Solutions to x 3 + y 3 + z3 + w 3 = (x + y + z + w)3: cubic surfaces, 27 lines, and the icosahedron.
Marcello Bernardara
Institut de Math´ematiquesde Toulouse
Marcello Bernardara (IMT) Cubic surfaces 1 / 19 One of the first paintings with modern perspective. The parallel lines on the floor converge to the same point L.B.Alberti: De pictura (1430s), Piero della Francesca De Prospectiva Pingendi (1470s)
Projective Geometry Perspective and infinite
A.Lorenzetti’s Annunciazione (1344)
Marcello Bernardara (IMT) Cubic surfaces 2 / 19 The parallel lines on the floor converge to the same point L.B.Alberti: De pictura (1430s), Piero della Francesca De Prospectiva Pingendi (1470s)
Projective Geometry Perspective and infinite
One of the first paintings with modern perspective.
A.Lorenzetti’s Annunciazione (1344)
Marcello Bernardara (IMT) Cubic surfaces 2 / 19 L.B.Alberti: De pictura (1430s), Piero della Francesca De Prospectiva Pingendi (1470s)
Projective Geometry Perspective and infinite
One of the first paintings with modern perspective. The parallel lines on the floor converge to the same point
A.Lorenzetti’s Annunciazione (1344)
Marcello Bernardara (IMT) Cubic surfaces 2 / 19 L.B.Alberti: De pictura (1430s), Piero della Francesca De Prospectiva Pingendi (1470s)
Projective Geometry Perspective and infinite
One of the first paintings with modern perspective. The parallel lines on the floor converge to the same point
A.Lorenzetti’s Annunciazione (1344)
Marcello Bernardara (IMT) Cubic surfaces 2 / 19 Projective Geometry Perspective and infinite
One of the first paintings with modern perspective. The parallel lines on the floor converge to the same point L.B.Alberti: De pictura (1430s), Piero della Francesca De Prospectiva Pingendi (1470s)
A.Lorenzetti’s Annunciazione (1344)
Marcello Bernardara (IMT) Cubic surfaces 2 / 19 A point P corresponds to a line through N. The ∞ point of li ’s corresponds to a line tangent at N with the same direction. {Points of projective plane} ↔ {Lines through origin in Euclidean space}
Projective Geometry The projective plane P2- stereographic projection
Idea: two parallel lines meet at an infinite point. How to add “infinite points” to the Euclidean plane?
Marcello Bernardara (IMT) Cubic surfaces 3 / 19 A point P corresponds to a line through N. The ∞ point of li ’s corresponds to a line tangent at N with the same direction. {Points of projective plane} ↔ {Lines through origin in Euclidean space}
Projective Geometry The projective plane P2- stereographic projection
Idea: two parallel lines meet at an infinite point. How to add “infinite points” to the Euclidean plane?
Marcello Bernardara (IMT) Cubic surfaces 3 / 19 The ∞ point of li ’s corresponds to a line tangent at N with the same direction. {Points of projective plane} ↔ {Lines through origin in Euclidean space}
Projective Geometry The projective plane P2- stereographic projection
Idea: two parallel lines meet at an infinite point. How to add “infinite points” to the Euclidean plane? A point P corresponds to a line through N.
Marcello Bernardara (IMT) Cubic surfaces 3 / 19 {Points of projective plane} ↔ {Lines through origin in Euclidean space}
Projective Geometry The projective plane P2- stereographic projection
Idea: two parallel lines meet at an infinite point. How to add “infinite points” to the Euclidean plane? A point P corresponds to a line through N. The ∞ point of li ’s corresponds to a line tangent at N with the same direction.
Marcello Bernardara (IMT) Cubic surfaces 3 / 19 Projective Geometry The projective plane P2- stereographic projection
Idea: two parallel lines meet at an infinite point. How to add “infinite points” to the Euclidean plane? A point P corresponds to a line through N. The ∞ point of li ’s corresponds to a line tangent at N with the same direction. {Points of projective plane} ↔ {Lines through origin in Euclidean space}
Marcello Bernardara (IMT) Cubic surfaces 3 / 19 2 {Points of P } ↔ {triples (x0, x1, x2) 6= 0}/(x0, x1, x2) = (λx0, λx1, λx2)
2 Homogeneous coordinates. A point of P is determined by an equivalence class of triples denoted (x0 : x1 : x2). Infinite points have coordinates (x1 : x2 : 0). Bezout’s Theorem (1779) 2 Two curves of degree m and n meet exactly in n · m points of P .
n The projective space P of dimension n has points with homogeneous coordinates (x0 : ... : xn)
Projective Geometry The projective plane P2: Mathematics
{Points of projective plane} ↔ {Lines through origin in Euclidean space}
Equation of such a line l: λ(x0, x1, x2), where P = (x0, x1, x2) is a point on l, and (x0, x1, x2) 6= (0, 0, 0).
Marcello Bernardara (IMT) Cubic surfaces 4 / 19 Bezout’s Theorem (1779) 2 Two curves of degree m and n meet exactly in n · m points of P .
n The projective space P of dimension n has points with homogeneous coordinates (x0 : ... : xn)
Projective Geometry The projective plane P2: Mathematics
{Points of projective plane} ↔ {Lines through origin in Euclidean space}
Equation of such a line l: λ(x0, x1, x2), where P = (x0, x1, x2) is a point on l, and (x0, x1, x2) 6= (0, 0, 0).
2 {Points of P } ↔ {triples (x0, x1, x2) 6= 0}/(x0, x1, x2) = (λx0, λx1, λx2)
2 Homogeneous coordinates. A point of P is determined by an equivalence class of triples denoted (x0 : x1 : x2). Infinite points have coordinates (x1 : x2 : 0).
Marcello Bernardara (IMT) Cubic surfaces 4 / 19 n The projective space P of dimension n has points with homogeneous coordinates (x0 : ... : xn)
Projective Geometry The projective plane P2: Mathematics
{Points of projective plane} ↔ {Lines through origin in Euclidean space}
Equation of such a line l: λ(x0, x1, x2), where P = (x0, x1, x2) is a point on l, and (x0, x1, x2) 6= (0, 0, 0).
2 {Points of P } ↔ {triples (x0, x1, x2) 6= 0}/(x0, x1, x2) = (λx0, λx1, λx2)
2 Homogeneous coordinates. A point of P is determined by an equivalence class of triples denoted (x0 : x1 : x2). Infinite points have coordinates (x1 : x2 : 0). Bezout’s Theorem (1779) 2 Two curves of degree m and n meet exactly in n · m points of P .
Marcello Bernardara (IMT) Cubic surfaces 4 / 19 Projective Geometry The projective plane P2: Mathematics
{Points of projective plane} ↔ {Lines through origin in Euclidean space}
Equation of such a line l: λ(x0, x1, x2), where P = (x0, x1, x2) is a point on l, and (x0, x1, x2) 6= (0, 0, 0).
2 {Points of P } ↔ {triples (x0, x1, x2) 6= 0}/(x0, x1, x2) = (λx0, λx1, λx2)
2 Homogeneous coordinates. A point of P is determined by an equivalence class of triples denoted (x0 : x1 : x2). Infinite points have coordinates (x1 : x2 : 0). Bezout’s Theorem (1779) 2 Two curves of degree m and n meet exactly in n · m points of P .
n The projective space P of dimension n has points with homogeneous coordinates (x0 : ... : xn)
Marcello Bernardara (IMT) Cubic surfaces 4 / 19 irreducible: F is irreducible.
conic + line: F = F2 · F1
3 lines: F = F1 · G1 · H1 2 2 lines, one double: F = F1 · G1 3 a triple line: F = F1
2 Example: Degree 3 curves in P
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 conic + line: F = F2 · F1
3 lines: F = F1 · G1 · H1 2 2 lines, one double: F = F1 · G1 3 a triple line: F = F1
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P irreducible: F is irreducible.
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 irreducible: F is irreducible.
3 lines: F = F1 · G1 · H1 2 2 lines, one double: F = F1 · G1 3 a triple line: F = F1
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P
conic + line: F = F2 · F1
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 irreducible: F is irreducible.
conic + line: F = F2 · F1
2 2 lines, one double: F = F1 · G1 3 a triple line: F = F1
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P
3 lines: F = F1 · G1 · H1
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 irreducible: F is irreducible.
conic + line: F = F2 · F1
3 lines: F = F1 · G1 · H1
3 a triple line: F = F1
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P
2 2 lines, one double: F = F1 · G1
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 irreducible: F is irreducible.
conic + line: F = F2 · F1
3 lines: F = F1 · G1 · H1 2 2 lines, one double: F = F1 · G1
Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P
3 a triple line: F = F1
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 Projective Geometry Projective varieties and Homogeneous equations
Homogeneous polynomials: polynomials F (x0,..., xn) whose monomials with nonzero coefficients all have the same total degree. If F (a0,..., an) = 0 , then F (λa0, . . . , λan) = 0 for all λ, so the Zero n Locus Z(F ) = {points of P where F vanishes} is well defined.
2 Example: Degree 3 curves in P irreducible: F is irreducible.
conic + line: F = F2 · F1
3 lines: F = F1 · G1 · H1 2 2 lines, one double: F = F1 · G1 3 a triple line: F = F1
Marcello Bernardara (IMT) Cubic surfaces 5 / 19 Cubic Surfaces Lines on cubic surfaces
3 A Cubic Surface in P is the zero locus S = Z(F ) of a degree 3 homogeneous polynomial. We suppose that S is smooth. Theorem (Cayley-Salmon 1849) There are exactly 27 lines on S.
Arthur Cayley (1821 - 1895) George Salmon (1819 - 1904)
Marcello Bernardara (IMT) Cubic surfaces 6 / 19 Cubic Surfaces
Marcello Bernardara (IMT) Cubic surfaces 7 / 19 Proof: wlog, let w = 0 be the equation of the plane Π, then Π ∩ S is a 2 degree three curve on P The line l :(z = w = 0) in Π is double if
2 F = z F1 + wF2
with Fi of degree i. But this would imply that S is nonsmooth Second Step There is at least a line on S.
Cubic Surfaces Proof of the Theorem
First step There are at most 3 lines in S through any point P of S. If there are 2 or 3, then they are coplanar. Every plane intersects S along either an irreducible cubic or a conic plus a line or three distinct lines.
Marcello Bernardara (IMT) Cubic surfaces 8 / 19 Second Step There is at least a line on S.
Cubic Surfaces Proof of the Theorem
First step There are at most 3 lines in S through any point P of S. If there are 2 or 3, then they are coplanar. Every plane intersects S along either an irreducible cubic or a conic plus a line or three distinct lines. Proof: wlog, let w = 0 be the equation of the plane Π, then Π ∩ S is a 2 degree three curve on P The line l :(z = w = 0) in Π is double if
2 F = z F1 + wF2
with Fi of degree i. But this would imply that S is nonsmooth
Marcello Bernardara (IMT) Cubic surfaces 8 / 19 Cubic Surfaces Proof of the Theorem
First step There are at most 3 lines in S through any point P of S. If there are 2 or 3, then they are coplanar. Every plane intersects S along either an irreducible cubic or a conic plus a line or three distinct lines. Proof: wlog, let w = 0 be the equation of the plane Π, then Π ∩ S is a 2 degree three curve on P The line l :(z = w = 0) in Π is double if
2 F = z F1 + wF2
with Fi of degree i. But this would imply that S is nonsmooth Second Step There is at least a line on S.
Marcello Bernardara (IMT) Cubic surfaces 8 / 19 PROOF: take l and a plane Π ⊃ l, Then Π ∩ S is one of the following:
We want exactly 5 planes giving a) or b). If l :(z = w = 0) then we can write F = Ax2 + Bxy + Cy 2 + Dx + Ey + G which is the equation of a conic in the plane (x, y) depending on z, w. Exactly 5 of them are reducible.
Cubic Surfaces Proof of the Theorem
Third Step 0 l a line on S. There are exactly 5 pairs (li , li ) of lines on S meeting l s.t. 0 for i = 1,... 5, l, li , and li are coplanar 0 0 for i 6= j, li ∪ li does not intersect lj ∪ lj .
Marcello Bernardara (IMT) Cubic surfaces 9 / 19 Cubic Surfaces Proof of the Theorem
Third Step 0 l a line on S. There are exactly 5 pairs (li , li ) of lines on S meeting l s.t. 0 for i = 1,... 5, l, li , and li are coplanar 0 0 for i 6= j, li ∪ li does not intersect lj ∪ lj .
PROOF: take l and a plane Π ⊃ l, Then Π ∩ S is one of the following:
We want exactly 5 planes giving a) or b). If l :(z = w = 0) then we can write F = Ax2 + Bxy + Cy 2 + Dx + Ey + G which is the equation of a conic in the plane (x, y) depending on z, w. Exactly 5 of them are reducible.
Marcello Bernardara (IMT) Cubic surfaces 9 / 19 Cubic Surfaces Proof of the Theorem
Third Step 0 l a line on S. There are exactly 5 pairs (li , li ) of lines on S meeting l s.t. 0 for i = 1,... 5, l, li , and li are coplanar 0 0 for i 6= j, li ∪ li does not intersect lj ∪ lj .
Marcello Bernardara (IMT) Cubic surfaces 10 / 19 Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ).
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i..
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i.
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅.
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 Cubic Surfaces
Fourth Step There exist two disjoint lines l and m on S.
Fifth Step Find 27 lines in term of the pairs configuration.
0 Consider l, m and the pairs (li , li ). 0 No double line on S =⇒ m cannot meet both li and li 3 0 Being in P =⇒ m meets either li or li for all i.. 0 Up to renaming: m ∩ l = m ∩ li = ∅ for all i. 0 0 0 Pairs (mi , mi ) are given by mi = li and mi ∩ li = ∅. Let n be any other line on S. If n meets more than 3 out of li ’s =⇒ n is double. Then n meets exactly 3 out of li , call n =: li,j,k
0 0 LINES l m li li mi lijk 1 + 1 + 5 + 5 + 5 + 10 = 27
Marcello Bernardara (IMT) Cubic surfaces 11 / 19 ai ∩ aj = bi ∩ bj = ∅ for i 6= j
ai ∩ bi = ∅, for all i = 1,... 6
ai ∩ bj is a point if i 6= j
a1 a2 a3 a4 a5 a6 b1 b2 b3 b4 b5 b6
Schl¨afli’sTheorem (1858) The 27 lines are completely determined by a double six. There are 36 double sixes for a given cubic surface.
Cubic Surfaces Schl¨afli’sdouble six
A double six is a configuration of 12 lines (a1,... a6, b1,..., b6) in the projective space such that
Marcello Bernardara (IMT) Cubic surfaces 12 / 19 Schl¨afli’sTheorem (1858) The 27 lines are completely determined by a double six. There are 36 double sixes for a given cubic surface.
Cubic Surfaces Schl¨afli’sdouble six
A double six is a configuration of 12 lines (a1,... a6, b1,..., b6) in the projective space such that
ai ∩ aj = bi ∩ bj = ∅ for i 6= j
ai ∩ bi = ∅, for all i = 1,... 6
ai ∩ bj is a point if i 6= j
a1 a2 a3 a4 a5 a6 b1 b2 b3 b4 b5 b6
Marcello Bernardara (IMT) Cubic surfaces 12 / 19 Cubic Surfaces Schl¨afli’sdouble six
A double six is a configuration of 12 lines (a1,... a6, b1,..., b6) in the projective space such that
ai ∩ aj = bi ∩ bj = ∅ for i 6= j
ai ∩ bi = ∅, for all i = 1,... 6
ai ∩ bj is a point if i 6= j
a1 a2 a3 a4 a5 a6 b1 b2 b3 b4 b5 b6
Schl¨afli’sTheorem (1858) The 27 lines are completely determined by a double six. There are 36 double sixes for a given cubic surface.
Marcello Bernardara (IMT) Cubic surfaces 12 / 19 Cubic Surfaces Schl¨afli’sdouble six
Marcello Bernardara (IMT) Cubic surfaces 13 / 19 Cubic Surfaces Schl¨afli’sdouble six
Marcello Bernardara (IMT) Cubic surfaces 13 / 19 Cubic Surfaces Schl¨afli’sdouble six
Marcello Bernardara (IMT) Cubic surfaces 13 / 19 Cubic Surfaces Schl¨afli’sdouble six
Marcello Bernardara (IMT) Cubic surfaces 13 / 19 From the plane to the cubic surface The blow-up of a point
Consider a point on a surface, and “replace it with a line”
The red line (the exceptional divisor) parameterizes all the lines through the red point (the center of the blow-up).
Marcello Bernardara (IMT) Cubic surfaces 14 / 19 Given a double six (a1,..., a6|b1,..., b6) the ai ’s are the exceptional divisors of such blow up.
2 There are 72 choices of 6 lines giving a map S → P .
From the plane to the cubic surface
Theorem Every cubic surface is the blow up of six points in general position in the 2 projective plane P .
Marcello Bernardara (IMT) Cubic surfaces 15 / 19 2 There are 72 choices of 6 lines giving a map S → P .
From the plane to the cubic surface
Theorem Every cubic surface is the blow up of six points in general position in the 2 projective plane P .
Given a double six (a1,..., a6|b1,..., b6) the ai ’s are the exceptional divisors of such blow up.
Marcello Bernardara (IMT) Cubic surfaces 15 / 19 From the plane to the cubic surface
Theorem Every cubic surface is the blow up of six points in general position in the 2 projective plane P .
Given a double six (a1,..., a6|b1,..., b6) the ai ’s are the exceptional divisors of such blow up.
2 There are 72 choices of 6 lines giving a map S → P .
Marcello Bernardara (IMT) Cubic surfaces 15 / 19 There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines) Each line through two points gives a line on the surface (15 of them)
From the plane to the cubic surface The 27 lines from this point of view 6 skew lines come from the 6 points
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 6 skew lines come from the 6 points
Each line through two points gives a line on the surface (15 of them)
From the plane to the cubic surface The 27 lines from this point of view
There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines)
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 6 skew lines come from the 6 points
Each line through two points gives a line on the surface (15 of them)
From the plane to the cubic surface The 27 lines from this point of view
There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines)
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 6 skew lines come from the 6 points There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines)
From the plane to the cubic surface The 27 lines from this point of view
Each line through two points gives a line on the surface (15 of them)
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 From the plane to the cubic surface The 27 lines from this point of view 6 skew lines come from the 6 points There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines) Each line through two points gives a line on the surface (15 of them)
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 From the plane to the cubic surface The 27 lines from this point of view 6 skew lines come from the 6 points There is a single conic for any choice of 5 points. Each of these conics give a line (6 skew lines) Each line through two points gives a line on the surface (15 of them)
Marcello Bernardara (IMT) Cubic surfaces 16 / 19 The Clebsch cubic surface The Clebsch cubic
Consider the homogeneous polynomial
F (x, y, z, w) = x3 + y 3 + z3 + w 3 − (x + y + z + w)3 3 the cubic surface in P defined as Z(F ) is the locus of points (x : y : w : z) satisfying
x3 + y 3 + z3 + w 3 = (x + y + z + w)3 This cubic is smooth and is called the Clebsch cubic
Marcello Bernardara (IMT) Cubic surfaces 17 / 19 The center of the icosahedron is the origin. √ 2 The 6 points in P have to be defined over Q[ −5]. There are 72 such icosahedra.
The Clebsch cubic surface
Theorem: Clebsch (1871), Klein (1873) The six points in the plane corresponding to the Clebsch cubic are given by the lines in the euclidean space joining opposite vertices of a regular icosahedron, with assigned vertex coordinates.
Marcello Bernardara (IMT) Cubic surfaces 18 / 19 √ 2 The 6 points in P have to be defined over Q[ −5]. There are 72 such icosahedra.
The Clebsch cubic surface
Theorem: Clebsch (1871), Klein (1873) The six points in the plane corresponding to the Clebsch cubic are given by the lines in the euclidean space joining opposite vertices of a regular icosahedron, with assigned vertex coordinates.
The center of the icosahedron is the origin.
Marcello Bernardara (IMT) Cubic surfaces 18 / 19 There are 72 such icosahedra.
The Clebsch cubic surface
Theorem: Clebsch (1871), Klein (1873) The six points in the plane corresponding to the Clebsch cubic are given by the lines in the euclidean space joining opposite vertices of a regular icosahedron, with assigned vertex coordinates.
The center of the icosahedron is the origin. √ 2 The 6 points in P have to be defined over Q[ −5].
Marcello Bernardara (IMT) Cubic surfaces 18 / 19 The Clebsch cubic surface
Theorem: Clebsch (1871), Klein (1873) The six points in the plane corresponding to the Clebsch cubic are given by the lines in the euclidean space joining opposite vertices of a regular icosahedron, with assigned vertex coordinates.
The center of the icosahedron is the origin. √ 2 The 6 points in P have to be defined over Q[ −5]. There are 72 such icosahedra.
Marcello Bernardara (IMT) Cubic surfaces 18 / 19 The Clebsch cubic surface The Icosahedron
Marcello Bernardara (IMT) Cubic surfaces 19 / 19