redefining the torus product

[PWrong]

Currently the torus product of A and B is defined as follows:

Construct an A.
Let u be the position vector of a point in A.
Construct a new B, with one axis pointing in the direction of u, and all other axes being perpendicular to the axes of A.
Let v be the position vector of a point in B.
The torus product of A and B is the set of all possible vectors u+v.

More succinctly, if A is a kD object, and B is a jD object, then A#B is a (k+j-1)D object, and
A#B = {u + v| u E A(x_1 ... x_k), v E B(u, x_k+1 ... x_k+j-1) }

where E is the membership symbol.

You may not have noticed this, but there is a huge problem with this definition- it creates lots of ugly shapes:
The product of a circle on the origin and another circle is a torus. But what if the first circle is not on the origin? If you work it out properly, you'll see that you don't get a doughnut. In fact you get an ugly looking shape that noone would ever dunk in their coffee.
It's fat at some points, thin at other points, because the position vector doesn't point from the centre.

This also means that the torus product is not associative.
circle#(circle#circle) puts a torus at every point on a circle. This is a nice, ordinary shape.

(circle#circle)#circle on the other hand, is ugly. It starts with a nice torus, and puts a circle at every point. If we take a cross section of the torus, on the xz plane, we see two displaced circles. So the final shape is actually the ugly shape, rotated through the yw plane.
So circle#(circle#circle) does not equal (circle#circle)#circle

Here is the simplest problem with the torus product as it stands.
What do you get if you spherate an arbitrary line (not through the origin) by a circle? A cylinder? No, you get a horrifying little thing who's radius decreases as you follow the line.
r = R/sqrt(1+y^2/x^2)
Where x is the shortest distance between the line and the origin, and y is the distance down the line you go. It looks like gamma from relativity, so I'm going to call it a "relativistic cylinder".

Anyway, the point is it's ugly. That's why we need a new definition for the torus product that can handle objects away from the origin. The key is to construct B so that one axis is parallel to the normal vector, not the position vector. The normal vector is simply the vector perpendicular to the surface.

So our new definition is this:

Construct an A.
Let u be the position vector of a point in A.
Let n be the normal vector of A at a point in A.
Construct a new B, with one axis pointing in the direction of n, and all other axes being perpendicular to the axes of A.
Let v be the position vector of a point in B.
The torus product of A and B is the set of all possible vectors u+v.

A#B = {u + v| u E A(x_1 ... x_k), v E B(n, x_k+1 ... x_k+j-1}

Lets investigate this new product.

Line#circle is always a cylinder.
circle#circle is always a torus.
(circle#circle)#circle = circle#(circle#circle).

Conjecture 1:
The new torus product is associative, that is (A#B)#C = A#(B#C)

I haven't yet found a construction using this product that leads to an ugly shape. I don't yet have a precise definition for "ugly", but you know an ugly shape when you see it.

Conjecture 2:
If A and B are not ugly, then A#B is not ugly.

One nice thing about this is we can now spherate the cylinder.
With the old definition, we'd get a relativistic torus. Now, we just get a torinder.

In my notation, we would write this as (2+1)*2 = 2*2 + 1
In yours, we would say (I think)
(2*1)+1 = (2+1)*1

Finally, there's a very important question we need to ask. What does this mean for the tiger? The tiger is supposed to be the 2D form of the duocylinder, spherated by a circle. But the 2D form of the duocylinder doesn't have a normal vector. A 3D object in 4D has a tangent realm and a normal vector, but a 2D object has a tangent plane and a normal plane. Does this mean the tiger will go extinct?

Actually, it just means the torus product is more complicated than I thought. Rather than spherating, we just place a circle onto the normal plane.

We can redefine the torus product again, to take into account that A may have a normal plane. If we using the torus product along with a subscript, we can can express different objects as the product of the same objects, but in a different dimension. For instance (circle#circle)₃ might be the torus, while (circle#circle)₄ is the duocylinder. More on this when I've worked it out entirely.

[wendy]

The torus product is a subclass of the comb product ie ##.

The actual torus is a fold downwards of the cartesian product of the surface, the shortnening happening in rays of the product direction. One can add the torus product in either a##b or b##a, by means of sock or hose joining.

I should need to look at your proposition, but what i have seen of the torus product, it is indeed associative.

Wendy

[wendy]

You have the torus-product wrong. The torus-product is a surface, that is the product of surfaces. That is, a circle ## circle ## circle is a 4d polytope, ie 2-1 + 2-1 + 2-1 = 4-1.

Line ## circle = two circles, not a cylinder.

Please note that the torus-product is a fold-down of a cartesian product of the surfaces of the bases. Removing all higher-dimensional and base-derived elements, one is left to the product of the surfaces only. This folds down in n! different ways,

Wendy

[PWrong]

I should need to look at your proposition, but what i have seen of the torus product, it is indeed associative.

I thought that too, and any correctly defined torus product should be associative. Unfortunately the old definition that I wrote wasn't correct, hence the new one.

You have the torus-product wrong. The torus-product is a surface, that is the product of surfaces. That is, a circle ## circle ## circle is a 4d polytope, ie 2-1 + 2-1 + 2-1 = 4-1.

I talked about surfaces and forms in the other thread. Every object comes in several forms, depending on how many of it's parameters are angles. This product, as I've defined it, takes two objects and returns another object. It makes no distinction between surfaces and solids.

Line ## circle = two circles, not a cylinder.

The cylinder comes in three forms. A solid, a hollow object with two cells, and a wireframe, which is just a pair of circles. I find it easiest to think of an object in terms of it's "minimal form", that is, with the smallest possible number of dimensions.

This folds down in n! different ways,

Does it? How did you work that out?

[wendy]

Hi

The definition given for the torus product does not match the results of
the torus product. Let's look at what's going on.

TORUS PRODUCT

We start out with a sheet of 100 squares, 10 rows, 10 columns. Each
row is numbered 00 to 90, each column 00 to 09.

We fold this into a decaognal cylinder. Row 0 and 5 are opposite each
other. We make the 0x squares shorter, and the 5x longer, so the
cylinder becomes a C - shape, and then a torus.

The square 55 is now on the outside layer, where the tread would be,
and 00 faces the hub. A ray through these two squares goes through

55, 50, 00, 05.

Now look at the second digits here, 5, 0, 0, 5. The sum of vectors
here would suggest that they ought be 5,0,5,0. This is where the
uglyness is comming from.

In general, we might note that the net of the torus-product is the
cartesian product of the nets of the bases, and it is the order
that we fold these that makes different figures.

For example, in 5D, there is a dodoecahedral-icosahedral torus.

The net of this consists of 240 pentagon-triangle pyramids, made
to make a dodecahedral net #* icosahedral net. If we fold up the
dodecahedral net, we end up with a 5d thing, made of 20 dodecahedra
#* triangles.

When we fold up the icosahedron, we have to shrink the dodecahedron.
To do this, we draw a ray through it, and points the centre of the
ray get contracted. We end up with a normal icosahedron, the surface
replaced by a dodecahedral prism-like layer, contracted to the centre
of the icosahedron.

TORUSES OF LINE + CIRCLE

The torus-product of a thing involving a line, is always two objects.

For example, consider the line ## circle product. We start with the
cartesian product of surfaces: two points by a ring. This gives only
the two circles of the cylinder. However, for this to bound, it must
be in 2d, so we have to 'fold' it.

One fold is to make line ## circle, ie the 'surface' of a line replaced
by a circle. This gives two separate circles.

The circle ## line gives the surface of a circle, replaced by that of
a line. This gives an anulus.

In terms of the cylinder, one could regard the 'outside' point as
being either the wall or one of the ends.

outside = wall -> inside = two circles = disjoint circles

outside = ends -> inside = vertical wall = annulus.

Note that we have

line-circle torus = two disjoint circles
circle-line torus = annulus
line-circle comb = two parallel circles in 3d + notional 2d interior.
line-circle prism = cylinder

The torus product is indeed associative, because it is a cumulation of
skewered polytopes. What the torus-product allows is only the forming
of sock and hose additions, as

sock [] hose.

Consider, the line-circle product. We draw the common nets as

line = 2 points. circle = line-segment.

We end up with in 2d, a set of parallel line segments.

If we suppose these are parrallel to y-axis (which is the leg), the
so that the lines go through 1,0 and -1,0. We end up with two circles
through 1,0 to 2,0 and -1,0 to -2,0.

In terms of the hose-join, we now take the same lines 1,y and -1,y
and stretch these around the point 2,0. This gives an annulus, that
crosses the x-axis at -1,0, 1,0, and again at 3,0 and 5,0. Note
here that a line along 0,y would now be inside the annulus, like
water in a pipe.

THE VECTOR SUM

The reason that the vector sum fails, is because the thing is a product
not a sum. That is, we pass a ray from 0,0 through the centre at eg
5,0, and the bits nearer 0,0 are scrunched up in the product.

But when we consider the product, the scaling of the "5,0" is set
by how faw away the next surface is. The direction changes.

For example, if we make one line of the circle-line product red, we
get either

line ## circle => red-circle + blue circle
circle ## line -> red circle around blue circle.

A line that passes through the 2nd product goes red-blue-blue-red,
and through the first as red-red-blue-blue. The vector sum would
suggest red-blue-red-blue, which never happens.

Therefore we can not express the torus-product as a sum of fixed
vectors. The directions vary.

Wendy

[PWrong]

The definition given for the torus product does not match the results of the torus product. Let's look at what's going on.

Which torus product are you talking about, the old one or the new one?

The torus-product of a thing involving a line, is always two objects.

Ah, I see what you mean. The minimal form of a line is two points. So what I said about the relativistic cylinder wasn't really about the torus product. However, what I said about the circle^3 object still applies.

Therefore we can not express the torus-product as a sum of fixed
vectors. The directions vary.

I never said fixed vectors. Are you sure you understood my definition? Under the old definition, B is rotated so that one axis is parallel to the position vector of A. Under the new definition, one axis is parallel to the normal vector of A.

At the moment the torus only works properly if the first object is in "surface form", that is, it's dimension is one less than the maximal, or solid, form. I'd like to extend it to work for arbitrary forms.

The nD object 'A' has a normal subspace and a tangent subspace at each point. These two subspaces are perpendicular, and their dimensions add to n. The varius forms of A determine the size of the tangent and normal subspace. The normal subspace is like a "leftover" space, which can be "filled up" with the dimensions of B. If the normal space isn't enough, extra dimensions are introduced.

The prism product doesn't use the normal subspace at all. All the dimensions of B are new. The torus product always uses a single normal vector. The tiger is not the torus product of a duocylinder and a circle, because the duocylinder (in 2D form) has a normal plane. We now define a third product that only takes a normal plane, and then creates new dimensions. So under this new product, duocylinder*circle = tiger is a 4D shape, but duocylinder*sphere is a 5D shape.

We could call the prism product A#₀B, the torus product A#₁B, and so on.

The dimension of the new object is easily found.
dim(A#₀B) = dim(A) + dim(B)
dim(A#₁B) = dim(A) + dim(B) - 1
dim(A#_kB) = dim(A) + dim(B) - k

I think (but I'm not sure), that codim(A#_kB) = codim(A) + codim(B), where codim is the dimension of the minimal form of the object.

I think I can work out a strict definition for the general product now.

A#₀B = {u + v| u E A(x_1 ... x_k), v E B(x_k+1 ... x_k+j) }

A#₁B = {u + v| u E A(x_1 ... x_k), v E B(n, x_k+1 ... x_k+j-1) }

A#₂B = {u + v| u E A(x_1 ... x_k), v E B(n₁, n₂, x_k+1 ... x_k+j-n) }

A#_nB = {u + v| u E A(x_1 ... x_k), v E B(N, x_k+1 ... x_k+j-n) }

where N is a basis for the normal subspace of A (that is, any set of independent vectors that span the subspace).