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Poincare Conjecture Proof
H. Vic Dannon
Poincare Conjecture Proof
H. Vic Dannon
[email protected]
September 2007
Revised Jan 2009
Abstract:
We prove that a 3-dimensional manifold Σ3 that is
Compact, Simply-Connected, and Borderless is homeomorphic to
the 3-dimensional sphere S 3 .
2000 Mathematics subject Classification: 57M40, 57N12.
Keywords: Poincare Conjecture, Topology, Manifold, Sphere, Compact,
Borderless, Simply-Connected.
Introduction
A 2-dimensional manifold Σ2 is a Topological Space with
l Hausdorff separation.
That is, distinct points have disjoint
neighborhoods,
l Each point x , has a 2-dimensional open neighborhood U x2 ,
homeomorphic to a disk Dx2 in the Euclidean plane E 2 .
The homeomorphism
hx : U x2 → Dx2 is a bicontinuous bijection
that endows each point x with a local coordinate system. The set
of all of these coordinate systems is called
manifold.
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an Atlas for the
Poincare Conjecture Proof
H. Vic Dannon
It is well-known [4], that a 2-manifold Σ2 that is Compact,
Simply-Connected, and Borderless
is Homeomorphic to the 2-
dimensional sphere S 2 .
For instance, the surface of an Ellipsoid is such a 2-dimensional
manifold.
On the other hand, the Torus is not Simply-Connected, and is not
homeomorphic to the 2-dimensional sphere.
In 1904, Poincare conjectured [7], that a three-dimensional
manifold Σ3 that is Compact, Simply-Connected, and Borderless,
may be homeomorphic to the three-dimensional sphere S 3 .
A 3-dimensional manifold Σ3 is a Topological space with
l Hausdorff separation.
That is, distinct points have disjoint
neighborhoods,
l Each point x , has a 3-dimensional open neighborhood U x3 ,
homeomorphic to a 3-dimensional ball Bx3 in the Euclidean
space E 3 .
The homeomorphism
hx : U x3 → Bx3 is a bicontinuous bijection
that endows each point x with a local coordinate system. The set
of all of the coordinate systems is called an Atlas for the manifold.
Denote the Poincare Manifold, by Σ3 , its topology by T , and an
open set in it by G .
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H. Vic Dannon
We first see that the Poincare manifold has a Metric.
1. Poincare Manifold and its Metric
1.1
The Poincare Manifold is Second Countable
Proof: A Topological Space is Second Countable if its topology
has a countable base. The 3-sphere
S3 =
{( x1, x 2, x 3, x 4 ) ∈ E 4 : x12 + x 22 + x 32 + x 42 = 1}
is Second Countable as a subspace of the closed 4-ball
B4 =
{( x1, x2, x 3, x 4 ) ∈ E 4 : x12 + x 22 + x 32 + x 42 ≤ 1} .
A countable base for the 3-sphere consists of 3-dimensional balls
with rational radius centered at points with rational components.
By [8], any compact manifold has a countable base.
Thus, Poincare Manifold has a countable base for its topology.
1.2
The Poincare Manifold has a Metric
Proof: By Uryson’s Metrization Theorem [3], if a topological space
is Second Countable, Hausdorff, and Compact, it is Metrizable.
The semi-metric is a metric because of the Hausdorff separation,
x ≠ y ⇒ d (x , y ) > 0 . Thus, the Poincare Manifold is a metric space.
2. Loops on the Poincare Manifold
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A manifold is Simply-Connected iff
any loop in it is contractible to a point.
For instance, on the 2-dimensional sphere, any loop is contractible
to a point.
On the other hand, the 2-dimensional Torus surface has loops that
can not be contracted to a point.
A non-self-intersecting curve homeomorphic to a loop, may not be
deformed into a loop, and we do not consider it a loop.
For instance, a non-self-intersecting figure-eight is homeomorphic
to a loop, but in the confinement of two dimensions, it cannot be
deformed into a loop.
Three dimensions are required to twist, unknot, and deform the
figure-eight into a loop.
Similarly, the following knot
is homeomorphic to a loop, but cannot be deformed into a loop in
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Poincare Conjecture Proof
H. Vic Dannon
the confinement of two dimensions.
Four dimensions are required to twist, unknot, and deform it into
a loop [6], [2].
3. Border of the Poincare Manifold
A 2-manifold is borderless iff it is
¾ Connected, and
¾ Its finite triangulation is such that each side of a triangle in
the triangulation is glued to one and only one side of another
triangle [1].
This
guarantees
no
border
points
with
neighborhoods
homeomorphic to semi-disks, as it happens with points on the
equator of a half-hemisphere.
The half-hemisphere has a non-closed triangulation, and the
equator is its border.
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Poincare Conjecture Proof
H. Vic Dannon
In contrast, the 2-dimensional sphere has a closed triangulation,
and is borderless.
Edwin Moise established in [5] that any 3-dimensional manifold
has a finite triangulation.
Thus, a 3-manifold is borderless iff it is
¾ Connected, and
¾ Its finite triangulation is such that each face of a
tetrahedron in the triangulation is glued to one and only one
face of another tetrahedron.
4. Pencil Bundle Structure on the Poincare Manifold
We construct the Poincare Homeomorphism between the Poincare
Manifold and the 3-sphere, as the union of homeomorphisms from
disjoint 2-maniflds that partition the Poincare Manifold, onto
respective disjoint 2-spheres that partition the 3-sphere.
We will show that each of the disjoint 2-manifolds is Compact,
Simply-Connected, and Borderless.
Then, by the theorem for 2-manifolds, each 2-manifold is
homeomorphic to its corresponding 2-sphere.
The Poincare Homeomorphism will be the union of these
homeomorphisms.
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Poincare Conjecture Proof
H. Vic Dannon
4.1 Circle Bundle, and Pencil Bundle
The 2-dimensional sphere
S 2 = {(x1, x 2, x 3 ) : x12 + x 22 + x 32 = 1 }
is a circle bundle [10], a fiber bundle whose fibers are circles, over
the 1-dimensional sphere
S 1 = {(x1, x 2 ) : x12 + x 22 = 1 } .
Each circle in the bundle is parametrized by the angle φ ∈ [0,2π ] .
As φ varies, the circle follows through a pencil [10] of 1dimensional spheres S 1 .
The figure depicts a pencil of 2-dimensional φ -angle planes.
Each is a Euclidean space E 2 that contains a 1-dimensional
sphere S 1 , and the x 3 axis.
The pencil is generated by turning the 1-sphere S 1 about the axis
x 3 that lies on the sphere’s diameter in the 3rd dimension.
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Poincare Conjecture Proof
H. Vic Dannon
4.2 The Pencil Bundle Structure of S 3 is induced on Σ3 and on E 4
Proof: As a compact set in a metric space, the Poincare Manifold
is bounded by a 3-dimensional sphere of radius a .
We’ll assume that a = 1 , and that the sphere is centered at the
origin of the 4-dimensional space that contains it. That is, the
sphere is S 3 . The four coordinate axes are x1, x 2, x 3, and x 4 .
The 3-dimensional sphere S 3 is a circle bundle [10], over the 2dimensional sphere S 2 .
Each circle is parametrized by the angle θ ∈ [0,2π ] .
As θ varies, the circle follows through a pencil [10] of 2dimensional spheres S 2 .
The figure depicts a pencil of 3-dimensional θ -angle hyperplanes.
Each is a Euclidean space E 3 that contains a 2-dimensional
sphere S 2 , and the x 4 axis.
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Poincare Conjecture Proof
H. Vic Dannon
The pencil is generated by turning the 2-sphere S 2 about the axis
x 4 that lies on the sphere’s diameter in the 4th dimension. Denote
S θ2 = S 3 ∩ ( the θ -angle hyperplane that contains the x 4 -axis ) .
Then, { S θ2 : θ ∈ [0,2π ]} is a partition of S 3 since
S θ2 ∩ S θ2 = ∅
1
for
2
θ1 ≠ θ2 , and S 3 =
∪
θ ∈[0,2 π ]
S θ2 .
Each S θ2 corresponds to
Σ2θ ≡ Σ3 ∩ ( the θ -angle hyperplane that contains the x 4 -axis ) ,
and
{ Σ2θ : θ ∈ [0,2π ]}
Σθ
1
∩ Σθ
2
is a partition of Σ3 since
= ∅ for θ1 ≠ θ2 , and
Σ3 =
∪
θ ∈[0,2 π ]
Σ2θ .
Each Σ2θ is embedded in the 3-dimensional Euclidean space
E θ3 ≡ E 4 ∩ ( the θ -angle hyperplane that contains the x 4 -axis ) ,
and
{ Eθ3 : θ ∈ [0,2π ]}
is a partition of E 4 since
E θ3 ∩ E θ3 = ∅ for θ1 ≠ θ2 , and E 4 =
1
4.3
2
∪
θ ∈[0,2 π ]
E θ3 .
T θ ≡ T ∩ ( the θ -angle hyperplane ) is a topology on Σ2θ
with open sets Gθ ≡ G ∩ ( the θ -angle hyperplane )
Proof: T θ is the relative topology of T on Σ2θ .
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Poincare Conjecture Proof
H. Vic Dannon
4.4 Each Σ2θ is Hausdorff
Proof: Every subspace of Hausdorff is Hausdorff.
4.5 Each Σ2θ is a 2-manifold with the topology T θ
Proof: Each point x ∈ Σ2θ , has a 2-dimensional open neighborhood
U x2 = U x3 ∩ ( the θ -angle hyperplane ) ,
homeomorphic
to
a
2-
dimensional disk Dx2 = Bx3 ∩ ( the θ -angle hyperplane ) .
4.6 Each Σ2θ is Compact
Proof:
Σ2θ is compact, as a subset of the compact Σ3 .
4.7 Each Σ2θ is Connected
Proof:
If Σ2θ is disconnected, then Σ3 is disconnected.
4.8 Each Σ2θ has a Closed Finite Triangulation
Proof:
Σ2θ inherits a finite triangulation from Σ3 .
Each side of a triangle in the induced triangulation on Σ2θ , is on
the face of some tetrahedron from the triangulation of Σ3 .
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Poincare Conjecture Proof
H. Vic Dannon
The tetrahedron shares its face with one and only one other
tetrahedron.
Consequently, the triangle side is glued to one and only one side of
another triangle, and the triangulation of Σ2θ is closed.
4.9 Each Σ2θ is Borderless
Proof: By 4.7 and 4.8, each Σ2θ is connected, and has a closed
finite triangulation. Therefore, Σ2θ has no boundary.
4.10 Each Σ2θ is Simply-Connected
If Σ2θ is not simply-connected, there is a loop
Proof:
0
2
θ0
in Σ2θ
0
that is prevented from contracting by a 2-disk Dθ2 that does not
0
belong to Σ2θ .
0
But Σ2θ
has no boundary,
and it cannot contain the disk
0
boundary.
Therefore, the loop
2
θ0
must be slipping along the surface of a 2-
Torus Tθ2 .
0
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Poincare Conjecture Proof
H. Vic Dannon
The Torus surface has no boundary and can be part of Σ2θ , but the
0
interior of the Torus does not belong to Σ2θ .
0
The interior lies in the 3-dimensional Euclidean space E θ3 that
0
contains Σ2θ .
0
Since the E θ3
loop
2
θ0
are disjoint from each other, the non-contracting
, and the Torus Tθ2 are embedded only in E θ3 , and in none
0
0
of the other E θ3 .
Therefore,
2
θ0
, does not contract in any other E θ3 , and in none of
the Σ2θ . Hence,
2
θ0
does not contract in
∪
θ ∈[0,2 π ]
Consequently, the Poincare Manifold
(Σ2θ , θ) = Σ3 .
has a loop that is not
contractible, and Σ3 is not Simply-Connected.
This says that if Σ3 is Simply-Connected, each Σ2θ is simplyconnected.
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Poincare Conjecture Proof
H. Vic Dannon
4.11 Each Σ2θ is homeomorphic to S θ2 with fθ : Σ2θ → S θ2
Proof:
By 4.5, 4.6, 4.9, and 4.10, each Σ2θ
is a 2-dimensional
manifold, which is Compact, Borderless, and Simply-Connected.
By the theorem for 2-dimensional manifolds [4], such manifold is
homeomorphic to the 2-dimensional sphere.
5. Constructing the Poincare Homeomorphism
We proceed to construct the Poincare Homeomorphism as the
union of the homeomorphisms fθ .
To that end, we will apply a theorem from [9].
5.1
¶ S is a subbase for a topology T on X
∙ S is a subbase for a topology T on X
¸ f : X → X is
¬ one-one function which induces a
Á one-one correspondence between the elements of
S , and the elements of S
Then, f is a homeomorphism between X , and X
Let τ be a topology on S 1 , with open sets Θ .
We proceed to write T in terms of T θ .
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Poincare Conjecture Proof
T =
5.2
∪
H. Vic Dannon
θ ∈[0,2 π ]
Tθ
has open sets G =
∪ Gθ
θ ∈Θ
Proof:
(⊇ )
For each θ ∈ [0,2π ] , T ⊇ T θ .
( ⊆ ) If x ∈ G ∈ T , then x belongs to some θ -angle hyperplane for
some θ ∈ [0,2π ] . Hence, x ∈ G ∩ ( the θ -angle hyperplane ) = Gθ ∈ T θ
by 4.3.
5.3 For each θ , fθ T θ is a topology on S θ2 , with open sets fθGθ
Proof: For each θ ∈ [0,2π ] , fθ is homeomorphism.
5.4 T =
∪
θ ∈[0,2 π ]
fθ T θ is a topology on S 3 with open sets
∪
θ ∈Θ∈ τ
fθGθ .
Proof:
∅ ∈ T θ ⇒ ∅ ∈ fθ T θ ⇒ ∅ ∈
Empty set
The Space Σ2θ ∈ T θ ⇒
S3 =
∪
θ ∈[0,2 π ]
(S θ2, θ) ∈
∪
fθ T θ = T .
∪
fθ T θ = T .
θ ∈[0,2 π ]
θ ∈[0,2 π ]
Infinite Union
For any i ∈ I , let G i ∈ T . We’ll show that
∪Gi ∈ T .
i ∈I
For each i ∈ I , there is Θi ∈ τ , so that for all θi ∈ Θi , there are
G i i ∈ T θi
θ
so that G i =
∪
θi ∈Θi
fθiG i i . Hence,
θ
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Poincare Conjecture Proof
H. Vic Dannon
∪Gi
=
i ∈I
∪ ∪
i ∈I θi ∈Θi
fθiG i i .
θ
Taking the union first over all i ∈ I , then over all θ ∈
∪ Θi ,
i ∈I
∪Gi
=
i ∈I
∪ ∪ fθGθi
θ ∈ ∪ Θi i ∈I
i ∈I
∪ Θi ∈ τ ,
Since
and
∪ fθGθi ∈ fθ Tθ
∪Gi ∈ T .
, we have
i ∈I
i ∈I
i ∈I
Finite intersection
Let G 1,G 2 ∈ T . We’ll show that G 1 ∩G 2 ∈ T .
Since G 1,G 2 ∈ T , there are Θ1, Θ2 ∈ τ so that for all θ1 ∈ Θ1 , and
for all θ 2 ∈ Θ2 , there are G 11 ∈ T θ1 , G 22 ∈ T θ2
θ
G1 =
∪
θ1 ∈Θ1
so that
θ
fθ1G 11 , and G 2 =
θ
∪
θ2 ∈Θ2
fθ2G 22 .
θ
Hence,
G 1 ∩G 2 =
∪
θ1 ∈Θ1
=
fθ1G 11 ∩
θ
∪
θ ∈Θ1 ∩ Θ2
∪
θ2 ∈Θ2
fθ2G 22
θ
fθGθ1 ∩ fθGθ2
Since Θ1 ∩ Θ2 ∈ τ , and fθGθ1 ∩ fθGθ2 ∈ fθ T θ , we have G 1 ∩G 2 ∈ T .
5.5
T is Sup Topology on Σ3, and
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{ Tθ
: θ ∈ [0,2π ]} partitions T
Poincare Conjecture Proof
H. Vic Dannon
is the smallest topology on Σ3 which is larger than each
Proof: T
T θ . Such topology is called sup topology [11].
In [11], each of the topologies is defined on the same space X , and
contains it. Here, for each θ , T θ ⊆ Σ2θ ⊆ E θ3 , and all the T θ are
disjoint from each other.
Tθ
1
∩ Tθ
2
{ Tθ
: θ ∈ [0,2π ]} partitions T since
= ∅ for θ1 ≠ θ2 , and T =
5.6 T is Sup Topology on S 3, and
Proof:
fθ T θ
1
1
∩ fθ Tθ
2
2
∪
θ ∈[0,2 π ]
{ fθ Tθ
Tθ .
: θ ∈ [0,2π ]} partitions T
= ∅ for θ1 ≠ θ2 , and T =
∪
θ ∈[0,2 π ]
fθ T θ .
5.7 For each θ , let Sθ be a Subbase for T θ on Σ2θ , with sets H θ .
Then,
∪
θ ∈[0,2 π ]
Sθ is a subbase for T , with subbase sets
∪
θ ∈Θ∈ τ
Hθ
Proof: If G ∈ T , there is Θ ∈ τ , and there are Gθ ∈ T θ , so that
∪ Gθ .
G =
θ ∈Θ
For each Gθ , there are base sets Bθj , j ∈ J , so that
Gθ =
∪ Bθj
j ∈J
For each Bθj , there are subbase sets H θj ,k ∈ Sθ , k = 1...n so that
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Poincare Conjecture Proof
H. Vic Dannon
n
Bθj
∩ H θj,k
=
k =1
Therefore,
n
G =
∪ ∪ ∩ H θj,k
θ∈Θ j ∈J k =1
n
=
∪ ∩ ∪ H θj,k .
j ∈J k =1 θ ∈Θ
Thus,
∪
θ ∈[0,2 π ]
Sθ is a subbase for T , with subbase sets
∪
θ ∈Θ∈ τ
Hθ .
5.8 For each θ , fθ Sθ is a subbase for fθ T θ with subbase sets fθH θ
Proof: For each θ ∈ [0,2π ] , fθ is homeomorphism.
5.9
∪
θ ∈[0,2 π ]
fθ Sθ is a subbase for T on S 3 with subbase sets
∪
θ ∈Θ∈ τ
fθH θ
Proof: If G ∈ T , there is Θ ∈ τ , and there are fθGθ ∈ fθ T θ , so that
G =
∪ fθGθ .
θ∈Θ
For each Gθ , there are base sets Bθj , j ∈ J , so that
fθGθ =
∪ fθBθj
j ∈J
For each Bθj , there are subbase sets H θj ,k ∈ Sθ , k = 1...n so that
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Poincare Conjecture Proof
H. Vic Dannon
n
fθBθj
∩ fθH θj,k
=
k =1
Therefore,
n
G =
∪ ∪ ∩ fθH θj,k
θ∈Θ j ∈J k =1
n
=
∪ ∩ ∪ fθH θj,k .
j ∈J k =1 θ ∈Θ
Thus,
∪
θ ∈[0,2π ]
fθ Sθ is a subbase for T , with subbase sets
∪
θ ∈Θ∈ τ
fθH θ .
To obtain the Poincare Homeomorphism by 5.1, we need to define
a function f : Σ3 → S 3 which is
¬ one-one
Á one-one correspondence between a subbase set for T ,
∪ Hθ ,
and a subbase set for T ,
θ∈Ω
5.10
∪ fθH θ .
θ∈Ω
Definition of f
For x ∈ Σ3 , there are unique θ ∈ [0,2π ] , and x θ ∈ Σ2θ so that
x = (x θ , θ) .
We define the mapping f from Σ3 onto S 3 by f (x θ , θ) = ( fθx θ , θ) .
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Poincare Conjecture Proof
H. Vic Dannon
f is one-one
5.11
Proof:
Let
x = (x θ , θ) , and y = (yφ , φ) .
Assume that
( fθx θ , θ) = ( fφyφ , φ) .
Since Σ2θ are disjoint,
θ = φ , and fθx θ = fθyθ .
Since fθ is a homeomorphism,
x θ = yθ .
Hence,
x = y.
f is one-one between the subbase set
5.12
and the subbase set
∪ fθH θ
in Σ
3
∪ Hθ
in S 3
θ ∈Θ
θ ∈Θ
Let H =
Proof:
∪ Hθ ,
and K =
θ∈Θ
∪ Kφ .
φ∈Φ
Assume that,
{(fθH θ , θ) : θ ∈ Θ, H θ
⊆ H } = { ( fφK φ , φ) : φ ∈ Φ, K φ ⊆ K } .
Since the Σ2θ are disjoint,
Θ = Φ,
and fθH θ = fθK θ , for any θ ∈ Θ .
Since fθ is a homeomorphism, for each θ ∈ Θ ,
H θ = Kθ .
Hence, H = K .
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Poincare Conjecture Proof
5.13
H. Vic Dannon
f is the Poincare Homeomorphism
Proof: By 5.1, 5.2, 5.4, 5.7, 5.9, 5.11, and 5.12, f is a homeomorphism. It is the Poincare Homeomorphism from Σ3 onto S 3 .
5.14 The Poincare Homeomorphism restriction to Σ2θ is fθ .
f
Σ2θ
= fθ .
5.15 The Poincare Homeomorphism is the union of all the fθ .
f = { fθ : θ ∈ [0,2π ]} =
20
∪
θ ∈[0,2 π ]
fθ .
Poincare Conjecture Proof
H. Vic Dannon
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