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Transcript of 4 Manifold
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Some examples of aspherical
homology 4-spheres
John Ratcliffe and Steven Tschantz
Vanderbilt University
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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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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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The regular Euclidean 24-cell
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 Euclidean 24-cell
A regular 24-cell is self-dual. It has 24 vertices, 96edges, 96 ridges, 24 sides, and dihedral angle 120 .
Some exam les of as herical homolo 4-s heres . 5/31
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The regular Euclidean 24-cell
A regular 24-cell is self-dual. It has 24 vertices, 96edges, 96 ridges, 24 sides, and dihedral angle 120 .
Some exam les of as herical homolo 4-s heres . 5/31
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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.
Some exam les of as herical homolo 4-s heres . 6/31
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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.
Some exam les of as herical homolo 4-s heres . 6/31
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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.
Some exam les of as herical homolo 4-s heres . 6/31
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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.
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H b li if ld
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Hyperbolic manifolds
A hyperbolic n-manifold M is a complete Riemanniann-manifold of constant sectional curvature 1.
Riemann proved that a simply connected hyperbolicn-manifold is isometric to hyperbolic n-space Hn.
Some exam les of as herical homolo 4-s heres . 8/31
H b li if ld
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Hyperbolic manifolds
A hyperbolic n-manifold M is a complete Riemanniann-manifold of constant sectional curvature 1.
Riemann proved that a simply connected hyperbolicn-manifold is isometric to hyperbolic n-space Hn.
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.
Some exam les of as herical homolo 4-s heres . 8/31
H b li if ld
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Hyperbolic manifolds
A hyperbolic n-manifold M is a complete Riemanniann-manifold of constant sectional curvature 1.
Riemann proved that a simply connected hyperbolicn-manifold is isometric to hyperbolic n-space Hn.
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.
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.
Some exam les of as herical homolo 4-s heres . 8/31
H b li if ld
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Hyperbolic manifolds
A hyperbolic n-manifold M is a complete Riemanniann-manifold of constant sectional curvature 1.
Riemann proved that a simply connected hyperbolicn-manifold is isometric to hyperbolic n-space Hn.
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.
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.
Some exam les of as herical homolo 4-s heres . 8/31
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.
Some exam les of as herical homolo 4-s heres . 9/31
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.
Some exam les of as herical homolo 4-s heres . 9/31
The reg lar ideal h perbolic 24 cell
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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.
Some exam les of as herical homolo 4-s heres . 9/31
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
.
Some exam les of as herical homolo 4-s heres . 10/31
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
1
1
1
3
.
Some exam les of as herical homolo 4-s heres . 10/31
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 =
2 0 0 2
2 0 0 2
0 0 0 0
0 0 0 0
2 2 0 0
.
Some exam les of as herical homolo 4-s heres . 10/31
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
.
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).
Some exam les of as herical homolo 4-s heres . 10/31
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).
Some exam les of as herical homolo 4-s heres . 11/31
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).
Let us try to construct hyperbolic 4-manifolds by gluing
together the sides of Q using just these 16 symmetries.
Some exam les of as herical homolo 4-s heres . 11/31
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).
Let us try to construct hyperbolic 4-manifolds by gluing
together the sides of Q using just these 16 symmetries.
It turns out that there are 137,075 such side-pairingsthat yield a hyperbolic 4-manifold.
Some exam les of as herical homolo 4-s heres . 11/31
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).
Let us try to construct hyperbolic 4-manifolds by gluing
together the sides of Q using just these 16 symmetries.
It turns out that there are 137,075 such side-pairingsthat yield a hyperbolic 4-manifold.
We classified these hyperbolic 4-manifolds up toisometry and found 1171 different isometry types.
Some exam les of as herical homolo 4-s heres . 11/31
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).
Let us try to construct hyperbolic 4-manifolds by gluing
together the sides of Q using just these 16 symmetries.
It turns out that there are 137,075 such side-pairingsthat yield a hyperbolic 4-manifold.
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.
Some exam les of as herical homolo 4-s heres . 11/31
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.
Some exam les of as herical homolo 4-s heres . 12/31
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.
Some exam les of as herical homolo 4-s heres . 12/31
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.
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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.
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David Hilbert
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David Hilbert
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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.
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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.
Some exam les of as herical homolo 4-s heres . 15/31
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.
Some exam les of as herical homolo 4-s heres . 16/31
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.
Some exam les of as herical homolo 4-s heres . 16/31
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.
Some exam les of as herical homolo 4-s heres . 16/31
A truncated hyperbolic 4-manifold
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Let M be the 4-manifold with boundary obtained from M byremoving open horocusp neighborhoods from each cusp.
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A truncated hyperbolic 4-manifold
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Let M be the 4-manifold with boundary obtained from M byremoving open horocusp neighborhoods from each cusp.
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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.
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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.
Hence the intrinsic geometry of the boundary of the
truncated manifold M is Euclidean (flat).
Some exam les of as herical homolo 4-s heres . 17/31
Carl Friedrich Gauss
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Some exam les of as herical homolo 4-s heres . 18/31
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 .
Some exam les of as herical homolo 4-s heres . 19/31
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
Some exam les of as herical homolo 4-s heres . 19/31
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.
Some exam les of as herical homolo 4-s heres . 20/31
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
Some exam les of as herical homolo 4-s heres . 20/31
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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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.
A circle S1 {1} D2 T2 is called a meridian of thesolid 4-torus D2 T2.
Some exam les of as herical homolo 4-s heres . 21/31
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.
A circle S1 {1} D2 T2 is called a meridian of thesolid 4-torus D2 T2.
The important property of a meridian is that it bounds a
2-disk.
Some exam les of as herical homolo 4-s heres . 21/31
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.
A circle S1 {1} D2 T2 is called a meridian of thesolid 4-torus D2 T2.
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 .
Some exam les of as herical homolo 4-s heres . 21/31
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.
Some exam les of as herical homolo 4-s heres . 22/31
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.
Let hi : T3i T
3i be an affine homeomorphism for each
i. Then h1, . . . , h5 determine an affine homeomorphismh : M M.
Some exam les of as herical homolo 4-s heres . 22/31
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.
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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Dehn filling the truncated manifold
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Some exam les of as herical homolo 4-s heres . 23/31
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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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.
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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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.
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
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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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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).
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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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).
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.
Some exam les of as herical homolo 4-s heres . 25/31
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.
Some exam les of as herical homolo 4-s heres . 26/31
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.
Some exam les of as herical homolo 4-s heres . 26/31
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 .
Some exam les of as herical homolo 4-s heres . 27/31
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.
Some exam les of as herical homolo 4-s heres . 27/31
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.
Some exam les of as herical homolo 4-s heres . 28/31
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.
Some exam les of as herical homolo 4-s heres . 28/31
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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.
Some exam les of as herical homolo 4-s heres . 29/31
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.
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.
Some exam les of as herical homolo 4-s heres . 29/31
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.
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.
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Albert Einstein
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