## non-cubic toratopes?

Discussion of shapes with curves and holes in various dimensions.

### Re: non-cubic toratopes?

wendy wrote:The may-13 piccie
ICN5D wrote: looks like some sort of partial hedral spheration of oo3ox3xo3oo, that is a pair of inverted rectified pentachora (tetrahedron || octahedron by Klitzing).

In fact, that animated picture shows clearly the dually arranged tetrahedral structure. And the dot = dodecateron = o3o3x3o3o can be given as rap || inv rap = oo3ox3xo3oo&#x. And as rap (rectified pentachoron = o3x3o3o) itself can be given as tet || oct = xo3ox3oo&#x we indeed get a lace city of type
Code: Select all
`    o3x3o   x3o3o                 o3o3x   o3x3o    `
here. Please note that this lace city indeed is slightly slanted! The octahedra are joined by lacing edges, thus an equatorial section of an octahedral prism can be obtained here. While the segmentochoron of 2 dually arranged tetrahedra usually would result in an hexadecachoron (x3o3o4o = xo3oo3ox&#x). But in here the respective vertices would ask for a q-sized lacing, if to be connected at all.
Thus your picture simply misses that equatorial octahedral prism in general. It only reaches that stage at the bifurcating stage, where the tubes connect temporarily: then you'll have 2 momentary octahedra atop each other.
What Wendy means there by partial hedral spheration obviously is that those tubes wander (in time) around the squares of the pseudo trips (triangular prisms) of that 5D structure.
--- rk
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### Re: non-cubic toratopes?

Here's something interesting:

Going off Marek's link to the cassini ovals,

Using 2-center bipolar coordinates :

r1*r2 = b^4

((x-a)^2 + y^2)((x+a)^2 + y^2) = b^4

Makes two cassini ovals

Trying for tripolar cassini ovals, using roots of unity solutions to place circles in triangular array on (x,y) plane:

Three roots of x^3 = 1:

x^3-1 == (x-1) (x + 1/2 - i sqrt(3)/2) (x + 1/2 + i sqrt(3)/2)

Converting (Re,Im) points to (x,y) coordinates:

x = a , -a/2 , -a/2

y = 0 , -a(sqrt(3)/2) , a(sqrt(3)/2)

Tripolar cassini ovals, by setting the center of three circles to (x,y) coord:

Three-center tripolar coordinates:

r1*r2*r3 = b^6

((x-a)^2 + y^2) ((x+a/2)^2 + (y-a(sqrt3/2))^2) ((x+a/2)^2 + (y+a(sqrt3/2))^2) = b^6

Product of three circles in unit triangle:

((x-a)^2 + y^2)((x+a/2)^2 + (y-a(sqrt3/2))^2)((x+a/2)^2 + (y+a(sqrt3/2))^2) = b^6

Desmos Script of above function, plus 2-center Cassini Ovals

Produces three cassini ovals, with deformation and 3-lobe lemniscate at a=b

I'm not sure how to unite the three circles into a 3-bar cage in 3D. The function,

((x-a)^2 + y^2)((x+a/2)^2 + (y-a(sqrt3/2))^2)((x+a/2)^2 + (y+a(sqrt3/2))^2) = b^6

will expand and simplify into

a^6 -2a^3x^3 +6a^3xy^2 +x^6 +3x^4y^2 +3x^2y^4 +y^6 = b^6

How can we fit the proper terms of z into this function, to unite as cage? We should see z^6 , at least ...
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### Re: non-cubic toratopes?

Well, I think that each of the three roots of unity should be multiplied by cos z -- this reflects that the three points are not the same ones at all z-levels -- they move on semicircles.
However, the b will then also need to be z-dependent; you can probably find out the necessary factors by studying how torus is sliced.
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### Re: non-cubic toratopes?

Comparing expanded torus equation with expanded Cassini Oval equation, looking for differences and missing terms:

Bipolar Cassini Ovals:

((x-a)^2 + y^2)((x+a)^2 + y^2) = b^4

x^4 +y^4 +a^4 -b^4 -2a^2x^2 +2a^2y^2 +2x^2y^2

Rewriting to,

((x-a)^2 + y^2 - b^2)((x+a)^2 + y^2 - b^2) = 0

x^4 +y^4 +a^4 +b^4 -2a^2b^2 -2a^2x^2 +2a^2y^2 -2b^2x^2 -2b^2y^2 +2x^2y^2

Expanded Torus

x^4 +y^4 +z^4 +a^4 +b^4 -2a^2b^2 -2a^2x^2 -2a^2y^2 +2a^2z^2 -2b^2x^2 -2b^2y^2 -2b^2z^2 +2x^2y^2 +2x^2z^2 +2y^2z^2 = 0

Missing terms from expanded, rewritten equation, [TorusEq] - [2ndCassniOvalEq] ==

z^4 -2a^2y^2 +2a^2z^2 -2b^2z^2 +2x^2z^2 +2y^2z^2

Organized Expanded Cassini oval equation
x^4
+y^4
+a^4
+b^4
---------
-2a^2b^2
-2a^2x^2
+2a^2y^2
-2b^2x^2
-2b^2y^2
+2x^2y^2

Missing Terms
+z^4
-------
-2a^2y^2 , just - sign
+2a^2z^2
-2b^2z^2
+2x^2z^2
+2y^2z^2

Organized Expanded Torus Equation
x^4
+y^4
+z^4
+a^4
+b^4
-----------
-2a^2b^2
-2a^2x^2
-2a^2y^2
+2a^2z^2
-2b^2x^2
-2b^2y^2
-2b^2z^2
+2x^2y^2
+2x^2z^2
+2y^2z^2

Tripolar Cassini Ovals

((x-a)^2 + y^2) ((x+a/2)^2 + (y-a(sqrt(3)/2))^2) ((x+a/2)^2 + (y+a(sqrt(3)/2))^2) = b^6

x^6 +y^6 + a^6 -2a^3x^3 +3x^4y^2 +3x^2y^4 +6a^3xy^2 = b^6

(I think it's amazing that it simplifies into this smaller equation, considering how complex it is)

Rewriting to,

((x-a)^2 + y^2 -b^2) ((x+a/2)^2 + (y-a(sqrt(3)/2))^2 -b^2) ((x+a/2)^2 + (y+a(sqrt(3)/2))^2 -b^2) = 0

x^6 +y^6 +a^6 -b^6 -3a^4b^2 +3a^2b^4 -2a^3x^3 +3b^4x^2 +3b^4y^2 -3b^2x^4 -3b^2y^4 +3x^4y^2 +3x^2y^4 -3a^2b^2x^2 -3a^2b^2y^2 +6a^3xy^2 -6b^2x^2y^2

Re-arranged and grouped like terms:

x^6
+y^6
+a^6
-b^6
===========
-3a^4b^2
+3a^2b^4
-2a^3x^3
+3b^4x^2
+3b^4y^2
-3b^2x^4
-3b^2y^4
----------
+3x^4y^2
+3x^2y^4
=============
-3a^2b^2x^2
-3a^2b^2y^2
--------------
+6a^3xy^2
-------------
-6b^2x^2y^2

Extrapolated guesses for theoretical z-terms:

+z^6
==========
-3b^4z^2
-3b^2z^4
------------
+3x^4z^2
+3z^4x^2
+3y^4z^2
+3z^4y^2
===========
+3a^2b^2z^2
------------
+6a^3zy^2
+6a^3yz^2
+6a^3xz^2
+6a^3zx^2
------------
-6b^2x^2z^2
-6b^2y^2z^2

z^6 -3b^4z^2 -3b^2z^4 +3x^4z^2 +3z^4x^2 +3y^4z^2 +3z^4y^2 +3a^2b^2z^2 +6a^3zy^2 +6a^3yz^2 +6a^3xz^2 +6a^3zx^2 +6b^2x^2z^2 +6b^2y^2z^2

Adding these with the expanded cassini oval equation makes strange object, not what we're looking for. The three circles will smoothly connect just by adding " -z^6 ", but with continuous surfaces . Not sure how to connect as a 3-bar cage. Maybe a bisecting rotation of the triangular array into 3D and 4D, then cut at some angle in between? We can build a torus with rotations of its row of two circles. And, tiger with bisecting rotations of the square array of circles. So, how about bisecting rotations of other arrays?

I tried semicircle equations, and spherating with a circle to get something close to the coordinate cut of xz. What I used was

(x-sqrt(a^2-z^2))^2+y^2-b^2

which makes a banana, with infinite cylinders at both ends, when a=3 , b=1 . But, I should probably assume it's a round-off error, since the half-circle will graph differently, depending on which program you use. Desmos will plot a perfect finite half circle, but calcplot will "stamp" the half circle into a 2D plane.

EDIT : corrected the expanded tripolar cassini oval equation
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### Re: non-cubic toratopes?

This is a general equation for a product of circles, in vertices of an equilateral polygon. Their centers are derived from the n-th roots of unity. For a triangle of three vertices, n=3 , leading to, in simplified form: in search of combinatorial objects of finite extent
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### Re: non-cubic toratopes?

And a cool explorer of the above equation (I'm glad desmos does the product function) :

https://www.desmos.com/calculator/y8u0iqodyp
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### Re: non-cubic toratopes?

Had some amazing insights today, after lots of hard work with these functions. So many experiments, so many equations, so much testing and extrapolating one result to another. But, I believe I found something very worthy : a factored out equation for the cut of mantis. It makes three inward facing tori, with some peculiar properties in the function. Elegant equations are one way of seeing yourself on the right track. I'll make pics when I can, got things to do tomorrow.

Third Roots of Unity
x^3 = 1

solve for x:
x = 1
x = -1/2 - i*sqrt(3)/2
x = -1/2 + i*sqrt(3)/2

Circle equation
(x^2 + y^2 -b^2) = 0

Applying coordinate translation of circles, based on third roots of 1 , from (Re,Im) to (x,y):
x = a
x = -a/2 , y = a sqrt(3)/2
x = -a/2 , y = -a sqrt(3)/2

equals,
((x-a)^2 + y^2 -b^2)((x+a/2)^2 + (y- a sqrt(3)/2)^2 -b^2) ((x+a/2)^2 + (y+ a sqrt(3)/2)^2 -b^2) = 0

* Produces three circles in vertices of triangle, whose centers are derived from x^3 = 1

Torus equation of ((I)I), two displaced circles in xy plane
((sqrt(y^2)-3)^2 + x^2 -1) = 0

Applying coordinate translations of torus cut, based on third roots of 1 , from (Re,Im) to (x,y):
x = a
x = -a/2 , y = a sqrt(3)/2
x = -a/2 , y = -a sqrt(3)/2

equals,
((sqrt(y^2)-3)^2 + (x-a)^2 -1)((sqrt((y- a sqrt(3)/2)^2)-3)^2 + (x+a/2)^2 -1)((sqrt((y+ a sqrt(3)/2)^2)-3)^2 + (x+a/2)^2 -1) = 0

* Produces three torus cuts of ((I)I) in triangle array, all facing same direction as principle root torus.

For rotating the two primitive root tori, so they face inwards, I set up an x,y rotate function, with n*pi/6 steps, for aligning the hexagonal array of circles:

Primitive in negative y region
x = (x*sin(pi/6)+y*cos(pi/6))
y = (x*cos(pi/6)-y*sin(pi/6))

Primitive in positive y region
x = (x*sin(5pi/6)+y*cos(5pi/6))
y = (x*cos(5pi/6)-y*sin(5pi/6))

Equals the product of six circles in hexagon array, from triangle array of 3 tori, of mantis cut:

((sqrt(y^2)-3)^2+(x-a)^2-1) = 0

((sqrt(((x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6))^2)-3)^2+((x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6))^2-1) = 0

((sqrt(((x+a/2)*cos(5pi/6)-(y-a*sqrt(3)/2)*sin(5pi/6))^2)-3)^2+((x+a/2)*sin(5pi/6)+(y-a*sqrt(3)/2)*cos(5pi/6))^2-1) = 0

Trying for 3D version of array. Apply +z^2 inside square root term, along with y^2:

Principle Root Torus
((sqrt(y^2+z^2)-b)^2+(x-a)^2) = c^2

Primitive Root Torus A
((sqrt(((x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6))^2+z^2)-b)^2+((x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6))^2) = c^2

Primitive Root Torus B
((sqrt(((x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6))^2+z^2)-b)^2+((x+a/2)*sin(5*pi/6)+(y-a*sqrt(3)/2)*cos(5*pi/6))^2) = c^2

Simplifying the forms of the primitive root tori, eliminating the sin/cos:

First Primitive root torus
((sqrt(((x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6))^2+z^2)-b)^2+((x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6))^2)

(x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6) == (sqrt(3)x -y)/2
(x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6) == (x +sqrt(3)y)/2 + a

equals,
((sqrt(((sqrt(3)x-y)/2)^2+z^2)-b)^2+((x+sqrt(3)y)/2 + a)^2)

Second Primitive root torus
((sqrt(((x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6))^2+z^2)-b)^2+((x+a/2)*sin(5*pi/6)+(y-a*sqrt(3)/2)*cos(5*pi/6))^2)

(x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6) == -(sqrt(3)x +y)/2
(x+a/2)*sin(5*pi/6)+(y-asqrt(3)/2)*cos(5*pi/6) == (x -sqrt(3)y)/2 + a

equals,
((sqrt((-(sqrt(3)x+y)/2)^2+z^2)-b)^2+((x-sqrt(3)y)/2 + a)^2)

Rewrite with simplified principle and primitive root tori,

Tri-toroidal Cassini Coordinate System
((sqrt(y^2+z^2)-b)^2+(x-a)^2)*((sqrt(((sqrt(3)x-y)/2)^2+z^2)-b)^2+((x+sqrt(3)y)/2+a)^2)*((sqrt((-(sqrt(3)x+y)/2)^2+z^2)-b)^2+((x-sqrt(3)y)/2+a)^2) = c^6
Symmetrical Three-Prong Multitorus at a=2 , b=5 , c=2.5

Mantis Cut, factored out as product of three tori.
Note the constants of x and y, in the root tori : the oblique plane equation terms are the third roots of unity! This part gave me the chills....

((sqrt(y^2+z^2)-b)^2+(x-a)^2-c^2)((sqrt(((sqrt(3)x-y)/2)^2+z^2)-b)^2+((x+sqrt(3)y)/2+a)^2-c^2)((sqrt((-(sqrt(3)x+y)/2)^2+z^2)-b)^2+((x-sqrt(3)y)/2+a)^2-c^2) = 0

Oblique terms pulled out:

Principle
x-a
y-0

Primitive A
(sqrt(3)x -y)/2
(x +sqrt(3)y)/2 +a

Primitive B
-(sqrt(3)x +y)/2
(x -sqrt(3)y)/2 +a

compare to x^3 = 1 :
x = 1
x = -1/2 - i*sqrt(3)/2
x = -1/2 + i*sqrt(3)/2

Next step is to take the relationship with x and y , and apply it to z with w, in the same equation. The aim would be for a mirror-image arrangement of tori, when flipping from z-cut to w-cut, as the equation is currently written. Not sure how to add in w in the root tori, may be just as simple as tiger terms...
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### Re: non-cubic toratopes?

Thinking a little more about it, we may be trying to define some kind of exotic rotation. Considering how a torus is made from a circle, we shift the circle along an axis by a certain value

x^2 + y^2 = r^2

(x-a)^2 + y^2 = r^2

then apply a rotation into 3D, that causes the circle to "puncture" through the original 2D plane in two places. This rotation is defined as the edge of a hollow circle,

x = sqrt(x^2 + z^2) ,

(sqrt(x^2 + z^2)-a)^2 + y^2 = r^2
circle over circle , or circle with 2-point rotation cycle.

A circle will have two intercepts on x, leading to two roots,

x^2 = a^2 , equal to (x-a)(x+a)

A 3-prong disk edge will have three roots,

x^3 = a^3, equal to (x-a) (x+a*(1-i√3)/2) (x+a*(1+i√3)/2)

where the imaginary values have to be tied in with the z-axis axis on the real plane.

So, what we want to do is define a hollow frame of three half-disks (exactly like you're thinking Marek). Embedding a circle over this frame will describe a rotation that causes an object to puncture in three places. The fact that a torus cut equation factors out into two equations of a circle is only an artifact of the circular frame that was embedded.

So, we're looking for a 1D object that's embedded in 3D. The three intercept points of this surface is merely an artifact of it's function. A circle has two half-disks, as the 2nd roots of unity. This 3-prong surface is a 3rd root version. The factored out mantis cut is mindblowing to look at, but doesn't show the full equation. It's merely an artifact of the 3-prong surface that a torus was stretched over.

So, if a simple circular rotation can be defined with sine and cosine for a 2-point cycle around the origin, how does one alter the pathway, into a more exotic 3-point cycle? It kind of does make me think about the surface of a sphere. Would we be looking for a single disk edge equation expressed in polar, with a change of periodicity in the complex plane?
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### Re: non-cubic toratopes?

ICN5D wrote:Would we be looking for a single disk edge equation expressed in polar, with a change of periodicity in the complex plane?

I think so.

I have tried to extend the toratopic notation to cover such objects. If (II) is a circle, [II]3 is a 3-edge circle in 3D and ([II]3 I) is a multitorus based on such object.

[II]3 is a 3D object, but as it only contains two I's, only one of its cuts is clearly defined in the notation. [I]3 are three points.

How would we define the "cuts" of multitorus ([II]3 I)? Well, ([I]3 I) is the ordinary cut with three circles, but ([II]3) is more complex. It's not a flat cut, it's three half-cuts of torus ((II)) stuck together, which translates to two concentric [II]3 shapes, one bigger and one smaller.

How would the cuts of ([II]3 (II)) look?

The 2D cut ([I]3 (I)) is six circles centered on (3 points) x (2 points)

([I]3 (II)) are three toruses: take two toruses ((I)(II)), split them horizontally to treat this as ([I]2 (II)) and then fit three to the axis of rotation.
([II]3 (I)) should be two multitoruses ([II]3 I) separated in 4th dimension.

This might not actually be the mantis, as I don't see an easy way to put the "other" set of three toruses in there... grumble...
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### Re: non-cubic toratopes?

Here's something:

Experiment with 4-prong multitorus with tiger cage as known surface. Tiger cage has two intersecting disk edges in 3D, a 1D object embedded in 3D.

Tiger

(sqrt(x^2 + y^2)-3)^2 + (sqrt(z^2 + w^2)-3)^2 = 1

Tiger cut of two tori, w = a

(sqrt(x^2 + y^2)-3)^2 + (sqrt(z^2 + a^2)-3)^2 = 1

Tiger cage between y-cut and w-cut

(sqrt(x^2 + (y*sin(pi/4)+a*cos(pi/4))^2)-3)^2 + (sqrt(z^2 + (y*cos(pi/4)+a*sin(pi/4))^2)-3)^2 = 1

Simplified form w/o sin/cos:

y*sin(pi/4)+a*cos(pi/4) == (y+a)/sqrt(2)
y*cos(pi/4)-a*sin(pi/4) == (y-a)/sqrt(2)

Equation for tiger cage in 3D:
(sqrt(x^2 + ((y+a)/sqrt(2))^2)-3)^2 + (sqrt(z^2 + ((y-a)/sqrt(2))^2)-3)^2 = 1

A 4-prong multitorus will have the circles on x and y axes, as the 4 root circles. Rotating the tiger cage, so circles in xy plane are in correct place for true 4-prong MT:

(sqrt(x^2 + ((z+a)/sqrt(2))^2)-3)^2 + (sqrt(y^2 + ((z-a)/sqrt(2))^2)-3)^2 = 1

When a=0, midsection and full tiger cage

(sqrt(x^2 + (z/sqrt(2))^2)-3)^2 + (sqrt(y^2 + (z/sqrt(2))^2)-3)^2 = 1

Apply rotation to x and y, for 1/4 turn

x = (x*sin(pi/4)+y*cos(pi/4)) == ((x+y)/sqrt(2))
y = (x*cos(pi/4)-y*sin(pi/4)) == ((x-y)/sqrt(2))

(sqrt(((x+y)/sqrt(2))^2 + (z/sqrt(2))^2)-3)^2 + (sqrt(((x-y)/sqrt(2))^2 + (z/sqrt(2))^2)-3)^2 = 1

Equation for 4-Prong Multitorus in 3D:

(sqrt(((x+y)/sqrt(2))^2 + (z/sqrt(2))^2)-a)^2 + (sqrt(((x-y)/sqrt(2))^2 + (z/sqrt(2))^2)-a)^2 = 1

How to build 4-Prong from a circle?

x^2 + y^2 = 1

shift by 'a' along x,

x = (x-a)

Principle root circle:
(x-a)^2 + y^2 = 1

Apply 4-point rotation, aka embed onto 1-frame of 4-edge

x = sqrt(((x+y)/sqrt(2))^2 + (z/sqrt(2))^2) == sqrt((x+y)^2+z^2)/sqrt(2)

y = (sqrt(((x-y)/sqrt(2))^2 + (z/sqrt(2))^2)-a) == (sqrt((x-y)^2+z^2)/sqrt(2)-a)

Resulting in,

(sqrt(((x+y)^2+z^2)/2)-a)^2 + (sqrt(((x-y)^2+z^2)/2)-a)^2 = 1
Last edited by ICN5D on Tue Jun 02, 2015 8:34 pm, edited 1 time in total.
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### Re: non-cubic toratopes?

Exploring a 4-prong multitorus in 2D : https://www.desmos.com/calculator/huspilqz6r

EDIT : Updated to better 4-prong genus-3
Last edited by ICN5D on Wed Jun 03, 2015 10:59 pm, edited 1 time in total.
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### Re: non-cubic toratopes?

So, we need to find how to define a 1D line of three half-disks, in 3D space,

full disk

x^2 + y^2 = a^2

half disks

x - sqrt(a^2-y^2) , right-hand half disk (principle)
x + sqrt(a^2-y^2) , left-hand half disk (primitive)

A 3-edge will have a principle root half disk, as the right handed. The two primitive root half edges will be left handed, turned into the complex plane of Z. Projecting from xy, into z, by the values of the 3rd roots of unity. A 3-edge will be product of them. Using 1/3 rotations from x into Z should do the trick.

Product of two half edges,
(x - sqrt(a^2-y^2))(x + sqrt(a^2-y^2)) == x^2 + y^2 -a^2

taking primitive root half disk, and setting Z projections by the values of 3rd roots

First primitive root half disk
x == (x+1/2)
z == (z-sqrt(3)/2)

Second primitive root half disk
x = (x+1/2)
z = (z+sqrt(3)/2)

Not sure how to incorporate z into a 2D circle equation, without it becoming a sphere. This will be the real trick.

But, maybe we SHOULD define it as a sphere, for only the primitives ...

x^2 + y^2 + z^2 = a^2

x -sqrt(a^2-y^2) and x +sqrt(a^2-y^2-z^2)

(x+1/2) +sqrt(a^2-y^2-(z-sqrt(3)/2)^2)

(x+1/2) +sqrt(a^2-y^2-(z+sqrt(3)/2)^2)

Now, express as full product of the three,

(x -sqrt(a^2-y^2))((x+1/2) +sqrt(a^2-y^2-(z-sqrt(3)/2)^2))((x+1/2) +sqrt(a^2-y^2-(z+sqrt(3)/2)^2)) = b^2

(x -sqrt(3^2-y^2))((x+1/2) +sqrt(3^2-y^2-(z-sqrt(3)/2)^2))((x+1/2) +sqrt(3^2-y^2-(z+sqrt(3)/2)^2))

Will make 3 half disk like object, but disconnected, and not perfect. Hmm....there must be some better way. I still like the idea of exploring rotations, that are more complex than a circle's edge. How can we do that? It's a rotation like a circle's edge, but continuously changing direction so the pathway has three locations of intersection, in the starting plane.
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### Re: non-cubic toratopes?

On the topic of exotic rotations, I have some ideas. I'm now pretty sure that we're looking for an extension of periodic functions, that are analogous to rotating in a circle. We can define rotations using the unit circle in the complex plane. In this case, we have one dimension in the real numbers, and one in the imaginary. It translates over to polar coordinates, which describes the circle. This works for generating shapes with perfect square roots, namely any toratope we've explored/defined so far. Take any equation, set any dimension(s) to zero, and the cross section equation can factor out into perfect square roots. This is why we get two, or multiples of two, objects in the intercept array. Toratopes are an extension to perfect square roots, where we get root circles, root spheres, root toruses, root tigers, etc.

In the case of Mantis, now we're looking at an equation with perfect cube roots. Using a degree-12 polynomial, set any dimension(s) to zero, and the cross section equation will factor out into three, or multiples of three, objects in the intercept array. If a periodic function on a circle makes perfect square roots, then we want a periodic function on a unit sphere in the bicomplex plane. That's a 1D real, and a 2D complex plane.

A rotation on this unit sphere will be more wave-like with two poles. I've been visualizing the rotation a torus would take to make mantis (or a circle to make 3-prong multitorus). The poles are where the 1st imaginary part is +i or -i , and 2nd complex component is 0. As is the case with cube roots, there will be a principle cycle, and two primitive cycles, which are the longitudinal lines inscribed onto the unit sphere, in 2nd complex dimension. A full cycle on a circle is 2*pi. In the case of unit sphere, a full cycle (for mantis or 3-prong) is 6*pi . Imagine the path a circle takes to make a 3-prong. Starting with (x-a)^2 + y^2 = b^2 , the circle moving along a wave, that goes through xy in three places. The object will undulate along a wave for three cycles of a circle. Note how the pathway for this rotation takes an object over/under the n-plane into n+1 in separate halves. A period of 3*pi will only trace out half of the object.

And that's where spherical coordinates come into play. I spent some time exploring the factored out mantis cut equation, and found something interesting. I'm glad I found it, since it holds the key to uncovering this difficult math problem. Or, at least it provides some information about the full equation, which points in the right direction to deriving it.

So, I took the terms from the factored equation and multiplied them together, as if they were the 3rd roots.

By analogy,

x^2 - 1 == (x-1)(x+1)

x^3 - 1 == (x-1)(x +1/2+(i sqrt(3))/2)(x +1/2-(i sqrt(3))/2)

In the actual equations, we often see:

x^2 - a^2 == (x-a)(x+a)

Which describes two locations displaced along x

This ultimately comes from

(sqrt(x^2)-a)^2 ,

which is exactly what we're after, but in cube root form.

Now, take the three terms from factored mantis cut,

(x-a)*((x +sqrt(3)y)/2 +a)*((x -sqrt(3)y)/2 +a) == ((3a+x)/4)x^2 +((3a-3x)/4)y^2 - a^3

Where the elegant polynome ((3a+x)/4)x^2 +((3a-3x)/4)y^2 - a^3 has perfect cube roots that define the spacing and oblique angle terms for the principle and primitive root toruses. There still exists the other reciprocal terms for y in the square root. A product of those should yield something else in the right direction:

y((sqrt(3)x -y)/2)(-(sqrt(3)x +y)/2) == ((y^2-3x^2)/4)y

In which ((y^2-3x^2)/4)y has perfect cube roots.

Obviously, we still should be aiming for a 3-prong multitorus, as generated from a circle. We can apply it to a torus when found and explored, to make mantis. This has all been extremely enlightening.
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### Re: non-cubic toratopes?

How exactly does the bicomplex plane work?

If you define bicomplex numbers have axes i^2 = -1 and j^2 = -1, what's i*j?
Let's presume you want the whole unit circle in ij plane to square to -1. In that case, we'll take something like sqrt(2)/2 i + sqrt(2)/2 j and its square should be -1.

1/2 i^2 + (i * j)/2 + (j * i)/2 + 1/2 j^2 = -1
-1/2 + (i * j)/2 + (j * i)/2 - 1/2 = -1
-1 + (i * j)/2 + (j * i)/2 = -1
(i * j)/2 + (j * i)/2 = 0

So either you have to go quaternion route and throw out commutativity, or you have to declare i * j equal to zero.
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### Re: non-cubic toratopes?

You just take a 2-tuple of 2 complex numbers, so you get (a+bi, c+di), you can write this as (z,a), with z and a complex numbers.
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### Re: non-cubic toratopes?

student91 wrote:You just take a 2-tuple of 2 complex numbers, so you get (a+bi, c+di), you can write this as (z,a), with z and a complex numbers.

Wouldn't it be a 3D number, like a + bi + cj ? Only one is real, with two orthogonal complex
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### Re: non-cubic toratopes?

Well, I think it may be just as simple as a simultaneous rotation of a rotation with a single variable, in two orthogonal directions. When they match up in perfect periodicity, we get things like 3-prong and mantis. The two oblique angle tori in mantis cut are simply rotated positions about an axis by pi/3, from the principle one. A tiger is made by full period of 2*pi from a torus. So, we're trying to rotate that rotation, as projected onto the 3-plane.
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### Re: non-cubic toratopes?

I found another, even more symmetrical mantis cut equation. This one has all -a terms , and identical oblique angle terms. It's exactly equal to the first one I posted.

((sqrt(y^2 + z^2)-3)^2 + (x - a)^2 - 1)

((sqrt(((sqrt(3)x + y)/2)^2 + z^2)-3)^2 + ((-x + sqrt(3)y)/2 - a)^2 - 1)

((sqrt(((-sqrt(3)x + y)/2)^2 + z^2)-3)^2 + ((-x - sqrt(3)y)/2 - a)^2 - 1)

which looks like this.

The equation will plot this, the mantis intercept array: in search of combinatorial objects of finite extent
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### Re: non-cubic toratopes?

You use pairs of complex numbers to effect the swirl of a three-dimensional figure.

Suppose you have a point x,y,z You now find the point x/2r, y/2r, z/2r, where r = x²+y²+z².

Add 0,0,½ to this lot, ie x/2r, y/2r, (z+r)/2r.

Draw a line from 0,0,1 through this point, and find the value for 0,0,0, giving, eg a = (x)(r-z)/4r² + i (y)(r-z)/4r²

This is the slope of a complex plane Y = a X. The relevant circle is X = r cis theta gives Y = (x+iy)(r-z) cis þ /4r , X = r cis þ

for all values of þ.

This plots onto four dimensions as X, Xi, Y, Yi.
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### Re: non-cubic toratopes?

Thanks wendy. I have another idea. They keep coming to me. As a general case, any multitorus is a spherated n-gonal hosohedron. The genus-2 has the frame of a trigonal hosohedron {2,3}. The real question is how do we define these edges algebraically? Is there a method to defining tesselated spherical lunes?
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### Re: non-cubic toratopes?

Using the same simplifying technique as the mantis cut above, I found a symmetric polynomial for the three circles in triangle cut, of a genus-2 multitorus:

-(x^2+y^2-b^2)^3 + 3a^2b^2(x^2+y^2-b^2) + 2a^3(x^3-3xy^2) + 3a^4b^2 = a^6

This equation factors out into real cube roots of 3 circles in equilateral triangle:

((x-a)^2 + y^2 - b^2) (((-x - sqrt(3)y)/2-a)^2 + ((-sqrt(3)x + y)/2)^2 - b^2) (((-x + sqrt(3)y)/2-a)^2 + ((sqrt(3)x + y)/2)^2 - b^2) = 0

Equations in LaTeX

This is a step closer to an unfactored equation of a genus-2. How to merge the 3 circles along a sphere into +z, -z is still a mystery. Setting z=0 leads to this unfactored equation, so it must go in there somehow!
Last edited by ICN5D on Fri Jun 05, 2015 6:03 am, edited 1 time in total.
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### Re: non-cubic toratopes?

Found something. I re-wrote the product of three circles in polar, and reduced to an unfactored equation. Then, I replaced r with ρ for spherical coordinates,

ρ^6 -3b^2(ρ^4 -ρ^2(b^2-a^2)) -(b^2-a^2)^3 -2a^3ρ^3*cos(3θ)

Which makes, Getting really close to a genus-2. The polar areas have always been an issue, but not so bad here. Not sure how to fatten up the stamped ends of the semi-torus legs. But, what's important are the poles do not have infinity errors, and the legs are perfect circles when z=0.
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### Re: non-cubic toratopes?

wendy wrote:You use pairs of complex numbers to effect the swirl of a three-dimensional figure.

Suppose you have a point x,y,z You now find the point x/2r, y/2r, z/2r, where r = x²+y²+z².

Add 0,0,½ to this lot, ie x/2r, y/2r, (z+r)/2r.

Draw a line from 0,0,1 through this point, and find the value for 0,0,0, giving, eg a = (x)(r-z)/4r² + i (y)(r-z)/4r²

This is the slope of a complex plane Y = a X. The relevant circle is X = r cis theta gives Y = (x+iy)(r-z) cis þ /4r , X = r cis þ

for all values of þ.

This plots onto four dimensions as X, Xi, Y, Yi.

That sounds interesting. How does one go about deriving an equation using that? If a circular path is defined by replacing x with (sqrt(x^2+y^2)-a) , then how would we define a spherical path, with periodicity of three half circles?

Starting with a circle x^2 + y^2 - a^2 , it would seem like using x = (sqrt(x^2+y^2+z^2)-a) , and then some additional terms with y. We want to restrict the rotation into +z, -z to just the surface of the sphere, but also rotate with respect to xy. The xy rotation should match up perfectly with z, to allow the traveling circle to intercept xy-plane at precise locations of a polygon vertex. And, somehow the equation becomes degree-6, probably by a product of a squared variable and 4th power one, i.e. x^2(x^2 + y^2)^2
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### Re: non-cubic toratopes?

I would not trust my algebra here, i think it's wrong.

But the description is spot on for making a swirlybob projection of 3d space.

1. derive the radius from 0 to the point P, and call it R.

2. derive a point on a unit (diameter) sphere as (x,y,z)/2R.

3. Add ½ to z, ie (x,y,z+r)/2R

4. Draw a line through (0,0,1) to (x,y,z+r)/2r until it strikes z=0. This gives S = A+Bi. (the slope) [this is where i made the mistake]

5. The desired projection is to project Y = X * S,

X = R cis(theta) and plots into w,x space. Y = R(A+Bi)cis(theta) and plots the y,z space.

The swirlybob of a torus centred at (0,0,0) is a tiger.
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### Re: non-cubic toratopes?

Found a much better equation for a 4-prong multitorus. This one is not made by turning a tiger cage 45 degrees. It's closer to what we're looking for, in terms of starting with a circle, and replacing x and y with other terms. It also defines a set of points over a sphere.

x^2 + y^2 = a^2

Shift center by +b along x

(x-b)^2 + y^2 = a^2

To make the circle's path symmetrical over a sphere, replace x with sqrt(x^2+y^2+z^2)

(sqrt(x^2+y^2+z^2)-b)^2 + y^2 = a^2

Equation makes a torus. To modify for multiple paths over the sphere, replace y with (x^2y^2)/(a^2b^2)

(sqrt(x^2+y^2+z^2)-b)^2 + (x^2y^2)/(a^2b^2) = a^2

Setting a=3 , b=1 makes a perfect 4-prong. Changing y to other variables makes other interesting surfaces. Trying it with a 3-prong is more difficult, though. It's the two semi-torus lobes that sit at the oblique angles that need more precise defining. Setting y = (x^3-3xy^2)/(2a^2b^2) will define a triangular arrangement of the lobes that branch from two poles, but highly skewed. It needs more precise oblique terms.

Using the equation (sqrt(x^2+y^2+z^2)-b)^2 + (x^3-3xy^2) = a^2 makes this : And, modifying the same way as 4-prong with (sqrt(x^2+y^2+z^2)-b)^2 + (x^3-3xy^2)/(2a^2b^2) = a^2 will make this: in search of combinatorial objects of finite extent
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### Re: non-cubic toratopes?

wendy wrote:I would not trust my algebra here, i think it's wrong.

But the description is spot on for making a swirlybob projection of 3d space.

1. derive the radius from 0 to the point P, and call it R.

2. derive a point on a unit (diameter) sphere as (x,y,z)/2R.

3. Add ½ to z, ie (x,y,z+r)/2R

4. Draw a line through (0,0,1) to (x,y,z+r)/2r until it strikes z=0. This gives S = A+Bi. (the slope) [this is where i made the mistake]

5. The desired projection is to project Y = X * S,

X = R cis(theta) and plots into w,x space. Y = R(A+Bi)cis(theta) and plots the y,z space.

The swirlybob of a torus centred at (0,0,0) is a tiger.

Looks promising! How would I apply that to a circle? If I'm only trying to rotate in xyz space, then would I use the same thing, like

X = R cis(theta) and plots into x,z space
Y = R(A+Bi)cis(theta) and plots the y,z space ?

I'm having some trouble trying to see how this works.
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### Re: non-cubic toratopes?

ICN5D wrote:Yes. Here's an animation of a=2 , b=2 , d=0 , t=0 , and animating c from 0 to 6.28. It's among the wilder things this equation makes. This is also the first gif I made using a screencapture program. The timing is a little tricky, but it's still much easier than manually one by one. What software did you use to create this visualization? Mathematica?
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### Re: non-cubic toratopes?

It's CalcPlot3D . You can have 5 adjustable parameters that you can animate.

I take a screenshot for each step, and make a gif out of them, using GifMaker.
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### Re: non-cubic toratopes?

Marek14 wrote:How would the cuts of ([II]3 (II)) look?

The 2D cut ([I]3 (I)) is six circles centered on (3 points) x (2 points)

([I]3 (II)) are three toruses: take two toruses ((I)(II)), split them horizontally to treat this as ([I]2 (II)) and then fit three to the axis of rotation.
([II]3 (I)) should be two multitoruses ([II]3 I) separated in 4th dimension.

This might not actually be the mantis, as I don't see an easy way to put the "other" set of three toruses in there... grumble...

Thinking about it some more, I think this is because we don't know what the equation symmetry is, for mantis. Some cuts work really well. After observing what a 3-prong looks like in 2D, I get the feeling that some other mantis sections might look really bizarre and unfamiliar, but still have the 3-lobe symmetry.

Or, do you think mantis would only have the two arrangements of 3 tori, in the fence?
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### Re: non-cubic toratopes?

ICN5D wrote:
Marek14 wrote:How would the cuts of ([II]3 (II)) look?

The 2D cut ([I]3 (I)) is six circles centered on (3 points) x (2 points)

([I]3 (II)) are three toruses: take two toruses ((I)(II)), split them horizontally to treat this as ([I]2 (II)) and then fit three to the axis of rotation.
([II]3 (I)) should be two multitoruses ([II]3 I) separated in 4th dimension.

This might not actually be the mantis, as I don't see an easy way to put the "other" set of three toruses in there... grumble...

Thinking about it some more, I think this is because we don't know what the equation symmetry is, for mantis. Some cuts work really well. After observing what a 3-prong looks like in 2D, I get the feeling that some other mantis sections might look really bizarre and unfamiliar, but still have the 3-lobe symmetry.

Or, do you think mantis would only have the two arrangements of 3 tori, in the fence?

Not sure...
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