Hunting for tigers in the parametric jungle

Discussion of shapes with curves and holes in various dimensions.

Hunting for tigers in the parametric jungle

Postby Marek14 » Sun Nov 20, 2005 12:56 pm

The recent toratopes discussion has inspired me to look at their parametric equations a bit closer.

First of all, you should notice that for spheres of dimension 4 and up, there are multiple perfectly good sets of parametric equation.

For a circle, we have
x = A * cos a
y = A * sin a

For a sphere, we have
x = A * cos a1 * cos a2
y = A * sin a1 * cos a2
z = A * sin a2

When we look at these equations we see that they are just combinations of two "circle" equations.

There are two different sets for glome:

x = A * cos a1 * cos a2 * cos a3
y = A * sin a1 * cos a2 * cos a3
z = A * sin a2 * cos a3
w = A * sin a3

and

x = A * cos a1 * cos a3
y = A * sin a1 * cos a3
z = A * cos a2 * sin a3
w = A * sin a2 * sin a3

Petaglome would have three different sets of equations. Exaglome would have six.

Now we see that the parametric equation for the tiger

x = A * cos a + C * cos a * cos c
y = A * sin a + C * sin a * cos c
z = B * cos b + C * cos b * sin c
w = B * sin b + C * sin b * sin c

is, basically, a sum of two circles and a glome.

Similarly, the parametric equations of five-dimensional sphere tiger a(x,y,z),b(w,v),c(a,b) are:

x = A * cos a1 * cos a2 + C * cos a1 * cos a2 * cos c
y = A * sin a1 * cos a2 + C * sin a1 * cos a2 * cos c
z = A * sin a2 + C * sin a2 * cos c
w = B * cos b + C * cos b * sin c
v = B * sin b + C * sin b * sin c

One way to classify the toratopes was in terms of unshrinkable curves and surfaces. Here, the middle-cuts are useful. It seems that the global rule is:

IF THE CUT WITH CERTAIN COORDINATE HYPERPLANE RESULTS IN TWO DISTINCT SURFACES OF LOWER DIMENSIONS, THOSE SURFACES CAN'T BE SHRUNK.

For example, the two distinct cuts of 3D torus both result in two circles. This leads directly to two categories of "main" circles which can't be shrunk to point on the surface of torus. On the other hand, cuts through sphere are always a SINGLE circle, which can be always shrunk.

So, in 4D, we have:

Glome a(x,y,z,w) - all cuts are single spheres - all of these can be shrunk into point. No shrinkables.

Sphere*circle a(x,y,z),b(a,w) - here the x,y, and z-cuts are single torii, but the w-cut are two spheres, leading to one type of sphere that can't be shrunk. Each of the cut torii contains two kinds of unshrinkable circles again - one of them is drawn on the unshrinkable sphere, and so it CAN be shrunk by drawing it along that sphere to a point. The other one is truly unshrinkable. a-spheres and b-circles

Circle^3 - a(x,y),b(a,z),c(b,w) - all 3 kinds of cuts result in two torii. So there are, altogether, three different unshrinkable torii - which can't be compressed in a circle. There are 3 types of unshrinkable circles, each common to two types of these torii. ab-torii, ac-torii, bc-torii, a-circles, b-circles, and c-circles.

Tiger - a(x,y),b(z,w),c(a,b) - each cut results in 2 torii. There are, in fact, 2 classes of cuts (xy and zw) which are symmetrical. So, there are 2 kinds of unshrinkable torii (ac and bc), and 3 kinds of unshrinkable circles (a, b, and c) altogether. ac-torii, bc-torii, a-circles, b-circles, and c-circles.

Circle*sphere - a(x,y),b(a,w,v) - The cuts are two displaced spheres b or torus ab. From first, we have an unshrinkable sphere b, the torus is shrinkable, as it's single. The b circle of ab torus is shrinkable, as it belongs to a sphere, the a circle is not. So, the topology of unshrinkable objects is the same as on surface of sphere*circle. b-spheres and a-circles

And in 5D:

Petaglome - a(x,y,z,w,v) - no unshrinkable objects.
Glome*circle - a(x,y,z,w),b(a,v) - a-glomes and b-circles
Sphere*circle*circle - a(x,y,z),b(a,w),c(b,v) - ab-sphere*circles, ac-sphere*circles, a-spheres, bc-torii, b-circles, c-circles
Circle^4 - a(x,y),b(a,z),c(b,w),d(c,v) - abc-circles^3, abd-circles^3, acd-circles^3, bcd-circles^3, ab-torii, ac-torii, ad-torii, bc-torii, bd-torii, cd-torii, a-circles, b-circles, c-circles, d-circles.
Tiger*circle - a(x,y),b(z,w),c(a,b),d(c,v) - abc-tigers, acd-circles^3, bcd-circles^3, ac-torii, ad-torii, bc-torii, bd-torii, cd-torii, a-circles, b-circles, c-circles, d-circles.
Circle*sphere*circle - a(x,y),b(a,z,w),c(b,v) - ab-circle*spheres, bc-sphere*circles, ac-torii, b-spheres, a-circles, c-circles.
Sphere tiger - a(x,y,z),b(w,v),c(a,b) - ac-sphere*circles, a-spheres, bc-torii, b-circles, c-circles.
Torus tiger - a(x,y),b(a,z),c(w,v),d(b,c) - abd-circles^3, acd-tigers, bcd-tigers, ab-torii, ad-torii, bd-torii, cd-torii, a-circles, b-circles, c-circles, d-circles.
Sphere*sphere - a(x,y,z),b(a,w,v) - a-spheres, b-spheres (note that all 4D toratopes in this can be shrunk!)
Circle*circle*sphere - a(x,y),b(a,z),c(b,w,v) - ac-circle*spheres, bc-circle*spheres, ab-torii, c-spheres, a-circles, b-circles
Circtiger - a(x,y),b(z,w),c(a,b,v) - ac-circle*spheres, bc-circle*spheres, c-spheres, a-circles, b-circles.
Circle*glome - a(x,y),b(a,z,w,v) - b-glomes, a-circles.

Is there a way to derive the hole information from these data?
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Postby PWrong » Mon Nov 21, 2005 2:20 pm

One way to classify the toratopes was in terms of unshrinkable curves and surfaces. Here, the middle-cuts are useful. It seems that the global rule is:

IF THE CUT WITH CERTAIN COORDINATE HYPERPLANE RESULTS IN TWO DISTINCT SURFACES OF LOWER DIMENSIONS, THOSE SURFACES CAN'T BE SHRUNK.


What about the torinder? You can wrap a string around the tube, but if you pull it all the way to the edge, it will come off and shrink.

Here's a recursive definition for the equation for an n-sphere, that might come in handy. I'll try to find more stuff like this later.

S_2 = (r cos a_1, r sin a_1)
S_n+1 = S_n sin a_n + r sin a_n e_n+1

You could also write this with a matrix.
Code: Select all
[b]S[/b]_n+1 =
[cos a_n       0    ] [S_n]
[    0       sin a_n ] [ 1  ]


I think we need some kind of transformation matrix or something, that will take a rotatope and "squeeze" it into a toratope, like the duocylinder into a torus.

Maybe I'll make a table of some data for each rotatope, like the dimensions of the actual shape, the dimension of the space it's embedded in, the number of parameters describing it and maybe the kind of holes it has.
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Postby Marek14 » Mon Nov 21, 2005 10:20 pm

PWrong wrote:
One way to classify the toratopes was in terms of unshrinkable curves and surfaces. Here, the middle-cuts are useful. It seems that the global rule is:

IF THE CUT WITH CERTAIN COORDINATE HYPERPLANE RESULTS IN TWO DISTINCT SURFACES OF LOWER DIMENSIONS, THOSE SURFACES CAN'T BE SHRUNK.


What about the torinder? You can wrap a string around the tube, but if you pull it all the way to the edge, it will come off and shrink.



Take note that my definition of "toratope" already excludes torinder, as it's not smooth. This might be another reason :)
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Postby jinydu » Thu Nov 24, 2005 9:36 pm

I asked a question about higher-dimensional spheres in another forum:

http://www.sosmath.com/CBB/viewtopic.ph ... ight=space
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Postby Marek14 » Fri Nov 25, 2005 10:53 am

There is a simple method to come up with parametric equations of sphere of any dimension. You just start with the most basic goniometric equality:

sin^2 x + cos^2 x = 1

This means that we can put
x = cos a
y = sin a
as parametric equations of circle, satisfying x^2+y^2=1. For higher dimensions, we start from this and then replace one of the coordinate by two: For example, we can replace cos a with:
cos b * cos a, sin b * cos a
Because of the basic equality, if you square these two terms and add them, you get (cos^2 b + sin^2 b)*cos^2 a = cos^2 * a. If the third coordinate has z = sin a, x^2+y^2+z^2 = 1.

For 4D, you can either replace one coordinate in the circle by sphere, or replace both by circles:

x = cos c * cos b * cos a
y = sin c * cos b * cos a
z = sin b * cos a
w = sin a

x = cos b * cos a
y = sin b * cos a
z = cos c * sin a
w = sin c * sin a

For 5D, you can replace one coordinate with 4-sphere (2 ways to do that) or replace one with 3-sphere and one with circle:

x = cos d * cos c * cos b * cos a
y = sin d * cos c * cos b * cos a
z = sin c * cos b * cos a
w = sin b * cos a
v = sin a

x = cos c * cos b * cos a
y = sin c * cos b * cos a
z = cos d * sin b * cos a
w = sin d * sin b * cos a
v = sin a

x = cos d * cos b * cos a
x = sin d * cos b * cos a
z = sin b * cos a
w = cos c * sin a
v = sin c * sin a

You can easily verify that the sum of squares of coordinates equals to 1.

In 6D, there are 3 possibilities for (5+1) split, 2 for (4+2) split and 1 for (3+3) split, 6 in total.

In 7D, there are 6+3+2 = 11, in 8 there are 11+6+3+(2*3/2) = 23, in 9D 23+11+6+3*2 = 46, and so on.
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Postby PWrong » Sun Nov 27, 2005 4:55 pm

This is a quote from Jinydu's forum
The new coordinate system must also be orthogonal; in other words, the dot product of a position vector and the partial derivative of that vector with respect to any of the angular coordinates must equal zero. We see this for spherical coordinates, for example.


This could be a useful test. We should find out if it's true for the other form of glomar coordinates, for toratopes, and for the tiger.

Marek, I think you've switched the sine and cos around in all your equations. I made the same mistake in another thread. The azimuthal angle is cos then sine, but all the polar angles are sine then cos. I didn't think it would make a difference at first, but all the boundaries are different. A link in Jinydu's forum has the correct method.

In 6D, there are 3 possibilities for (5+1) split, 2 for (4+2) split and 1 for (3+3) split, 6 in total.

In 7D, there are 6+3+2 = 11, in 8 there are 11+6+3+(2*3/2) = 23, in 9D 23+11+6+3*2 = 46, and so on.


Ok, so let a(n) be the number of coordinate systems in nD. The number of splits is the integer part of n/2. If we split into a k-sphere and an (n-k)-sphere, there are a(k)*a(n-k) possibilities. So we have this recurrence relation:

a(n+1) = sum ( a(k)*a(n-k), from k=1 to [n/2])
where [n/2] is the integer part.

Is that right? I couldn't find the sequence 1,1,2,3,3,6,11,23,46 at Sloanes. Maybe one of these is wrong?
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Postby PWrong » Mon Nov 28, 2005 8:32 am

My formula gives 1, 1, 1, 2, 3, 6, 11, 24, 47, which is Sloane's A000992

Actually, I'm not sure if this sequence is right. We should look more closely at 8D.

The important split is (4+4). Here there are two sets of glomar coordinates, and two choices for each one. But since order isn't important, the number is not 2*2, but the 2nd triangular number, 3.
However, if we consider the (5+4) split of the 9D sphere, the order is important, so there are 3*2 choices.

For an even dimension (2n-D), we can split into two identical n-spheres. The number of possibilities for this split is the a(n)-th triangular number, which is a(n)*(a(n)+1)/2.

We end up with a different recurrence relation for odd and even dimensions.

a(2n) = sum (1 to n-1, a(2n-k)*a(k)) + a(n)*(a(n)+1)/2
a(2n+1) = sum (1 to n, a(2n+1-k)*a(k))

It's not surprising I couldn't find this on Sloanes. It's the same sequence as yours, so you were right.
1, 1, 1, 2, 3, 6, 11, 23, 46, 98
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Postby Marek14 » Mon Nov 28, 2005 8:34 am

What exactly is the result of switching sin and cos? The splitting of coordinates I described can be used equally well on the sin AND cos parts.

For example, putting sphere as
x = cos a cos b
y = sin a cos b
z = sin b

gives x^2+y^2+z^2=1, the same as

x = cos b
y = cos a sin b
z = sin a sin b

(plus, of course, in both cases we may do arbitrary permutation of the equations)

Your formula for number of representation seem to be a bit wrong, because of case where n is even and you split it to two sub-spheres of equal dimension. If k if the number of sphere representation in that dimension, then the correct value for this term is not k*k, but rather k*(k+1)/2, to avoid the double countings of terms that have the same two representations in different order. This might make it harder to find the general formula.

Edit: I see you found this out too.
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Postby PWrong » Wed Nov 30, 2005 5:26 pm

The difference is the values the angles range between. If you use the conventional coordinate system, the first angle (azimuthal, east-west) runs from 0 to 2 pi, and all the other angles (polar angles, like north-south and garp-marp) from 0 to pi. The north pole is at 0, and the south pole is at pi.

Under your system, the polar angles go from -pi/2 to pi/2, with the equator at 0. I'm not sure what's wrong with this, but when I was trying to integrate something once with the wrong coordinates it didn't work properly. Anyway, this stuff has already been defined, there's no point changing it now. http://planetmath.org/encyclopedia/HypersphericalCoordinates.html

Let's get back to the tiger. The equations for the tiger are the sum of two circles and a glome. But this means the tiger probably has two sets of equations, like the glome. How can we find the other set?

On the subject of cutting objects, what happens if we play with the radii and angles, as well as the cartesian coordinates? For instance, say we take a radius from the tiger and let it go to zero.

x = A * cos a + C * cos a * cos c
y = A * sin a + C * sin a * cos c
z = B * cos b + C * cos b * sin c
w = B * sin b + C * sin b * sin c

If A=0 or B=0, we have the equations for a circle plus a glome. This is the circle*sphere torus.
If we let C=0, we get a duocylinder.
If we hold the angle c fixed at any value, we also get a duocylinder, with
r1= (A + C cos c), and r2 = (B + C sin c)
If we let a = 0, we get a torus, with its centre on (A, 0, 0)
This suggests that if we hold either a or b fixed, we'll always get a rotated torus.

The only unexpected shape here is the circle*sphere torus. I didn't think that would be related to the tiger much.
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Postby Marek14 » Thu Dec 01, 2005 6:45 am

PWrong wrote:The difference is the values the angles range between. If you use the conventional coordinate system, the first angle (azimuthal, east-west) runs from 0 to 2 pi, and all the other angles (polar angles, like north-south and garp-marp) from 0 to pi. The north pole is at 0, and the south pole is at pi.

Under your system, the polar angles go from -pi/2 to pi/2, with the equator at 0.

Yes, but isn't that practically the same? This is the same way, for example, how en equator is defined on Earth. Plus, if we let all the coordinates just run from 0 to 2*pi, we get the same results, albeit with multiple covering. (which can be dealt with through some simple identities).
I'm not sure what's wrong with this, but when I was trying to integrate something once with the wrong coordinates it didn't work properly. Anyway, this stuff has already been defined, there's no point changing it now. http://planetmath.org/encyclopedia/HypersphericalCoordinates.html

The thing is that my approach requires to treat both sin and cos the same, as the hypersphere equation used in tiger equations, for example, require splitting both cos and sin parts.
The page you refer to shows merely a convention, and it even mentions that mathematicians and physicians don't agree on it.

Let's get back to the tiger. The equations for the tiger are the sum of two circles and a glome. But this means the tiger probably has two sets of equations, like the glome. How can we find the other set?

Nope, it doesn't mean that at all. The equation of the glome must be "split" a certain way, otherwise you can't add those two circles. Basically, the glome can be expressed (in parametric equations) as having (3+1) dimensions or (2+2) dimensions. if you have two 2's (the circles) already defined, then (2+2) split is the only one possible.

On the subject of cutting objects, what happens if we play with the radii and angles, as well as the cartesian coordinates? For instance, say we take a radius from the tiger and let it go to zero.

x = A * cos a + C * cos a * cos c
y = A * sin a + C * sin a * cos c
z = B * cos b + C * cos b * sin c
w = B * sin b + C * sin b * sin c

If A=0 or B=0, we have the equations for a circle plus a glome. This is the circle*sphere torus.
If we let C=0, we get a duocylinder.
If we hold the angle c fixed at any value, we also get a duocylinder, with
r1= (A + C cos c), and r2 = (B + C sin c)
If we let a = 0, we get a torus, with its centre on (A, 0, 0)
This suggests that if we hold either a or b fixed, we'll always get a rotated torus.

The only unexpected shape here is the circle*sphere torus. I didn't think that would be related to the tiger much.

It looks correct. Tiger is 2+2. Letting C go to zero leads to 2*2 (duocylinder). Letting A or B to zero splits 2 into 1+1, leading into 2+1+1 - circle*sphere.
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Postby PWrong » Fri Dec 02, 2005 11:36 am

Plus, if we let all the coordinates just run from 0 to 2*pi, we get the same results, albeit with multiple covering. (which can be dealt with through some simple identities).


With your coordinate system, the jacobian of the sphere is R^2 sin a.
If you let 'a' run from 0 to pi, or 0 to 2*pi, then the volume of a sphere is zero. Is that what you want?

The page you refer to shows merely a convention, and it even mentions that mathematicians and physicians don't agree on it.

It's easy for you to say, but when you're finding scale factors and integrating stuff, you want to be able to look at existing work and understand it, without changing their notation. Mathematicians and physicists only disagree on whether to use phi or theta for the azimuthal and polar coordinates, which doesn't make much difference. Your coordinate system would change nearly everything we know about the sphere. Scale factors, jacobians, unit vectors...

The thing is that my approach requires to treat both sin and cos the same, as the hypersphere equation used in tiger equations, for example, require splitting both cos and sin parts.


You don't have to change the equation for the tiger at all. The tiger has three sets of azimuthal coordinates, like the second kind of glome. The standard system for glome has one azimuthal and three polar angles.

Azimuthal angles go cos, sine, in that order. Polar angles go sine, cos. Under your system, polar angles would go cos sine. There are important differences between azimuthal and polar angles, and you can't make them the same just by switching the cos and sine around.
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Postby Marek14 » Fri Dec 02, 2005 2:01 pm

Ah, I see. Well, it doesn't really matter - all that is needed is to switch few sin's and cos's in the parametric equations. However, I still find it a bit simpler to just split under equal rules.

So, I guess that a "polar" coordinate is such where only one part is split, and the split part is (by definition) sin. Azimuthal coordinate has split neither cos nor sin, OR both.

So, the equations for petaglome should be:

x = cos a*sin b*sin c*sin d
y = sin a*sin b*sin c*sin d
z = cos b*sin c*sin d
w = cos c*sin d
v = cos d

x = cos a*cos c*sin d
y = sin a*cos c*sin d
z = cos b*sin c*sin d
w = sin b*sin c*sin d
v = cos d

x = cos a*sin b*cos d
y = sin a*sin b*cos d
z = cos b*cos d
w = cos c*sin d
b = sin c*sin d

Is this correct?
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Postby PWrong » Sun Dec 04, 2005 10:01 am

Yep, that's right. :)
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Postby Marek14 » Sun Dec 04, 2005 7:00 pm

So, for the third set of equations (3+2 split), how do you determine whether the 3 should be paired with cos or sin?
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Postby PWrong » Sat Dec 10, 2005 12:48 pm

I'm not sure what you mean. Wouldn't you get the same thing either way?

I'm more worried about which angles are azimuthal now. There's a 6D sphere, ((2+2)+2) that can be made in two ways. You can take a circle and replace one component with a (2+2) glome. Or you can take a sphere, and replace each component with a circle. We could label this as (2+2+2). Either way you get the same set of equations, but the first set has five azimuthal angles, and the other has one polar angle.

The good news is there's only one correct answer. The volume of ((2+2)+2) is twice that of (2+2+2), so one of them is an imposter.
The bad news is, to find the volume we have to find the jacobian, which is a 6*6 matrix, work out its determinant, and then integrate it. :(
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Postby Marek14 » Sat Dec 10, 2005 9:49 pm

PWrong wrote:I'm not sure what you mean. Wouldn't you get the same thing either way?

I'm more worried about which angles are azimuthal now. There's a 6D sphere, ((2+2)+2) that can be made in two ways. You can take a circle and replace one component with a (2+2) glome. Or you can take a sphere, and replace each component with a circle. We could label this as (2+2+2). Either way you get the same set of equations, but the first set has five azimuthal angles, and the other has one polar angle.

The good news is there's only one correct answer. The volume of ((2+2)+2) is twice that of (2+2+2), so one of them is an imposter.
The bad news is, to find the volume we have to find the jacobian, which is a 6*6 matrix, work out its determinant, and then integrate it. :(


Isn't there a way to redo the integral so it would give the same result for my original notation? (which made no distinction between azimuthal and polar angle)
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Postby PWrong » Mon Dec 12, 2005 9:16 am

(which made no distinction between azimuthal and polar angle)

Actually, it does make a distinction, we just don't know which is which. An azimuthal angle runs all the way round the circle, while a polar angle only goes half way. In conventional notation, the polar angles run from 0 to pi. Your polar angles would run from -pi/2 to pi/2. You can't have a spherical coordinate system where both angles have the same range, without doubling up on certain points.

By the way, we might not have to find the determinant of the jacobian after all. The jacobian happens to be the product of the scale factors in many cases. Each scale factor is the magnitude of a column of the matrix.
I haven't seen it stated anywhere, and it's obviously not true for all matrices. However, I can prove that for a 2*2 matrix, the rule applies iff the column vectors are perpendicular. That's good because we already know the coordinates have to be orthogonal.
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