Stony Brook University Seminario de Geometr a ICMAT ...De nition.Complete, non-compact n-manifold...
Transcript of Stony Brook University Seminario de Geometr a ICMAT ...De nition.Complete, non-compact n-manifold...
Mass, Scalar Curvature, &
Kahler Geometry, II
Claude LeBrunStony Brook University
Seminario de GeometrıaICMAT, November 5, 2018
1
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(R
n −Dn)/Γi, where Γi ⊂ O(n), such that
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Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(R
n −Dn)/Γi, where Γi ⊂ O(n), such that
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Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(R
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4
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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5
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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6
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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7
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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8
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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gjk = δjk + O(|x|1−n2−ε)
gjk,` = O(|x|−n2−ε), s ∈ L1
9
Why consider ALE spaces?
10
Key examples:
11
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
12
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
13
vv
vvv
Data: ` points in R3. =⇒ V with ∆V = 0
14
vv
vvv
Data: ` points in R3. =⇒ V with ∆V = 0
V =∑j=1
1
2%j
15
vv
vvv
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Data: ` points in R3. =⇒ V with ∆V = 0
V =∑j=1
1
2%j
F = ?dV curvature θ on P → R3 − {pts}.
16
vv
vvv
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
F = ?dV curvature θ on P → R3 − {pts}.
17
vv
vvv
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V (dx2 + dy2 + dz2) + V −1θ2
F = ?dV curvature θ on P → R3 − {pts}.
18
vv
vvv
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........................
Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
F = ?dV curvature θ on P → R3 − {pts}.
19
vv
vvv
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
on P . Then take M4 = Riemannian completion.
20
vv
vvv
.....................................................................................................................................................................................
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
on P . Then take M4 = Riemannian completion.
21
vv
vvv
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.
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
on P . Then take M4 = Riemannian completion.
22
vv
vvv
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............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .............
Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
on P . Then take M4 = Riemannian completion.
23
vv
vvv
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Data: ` points in R3. =⇒ V with ∆V = 0
g = V h + V −1θ2
on P . Then take M4 = Riemannian completion.
24
Deform retracts to k = `− 1 copies of S2,
25
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,
26
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
27
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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28
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
Configuration dual to Dynkin diagram Ak:
29
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
Configuration dual to Dynkin diagram Ak:
• • • •.............................................................................................................................................................................................................................................................................................................
30
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
Configuration dual to Dynkin diagram Ak:
• • • •.............................................................................................................................................................................................................................................................................................................
Diffeotype:
31
Deform retracts to k = `− 1 copies of S2,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
Configuration dual to Dynkin diagram Ak:
• • • •.............................................................................................................................................................................................................................................................................................................
Diffeotype:
Plumb together k copies of T ∗S2
according to diagram.
32
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
33
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
34
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
The G-H metrics are hyper-Kahler, and were soonindependently rediscovered by Hitchin.
35
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
36
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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37
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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38
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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39
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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40
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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41
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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42
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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43
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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44
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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45
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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46
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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47
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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48
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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49
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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50
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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51
Kahler metrics:
(Mn, g): Kahler ⇐⇒ holonomy ⊂ O(n)
s
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52
Kahler metrics:
(M2m, g): Kahler ⇐⇒ holonomy ⊂ U(m)
s
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53
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
s
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54
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
s
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U(m) := O(2m) ∩GL(m,C)
55
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
s
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Makes tangent space a complex vector space!
56
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
s
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Makes tangent space a complex vector space!
J : TM → TM, J2 = −identity
“almost-complex structure”
57
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
s
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Makes tangent space a complex vector space!
Invariant under parallel transport!
58
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
59
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
60
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
dω = 0
61
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
[ω] ∈ H2(M)
“Kahler class”
62
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
63
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
⇐⇒ In local complex coordinates (z1, . . . , zm), ∃f (z)
g = −m∑
j,k=1
∂2f
∂zj∂zk
[dzj ⊗ dzk + dzk ⊗ dzj
]
64
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
⇐⇒ In local complex coordinates (z1, . . . , zm), ∃f (z)
g = −m∑
j,k=1
∂2f
∂zj∂zk
[dzj ⊗ dzk + dzk ⊗ dzj
]
65
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
⇐⇒ In local complex coordinates (z1, . . . , zm), ∃f (z)
ω = im∑
j,k=1
∂2f
∂zj∂zkdzj ∧ dzk
66
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
⇐⇒ In local complex coordinates (z1, . . . , zm), ∃f (z)
g = −m∑
j,k=1
∂2f
∂zj∂zk
[dzj ⊗ dzk + dzk ⊗ dzj
]
67
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
Kahler magic:
r = −m∑
j,k=1
∂2
∂zj∂zklog det[gpq]
[dzj⊗dzk+dzk⊗dzj
]
68
Kahler metrics:
(M2m, g) Kahler ⇐⇒ holonomy ⊂ U(m)
⇐⇒ ∃ almost complex-structure J with ∇J = 0and g(J ·, J ·) = g.
⇐⇒ (M2m, g) is a complex manifold & ∃ J -invariantclosed 2-form ω such that g = ω(·, J ·).
Kahler magic:
If we define the Ricci form by
ρ = r(J ·, ·)then iρ is curvature of canonical line bundle Λm,0.
69
Kahler metrics:
(M2m, g): Ricci-flat Kahler⇐= holonomy⊂ SU(m)
s
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70
Kahler metrics:
(M2m, g): Ricci-flat Kahler⇐= holonomy⊂ SU(m)
s
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71
Kahler metrics:
(M2m, g): Ricci-flat Kahler⇐= holonomy⊂ SU(m)
s
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SU(m) ⊂ U(m) : {A | detA = 1}
72
Kahler metrics:
(M2m, g): Ricci-flat Kahler⇐= holonomy⊂ SU(m)
s
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73
Kahler metrics:
(M2m, g): Ricci-flat Kahler⇐⇒ holonomy⊂ SU(m)
s
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if M is simply connected.
74
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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75
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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76
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) := O(4`) ∩GL(`,H)
77
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) ⊂ SU(2`)
78
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) ⊂ SU(2`)
in many ways! (For example, permute i, j, k. . . )
79
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) ⊂ SU(2`)
in many ways! (For example, permute i, j, k. . . )
80
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) ⊂ SU(2`)
Ricci-flat and Kahler,
for many different complex structures!
81
Hyper-Kahler metrics:
(M4`, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(`)
s
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Sp(`) ⊂ SU(2`)
82
Hyper-Kahler metrics:
(M4, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(1)
s
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Sp(1) = SU(2)
83
Hyper-Kahler metrics:
(M4, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(1)
s
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Sp(1) = SU(2)
When (M4, g) simply connected:
hyper-Kahler ⇐⇒ Ricci-flat Kahler.
84
Hyper-Kahler metrics:
(M4, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(1)
s
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Sp(1) = SU(2)
85
Hyper-Kahler metrics:
(M4, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(1)
s
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Sp(1) = SU(2)
Ricci-flat and Kahler,
for many different complex structures!
86
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold
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87
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold
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88
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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89
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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Complex structure faithfully encodes the metric.
90
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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Complex structure encodes metric mod homothety.
91
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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Complex structure faithfully encodes the metric.
92
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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Complex structure faithfully encodes the metric.
Constructing twistor space suffices for existence.
93
Hitchin’s Twistor Spaces:
94
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3⊃ R3.
95
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
96
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
vv
vvv
So ` points determine P1, . . . , P` ∈ H0(CP1,O(2)).
97
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
vv
vvv
So ` points determine P1, . . . , P` ∈ H0(CP1,O(2)).
Small resolution Z of Z ⊂ O(`)⊕O(`)⊕O(2)
98
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
vv
vvv
So ` points determine P1, . . . , P` ∈ H0(CP1,O(2)).
Small resolution Z of Z ⊂ O(`)⊕O(`)⊕O(2)
xy = (z − P1) · · · (z − P`)
99
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
vv
vvv
So ` points determine P1, . . . , P` ∈ H0(CP1,O(2)).
Small resolution Z of Z ⊂ O(`)⊕O(`)⊕O(2)
xy = (z − P1) · · · (z − P`)
100
Hitchin’s Twistor Spaces:
H0(CP1,O(2)) = C3 ⊃ R3.
vv
vvv
So ` points determine P1, . . . , P` ∈ H0(CP1,O(2)).
Small resolution Z of Z ⊂ O(`)⊕O(`)⊕O(2)
xy = (z − P1) · · · (z − P`)is the twistor space of a Gibbons-Hawking metric.
101
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
The G-H metrics are hyper-Kahler, and were soonindependently rediscovered by Hitchin.
102
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
The G-H metrics are hyper-Kahler, and were soonindependently rediscovered by Hitchin.
Hitchin conjectured that similar metrics would existfor each finite Γ ⊂ SU(2).
103
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
The G-H metrics are hyper-Kahler, and were soonindependently rediscovered by Hitchin.
Hitchin conjectured that similar metrics would existfor each finite Γ ⊂ SU(2).
This conjecture was proved by Kronheimer, 1986.
104
Given Γ ⊂ SU(2) finite subgroup,
105
Given Γ ⊂ SU(2) finite subgroup,
106
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
107
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ mathbbC3 by
108
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
109
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
110
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
Example.
[e2πi/m
e−2πi/m
]∈ SU(2)
111
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
Example.
[e2πi/m
e−2πi/m
]∈ SU(2)
generates Γ ∼= Zm.
112
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
Example.
[e2πi/m
e−2πi/m
]∈ SU(2)
generates Γ ∼= Zm. Setting
u = zm1 , v = zm2 , y = z1z2,
113
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
Example.
[e2πi/m
e−2πi/m
]∈ SU(2)
generates Γ ∼= Zm. Setting
u = zm1 , v = zm2 , y = z1z2,
then identifies C2/Γ with
uv = ym.
114
Given Γ ⊂ SU(2) finite subgroup,
the orbifold C2/Γ can be viewed as
singular complex surface ⊂ C3 by
choosing 3 generators of Γ-invariant polynomials.
Example.
[e2πi/m
e−2πi/m
]∈ SU(2)
generates Γ ∼= Zm. Setting
w =1
2(zm1 −z
m2 ), x =
i
2(zm1 +zm2 ), y = z1z2,
then identifies C2/Γ with
w2 + x2 + ym = 0.
115
Felix Klein’s Table of Singularities (1884)
116
Felix Klein’s Table of Singularities (1884)
Zm ←→ w2 + x2 + ym = 0
117
Felix Klein’s Table of Singularities (1884)
Zm ←→ w2 + x2 + ym = 0
Dih∗m ←→ w2 + y(x2 + ym) = 0
118
Felix Klein’s Table of Singularities (1884)
Zm ←→ w2 + x2 + ym = 0
Dih∗m ←→ w2 + y(x2 + ym) = 0
T ∗ ←→ w2 + x3 + y4 = 0
119
Felix Klein’s Table of Singularities (1884)
Zm ←→ w2 + x2 + ym = 0
Dih∗m ←→ w2 + y(x2 + ym) = 0
T ∗ ←→ w2 + x3 + y4 = 0
O∗ ←→ w2 + x3 + xy3 = 0
120
Felix Klein’s Table of Singularities (1884)
Zm ←→ w2 + x2 + ym = 0
Dih∗m ←→ w2 + y(x2 + ym) = 0
T ∗ ←→ w2 + x3 + y4 = 0
O∗ ←→ w2 + x3 + xy3 = 0
I∗ ←→ w2 + x3 + y5 = 0
121
Prototypical Klein singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = 0
122
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = 0
123
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = 0
124
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
125
126
127
128
129
130
131
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
132
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
133
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
134
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
135
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
136
137
138
139
140
141
142
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
143
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
Usually these are topologically different.
144
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
Usually these are topologically different.
But for Klein singularities, they are diffeomorphic!
145
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
Usually these are topologically different.
But for Klein singularities, they are diffeomorphic!
Gorenstein singularities. Crepant Resolutions.
146
Two ways to get rid of a singularity:
• Smooth it, by deformation:
w2 + x2 + y2 = ε
• Resolve it, by blowing up, iteratively:
w2 + x2 + y2 = 0
O(−2) → O(−1)↓ ↓
CP1 ↪→ CP2
Usually these are topologically different.
But for Klein singularities, they are diffeomorphic!
Gorenstein singularities. Crepant Resolutions.
147
Resolutions of Klein Singularities:
148
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
149
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
150
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
151
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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...........................................................................................................................................................................................................................................................
152
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
153
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
154
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
155
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
Intersection pattern dual to Dynkin diagram!
• • • • •••
........................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................
156
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
Intersection pattern dual to Dynkin diagram!
• • • • •••
........................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................
157
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
•.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
Intersection pattern dual to Dynkin diagram!
• • • • •••
........................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................
158
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
Intersection pattern dual to Dynkin diagram!
• • • • •••
........................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................
159
Resolutions of Klein Singularities:
∀ Klein singularity V ⊂ C3, ∃! resolution
V → V
with c1(T 1,0V ) = 0.
Replaces origin with a union of CP1’s,each with self-intersection −2,meeting transversely, & forming connected set:
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................
Intersection pattern dual to Dynkin diagram!
• • • • •••
........................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................
160
Zk+1←→Ak • • • •.......................................................................................................................................................
161
Zk+1←→Ak • • • •.......................................................................................................................................................
Dih∗k−2←→Dk • • • ••.................................................................................................................
................................................
...........................................................
162
Zk+1←→Ak • • • •.......................................................................................................................................................
Dih∗k−2←→Dk • • • ••.................................................................................................................
................................................
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T ∗←→E6 • • • •••................................................................................................................................................................................................................
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163
Zk+1←→Ak • • • •.......................................................................................................................................................
Dih∗k−2←→Dk • • • ••.................................................................................................................
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T ∗←→E6 • • • •••................................................................................................................................................................................................................
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O∗←→E7 • • • ••• •..................................................................................................................................................................................................................................................................
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164
Zk+1←→Ak • • • •.......................................................................................................................................................
Dih∗k−2←→Dk • • • ••.................................................................................................................
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T ∗←→E6 • • • •••................................................................................................................................................................................................................
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O∗←→E7 • • • ••• •..................................................................................................................................................................................................................................................................
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I∗←→E8 • • • ••• • •....................................................................................................................................................................................................................................................................................................................
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165
Key examples:
Term ALE coined by Gibbons & Hawking, 1979.
They wrote down various explicit Ricci-flat ALE4-manifolds they called gravitational instantons.
Their examples have just one end, with
Γ ∼= Z` ⊂ SU(2) ⊂ O(4).
The G-H metrics are hyper-Kahler, and were soonindependently rediscovered by Hitchin.
Hitchin conjectured that similar metrics would existfor each finite Γ ⊂ SU(2).
Proved by Kronheimer, who also showed (1989) thisgives complete classification of ALE hyper-Kahlers.
166
Hyper-Kahler metrics:
(M4, g) hyper-Kahler ⇐⇒ holonomy ⊂ Sp(1)
s
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Sp(1) = SU(2)
Ricci-flat and Kahler,
for many different complex structures!
167
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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168
All these complex structures can be repackaged as
Penrose Twistor Space (Z6, J),
which is a complex 3-manifold.
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But similar for scalar-flat Kahler surfaces (M4, g, J)!
169
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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170
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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Integrability condition for twistor space: W+ ≡ 0.
171
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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Integrability condition for twistor space: W+ ≡ 0.
For Kahler surfaces, |W+|2 = s2/24.
172
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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Integrability condition for twistor space: W+ ≡ 0.
For Kahler surfaces, integrable ⇐⇒ scalar-flat!
173
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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Integrability condition for twistor space: W+ ≡ 0.
For Kahler surfaces, integrable ⇐⇒ scalar-flat!
Leads to constructions of explicit examples.
174
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z6, J),
which is once again a complex 3-manifold.
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Integrability condition for twistor space: W+ ≡ 0.
For Kahler surfaces, integrable ⇐⇒ scalar-flat!
Many simple examples are AE or ALE.
175
Some AE Scalar-Flat Kahler Surfaces:
176
Some AE Scalar-Flat Kahler Surfaces:
(L ’91)
177
Some AE Scalar-Flat Kahler Surfaces:
f
vv
vvv
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.............
178
Some AE Scalar-Flat Kahler Surfaces:
f
vv
vvv
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Data: k points inH3 and one point at infinity. =⇒V with ∆V = 0
179
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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Data: k points in H3 = upper half-space model.
180
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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Data: k points in H3. =⇒ V with ∆V = 0
V = 1 +
k∑j=1
Gj
181
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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Data: k points in H3. =⇒ V with ∆V = 0
V = 1 +
k∑j=1
1
e2%j − 1
182
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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Data: k points in H3. =⇒ V with ∆V = 0
V = 1 +
k∑j=1
Gj
183
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
.......
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Data: k points in H3. =⇒ V with ∆V = 0
F = ?dV curvature θ on P → H3 − {pts}.
184
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
185
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
.......
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Data: k points in H3. =⇒ V with ∆V = 0
g = z2
(Vdx2 + dy2 + dz2
z2+ V −1θ2
)
186
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
.......
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Data: k points in H3. =⇒ V with ∆V = 0
g = V (dx2 + dy2 + dz2) + z2V −1θ2
187
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
.......
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Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
188
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
.......
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Riemannian completion is AE scalar-flat Kahler.
g = z2(V h + V −1θ2
)
189
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
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............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
190
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
C×× ×××
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Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
191
Some AE Scalar-Flat Kahler Surfaces:
C2
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.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.............................
......................
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............. C×× ×××
Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
192
193
Some AE Scalar-Flat Kahler Surfaces:
C2
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.............................
......................
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............. C×× ×××
Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
194
Some AE Scalar-Flat Kahler Surfaces:
C2
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.............................
......................
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.............
−1−1 −1−1−1
........................................................................................................
........................................................................................................
........................................................................................................
........................................................................................................
........................................................................................................
Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
195
Some AE Scalar-Flat Kahler Surfaces:
vv
vvv
C×× ×××
.................................................................................................................................................................
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
......
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
.............
Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
196
Some AE Scalar-Flat Kahler Surfaces:
C2
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.............................
......................
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.............
−1−1 −1−1−1
........................................................................................................
........................................................................................................
........................................................................................................
........................................................................................................
........................................................................................................
Data: k points in H3. =⇒ V with ∆V = 0
g = z2(V h + V −1θ2
)
197
Some ALE Scalar-Flat Kahler Surfaces:
198
Some ALE Scalar-Flat Kahler Surfaces:
(L ’91)
199
Some ALE Scalar-Flat Kahler Surfaces:
200
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
201
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
V = 1 + `G0 +
k∑j=1
Gj
202
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
203
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
V = 1 + `G0 +
k∑j=1
Gj
204
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.......
........................................
..............................................................................................................................................................................................................................................................................................................................................................................
..........
.......
........................................
..............................................................................................................................................................................................................................................................................................................................................................................
..........
.......
........................................
..............................................................................................................................................................................................................................................................................................................................................................................
..........
.......
........................................
..............................................................................................................................................................................................................................................................................................................................................................................
..........
.......
....................................
..............................................................................................................................................................................................................................................................................
....
.......
........................................
..............................................................................................................................................................................................................................................................................................................................................................................
..........
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
F = ?dV curvature θ on P → H3 − {pts}.
205
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Data: k + 1 points in H3. =⇒ V with ∆V = 0
g =1
4 sinh2 %0
(V h + V −1θ2
)
206
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Riemannian completion is ALE scalar-flat Kahler.
g =1
4 sinh2 %0
(V h + V −1θ2
)
207
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
Riemannian completion is ALE with Γ = Z`.
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
208
Some ALE Scalar-Flat Kahler Surfaces:
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
.................................................................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ......
..........................
..........................
............................................
.................................................................
×
×
×
×
×
Riemannian completion is ALE with Γ = Z`.
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
209
Some ALE Scalar-Flat Kahler Surfaces:
Blow up of Chern-class −` line bundle over CP1 atk points on zero section Σ.
......................................................................................................................................................................................................................................................................................
..................................
................
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
Σ−`
...................................................................................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................................
Riemannian completion is ALE with Γ = Z`.
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
210
Some ALE Scalar-Flat Kahler Surfaces:
Blow up of Chern-class −` line bundle over CP1 atk points on zero section Σ.
......................................................................................................................................................................................................................................................................................
..................................
................
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
×× × Σ−`
...................................................................................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................................
Riemannian completion is ALE with Γ = Z`.
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
211
Some ALE Scalar-Flat Kahler Surfaces:
Blow up of Chern-class −` line bundle over CP1 atk points on zero section Σ.
......................................................................................................................................................................................................................................................................................
..................................
................
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
............................................................................................................................................................................................................................................................................................................................................................
............................................
........................
.............
E1
−1−1−1
E2Ek
Σ−`− k
...................................................................................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................................
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
......
.......
.......
.......
.......
.......
.......
.......
.......
.......
.......
......
Riemannian completion is ALE with Γ = Z`.
V = 1 +`
e2%0 − 1+
k∑j=1
1
e2%j − 1
212
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z, J),
which is once again a complex 3-manifold.
.....................................................................................................................................................................................................................................................................................................................................................................................................................................
............................
.......................
.........................
.......................................................................................................................................................
.................
.............
..........
......................................
...................................
........................
.................................................................
....................................
r
rr Z
M 4
↓
.......................................................................
..................
..................
..................
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.............................................................................................................................................................
..................
..................
..................
................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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............................... ...............................
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......................................................................
...................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.........................................................................................................
.............................................................................
...............................................................................................................................................................................................................................................................................................................................................................................................................
213
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4⊃ R1,3 ⊃ H3
214
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3⊃ H3
215
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
216
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...................
....................
......................
.......................
.........................
...........................
...............................
......................................
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..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................
...............................................................
..............................................
....................................
....................................................................................
...................
............. .............
..........................
............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ............. ..........................
.............
So k + 1 points in H3 give rise to
217
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
f
vv
vvv
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
............................................
...................................
..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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......................................
....................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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.............
So k + 1 points in H3 give rise to
P0, P1, . . . , Pk ∈ H0(CP1 × CP1,O(1, 1)).
218
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
219
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
220
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
let Z be the hypersurface
xy = P `0 P1 · · · Pk.
221
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
let Z be the hypersurface
xy = P `0 P1 · · · Pk.
Then twistor space Z obtained from Z by
222
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
let Z be the hypersurface
xy = P `0 P1 · · · Pk.
Then twistor space Z obtained from Z by
• removing curve in zero section cut out by P0,
223
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
let Z be the hypersurface
xy = P `0 P1 · · · Pk.
Then twistor space Z obtained from Z by
• removing curve in zero section cut out by P0,
• adding two rational curves at infinity, and
224
Twistor Spaces for These Metrics:
H0(CP1 × CP1,O(1, 1)) = C4 ⊃ R1,3 ⊃ H3
InO(k + `− 1, 1)⊕O(1, k + `− 1)→ CP1×CP1,
let Z be the hypersurface
xy = P `0 P1 · · · Pk.
Then twistor space Z obtained from Z by
• removing curve in zero section cut out by P0,
• adding two rational curves at infinity, and
•making small resolutions of isolated singularities.
225
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z, J),
which is once again a complex 3-manifold.
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226
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z, J),
which is once again a complex 3-manifold.
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rr Z
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Lots more ALE scalar-flat Kahler surfaces now known:
227
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z, J),
which is once again a complex 3-manifold.
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Lots more ALE scalar-flat Kahler surfaces now known:
Joyce, Calderbank-Singer, Lock-Viaclovsky. . .
228
Any scalar-flat Kahler surface (M4, g, J) has a
Penrose Twistor Space (Z, J),
which is once again a complex 3-manifold.
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rr Z
M 4
↓
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Lots more ALE scalar-flat Kahler surfaces now known:
Joyce, Calderbank-Singer, Lock-Viaclovsky. . .
But full classification remains an open problem.
229
Definition. Complete, non-compact n-manifold(Mn, g) is asymptotically locally Euclidean (ALE)if ∃ compact set K ⊂ M such that M − K ≈∐i(Rn −Dn)/Γi, where Γi ⊂ O(n), such that
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gjk = δjk + O(|x|1−n2−ε)
gjk,` = O(|x|−n2−ε), s ∈ L1
230
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
231
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
232
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
233
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
234
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
235
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
236
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
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237
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
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238
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
239
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
240
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Bartnik/Chrusciel (1986): With weak fall-offconditions, the mass is well-defined & coordinateindependent.
241
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Bartnik/Chrusciel (1986): With weak fall-offconditions, the mass is well-defined & coordinateindependent.
242
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
gjk = δjk + O(|x|1−n2−ε)
gjk,` = O(|x|−n2−ε), s ∈ L1
243
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Bartnik/Chrusciel (1986): With weak fall-offconditions, the mass is well-defined & coordinateindependent.
244
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Bartnik/Chrusciel (1986): With weak fall-offconditions, the mass is well-defined & coordinateindependent.
245
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Chrusciel-type fall-off:
gjk − δjk ∈ C1−τ , τ >
n− 2
2
246
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
Bartnik/Chrusciel (1986): With weak fall-offconditions, the mass is well-defined & coordinateindependent.
247
Definition. The mass (at a given end) of anALE n-manifold is defined to be
m(M, g) := lim%→∞
Γ(n2)
4(n− 1)πn/2
∫Σ(%)
[gij,i − gii,j
]νjαE
where
• Σ(%) ≈ Sn−1/Γi is given by |~x| = %;
• ν is the outpointing Euclidean unit normal;and
• αE is the volume (n− 1)-from induced by theEuclidean metric.
We’ll see a new proof of this in the Kahler case.
248
In fact, we’ll eventually prove:
249
In fact, we’ll eventually prove:
Theorem C. Any ALE Kahler manifold (M, g, J)of complex dimension m has mass given by
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
250
In fact, we’ll eventually prove:
Theorem C. Any ALE Kahler manifold (M, g, J)of complex dimension m has mass given by
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
where
• s = scalar curvature;
• dµ = metric volume form;
• c1 = c1(M,J) ∈ H2(M) is first Chern class;
• [ω] ∈ H2(M) is Kahler class of (g, J); and
• 〈 , 〉 is pairing between H2c (M) and H2m−2(M).
251
In fact, we’ll eventually prove:
Theorem C. Any ALE Kahler manifold (M, g, J)of complex dimension m has mass given by
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
where
• s = scalar curvature;
• dµ = metric volume form;
• c1 = c1(M,J) ∈ H2(M) is first Chern class;
• [ω] ∈ H2(M) is Kahler class of (g, J); and
• 〈 , 〉 is pairing between H2c (M) and H2m−2(M).
• ♣ : H2(M)∼=−→ H2
c (M) inverse of natural map.
252
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
253
Scalar-flat Kahler case:
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
254
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
255
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Today: What does this mean in practice?
256
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
257
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
258
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
259
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
Ricci flat! =⇒ c1 = 0.
260
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
Ricci flat! =⇒ c1 = 0.
261
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
Ricci flat! =⇒ c1 = 0.
Mass automatically vanishes!
262
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
Ricci flat! =⇒ c1 = 0.
Mass automatically vanishes!
Bartnik: Ricci-flat =⇒ faster fall-off of metric!
263
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Gravitational instantons?
Ricci flat! =⇒ c1 = 0.
Mass automatically vanishes!
Bartnik: Ricci-flat =⇒ mass vanishes!
264
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Today: What does this mean in practice?
265
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Today: Exploit Poincare duality. . .
266
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by
267
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by
268
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by
269
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by
270
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
271
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
272
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
273
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
where [ω] denotes the Kahler class of (M, g, J).
274
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
275
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
276
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
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.............×× ×××
277
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
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×××××
278
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
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279
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
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.............
−1−1 −1−1−1
E4E3 E5E2E1 ........................................................................................................
........................................................................................................
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280
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
where [ω] denotes the Kahler class of (M, g, J).
281
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
m(M, g) =1
3π
∑j=1
∫Ej
[ω]
282
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
m(M, g) =1
3π
∑j=1
∫Ej
[ω]
Always positive! (AE): Positive mass theorem.
283
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up C2 at k points.
m(M, g) =1
3π
∑j=1
∫Ej
[ω]
Always positive! (AE): Positive mass theorem.
284
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
285
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
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Σ−`
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286
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
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×× × Σ−`
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287
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
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E1
−1−1−1
E2Ek
Σ−`− k
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288
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
where [ω] denotes the Kahler class of (M, g, J).
289
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
m(M, g) =
290
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
m(M, g) =1
3π`
(2− `)∫
Σω+2
k∑j=1
∫Ej
ω
.
291
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
Example. Blow up Chern-class −` line bundleover CP1 at k points on zero section Σ.
m(M, g) =1
3π`
(2− `)∫
Σω + 2
k∑j=1
∫Ej
ω
.
292
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
where [ω] denotes the Kahler class of (M, g, J).
293
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
294
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
295
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
Examples: Hirzebruch-Jung resolution of C2/Z`.
296
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
Examples: Hirzebruch-Jung resolution of C2/Z`.
(z1, z2) 7→ (e2πi/`z1, e2πik/`z2)
297
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
Examples: Hirzebruch-Jung resolution of C2/Z`.
Calderbank-Singer metrics generalize for k 6= ±1.
298
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then zapm(M, g) ≤ 0, with = iff g is Ricci-flat.
299
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then m(M, g) ≤ 0, with = iff g is Ricci-flat.
300
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then m(M, g) ≤ 0, with = iff g is Ricci-flat.
301
Proposition. Let (M, g, J) be an ALE scalar-flat Kahler surface. Let E1, . . . E` be a basis forH2(M,R), and let Q = [Qjk] = [Ej · Ek] be thecorresponding intersection matrix. If we definea1, . . . , a` by a1
...a`
= Q−1∫E1c1
...∫E`c1
then the mass of (M, g) is given by
m(M, g) = − 1
3π
∑j=1
aj
∫Ej
[ω]
where [ω] denotes the Kahler class of (M, g, J).
302
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then m(M, g) ≤ 0, with = iff g is Ricci-flat.
V. Alexeev: Q−1 term-by-term ≤ 0 for these.
303
Theorem B. Let (M4, g, J) be an ALE scalar-flat Kahler surface, and suppose that (M,J) isthe minimal resolution of a surface singularity.Then m(M, g) ≤ 0, with = iff g is Ricci-flat.
V. Alexeev: Q−1 term-by-term ≤ 0 for these.
Brought to our attention by C. Spotti.
304
Scalar-flat Kahler surface:
m(M, g) = − 1
3π〈♣(c1), [ω]〉
305
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
306
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
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..................................
..................................
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.....................
...............................
307
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
308
m(M, g) = −〈♣(c1), [ω]m−1〉(2m− 1)πm−1
+(m− 1)!
4(2m− 1)πm
∫Msgdµg
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309
End, Part II
310