bfff”.

In[]:=

halfAngleSubs={xTan[ξ/2],yTan[η/2]};

In[]:=

halfAngleBackSubs={Tan[ξ/2]x,Tan[η/2]->y,Cot[ξ/2]1/x,Cot[η/2]1/y};

We can set the five phases using either compatible pair of solutions for

and

In[]:=

θ[3]=0;

In[]:=

θ[4]=2ArcTan[

]/.t3vals[[1]]/.halfAngleSubs;

In[]:=

θ[5]=θ[4]+η;

In[]:=

θ[2]=-2ArcTan[

]/.t2vals[[1]]/.halfAngleSubs;

In[]:=

θ[1]=θ[2]-ξ;

We compute “bfff” by plugging those phases into the definition of the bag-of-edges metric in (1). That computation itself is straightforward; but we don’t print the result, since it is lengthy and opaque.

In[]:=

bfffRaw=Sum[(Dt[θ[j]]-Dt[θ[i]])^2/25,{i,1,5},{j,i+1,5}];

How can we simplify that result enough to be useful? Invoking Simplify or FullSimplify in the obvious way doesn’t help, since they keep executing indefinitely — apparently exploring countless ways to exploit trigonometric identities. Fortunately, FullSimplify does well if we outlaw trigonometry by specifying

TrigFalse

One tricky point, though: Differentiation has introduced factors of

Sec[ξ/2]

and

Sec[η/2]

into bfffRaw. We would prefer to stick entirely with tangents, and we can’t leave that chore to FullSimplify, since we are forbidding it to use trigonometry. So we apply our own substitution rule that replaces

Sec[z]

with

Tan[z]

In[]:=

secToTan=Sec[arg_]^(n_/;EvenQ[n])(Tan[arg]^2+1)^(n/2);

The following line then takes perhaps 15 seconds to do the crucial simplification:

In[]:=

bfff=FullSimplify[bfffRaw/.secToTan,TrigFalse]

Out[]=

Dt[ξ]

5-3+

Tan



5+9

Tan



-5+9

Tan



45+17

Tan



Tan



--3+

Tan



5+9

Tan



Tan



-3

-3+

Tan



Tan



+4Dt[η]Dt[ξ]Tan

Tan

55+50

Tan



+27

Tan



Tan



-3+

Tan



Tan



-2

Tan



Tan



-5+3

Tan



Dt[η]

-3

Tan



-3+

Tan



+5-3+

Tan



5+9

Tan



Tan



5+9

Tan



45+17

Tan



Tan



15+22

Tan



-9

Tan



50

Tan



-3+

Tan



+-3+

Tan



5+9

Tan



-2

Tan



11+30

Tan



Tan



We partition the first fundamental form into its three pieces

, and

In[]:=

bE=Coefficient[bfff,Dt[ξ]^2];

In[]:=

bF=Coefficient[bfff,Dt[ξ]Dt[η]]/2;

In[]:=

bG=Coefficient[bfff,Dt[η]^2];

To quote those formulas in the paper as (8), we use halfAngleBackSubs to substitute in the Weierstrass variables

and

. To make the symmetries more apparent, we also call Factor to expand out the Horner-like nested structure that FullSimplify comes up with for polynomials.

In[]:=

bE/.halfAngleBackSubs//Factor

Out[]=

75+225

-15

+27

+110

+490

-22

-18

-45

+153

50(-15-22

-22

-60

-6

)

In[]:=

bF/.halfAngleBackSubs//Factor

Out[]

The Bring sextic of equilateral pentagons

[A1]: The singular points of E4

[A2]: Figures 2 depicts +E5

[A3]: Relations among the vertex angles

[A4]: Compatible values for t2 and t3

[A5]: The first fundamental form of (E5,e5)

[A1]: The singular points of
E
4

[A2]: Figures 2 depicts
+
E
5

[A4]: Compatible values for
t
2
and
t
3

[A5]: The first fundamental form of
(
E
5
,
e
5
)