## Monday, December 31, 2007

### Overlap Examples

Here are three tetrahedron overlaps. The right two differ from the disjoint one by single migrations. The middle one is a point migration since it is a point jumping across a boundary. The rightmost one is a segment migration since it is a segment jumping across another segment.

## Saturday, December 8, 2007

### Tetrahedron Overlaps

Linear Space Equivalence

Two linear spaces are equivalent iff they are isomorphic to the same finite set of nonparallel noncoincidental linear surfaces. Two finite sets of nonparallel noncoincidental linear surfaces are equivalent iff they are isomorphic to the same linear space. A finite set of nonparallel noncoincidental linear surfaces are said to be equivalent to a linear space iff they are isomorphic.

Migration

In space ** S**, regions

^{n}= (B,P,S,n)

**and**R

_{1}

**are neighbors with respect to**R

_{2}

**iff**b

_{0}∈B

**and**R = R

_{1}(b)

_{2}(b)

**, for**R ≠ R

_{1}(b

_{0})

_{2}(b

_{0})

**, where**b∈B

_{0}

**. In space**B = B

_{0}∪{b

_{0}}

**, region**S = (B,P,S,n)

^{n}

**is the opposite of region**R

_{2}

**iff for**R

_{1}

**,**b∈B

**has a nonempty neighbor with respect to**R ≠ R ⇔ R

_{1}(b)

_{2}(b)

_{1}

**. Two spaces**b

**and**S = (B,P

_{1}

^{n}

_{1},S,n)

**are migrations of each other iff for each region**S = (B,P

_{2}

^{n}

_{2},S,n)

**in**R

**and**S

_{1}

^{n}

**, one or both of the following is true.**S

_{2}

^{n}

**is empty in**R

**is empty in**S ⇔ R

_{1}

^{n}

**. Or,**S

_{2}

^{n}

**is empty**R

**⇔**the opposite of

**is not empty.**R

Example

Regions ** A**,

**, and**B

**have empty opposites. Region**C

**is the opposite of region**A

**. Region**E

**is the opposite of region**E

**.**D

Migration Theorem

If ** S** and

_{1}

^{n}

**are migrations of each other then,**S

_{2}

^{n}

**is linear**S

_{1}

^{n}

**⇔**S is linear.

_{2}

^{n}

Without loss of generality, assume ** S** is linear,

_{1}

^{n}

**is empty in**R

**,**S

_{2}

^{n}

and its opposite is not empty in

**. Suppose**S

_{2}

^{n}

**is any subspace of**S

_{3}

^{n}

**. If**S

_{2}

^{n}

**is not a region of**R

**, then it is linear because its identical subspace in**S

_{3}

^{n}

**is linear. If**S

_{1}

^{n}

**is a region of**R

**, then so is its opposite, so the number of nonempty regions in**S

_{3}

^{n}

**is unaltered, so it is linear, and the theorem is proved.**S

_{3}

^{n}

Continuity Theorem

There exists continuous linear surfaces in

*L*:

*R*→{m

**, such that**

*R*

^{n}}

**is nonparallel and nonconicidental for**

*L*(

*r*)

**, only if there exist** ≠

*r*

*r*

_{0}∈

*R*

**,**

*r*

_{1}

**, and linear spaces**

*r*

_{2}∈

*R*

**and**S

_{1}

^{n}

**, such that**S

_{2}

^{n}

**,**

*r*

_{1}<

*r*

_{0}<

*r*

_{2}

**,**

*L*(

*r*

_{1})≡S

_{1}

^{n}

**, and**

*L*(

*r*

_{2})≡S

_{2}

^{n}

**and**S

_{1}

^{n}

**are migrations of each other.**S

_{2}

^{n}

As a set of linear surfaces changes continuously, its regions cannot suddenly become empty. Before becoming empty, a region must become infinitesimally small. As a region becomes infinitesimally small, so does its opposite. Thus, as a region becomes empty, its opposite becomes nonempty.

Simplex Overlap

A simplex overlap is two disjoint sets of boundaries, each of which is a simplex. The inside of a simplex overlap ^{n}_{1}^{n}_{1},S,n)_{1} is the union of the insides of the simplices. Two simplex overlaps, ** (M**, and

_{1},M

_{2})

**, where**(N

_{1},N

_{2})

**,**M

_{1}

**,**M

_{2}

**,**N

_{1}

**are the simplices of the overlaps, are equivalent iff there exists one to one mapping**N

_{2}

**, such that**g:M

_{1}∪M

_{2}→N

_{1}∪N

_{2}

**for some**g(M = N

_{1})

_{i}

**, and for each**i∈{1,2}

**M**b∈

_{1}∪M

_{2}the section of the inside of (M

_{1},M

_{2}) by

**is equivalent to the section of the inside of (N**b

_{1},N

_{2}) by

**. There are 11 equivalence classes of triangle overlap. How many equivalence classes of tetrahedron overlap are there?**g(b)

## Sunday, November 11, 2007

### Sidedness Geometry

##
Linear Space

Take points and boundaries as undefined. Define a sidedness relation **, where**S:B×P→{0,1}

**is a finite set of boundaries and**B

**is a finite set of points. Two points**P

**,**p

_{0}

**are said to be on the same side of a boundary**p

_{1}

**iff**b

**, and they are said to be opposite**S(b,p = S(b,p

_{0})

_{1})

**iff**b

**. Define a space as**S(b,p ≠ S(b,p

_{0})

_{1})

**, where**S = (B,P,S,n)

^{n}

**is a sidedness relation on**S

**, and integer**B×P

**is called the dimension of the space. Define a region as**n>0

**. Note that there are**R:B→{0,1}

**regions in a space with**2

^{m}

**boundaries. A point**m

**is said to be in region**p∈P

**iff**R

**for each**S(b,p) = R(b)

**. A region is said to be empty iff no points are in it. Define a subspace of**b∈B

**to be**S = (B,P,S,n)

^{n}

**such that**S = (B

_{0}

^{n}

_{0},P,S

_{0},n)

**, and**B

_{0}⊂B

**for each**S = S(b,p)

_{0}(b,p)

**, and**b∈B

_{0}

**. A space**p∈P

**is said to be linear iff in each subspace the number of nonempty regions is equal to**S

^{n}

**, where**f(n,m)

**is the number of boundaries in the subspace,**m

**,**f(1,m) = m+1

**, and**f(n,0) = 1

**. A sidedness relation**f(n+1,m+1) = f(n+1,m)+f(n,m)

**, where**

*S*:

*B*×

*P*→{0,1}

**is a finite set of linear surfaces in Euclidean space**

*B*

**and**

*R*

^{n}

**is a finite set of points not on the surfaces in**

*P*⊂

*R*

^{n}

**, is given by**

*B*

**, where**

*n*

_{b}∙(

*p*-

*p*

_{b})>0

**is a normal to**

*n*

_{b}

**,**

*b*∈

*B*

**is a point on**

*p*

_{b}

**, and**

*b*

**. A set of**

*p*∈

*P*

**linear surfaces in**n

**determines an intersection point. A set of linear surfaces in**

*R*

^{n}

**is associated with a set of intersection points determined by all subsets of**

*R*

^{n}

**linear surfaces. A finite set of linear surfaces in**n

**is noncoincidental iff no two surfaces in it are parallel, no**

*R*

^{n}

**surfaces in it share a point, and no linear surface goes through more than**n+1

**intersection points. A space**n

**is isomorphic with a finite set of noncoincidental linear surfaces in**S

^{n}

**iff there exist one to one mappings**

*R*

^{n}

**,**g:B→

*B*

**, and**h:P→

*P*

**, such that**s

_{b}:{0,1}→{0,1}

**for each**S(b,p) = s

_{b}(

*S*(g(b),h(p)))

**and**b∈B

**.**p∈P

##
Linear Space Theorem

A finite set of noncoincidental linear surfaces in **exists if and only if isomorphic linear space**

*R*

^{n}

**exists.**S

^{n}

First I will prove that a finite set of linear surfaces in

**exists only if isomorphic linear space**

*R*

^{n}

**exists. Each point in**S

^{n}

**divides one of a finite set of regions spanning**

*R*

**, so the number of regions in**

*R*

**is**

*R*

**, where**f(1,m) = m+1

**is the number of points dividing**m

**. Thus, the "only if" part of the theorem is true for**

*R*

**. There exist any number of points in**n = 1

**on either side of a linear surface, so**

*R*

^{n}

**, for all**f(n,1) = 2

**. Thus, the "only if" part of the theorem is true for**n

**. Suppose the "only if" part of the theorem is true for**m = 1

**. Also suppose that if**n < i

**, then the "only if" part of the theorem is true for**n = i

**. I must show that the "only if" part of the theorem is true for**m < j

**and**n = i

**. Let**m = j

**be a set of**

*L*

**noncoincidental linear surfaces in**j-1

**, and let**

*R*

^{i}

**be a linear surface in**

*l*

**, noncoincidental with**

*R*

^{i}

**. By supposition,**

*L*

**divides**

*L*

**into**

*R*

^{i}

**regions. The intersections of**f(i,j-1)

**with the surfaces in**

*l*

**are isomorphic with linear surfaces in**

*L*

**. By supposition,**

*R*

^{i-1}

**divides**

*l*

**of the regions formed by**f(i-1,j-1)

**in**

*L*

**. Therefore, there are**

*R*

^{i}

**regions formed by**f(i,j-1)+f(i-1,j-1) = f(i,j)

**in**

*L*∪{

*l*}

**, and the "only if" part of the theorem is proved. The "if" part I will prove below after proving some supportive theorems.**

*R*

^{i}

##
Section

A boundary **is said to extend**b

**to**S = (B

_{0}

^{n}

_{0},P

_{0},S

_{0},n)

**iff**S = (B

_{1}

^{n}

_{1},P

_{1},S

_{1},n)

**is a subspace of**S

_{0}

^{n}

**, and**S

_{1}

^{n}

**. If**B = B

_{1}

_{0}∪{b}

**extends**b

**to**S = (B

_{0}

^{n}

_{0},P

_{0},S

_{0},n)

**, then**S

_{1}

^{n}

**divides region**b

**in**R

_{0}

**into regions**S

_{0}

^{n}

**and**R

_{1a}

**in**R

_{1b}

**iff**S

_{1}

^{n}

**and**R

_{1a}

**are not empty,**R

_{1b}

**, and**R ≠ R

_{1a}

_{1b}

**for each**R = R = R

_{1a}(b

_{0})

_{1b}(b

_{0})

_{0}(b

_{0})

**. If**b

_{0}∈B

_{0}

**extends**b

**, then**S = (B

_{0}

^{n}

_{0},P

_{0},S

_{0},n)

**is the section of**S

_{1}

^{n-1}

**by**S

_{0}

^{n}

**iff**b

**is isomorphic with**S

_{1}

^{n-1}

**, where**S = (B

_{1}

^{n}

_{0},P⊂P

_{0},S

_{0},n)

**iff**p∈P

**is in a region divided by**p

**.**b

##
Section Theorem

If **is a section of**S

^{n-1}

**by**S

_{0}

^{n}

**,**b

**extends**b

**to**S

_{0}

^{n}

**,**S

_{1}

^{n}

**is linear, and**S

_{0}

^{n}

**is linear, then**S

_{1}

^{n}

**is linear.**S

^{n-1}

The number of regions in

**is the number of regions divided by**S

^{n-1}

**. The number of regions divided by**b

**is the number of regions in**b

**subtracted from the number of regions in**S

_{0}

^{n}

**. Since**S

_{1}

^{n}

**and**S

_{0}

^{n}

**are linear, the number of regions in**S

_{1}

^{n}

**is**S

^{n-1}

**. The same reasoning applies to each subspace of**f(n,m+1)-f(n,m) = (f(n,m)+f(n-1,m))-f(n,m) = f(n-1,m)

**, because each subspace of**S

^{n-1}

**is the section of a subspace of**S

^{n-1}

**by**S

_{0}

^{n}

**. Therefore,**b

**is linear.**S

^{n-1}

##
Antisection Theorem

If **is a section of**S

^{n-1}

**by**S

_{0}

^{n}

**,**b

**extends**b

**to**S

_{0}

^{n}

**,**S

_{1}

^{n}

**is linear, and**S

^{n-1}

**is linear, then**S

_{0}

^{n}

**is linear.**S

_{1}

^{n}

The number of regions in

**is the number of regions divided by**S

^{n-1}

**. The number of regions in**b

**is the number of regions divided by**S

_{1}

^{n}

**added to the number of regions in**b

**. Since**S

_{0}

^{n}

**and**S

^{n-1}

**are linear, the number of regions in**S

_{0}

^{n}

**is**S

_{1}

^{n}

**The same reasoning applies to each subspace of**f(n-1,m)+f(n,m) = f(n,m+1).

**, because each subspace of**S

_{1}

^{n}

**is the extension of a subspace of**S

_{1}

^{n}

**by**S

_{0}

^{n}

**. Therefore,**b

**is linear.**S

_{1}

^{n}

##
Simplex

A simplex in linear space **is any**S

^{n}

**boundaries in**n+1

**. Note that because**S

^{n}

**linear surfaces are linearly independent in**n

**, and the "only if" part of the Linear Space Theorem is true,**

*R*

^{n}

**. By definition of linear space,**f(n,n) = 2

^{n}

**, and if**f(1,2) = 2

^{2}-1

**, then**f(n-1,n) = 2

^{n}-1

**, so the number of nonempty regions in a simplex is**f(n,n+1) = f(n,n)+f(n-1,n) = 2 = 2

^{n}+2

^{n}-1

^{n+1}-1

**. Consider the empty region of a simplex in space**2

^{n+1}-1

**. Reverse any one sidedness of the empty region, and call the resultant region a vertex region of the simplex. Since there are**S

^{n}

**boundaries in a simplex, there are**n+1

**vertex regions in it. Reverse all of the sidednesses of the empty region and call the resultant region the inside of the simplex. Since there is only one empty region in a simplex, there is only one inside in it.**n+1

##
Simplex Theorem

There exists no boundary dividing all vertex regions of a simplex. Suppose there does exist boundary

**dividing all**b

**vertex regions of simplex**n+1

**,**b

_{1}

**, ...**b

_{2}

**. Consider the section of**b

_{n+1}

**,**b

_{1}

**, ...**b

_{2}

**by**b

_{n}

**. That section is a simplex with vertices sectioned from the vertices opposite**b

**,**b

_{1}

**, ...**b

_{2}

**. Since the vertex region opposite**b

_{n}

**is opposite one of**b

_{n+1}

**,**b

_{1}

**, ...**b

_{2}

**from each of the sectioned vertex regions, it is not sectioned to any one region of the sectioned simplex. This contradicts the supposition of**b

_{n}

**, so no such**b

**exists, and the theorem is proved.**b

##
Independent Boundary Theorem

If **extended by**S

^{n}

**is linear, and**b

_{0}

**extended by**S

^{n}

**is linear, then there exists a linear superspace of**b

_{1}

**extended by**S

^{n}

**and**b

_{0}

**extended by**S

^{n}

**.**b

_{1}

(This proof attempt is wrong. I am working on a correction.)

**it is true because**S

^{1}

**and**b

_{0}

**divide different regions, so points on one or the other side of**b

_{1}

**in the region it divides will all be on one side of**b

_{0}

**. Suppose it is true for**b

_{1}

**. I must prove it for**S

^{n-1}

**. Suppose**S

^{n}

**is not linear if it is extended by both**S

^{n}

**and**b

_{0}

**together, but**b

_{1}

**is linear if it is extended by**S

^{n}

**or**b

_{0}

**individually. Discard boundaries from**b

_{1}

**until the reduced**S

^{n}

**is linear when extended by both**S

^{n}

**and**b

_{0}

**. This is possible because**b

_{1}

**and**b

_{0}

**alone in**b

_{1}

**is a linear space. Let**S

^{n}

**be the last boundary discarded from**b

_{2}

**. Consider the section by**S

^{n}

**,**b

_{2}

**. By the Section Theorem,**S

^{n-1}

**is linear because**S

^{n-1}

**is linear and**S

^{n}

**extended by**S

^{n}

**is linear. Similarly,**b

_{2}

**extended by the section of**S

^{n-1}

**, or by the section of**b

_{0}

**, is linear. Thus, by supposition,**b

_{1}

**extended by the sections of both**S

^{n-1}

**and**b

_{0}

**is linear. By the Antisection Theorem, since**b

_{1}

**extended by**S

^{n}

**and**b

_{0}

**is linear, and**b

_{1}

**extended by the sections of**S

^{n-1}

**and**b

_{0}

**is linear,**b

_{1}

**extended by**S

^{n}

**,**b

_{0}

**, and**b

_{1}

**is linear. This contradicts the supposition, so the theorem is proved.**b

_{2}

##
"If" Part of Linear Space Theorem

If linear space **exists, then a finite set of linear surfaces in**S

^{n}

**isomorphic with**

*R*

^{n}

**exist.**S

^{n}

(This proof attempt, in particular the struckout text, is wrong. I am working on a correction.) Suppose no

**exits with boundaries and sidednesses isomorphic to some linear space**

*R*

_{0}

^{n}

**. Let**S

_{0}

^{n}

**be boundaries of**S

_{1}

^{n}

**which are isomorphic with the boundaries and sidednesses of some**S

_{0}

^{n}

**. Until impossible, extend**

*R*

_{1}

^{n}

**by boundaries from**S

_{1}

^{n}

**such that**S

_{0}

^{n}

**is still isomorphic with an extension of**S

_{1}

^{n}

**. By supposition, there exists boundary**

*R*

_{1}

^{n}

**in**b

_{0}

**such that**S

_{0}

^{n}

**is not in**b

_{0}

**.**S

_{1}

^{n}

**,**R

_{1}

**, ...**R

_{2}

**be the**R

_{n+1}

**regions of**n+1

**divided by**S

_{1}

^{n}

**such that no**b

_{0}

**exists which divides those regions in**

*b*

_{0}

**. Let**

*R*

_{1}

^{n}

**and**p

_{1}

**be in**q

_{1}

**, opposite**R

_{1}

**, and similarly for**b

_{0}

**, ...**(p

_{2},q

_{2})

**. Construct**(p

_{n+1},q

_{n+1})

**,**

*b*

_{1}

**, ...**

*b*

_{2}

**such that**

*b*

_{n+1}

**and**p

_{1}

**are in one vertex region of the simplex formed by**q

_{1}

**,**

*b*

_{1}

**, ...**

*b*

_{2}

**, and**

*b*

_{n+1}

**, ...**(p

_{2},q

_{2})

**are in the other vertex regions. The prior construction is possible because, given**(p

_{n+1},q

_{n+1})

**points in**n+1

**a simplex can be constructed with those points as vertices; a point can be chosen in each of**

*R*

^{n}

**,**R

_{1}

**, ...**R

_{2}

**; the vertex region of each point at least partially overlaps the region the point is in; and without loss of generality, we may assume**R

_{n+1}

**and**p

_{i}

**are in the portion of**q

_{i}

**which overlaps the vertex region.**R

_{i}

**by**

*R*

_{1}

^{n}

**,**

*b*

_{1}

**, ...**

*b*

_{2}

**to**

*b*

_{n+1}

**. Since**

*R*

_{2}

^{n}

**obeys the "only if" part of the Linear Space Theorem,**

*R*

_{2}

^{n}

**can be similarly extended to**S

_{1}

^{n}

**such that**S

_{2}

^{n}

**is isomorphic to**S

_{2}

^{n}

**. Since**

*R*

_{2}

^{n}

**can be extended by**S

_{1}

^{n}

**, and**b

_{0}

**divides different regions from**b

_{0}

**,**b

_{1}

**, ...**b

_{2}

**,**b

_{n+1}

**can be extended by**S

_{2}

^{n}

**to**b

_{0}

**because of the Independent Boundary Theorem. Now,**S

_{3}

^{n}

**goes through each of the vertex regions of**b

_{0}

**,**b

_{1}

**, ...**b

_{2}

**in**b

_{n+1}

**. This violates the Simplex Theorem. Therefore the supposition was false, and there does exist**S

_{3}

^{n}

**with boundaries and sidednesses isomorphic with**

*R*

_{0}

^{n}

**.**S

_{0}

^{n}