4 Manifold

7/30/2019 4 Manifold

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Some examples of aspherical

homology 4-spheres

John Ratcliffe and Steven Tschantz

Vanderbilt University

Some exam les of as herical homolo 4-s heres . 1/31


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Thank hosts, Brent Everitt, and LMS



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Quotation

The mathematical universe is inhabited not only byimportant species but also by interesting individuals.

Carl Ludwig Siegel



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The regular Euclidean 24-cell

A regular Euclidean 24-cell is a 4-dimensional, regular,

convex polytope in Euclidean 4-space R4 with 24 sideseach a regular octahedron.



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A regular Euclidean 24-cell is a 4-dimensional, regular,

convex polytope in Euclidean 4-space R4 with 24 sideseach a regular octahedron.



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A regular 24-cell is self-dual. It has 24 vertices, 96edges, 96 ridges, 24 sides, and dihedral angle 120 .



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The regular hyperbolic ideal 24-cell

Consider a regular 24-cell Q inscribed in S3 with

vertices e1, . . . ,e4 and (1

2,1

2,1

2,1

2). Then Q is a

right-angled, regular, hyperbolic, ideal, convex polytope

in the projective ball model of hyperbolic 4-space.



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2,1

2,1

2,1

2). Then Q is a





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2,1

2,1

2,1

2). Then Q is a





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The regular ideal 24-cell tessellation

A regular hyperbolic ideal 24-cell Q is a Coxeter polytope,since Q is right-angled. Hence hyperbolic 4-space istessellated by Q and all its images under repeated

reflections in sides.



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The regular ideal 24-cell tessellation

A regular hyperbolic ideal 24-cell Q is a Coxeter polytope,since Q is right-angled. Hence hyperbolic 4-space istessellated by Q and all its images under repeated

reflections in sides.



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Hyperbolic manifolds

A hyperbolic n-manifold M is a complete Riemanniann-manifold of constant sectional curvature 1.


H b li if ld


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Riemann proved that a simply connected hyperbolicn-manifold is isometric to hyperbolic n-space Hn.


H b li if ld


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The fact that the regular ideal 24-cell Q tessellates H4

suggests that we can construct hyperbolic 4-manifolds

by gluing each side Sof Q to another side S of Q with asymmetry of Q of which there are 1152.


H b li if ld


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The problem with this idea is that there are about

4.7 1031 different ways of pairing off the sides of Q withsymmetries of Q.


H b li if ld


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The problem with this idea is that there are about

4.7 1031 different ways of pairing off the sides of Q withsymmetries of Q.

Let us be optimistic and hope for a simple way of

choosing side-pairings.


Th l id l h b li 24 ll


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The regular ideal hyperbolic 24-cell

Consider the 24-cell Q in the hyperboloid model ofhyperbolic 4-space

H4

= {x R5

: x2

1 + + x2

4 x2

5 = 1,and x5 > 0}.

Then Q has 24 ideal vertices represented by the 8points e1 + e5, . . . ,e4 + e5, and the 16 points

(1,1,1,1, 2) all on the light cone.


Th l id l h b li 24 ll


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H4

= {x R5

: x2

1 + + x2

4 x2

5 = 1,and x5 > 0}.




The reg lar ideal h perbolic 24 cell


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H4

= {x R5

: x2

1 + + x2

4 x2

5 = 1,and x5 > 0}.




The regular ideal 24 cell


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The regular ideal 24-cell

The isometries of H4 are represented by Lorentzian5 5-matrices. The reflections in the sides of Q arerepresented by the 24 matrices obtained by permuting

the first four rows and columns of the matrix

R =

1 2 0 0 2

2 1 0 0 2

0 0 1 0 0

0 0 0 1 0

2 2 0 0 3

.




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R =

1

1

1

1

3

.




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R =

2 0 0 2

2 0 0 2

0 0 0 0

0 0 0 0

2 2 0 0

.




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R =

1 2 0 0 2

2 1 0 0 2

0 0 1 0 0

0 0 0 1 0

2 2 0 0 3

.

We have that R Imod 2. Hence the reflections in thesides of Q are in the congruence two subgroup of the

integral Lorentz group PO(4, 1;Z).


24 cell hyperbolic 4 manifolds


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24-cell hyperbolic 4-manifolds

The symmetries of Q that are in the congruence twosubgroup are of the form Diag(1,1,1,1, 1).




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Let us try to construct hyperbolic 4-manifolds by gluing

together the sides of Q using just these 16 symmetries.




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It turns out that there are 137,075 such side-pairingsthat yield a hyperbolic 4-manifold.




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We classified these hyperbolic 4-manifolds up toisometry and found 1171 different isometry types.




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24 cell hyperbolic 4 manifolds





We classified these hyperbolic 4-manifolds up toisometry and found 1171 different isometry types.

Each of these manifolds M has minimum volume

among all hyperbolic 4-manifolds, since

Vol(M) = Vol(Q) = 432,

and so (M) = 1 by the Gauss-Bonnet Theorem.


The most symmetric 24-cell manifold


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The most symmetric 24 cell manifold

The most symmetric of these manifolds N has asymmetry group of order 320. In particular, N has anorder 5 symmetry that cyclically permutes its 5 cusps.


The most symmetric 24-cell manifold


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The most symmetric 24 cell manifold

The most symmetric of these manifolds N has asymmetry group of order 320. In particular, N has anorder 5 symmetry that cyclically permutes its 5 cusps.


David Hilberts theorem


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David Hilbert s theorem

Our illustration of a hyperbolic manifold is not correctlydrawn, since it obviously has some positive curvature.


David Hilberts theorem


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David Hilbert s theorem

Our illustration of a hyperbolic manifold is not correctlydrawn, since it obviously has some positive curvature.

In fact, it is impossible to correctly illustrate a hyperbolicmanifold, since a hyperbolic surface cannot beisometrically embedded in Euclidean 3-space by atheorem of David Hilbert.


David Hilbert


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David Hilbert


A symmetric hyperbolic 4-manifold


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y yp

The manifold N is nonorientable. Let M be theorientable double cover of N. Then (M) = 2. All thesymmetries of N lift to symmetries of M, and M has an

order 5 symmetry that cyclically permutes its 5 cusps.


A symmetric hyperbolic 4-manifold


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y yp

The manifold N is nonorientable. Let M be theorientable double cover of N. Then (M) = 2. All thesymmetries of N lift to symmetries of M, and M has an

order 5 symmetry that cyclically permutes its 5 cusps.


A truncated hyperbolic 4-manifold


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yp

Let M be the 4-manifold with boundary obtained from M byremoving open horocusp neighborhoods from each cusp.




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yp





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yp





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Horospherical geometry


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Gauss and his student F. L. Wachter observed in 1816that the intrinsic geometry on a horosphere inhyperbolic space is Euclidean.


Horospherical geometry


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Gauss and his student F. L. Wachter observed in 1816that the intrinsic geometry on a horosphere inhyperbolic space is Euclidean.

Hence the intrinsic geometry of the boundary of the

truncated manifold M is Euclidean (flat).


Carl Friedrich Gauss


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The boundary tori of the manifold


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Now M = T31 T35 where T3i is a flat 3-torus. Eachtorus T3i has a fundamental domain which is an

s 2s 2s rectangular solid R3i .


The boundary tori of the manifold


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Now M = T31 T35 where T3i is a flat 3-torus. Eachtorus T3i has a fundamental domain which is an

s 2s 2s rectangular solid R3i .

s

2s

2s


The first homology of the manifold


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Now M is a deformation retract of M, and so (M) = 2.In fact H1(M) = Z

5 and H1(M) is generated by thehomology classes [c1], . . . , [c5] where ci is a cycle

represented by a short edge of the box R3i for each i.


The first homology of the manifold


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Now M is a deformation retract of M, and so (M) = 2.In fact H1(M) = Z

5 and H1(M) is generated by thehomology classes [c1], . . . , [c5] where ci is a cycle

represented by a short edge of the box R3i for each i.

s

2s

2s


Meridians of a solid 4-torus


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Consider a flat, solid, 4-dimensional torus D2 T2.Then

(D2 T2) = S1 T2 = T3.




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(D2 T2) = S1 T2 = T3.

A circle S1 {1} D2 T2 is called a meridian of thesolid 4-torus D2 T2.




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(D2 T2) = S1 T2 = T3.


The important property of a meridian is that it bounds a

2-disk.




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(D2 T2) = S1 T2 = T3.


The important property of a meridian is that it bounds a

2-disk.

For example, the meridian S1 {1} of D2 T2 bounds

the 2-disk D2 {1} in D2 T2i .


Dehn filling the truncated manifold


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Consider a flat solid 4-torus D2 T2i for eachi = 1, . . . , 5. By rescaling the solid 4-torus D2 T2i , we

can identify its boundary 3-torus T3i with the ith

boundary component of M, and identify a meridian inT3i with the cycle ci for each i = 1, . . . , 5.




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Let hi : T3i T

3i be an affine homeomorphism for each

i. Then h1, . . . , h5 determine an affine homeomorphismh : M M.




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Let hi : T3i T

3i be an affine homeomorphism for each

i. Then h1, . . . , h5 determine an affine homeomorphismh : M M.

The closed 4-manifold M obtained by Dehn filling each

boundary component of M by h is the attaching space

M = 5

i=1

D2 T2i h M.Some exam les of as herical homolo 4-s heres . 22/31


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Homology of the Dehn filled manifold


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By a Mayer-Vietoris argument (M) = 2 and H1(M) isobtained from H1(M) by adding the relation

jh([ci]) = 0 for each i where j : MM.




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Now H1(T3i ) = Z3 with basis i, i, i with i = [ci] and

i, i the homology classes of cycles represented by

the equal length edges of the box R3i for each i.




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Now H1(T3i ) = Z3 with basis i, i, i with i = [ci] and

i, i the homology classes of cycles represented by

the equal length edges of the box R3i for each i.

s

2s

2s


Constructing homology 4-spheres


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The homology classes i and i, for i = 1, . . . , 5, are inthe kernel of j : H1(M) H1(M) and the homology

classes j(1), . . . , j(5) form a basis of H1(M).




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Hence H1(M) = 0 if and only if

h([ci]) = aii + bii i

for arbitrary integers ai and bi for each i = 1, . . . , 5, since

jh([ci]) = j(i).




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Hence H1(M) = 0 if and only if

h([ci]) = aii + bii i

for arbitrary integers ai and bi for each i = 1, . . . , 5, since

jh([ci]) = j(i).

Suppose we Dehn fill the truncated 4-manifold M so

that H1(M) = 0. Then Poincar duality and the fact that

(M) = 2 imply that M is a homology 4-sphere.


A hyperbolic link complement in S4


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For example, when ai = bi = 0 for each i, DubravkoIvanic proved that the Dehn filled 4-manifold M is

homeomorphic to the 4-sphere S4. Ivanic also verified

by Kirby calculus that M is diffeomorphic to S4

. This isinteresting, since it is unknown whether S4 has a uniquedifferentiable structure.


A hyperbolic link complement in S4


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For example, when ai = bi = 0 for each i, DubravkoIvanic proved that the Dehn filled 4-manifold M is

homeomorphic to the 4-sphere S4. Ivanic also verified

by Kirby calculus that M is diffeomorphic to S4

. This isinteresting, since it is unknown whether S4 has a uniquedifferentiable structure.

There are infinitely many classical knots and linksL S3 whose complement S3 L has a hyperbolicstructure (complete Riemannian metric of constant

sectional curvature 1). Ivanics identification ofMwith S4 gives the first example of a set of linked 2-tori

T S4 whose complement S4 T has a hyperbolicstructure, since S4 T is diffeomorphic to M.


Aspherical homology 4-spheres


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By the Gromov-Thurston 2 theorem, if

Lengthh(ci) > 2 for each i,

the Riemannian metric of constant curvature 1 on Mcan be extended to a Riemannian metric on M ofnonpositive curvature. Recall that we are attaching

solid flat 4-tori of the form D2

T2

i .


Aspherical homology 4-spheres


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By the Gromov-Thurston 2 theorem, if

Lengthh(ci) > 2 for each i,

the Riemannian metric of constant curvature 1 on Mcan be extended to a Riemannian metric on M ofnonpositive curvature. Recall that we are attaching

solid flat 4-tori of the form D2

T2

i .If we Dehn fill so that a2i + b

2i 5 for each i, we can

truncate M so that Lengthh(ci) > 2 for each i. Then M

is a homology 4-sphere with a Riemannian metric ofnonpositive curvature. The universal cover of M is R4

by Cartans Theorem, and so M is an aspherical

homology 4-sphere.


Aspherical Einstein 4-Manifolds


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The construction of these examples answers a questionof William Thurston as to whether aspherical homology4-spheres exist and solves Problem 4.17 on Kirbys1977 4-manifold problem list.


Aspherical Einstein 4-Manifolds


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The construction of these examples answers a questionof William Thurston as to whether aspherical homology4-spheres exist and solves Problem 4.17 on Kirbys1977 4-manifold problem list.

Michael Anderson proved, with PDE techniques, that ifall the curves h(ci) are long enough, then the

Riemannian metric on M can be deformed into aRiemannian metric g such that

Ricg = 3g and (M, g) (M,hyp) as |h(ci)| .

Moreover Anderson proved that infinitely many of ouraspherical homology 4-spheres are closed orientableEinstein 4-manifolds.



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Closed Einstein 4-manifolds


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Now let M be a closed, orientable, Einstein 4-manifold.By a theorem of Berger, (M) 0 with equality if andonly if M is flat.

Let (M) be the signature of M. By Poincar duality,(M) (M) mod 2. Hence if (M) = 0, then (M) iseven.




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By the Hitchin-Thorpe inequality, (M) 3

2 |(M)|.Hence if |(M)| 1, then (M) 3. Thus there are noclosed orientable Einstein 4-manifolds M with(M) = 1.




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By the Hitchin-Thorpe inequality, (M) 3

2 |(M)|.Hence if |(M)| 1, then (M) 3. Thus there are noclosed orientable Einstein 4-manifolds Mwith (M) = 1.

In conclusion, our Einstein 4-manifolds are examples ofnonflat, closed, orientable, Einstein 4-manifolds withminimal Euler characteristic (M) = 2 and (M) = 0.


Albert Einstein


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