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Cubic 4-folds with an Eckardt point
R. Laza
Stony Brook University
September 13, 2018
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 1 / 21
Outline
1 1. Cubic fourfolds1.A. Rationality Questions1.B. Cubics and Hyperkahlers
2 2. Cubics with an Eckardt point [LPZ17]2.A. Generalities2.B. The GIT model2.C. The D/Γ model2.D. The main result of [LPZ17]
3 3. Eckardt points and rationality [L18]
4 4. Back to hyper-Kahlers
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 2 / 21
1. Cubic Fourfolds
ConsiderX ⊂ P5
a smooth cubic hypersurface (/C).
The middle cohomology H4(X,C) is a Hodge structure of K3 type
0 1 21 1 0
Consequently, cubic 4-folds are very relevant for
Rationality QuestionsConstruction of HK manifolds
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 3 / 21
1. Cubic Fourfolds
ConsiderX ⊂ P5
a smooth cubic hypersurface (/C).
The middle cohomology H4(X,C) is a Hodge structure of K3 type
0 1 21 1 0
Consequently, cubic 4-folds are very relevant for
Rationality QuestionsConstruction of HK manifolds
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 3 / 21
1. Cubic Fourfolds
ConsiderX ⊂ P5
a smooth cubic hypersurface (/C).
The middle cohomology H4(X,C) is a Hodge structure of K3 type
0 1 21 1 0
Consequently, cubic 4-folds are very relevant for
Rationality QuestionsConstruction of HK manifolds
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 3 / 21
1.A. Rationality Questions
cubic surfaces S are rational
cubic threefolds Y are irrational (Clemens-Griffiths’ 72)
some cubic fourfolds are rational (Morin, Iskovskikh, Tregub,Hassett...)
but conjecturally, the very general cubic fourfold is not rational.
A lot of progress on rationality questions (Voisin, Colliot-Thelene,Totaro, Hassett, Tschinkel, Schreieder, etc.), but one case thatremains open is cubic fourfolds.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 4 / 21
1.A. Rationality Questions
cubic surfaces S are rational
cubic threefolds Y are irrational (Clemens-Griffiths’ 72)
some cubic fourfolds are rational (Morin, Iskovskikh, Tregub,Hassett...)
but conjecturally, the very general cubic fourfold is not rational.
A lot of progress on rationality questions (Voisin, Colliot-Thelene,Totaro, Hassett, Tschinkel, Schreieder, etc.), but one case thatremains open is cubic fourfolds.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 4 / 21
1.A. Rationality Questions
cubic surfaces S are rational
cubic threefolds Y are irrational (Clemens-Griffiths’ 72)
some cubic fourfolds are rational (Morin, Iskovskikh, Tregub,Hassett...)
but conjecturally, the very general cubic fourfold is not rational.
A lot of progress on rationality questions (Voisin, Colliot-Thelene,Totaro, Hassett, Tschinkel, Schreieder, etc.), but one case thatremains open is cubic fourfolds.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 4 / 21
1.A. Harris, Hassett, Kuznetsov, Kulikov, etc.
Modeled by the Clemens-Griffiths proof, it is natural to expect
Rationality Conjecture
Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice (e.g. Hodge general cubic, or Hodge general cubiccontaining a plane), then X is not rational.
For convenience
Definition (Potentially irrational)
Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice, we call X potentially irrational.
All known rational examples are not potentially irrational.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 5 / 21
1.A. Harris, Hassett, Kuznetsov, Kulikov, etc.
Modeled by the Clemens-Griffiths proof, it is natural to expect
Rationality Conjecture
Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice (e.g. Hodge general cubic, or Hodge general cubiccontaining a plane), then X is not rational.
For convenience
Definition (Potentially irrational)
Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice, we call X potentially irrational.
All known rational examples are not potentially irrational.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 5 / 21
1.B. HK Manifolds
A hyper-Kahler (HK) manifold (or IHSM) Z is a simply connectedKahler manifold admitting a holomorphic 2-form ω which isnon-degenerate, and unique
H2,0(Z) ∼= C[ω]
Example: K3 surfaces.
They are building blocks in AG (KZ ≡ 0, BB decomposition:abelian varieties, Calabi-Yau, or HK)
Few deformation classes known (Beauville, Mukai, O’Grady):
2 infinite series: K3[n], Kumn (dim = 2n)2 exceptional cases (O’Grady): OG10, OG6 (dim = 10, 6 resp.)
Are there any other examples?
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 6 / 21
1.B. HK Manifolds
A hyper-Kahler (HK) manifold (or IHSM) Z is a simply connectedKahler manifold admitting a holomorphic 2-form ω which isnon-degenerate, and unique
H2,0(Z) ∼= C[ω]
Example: K3 surfaces.
They are building blocks in AG (KZ ≡ 0, BB decomposition:abelian varieties, Calabi-Yau, or HK)
Few deformation classes known (Beauville, Mukai, O’Grady):
2 infinite series: K3[n], Kumn (dim = 2n)2 exceptional cases (O’Grady): OG10, OG6 (dim = 10, 6 resp.)
Are there any other examples?
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 6 / 21
1.B. HK Manifolds
A hyper-Kahler (HK) manifold (or IHSM) Z is a simply connectedKahler manifold admitting a holomorphic 2-form ω which isnon-degenerate, and unique
H2,0(Z) ∼= C[ω]
Example: K3 surfaces.
They are building blocks in AG (KZ ≡ 0, BB decomposition:abelian varieties, Calabi-Yau, or HK)
Few deformation classes known (Beauville, Mukai, O’Grady):
2 infinite series: K3[n], Kumn (dim = 2n)2 exceptional cases (O’Grady): OG10, OG6 (dim = 10, 6 resp.)
Are there any other examples?
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 6 / 21
1.B. HK Manifolds
A hyper-Kahler (HK) manifold (or IHSM) Z is a simply connectedKahler manifold admitting a holomorphic 2-form ω which isnon-degenerate, and unique
H2,0(Z) ∼= C[ω]
Example: K3 surfaces.
They are building blocks in AG (KZ ≡ 0, BB decomposition:abelian varieties, Calabi-Yau, or HK)
Few deformation classes known (Beauville, Mukai, O’Grady):
2 infinite series: K3[n], Kumn (dim = 2n)2 exceptional cases (O’Grady): OG10, OG6 (dim = 10, 6 resp.)
Are there any other examples?
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 6 / 21
1.B. Cubics and hyper-Kahlers
Cubic 4-folds give many interesting examples of hyper-Kahlermanifolds
Beauville-Donagi’ 85: X cubic fourfold =⇒ F (X) is HK 4-fold(deformation equivalent to K3[2])
[LLSvS’17] - twisted cubics on X =⇒ HK 8-fold (deformationequivalent to K3[4])
cubics are better than K3s - 20 dimensional moduli, leads tolocally complete examples of families of HKs.
recently, we (L–Sacca–Voisin [LSV17]) gave a new construction ofthe OG10 exceptional example
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 7 / 21
1.B. Cubics and hyper-Kahlers
Cubic 4-folds give many interesting examples of hyper-Kahlermanifolds
Beauville-Donagi’ 85: X cubic fourfold =⇒ F (X) is HK 4-fold(deformation equivalent to K3[2])
[LLSvS’17] - twisted cubics on X =⇒ HK 8-fold (deformationequivalent to K3[4])
cubics are better than K3s - 20 dimensional moduli, leads tolocally complete examples of families of HKs.
recently, we (L–Sacca–Voisin [LSV17]) gave a new construction ofthe OG10 exceptional example
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 7 / 21
1.B. Cubics and hyper-Kahlers
Cubic 4-folds give many interesting examples of hyper-Kahlermanifolds
Beauville-Donagi’ 85: X cubic fourfold =⇒ F (X) is HK 4-fold(deformation equivalent to K3[2])
[LLSvS’17] - twisted cubics on X =⇒ HK 8-fold (deformationequivalent to K3[4])
cubics are better than K3s - 20 dimensional moduli, leads tolocally complete examples of families of HKs.
recently, we (L–Sacca–Voisin [LSV17]) gave a new construction ofthe OG10 exceptional example
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 7 / 21
1.B. Cubics and hyper-Kahlers
Cubic 4-folds give many interesting examples of hyper-Kahlermanifolds
Beauville-Donagi’ 85: X cubic fourfold =⇒ F (X) is HK 4-fold(deformation equivalent to K3[2])
[LLSvS’17] - twisted cubics on X =⇒ HK 8-fold (deformationequivalent to K3[4])
cubics are better than K3s - 20 dimensional moduli, leads tolocally complete examples of families of HKs.
recently, we (L–Sacca–Voisin [LSV17]) gave a new construction ofthe OG10 exceptional example
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 7 / 21
1.B. LSV’s Construction of OG10 example
Theorem [LSV17]
Let X ⊂ P5 be a Hodge general cubic fourfold, and B = (P5)∨. LetY/B the universal hyperplane section, with J /U the associatedrelative intermediate Jacobian bundle (with U = B \X∨). Then
i) There exists a compactification Z/B of J /U such that Z issmooth, and Z/B is flat.
ii) Z is a 10-dimensional HK manifold (and Z/B is a Lagrangianfibration).
iii) Z is deformation equivalent to OG10.
(A lot of previous work: Donagi-Markman, O’Grady–Rapagnetta,Markushevich, etc.)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 8 / 21
1.B. LSV’s Construction of OG10 example
Theorem [LSV17]
Let X ⊂ P5 be a Hodge general cubic fourfold, and B = (P5)∨. LetY/B the universal hyperplane section, with J /U the associatedrelative intermediate Jacobian bundle (with U = B \X∨). Then
i) There exists a compactification Z/B of J /U such that Z issmooth, and Z/B is flat.
ii) Z is a 10-dimensional HK manifold (and Z/B is a Lagrangianfibration).
iii) Z is deformation equivalent to OG10.
(A lot of previous work: Donagi-Markman, O’Grady–Rapagnetta,Markushevich, etc.)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 8 / 21
1.B. LSV’s Construction of OG10 example
Theorem [LSV17]
Let X ⊂ P5 be a Hodge general cubic fourfold, and B = (P5)∨. LetY/B the universal hyperplane section, with J /U the associatedrelative intermediate Jacobian bundle (with U = B \X∨). Then
i) There exists a compactification Z/B of J /U such that Z issmooth, and Z/B is flat.
ii) Z is a 10-dimensional HK manifold (and Z/B is a Lagrangianfibration).
iii) Z is deformation equivalent to OG10.
(A lot of previous work: Donagi-Markman, O’Grady–Rapagnetta,Markushevich, etc.)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 8 / 21
1.B. LSV’s Construction of OG10 example
Theorem [LSV17]
Let X ⊂ P5 be a Hodge general cubic fourfold, and B = (P5)∨. LetY/B the universal hyperplane section, with J /U the associatedrelative intermediate Jacobian bundle (with U = B \X∨). Then
i) There exists a compactification Z/B of J /U such that Z issmooth, and Z/B is flat.
ii) Z is a 10-dimensional HK manifold (and Z/B is a Lagrangianfibration).
iii) Z is deformation equivalent to OG10.
(A lot of previous work: Donagi-Markman, O’Grady–Rapagnetta,Markushevich, etc.)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 8 / 21
2. New title - “Pseudo-cubics”
Question
Are there any other cubic like fourfolds?
joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)
what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin
inspired by M. Reid list of 95 weighted K3 surfaces
Specifically, we wantX ⊂WP4
quasi-smooth with H4(X) of K3 type.
Answer - “Pseudo-Cubics”
17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 9 / 21
2. New title - “Pseudo-cubics”
Question
Are there any other cubic like fourfolds?
joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)
what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin
inspired by M. Reid list of 95 weighted K3 surfaces
Specifically, we wantX ⊂WP4
quasi-smooth with H4(X) of K3 type.
Answer - “Pseudo-Cubics”
17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 9 / 21
2. New title - “Pseudo-cubics”
Question
Are there any other cubic like fourfolds?
joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)
what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin
inspired by M. Reid list of 95 weighted K3 surfaces
Specifically, we wantX ⊂WP4
quasi-smooth with H4(X) of K3 type.
Answer - “Pseudo-Cubics”
17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 9 / 21
2. New title - “Pseudo-cubics”
Question
Are there any other cubic like fourfolds?
joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)
what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin
inspired by M. Reid list of 95 weighted K3 surfaces
Specifically, we wantX ⊂WP4
quasi-smooth with H4(X) of K3 type.
Answer - “Pseudo-Cubics”
17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 9 / 21
2. New title - “Pseudo-cubics”
Question
Are there any other cubic like fourfolds?
joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)
what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin
inspired by M. Reid list of 95 weighted K3 surfaces
Specifically, we wantX ⊂WP4
quasi-smooth with H4(X) of K3 type.
Answer - “Pseudo-Cubics”
17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 9 / 21
2.A. Cubics with an Eckardt point
Main example: cubic 4-folds (20 moduli),
but with 14 moduli, weget another new(ish) example:
X6 ⊂ P(1, 2, 2, 2, 2, 3)
[Un]fortunately, X6 is birational to a cubic with an Eckardt point
X = V(f(x0, . . . , x4) + x0x
25
)⊂ P5
Question
A smooth cubic hypersurface X (of dimension n) has an Eckardt pointp ∈ X if TpX ∩X = CS , where CS is the cone over an (n− 2)dimensional cubic S.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 10 / 21
2.A. Cubics with an Eckardt point
Main example: cubic 4-folds (20 moduli),but with 14 moduli, weget another new(ish) example:
X6 ⊂ P(1, 2, 2, 2, 2, 3)
[Un]fortunately, X6 is birational to a cubic with an Eckardt point
X = V(f(x0, . . . , x4) + x0x
25
)⊂ P5
Question
A smooth cubic hypersurface X (of dimension n) has an Eckardt pointp ∈ X if TpX ∩X = CS , where CS is the cone over an (n− 2)dimensional cubic S.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 10 / 21
2.A. Cubics with an Eckardt point
Main example: cubic 4-folds (20 moduli),but with 14 moduli, weget another new(ish) example:
X6 ⊂ P(1, 2, 2, 2, 2, 3)
[Un]fortunately, X6 is birational to a cubic with an Eckardt point
X = V(f(x0, . . . , x4) + x0x
25
)⊂ P5
Question
A smooth cubic hypersurface X (of dimension n) has an Eckardt pointp ∈ X if TpX ∩X = CS , where CS is the cone over an (n− 2)dimensional cubic S.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 10 / 21
2.A. Cubics with an Eckardt point
Main example: cubic 4-folds (20 moduli),but with 14 moduli, weget another new(ish) example:
X6 ⊂ P(1, 2, 2, 2, 2, 3)
[Un]fortunately, X6 is birational to a cubic with an Eckardt point
X = V(f(x0, . . . , x4) + x0x
25
)⊂ P5
Question
A smooth cubic hypersurface X (of dimension n) has an Eckardt pointp ∈ X if TpX ∩X = CS , where CS is the cone over an (n− 2)dimensional cubic S.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 10 / 21
2.A. Cubics with an Eckardt point (2)
Two other natural characterization:
X has an Eckardt point iff X admits an involution fixing ahyperplane H0.
ι : x5 → −x5(fixes H0 = V (x5) and p = (0 : · · · : 0 : 1) the Eckardt point)
(X, p) ⇐⇒ (Y, S) a n− 1 dimensional cubic pair (withY = X ∩H0, S = X ∩H0 ∩ TpX)
going from (Y, S) to (X, p) was inspired by Allcock-Carlson-Toledo:
Y = V (f3(x0, . . . , x4)), S = V (f3, `) =⇒ X = V (f3 + x25`)
(Recall ACT: Y = V (f3) =⇒ X = V (f3 + x35))
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 11 / 21
2.A. Cubics with an Eckardt point (2)
Two other natural characterization:
X has an Eckardt point iff X admits an involution fixing ahyperplane H0.
ι : x5 → −x5(fixes H0 = V (x5) and p = (0 : · · · : 0 : 1) the Eckardt point)
(X, p) ⇐⇒ (Y, S) a n− 1 dimensional cubic pair (withY = X ∩H0, S = X ∩H0 ∩ TpX)
going from (Y, S) to (X, p) was inspired by Allcock-Carlson-Toledo:
Y = V (f3(x0, . . . , x4)), S = V (f3, `) =⇒ X = V (f3 + x25`)
(Recall ACT: Y = V (f3) =⇒ X = V (f3 + x35))
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 11 / 21
2.B. Moduli - The GIT Model
GIT model: X6 ⊂WP5 ??? (non-reductive group)
We have a GIT model for pairs
Mpairs = P55 × (P5)∨//SL(6)
rk Pic(P55 × (P5)∨) = 2 =⇒ choice of linearization, VGIT(Thaddeus, Dolgachev-Hu)...
Family of compactifications for pairs
M(t) = P55 × (P5)∨//O(1,t) SL(6), t ∈ Q+.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 12 / 21
2.B. Moduli - The GIT Model
GIT model: X6 ⊂WP5 ??? (non-reductive group)
We have a GIT model for pairs
Mpairs = P55 × (P5)∨//SL(6)
rk Pic(P55 × (P5)∨) = 2 =⇒ choice of linearization, VGIT(Thaddeus, Dolgachev-Hu)...
Family of compactifications for pairs
M(t) = P55 × (P5)∨//O(1,t) SL(6), t ∈ Q+.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 12 / 21
2.B. Moduli - The GIT Model (2)
There are two forgetful maps:
M(ε) → M3,3
(Y, S) → Y
and at the opposite end
M(3/4− ε) → M3,2
(Y, S) → S
and flips M(ε) 99KM(3/4− ε) giving a wall crossing situation.
see my thesis [L09] (related to deformations of singularities, N16,O16)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 13 / 21
2.B. Moduli - The GIT Model (2)
There are two forgetful maps:
M(ε) → M3,3
(Y, S) → Y
and at the opposite end
M(3/4− ε) → M3,2
(Y, S) → S
and flips M(ε) 99KM(3/4− ε) giving a wall crossing situation.
see my thesis [L09] (related to deformations of singularities, N16,O16)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 13 / 21
2.B. Moduli - The GIT Model (2)
There are two forgetful maps:
M(ε) → M3,3
(Y, S) → Y
and at the opposite end
M(3/4− ε) → M3,2
(Y, S) → S
and flips M(ε) 99KM(3/4− ε) giving a wall crossing situation.
see my thesis [L09] (related to deformations of singularities, N16,O16)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 13 / 21
2.C. Moduli (2) - The D/Γ model
A cubic surface S contains 27 lines, thus the cone CS contains 27planes.
If X has an Eckardt point p, CS ⊂ X, and then there are 27planes on X
X is an M -polarized cubic fourfold in the sense of Dolgachev, withM a rank 7 lattice (=⇒ 14 moduli)
In fact,
M =
7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3
with M ⊂ H2,2 ∩H4(X,Z).
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 14 / 21
2.C. Moduli (2) - The D/Γ model
A cubic surface S contains 27 lines, thus the cone CS contains 27planes.
If X has an Eckardt point p, CS ⊂ X, and then there are 27planes on X
X is an M -polarized cubic fourfold in the sense of Dolgachev, withM a rank 7 lattice (=⇒ 14 moduli)
In fact,
M =
7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3
with M ⊂ H2,2 ∩H4(X,Z).
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 14 / 21
2.C. Moduli (2) - The D/Γ model
A cubic surface S contains 27 lines, thus the cone CS contains 27planes.
If X has an Eckardt point p, CS ⊂ X, and then there are 27planes on X
X is an M -polarized cubic fourfold in the sense of Dolgachev, withM a rank 7 lattice (=⇒ 14 moduli)
In fact,
M =
7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3
with M ⊂ H2,2 ∩H4(X,Z).
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 14 / 21
2.C. The D/Γ model (2)
Easier to remember, the primitive algebraic part:
Mprim∼= E6(2)
(corresponding to H2(S)prim ∼= E6)
Giving the transcendental lattice
T ∼= (D4)3 ⊕ U2
signature (14, 2)
As usually (e.g. see M -polarized K3s), we get a 14 dimensionallocally symmetric variety D/Γ as model for the moduli space,where
D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦
Γ ∼ O(T )
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 15 / 21
2.C. The D/Γ model (2)
Easier to remember, the primitive algebraic part:
Mprim∼= E6(2)
(corresponding to H2(S)prim ∼= E6)
Giving the transcendental lattice
T ∼= (D4)3 ⊕ U2
signature (14, 2)
As usually (e.g. see M -polarized K3s), we get a 14 dimensionallocally symmetric variety D/Γ as model for the moduli space,where
D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦
Γ ∼ O(T )
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 15 / 21
2.C. The D/Γ model (2)
Easier to remember, the primitive algebraic part:
Mprim∼= E6(2)
(corresponding to H2(S)prim ∼= E6)
Giving the transcendental lattice
T ∼= (D4)3 ⊕ U2
signature (14, 2)
As usually (e.g. see M -polarized K3s), we get a 14 dimensionallocally symmetric variety D/Γ as model for the moduli space,where
D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦
Γ ∼ O(T )
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 15 / 21
2.D. Main result of [LPZ17]
Theorem [LPZ17]
With notations as above
M(1/3) ∼= (D/Γ)∗
compare
Theorem (L./Looijenga 2017; Voisin ’86, Hassett ’96)
M3,4∼= (D20 \ H2 \ H6)/Γ
Furthermore
M3,4
K−blow−up��
flip// D20/Γ
L−factorialization��
M3,4GIT P // (D20/Γ)∗
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 16 / 21
2.D. Main result of [LPZ17]
Theorem [LPZ17]
With notations as above
M(1/3) ∼= (D/Γ)∗
compare
Theorem (L./Looijenga 2017; Voisin ’86, Hassett ’96)
M3,4∼= (D20 \ H2 \ H6)/Γ
Furthermore
M3,4
K−blow−up��
flip// D20/Γ
L−factorialization��
M3,4GIT P // (D20/Γ)∗
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 16 / 21
2.D. Main result of [LPZ17]
Theorem [LPZ17]
With notations as above
M(1/3) ∼= (D/Γ)∗
compare
Theorem (L./Looijenga 2017; Voisin ’86, Hassett ’96)
M3,4∼= (D20 \ H2 \ H6)/Γ
Furthermore
M3,4
K−blow−up��
flip// D20/Γ
L−factorialization��
M3,4GIT P // (D20/Γ)∗
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 16 / 21
2.X. Commercial Break “HKL Program”
with K. O’Grady, we have a “Hassett–Keel–LooijengaProgram” to systematically understand
P :M 99K (D/Γ)∗
GIT vs. Baily-Borel compactifications (Type IV case, or ballquotients)
Thus, we understand well the “compare” above.
“Modularity principle”: The tautological models of D/Γ areobtained by arithmetic modifications of the Baily-Borelcompactification.
The upshot, the complexity of P is determined by the complexityof the “discriminant” hyperplane arrangement ∆
deg 2 K3 surfaces, cubic fourfolds ... Easy.deg 4 K3 surfaces, EPW sextics, DV ... Hard
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 17 / 21
2.X. Commercial Break “HKL Program”
with K. O’Grady, we have a “Hassett–Keel–LooijengaProgram” to systematically understand
P :M 99K (D/Γ)∗
GIT vs. Baily-Borel compactifications (Type IV case, or ballquotients)
Thus, we understand well the “compare” above.
“Modularity principle”: The tautological models of D/Γ areobtained by arithmetic modifications of the Baily-Borelcompactification.
The upshot, the complexity of P is determined by the complexityof the “discriminant” hyperplane arrangement ∆
deg 2 K3 surfaces, cubic fourfolds ... Easy.deg 4 K3 surfaces, EPW sextics, DV ... Hard
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 17 / 21
2.X. Commercial Break “HKL Program”
with K. O’Grady, we have a “Hassett–Keel–LooijengaProgram” to systematically understand
P :M 99K (D/Γ)∗
GIT vs. Baily-Borel compactifications (Type IV case, or ballquotients)
Thus, we understand well the “compare” above.
“Modularity principle”: The tautological models of D/Γ areobtained by arithmetic modifications of the Baily-Borelcompactification.
The upshot, the complexity of P is determined by the complexityof the “discriminant” hyperplane arrangement ∆
deg 2 K3 surfaces, cubic fourfolds ... Easy.deg 4 K3 surfaces, EPW sextics, DV ... Hard
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 17 / 21
2.X. Commercial Break “HKL Program”
with K. O’Grady, we have a “Hassett–Keel–LooijengaProgram” to systematically understand
P :M 99K (D/Γ)∗
GIT vs. Baily-Borel compactifications (Type IV case, or ballquotients)
Thus, we understand well the “compare” above.
“Modularity principle”: The tautological models of D/Γ areobtained by arithmetic modifications of the Baily-Borelcompactification.
The upshot, the complexity of P is determined by the complexityof the “discriminant” hyperplane arrangement ∆
deg 2 K3 surfaces, cubic fourfolds ... Easy.deg 4 K3 surfaces, EPW sextics, DV ... Hard
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 17 / 21
3. Rationality of cubics revisited
It is expected that the Hodge general cubic fourfold is not rational.
In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.
But a cubic fourfold containing two disjoint planes is rational.
Natural to ask
Question (Most algebraic irrational cubic)
Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?
Two issues
define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 18 / 21
3. Rationality of cubics revisited
It is expected that the Hodge general cubic fourfold is not rational.
In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.
But a cubic fourfold containing two disjoint planes is rational.
Natural to ask
Question (Most algebraic irrational cubic)
Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?
Two issues
define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 18 / 21
3. Rationality of cubics revisited
It is expected that the Hodge general cubic fourfold is not rational.
In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.
But a cubic fourfold containing two disjoint planes is rational.
Natural to ask
Question (Most algebraic irrational cubic)
Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?
Two issues
define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 18 / 21
3. Rationality of cubics revisited
It is expected that the Hodge general cubic fourfold is not rational.
In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.
But a cubic fourfold containing two disjoint planes is rational.
Natural to ask
Question (Most algebraic irrational cubic)
Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?
Two issues
define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 18 / 21
3. Rationality of cubics revisited
It is expected that the Hodge general cubic fourfold is not rational.
In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.
But a cubic fourfold containing two disjoint planes is rational.
Natural to ask
Question (Most algebraic irrational cubic)
Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?
Two issues
define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 18 / 21
3. Maximally algebraic Cubics
Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)
Clearly, we want
ρX := rank(H2,2(X) ∩H4(X,Z)
)to be as large as possible
but we also want
dX := det(H2,2(X) ∩H4(X,Z)
)to be as small as possible
combine them in a single index
κX :=2dX
ρX
The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221
3 or 219 resp.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 19 / 21
3. Maximally algebraic Cubics
Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)
Clearly, we want
ρX := rank(H2,2(X) ∩H4(X,Z)
)to be as large as possible
but we also want
dX := det(H2,2(X) ∩H4(X,Z)
)to be as small as possible
combine them in a single index
κX :=2dX
ρX
The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221
3 or 219 resp.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 19 / 21
3. Maximally algebraic Cubics
Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)
Clearly, we want
ρX := rank(H2,2(X) ∩H4(X,Z)
)to be as large as possible
but we also want
dX := det(H2,2(X) ∩H4(X,Z)
)to be as small as possible
combine them in a single index
κX :=2dX
ρX
The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221
3 or 219 resp.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 19 / 21
3. Maximally algebraic Cubics
Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)
Clearly, we want
ρX := rank(H2,2(X) ∩H4(X,Z)
)to be as large as possible
but we also want
dX := det(H2,2(X) ∩H4(X,Z)
)to be as small as possible
combine them in a single index
κX :=2dX
ρX
The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221
3 or 219 resp.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 19 / 21
3. Maximally algebraic Cubics
Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)
Clearly, we want
ρX := rank(H2,2(X) ∩H4(X,Z)
)to be as large as possible
but we also want
dX := det(H2,2(X) ∩H4(X,Z)
)to be as small as possible
combine them in a single index
κX :=2dX
ρX
The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221
3 or 219 resp.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 19 / 21
3. Maximally algebraic potentially irrational Cubics
Theorem [L18]
Let X be a cubic fourfold.
(0) If X is potentially irrational, then κX ≤ 1.
(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.
(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.
If X has two Eckardt points, then X is rational.
Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 20 / 21
3. Maximally algebraic potentially irrational Cubics
Theorem [L18]
Let X be a cubic fourfold.
(0) If X is potentially irrational, then κX ≤ 1.
(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.
(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.
If X has two Eckardt points, then X is rational.
Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 20 / 21
3. Maximally algebraic potentially irrational Cubics
Theorem [L18]
Let X be a cubic fourfold.
(0) If X is potentially irrational, then κX ≤ 1.
(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.
(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.
If X has two Eckardt points, then X is rational.
Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 20 / 21
3. Maximally algebraic potentially irrational Cubics
Theorem [L18]
Let X be a cubic fourfold.
(0) If X is potentially irrational, then κX ≤ 1.
(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.
(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.
If X has two Eckardt points, then X is rational.
Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 20 / 21
3. Maximally algebraic potentially irrational Cubics
Theorem [L18]
Let X be a cubic fourfold.
(0) If X is potentially irrational, then κX ≤ 1.
(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.
(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.
If X has two Eckardt points, then X is rational.
Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 20 / 21
4. Cubics with an Eckardt point and HK Manifolds
Question
Is there an exotic hyper-Kahler 8-fold?
[LPZ18+] – Maybe?
R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 21 / 21