Classical Enumerative Geometry and Quantum Cohomology
James McKernan
UCSB
Classical Enumerative Geometry and Quantum Cohomology – p.1 Amazing Equation
It is the purpose of this talk to convince the listener that the following formula is truly amazing. end
a − a − b c 2 2 3 − 3 3 Nd(a, b, c) = Nd1 (a1, b1, c1)Nd2 (a2, b2, c2) d1d2 d1d2 a1 − 1 a1 b1 c1 d1+Xd2=d a1+a2=a−1 b1+b2=b c1+c2=c a − a − b c · 2 3 − 3 3 + 2 Nd1 (a1, b1, c1)Nd2 (a2, b2, c2) d1d2 d1 a1 − 1 a1 b1 b2 1 c1 d1+Xd2=d a1+a2=a b1+b2=b−1 c1+c2=c a − a − b c · 3 − 2 3 + 4 Nd1 (a1, b1, c1)Nd2 (a2, b2, c2) d1d2 d1 a1 − 2 a1 − 1 b1 b2 2 c1 d1+Xd2=d a1+a2=a+1 b1+b2=b−2 c1+c2=c a − a − b c · 3 − 2 3 + 2 Nd1 (a1, b1, c1)Nd2 (a2, b2, c2) d1d2 d1 . a1 − 2 a1 − 1 b1 c1 c2 1 d1+Xd2=d a1+a2=a+1 b1+b2=b Classical Enumerative Geometry and Quantum Cohomology – p.2 c1+c2=c−1 Suppose n = 1. Pick p = (x, y) ∈ C2. The line through p is represented by its slope, that is the ratio z = y/x. We get the classical Riemann sphere, C compactified by adding a point at infinity.
z ∈ C ∪ {∞}
Projective Space
Pn is the set of lines in Cn+1.
Classical Enumerative Geometry and Quantum Cohomology – p.3 We get the classical Riemann sphere, C compactified by adding a point at infinity.
z ∈ C ∪ {∞}
Projective Space
Pn is the set of lines in Cn+1. Suppose n = 1. Pick p = (x, y) ∈ C2. The line through p is represented by its slope, that is the ratio z = y/x.
Classical Enumerative Geometry and Quantum Cohomology – p.3 Projective Space
Pn is the set of lines in Cn+1. Suppose n = 1. Pick p = (x, y) ∈ C2. The line through p is represented by its slope, that is the ratio z = y/x. We get the classical Riemann sphere, C compactified by adding a point at infinity.
z ∈ C ∪ {∞}
Classical Enumerative Geometry and Quantum Cohomology – p.3 Suppose we are given four points p, q, r and s. Then there is a unique Möbius transformation, sending p −→ 0, q −→ 1, r −→ ∞. The image s −→ λ ∈ C − {0, 1} is called the cross-ratio. link. In fact (r − q)(p − s) λ = . (p − q)(r − s)
Automorphisms of P1
P1 az+b Aut( ) = cz+d, the group of Möbius transformations.
Classical Enumerative Geometry and Quantum Cohomology – p.4 The image s −→ λ ∈ C − {0, 1} is called the cross-ratio. link. In fact (r − q)(p − s) λ = . (p − q)(r − s)
Automorphisms of P1
P1 az+b Aut( ) = cz+d, the group of Möbius transformations. Suppose we are given four points p, q, r and s. Then there is a unique Möbius transformation, sending p −→ 0, q −→ 1, r −→ ∞.
Classical Enumerative Geometry and Quantum Cohomology – p.4 link. In fact (r − q)(p − s) λ = . (p − q)(r − s)
Automorphisms of P1
P1 az+b Aut( ) = cz+d, the group of Möbius transformations. Suppose we are given four points p, q, r and s. Then there is a unique Möbius transformation, sending p −→ 0, q −→ 1, r −→ ∞. The image s −→ λ ∈ C − {0, 1} is called the cross-ratio.
Classical Enumerative Geometry and Quantum Cohomology – p.4 Automorphisms of P1
P1 az+b Aut( ) = cz+d, the group of Möbius transformations. Suppose we are given four points p, q, r and s. Then there is a unique Möbius transformation, sending p −→ 0, q −→ 1, r −→ ∞. The image s −→ λ ∈ C − {0, 1} is called the cross-ratio. link. In fact (r − q)(p − s) λ = . (p − q)(r − s)
Classical Enumerative Geometry and Quantum Cohomology – p.4 In fact these two curves intersect at two points and these two lines
meet at a point. Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Some Consequences
Our definition of Pn has some interesting consequences.
Classical Enumerative Geometry and Quantum Cohomology – p.5 intersect at two points and these two lines
meet at a point. Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Some Consequences
Our definition of Pn has some interesting consequences. In fact these two curves
Classical Enumerative Geometry and Quantum Cohomology – p.5 and these two lines
meet at a point. Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Some Consequences
Our definition of Pn has some interesting consequences. In fact these two curves
intersect at two points
Classical Enumerative Geometry and Quantum Cohomology – p.5 meet at a point. Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Some Consequences
Our definition of Pn has some interesting consequences. In fact these two curves
intersect at two points and these two lines
Classical Enumerative Geometry and Quantum Cohomology – p.5 Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Some Consequences
Our definition of Pn has some interesting consequences. In fact these two curves
intersect at two points and these two lines
meet at a point.
Classical Enumerative Geometry and Quantum Cohomology – p.5 Some Consequences
Our definition of Pn has some interesting consequences. In fact these two curves
intersect at two points and these two lines
meet at a point. Principle of Continuity The number of intersection points is an invariant of a continuous family of curves.
Classical Enumerative Geometry and Quantum Cohomology – p.5 Indeed the result is obviously true when our two curves are the union of d and e lines
Proof. Let F∞ and G∞ be the product of linear forms. Then F + tF∞ and G + tG∞ define continuous families and when t = ∞, the answer is obviously de.
Bézout’s Theorem
Theorem. (Bezout’s Theorem) Given two plane curves C and D defined by polynomials F and G of degree d and e, then #C ∩ D = de.
Classical Enumerative Geometry and Quantum Cohomology – p.6 Proof. Let F∞ and G∞ be the product of linear forms. Then F + tF∞ and G + tG∞ define continuous families and when t = ∞, the answer is obviously de.
Bézout’s Theorem
Theorem. (Bezout’s Theorem) Given two plane curves C and D defined by polynomials F and G of degree d and e, then #C ∩ D = de.
Indeed the result is obviously true when our two curves are the union of d and e lines
Classical Enumerative Geometry and Quantum Cohomology – p.6 Bézout’s Theorem
Theorem. (Bezout’s Theorem) Given two plane curves C and D defined by polynomials F and G of degree d and e, then #C ∩ D = de.
Indeed the result is obviously true when our two curves are the union of d and e lines
Proof. Let F∞ and G∞ be the product of linear forms. Then F + tF∞ and G + tG∞ define continuous families and when t = ∞, the answer is obviously de.
Classical Enumerative Geometry and Quantum Cohomology – p.6 Bézout’s Theorem: If we have n hypersurfaces X1, X2, . . . , Xn of degrees d1, d2, . . . , dn, then the number of common intersection points is d1d2 . . . dn.
Indeed, the same proof applies.
The degree
The degree d of a variety Xk ⊂ Pn is the number of points X ∩ Λ, where Λ is a general linear subspace of dimension n − k.
Classical Enumerative Geometry and Quantum Cohomology – p.7 Indeed, the same proof applies.
The degree
The degree d of a variety Xk ⊂ Pn is the number of points X ∩ Λ, where Λ is a general linear subspace of dimension n − k. Bézout’s Theorem: If we have n hypersurfaces X1, X2, . . . , Xn of degrees d1, d2, . . . , dn, then the number of common intersection points is
d1d2 . . . dn.
Classical Enumerative Geometry and Quantum Cohomology – p.7 The degree
The degree d of a variety Xk ⊂ Pn is the number of points X ∩ Λ, where Λ is a general linear subspace of dimension n − k. Bézout’s Theorem: If we have n hypersurfaces X1, X2, . . . , Xn of degrees d1, d2, . . . , dn, then the number of common intersection points is
d1d2 . . . dn. Indeed, the same proof applies.
Classical Enumerative Geometry and Quantum Cohomology – p.7 How many conics pass through five points, p1, p2, p3, p4 and p5? The condition that a conic contains a point p is a linear condition on the coefficients.
Conics in P2
A conic is given as the zero locus of
aX2 + bY 2 + cZ2 + dXY + eY Z + fXZ, where [a : b : c : d : e : f] ∈ P5.
Classical Enumerative Geometry and Quantum Cohomology – p.8 The condition that a conic contains a point p is a linear condition on the coefficients.
Conics in P2
A conic is given as the zero locus of
aX2 + bY 2 + cZ2 + dXY + eY Z + fXZ, where [a : b : c : d : e : f] ∈ P5. How many conics pass through five points, p1, p2, p3, p4 and p5?
Classical Enumerative Geometry and Quantum Cohomology – p.8 Conics in P2
A conic is given as the zero locus of
aX2 + bY 2 + cZ2 + dXY + eY Z + fXZ, where [a : b : c : d : e : f] ∈ P5. How many conics pass through five points, p1, p2, p3, p4 and p5? The condition that a conic contains a point p is a linear condition on the coefficients.
Classical Enumerative Geometry and Quantum Cohomology – p.8 Thus the set of conics passing through five points, corresponds to the intersection of five hyperplanes. So the answer is one.
Correspondence
So we have a correspondence between
5 { C | C contains pi } and Hi ⊂ P ,
where Hi is a hyperplane.
Classical Enumerative Geometry and Quantum Cohomology – p.9 So the answer is one.
Correspondence
So we have a correspondence between
5 { C | C contains pi } and Hi ⊂ P ,
where Hi is a hyperplane. Thus the set of conics passing through five points, corresponds to the intersection of five hyperplanes.
Classical Enumerative Geometry and Quantum Cohomology – p.9 Correspondence
So we have a correspondence between
5 { C | C contains pi } and Hi ⊂ P ,
where Hi is a hyperplane. Thus the set of conics passing through five points, corresponds to the intersection of five hyperplanes. So the answer is one.
Classical Enumerative Geometry and Quantum Cohomology – p.9 Schubert Calculus: Let gl represent the condition to meet a line, gP be the condition to meet a point, and gπ be the condition to meet a plane. I claim 2 gl = gP + gπ. Indeed, fix two lines l and m, degenerate them until they are concurrent, and use the principle of continuity.
Lines in P3
How many lines in P3 meet four given lines?
Classical Enumerative Geometry and Quantum Cohomology – p.10 I claim 2 gl = gP + gπ. Indeed, fix two lines l and m, degenerate them until they are concurrent, and use the principle of continuity.
Lines in P3
How many lines in P3 meet four given lines? Schubert Calculus: Let
gl represent the condition to meet a line,
gP be the condition to meet a point, and
gπ be the condition to meet a plane.
Classical Enumerative Geometry and Quantum Cohomology – p.10 Indeed, fix two lines l and m, degenerate them until they are concurrent, and use the principle of continuity.
Lines in P3
How many lines in P3 meet four given lines? Schubert Calculus: Let
gl represent the condition to meet a line,
gP be the condition to meet a point, and
gπ be the condition to meet a plane. I claim 2 gl = gP + gπ.
Classical Enumerative Geometry and Quantum Cohomology – p.10 Lines in P3
How many lines in P3 meet four given lines? Schubert Calculus: Let
gl represent the condition to meet a line,
gP be the condition to meet a point, and
gπ be the condition to meet a plane. I claim 2 gl = gP + gπ. Indeed, fix two lines l and m, degenerate them until they are concurrent, and use the principle of continuity.
Classical Enumerative Geometry and Quantum Cohomology – p.10 In fact we are working in the cohomology ring. Z[x] H∗(P2) = hx3i where x is the class of a line, and C ∼ dx and D ∼ ex, so that C · D = dex2 = de.
A little Algebra
So 4 2 gl = (gP + gπ) 2 2 = gP + 2gP gπ + gπ = 1 + 2 · 0 + 1 = 2.
Classical Enumerative Geometry and Quantum Cohomology – p.11 Z[x] H∗(P2) = hx3i where x is the class of a line, and C ∼ dx and D ∼ ex, so that C · D = dex2 = de.
A little Algebra
So 4 2 gl = (gP + gπ) 2 2 = gP + 2gP gπ + gπ = 1 + 2 · 0 + 1 = 2.
In fact we are working in the cohomology ring.
Classical Enumerative Geometry and Quantum Cohomology – p.11 A little Algebra
So 4 2 gl = (gP + gπ) 2 2 = gP + 2gP gπ + gπ = 1 + 2 · 0 + 1 = 2.
In fact we are working in the cohomology ring. Z[x] H∗(P2) = hx3i where x is the class of a line, and C ∼ dx and D ∼ ex, so that C · D = dex2 = de.
Classical Enumerative Geometry and Quantum Cohomology – p.11 Fact: The set of conics tangent to a given conic corresponds to a hypersurface of degree 6, so by Bézout the answer ought to be 65. Wrong! Conics tangent to five lines? The set of conics tangent to one line corresponds to a hypersurface of degree two. Bézout predicts the answer is 25. But the actual answer is 1.
A classical Problem
How many conics tangent to five given conics?
Classical Enumerative Geometry and Quantum Cohomology – p.12 Wrong! Conics tangent to five lines? The set of conics tangent to one line corresponds to a hypersurface of degree two. Bézout predicts the answer is 25. But the actual answer is 1.
A classical Problem
How many conics tangent to five given conics? Fact: The set of conics tangent to a given conic corresponds to a hypersurface of degree 6, so by Bézout the answer ought to be 65.
Classical Enumerative Geometry and Quantum Cohomology – p.12 Conics tangent to five lines? The set of conics tangent to one line corresponds to a hypersurface of degree two. Bézout predicts the answer is 25. But the actual answer is 1.
A classical Problem
How many conics tangent to five given conics? Fact: The set of conics tangent to a given conic corresponds to a hypersurface of degree 6, so by Bézout the answer ought to be 65. Wrong!
Classical Enumerative Geometry and Quantum Cohomology – p.12 A classical Problem
How many conics tangent to five given conics? Fact: The set of conics tangent to a given conic corresponds to a hypersurface of degree 6, so by Bézout the answer ought to be 65. Wrong! Conics tangent to five lines? The set of conics tangent to one line corresponds to a hypersurface of degree two. Bézout predicts the answer is 25. But the actual answer is 1.
Classical Enumerative Geometry and Quantum Cohomology – p.12 Given a conic C in P2, we associate a dual conic Cˆ in Pˆ2, simply by sending a point of p of C to its tangent line, a point of Pˆ2. So the number of conics tangent to five given lines, is equal, by duality, to the number of conics through five given points which, as we have seen, is one. (Hwk) What is wrong?
Use duality
Pˆ2 is the dual space to P2. It consists of the set of lines in P2.
Classical Enumerative Geometry and Quantum Cohomology – p.13 So the number of conics tangent to five given lines, is equal, by duality, to the number of conics through five given points which, as we have seen, is one. (Hwk) What is wrong?
Use duality
Pˆ2 is the dual space to P2. It consists of the set of lines in P2. Given a conic C in P2, we associate a dual conic Cˆ in Pˆ2, simply by sending a point of p of C to its tangent line, a point of Pˆ2.
Classical Enumerative Geometry and Quantum Cohomology – p.13 (Hwk) What is wrong?
Use duality
Pˆ2 is the dual space to P2. It consists of the set of lines in P2. Given a conic C in P2, we associate a dual conic Cˆ in Pˆ2, simply by sending a point of p of C to its tangent line, a point of Pˆ2. So the number of conics tangent to five given lines, is equal, by duality, to the number of conics through five given points which, as we have seen, is one.
Classical Enumerative Geometry and Quantum Cohomology – p.13 Use duality
Pˆ2 is the dual space to P2. It consists of the set of lines in P2. Given a conic C in P2, we associate a dual conic Cˆ in Pˆ2, simply by sending a point of p of C to its tangent line, a point of Pˆ2. So the number of conics tangent to five given lines, is equal, by duality, to the number of conics through five given points which, as we have seen, is one. (Hwk) What is wrong?
Classical Enumerative Geometry and Quantum Cohomology – p.13 Physics
Here is a typical Feynmann diagram.
Feynmann diagrams are used to encode the complicated interactions which particles undergo.
Classical Enumerative Geometry and Quantum Cohomology – p.14 String Theory
One of the ideas of string theory, is that a string is the basic object and not particles. Replacing a point by a string, means replacing a line by a tube and our Feynmann diagram becomes:
Classical Enumerative Geometry and Quantum Cohomology – p.15 Points at Infinity
Now take this picture and extend the tubes to infinity. Adding the points at infinity is topologically equivalent to adding caps.
p r
q s
Classical Enumerative Geometry and Quantum Cohomology – p.16 So now we have the Riemann sphere, that is a copy of P1, with four marked points. More generally, we will get a Riemann surface, together with a collection of marked points.
Back to the Riemann sphere
Topologically, the resulting surface is a sphere.
Classical Enumerative Geometry and Quantum Cohomology – p.17 More generally, we will get a Riemann surface, together with a collection of marked points.
Back to the Riemann sphere
Topologically, the resulting surface is a sphere. So now we have the Riemann sphere, that is a copy of P1, with four marked points.
Classical Enumerative Geometry and Quantum Cohomology – p.17 Back to the Riemann sphere
Topologically, the resulting surface is a sphere. So now we have the Riemann sphere, that is a copy of P1, with four marked points. More generally, we will get a Riemann surface, together with a collection of marked points.
Classical Enumerative Geometry and Quantum Cohomology – p.17 For example, if d = 1 we get N = 2 (in fact Pˆ2) and if d = 2, we get N = 5, P5. Some plane curves are rational, that is to say there is a map P1 −→ P2, [S : T ] −→ [F : G : H], where F , G and H are polynomials of degree d, in S and T . Let X ⊂ PN , be the locus of these rational curves. Basic question: what is the degree of X ⊂ PN ?
Rational curves in P2
Note that plane curves of degree d correspond to polynomials of degree d, modulo scalars, which in turn corresponds to a PN , for some N.
Classical Enumerative Geometry and Quantum Cohomology – p.18 Some plane curves are rational, that is to say there is a map P1 −→ P2, [S : T ] −→ [F : G : H], where F , G and H are polynomials of degree d, in S and T . Let X ⊂ PN , be the locus of these rational curves. Basic question: what is the degree of X ⊂ PN ?
Rational curves in P2
Note that plane curves of degree d correspond to polynomials of degree d, modulo scalars, which in turn corresponds to a PN , for some N. For example, if d = 1 we get N = 2 (in fact Pˆ2) and if d = 2, we get N = 5, P5.
Classical Enumerative Geometry and Quantum Cohomology – p.18 Let X ⊂ PN , be the locus of these rational curves. Basic question: what is the degree of X ⊂ PN ?
Rational curves in P2
Note that plane curves of degree d correspond to polynomials of degree d, modulo scalars, which in turn corresponds to a PN , for some N. For example, if d = 1 we get N = 2 (in fact Pˆ2) and if d = 2, we get N = 5, P5. Some plane curves are rational, that is to say there is a map P1 −→ P2, [S : T ] −→ [F : G : H], where F , G and H are polynomials of degree d, in S and T .
Classical Enumerative Geometry and Quantum Cohomology – p.18 Rational curves in P2
Note that plane curves of degree d correspond to polynomials of degree d, modulo scalars, which in turn corresponds to a PN , for some N. For example, if d = 1 we get N = 2 (in fact Pˆ2) and if d = 2, we get N = 5, P5. Some plane curves are rational, that is to say there is a map P1 −→ P2, [S : T ] −→ [F : G : H], where F , G and H are polynomials of degree d, in S and T . Let X ⊂ PN , be the locus of these rational curves. Basic question: what is the degree of X ⊂ PN ?
Classical Enumerative Geometry and Quantum Cohomology – p.18 2 N1 = 1. Indeed X1 = P , as every line is rational. 5 Similarly every conic is rational, so X2 = P and N2 = 1.
In general to calculate the degree of Xd we need to cut by hyperplanes. As the dimension of Xd is 3d − 1, we want to cut by 3d − 1 hyperplanes. Now imposing the condition that a curve passes through a point is one linear condition. So we want to count the number of rational curves of degree d that pass through 3d − 1 points.
The Grade
Call this degree Nd. Classically this number is known as the grade.
Classical Enumerative Geometry and Quantum Cohomology – p.19 In general to calculate the degree of Xd we need to cut by hyperplanes. As the dimension of Xd is 3d − 1, we want to cut by 3d − 1 hyperplanes. Now imposing the condition that a curve passes through a point is one linear condition. So we want to count the number of rational curves of degree d that pass through 3d − 1 points.
The Grade
Call this degree Nd. Classically this number is known as the grade. 2 N1 = 1. Indeed X1 = P , as every line is rational. 5 Similarly every conic is rational, so X2 = P and N2 = 1.
Classical Enumerative Geometry and Quantum Cohomology – p.19 Now imposing the condition that a curve passes through a point is one linear condition. So we want to count the number of rational curves of degree d that pass through 3d − 1 points.
The Grade
Call this degree Nd. Classically this number is known as the grade. 2 N1 = 1. Indeed X1 = P , as every line is rational. 5 Similarly every conic is rational, so X2 = P and N2 = 1.
In general to calculate the degree of Xd we need to cut by hyperplanes. As the dimension of Xd is 3d − 1, we want to cut by 3d − 1 hyperplanes.
Classical Enumerative Geometry and Quantum Cohomology – p.19 The Grade
Call this degree Nd. Classically this number is known as the grade. 2 N1 = 1. Indeed X1 = P , as every line is rational. 5 Similarly every conic is rational, so X2 = P and N2 = 1.
In general to calculate the degree of Xd we need to cut by hyperplanes. As the dimension of Xd is 3d − 1, we want to cut by 3d − 1 hyperplanes. Now imposing the condition that a curve passes through a point is one linear condition. So we want to count the number of rational curves of degree d that pass through 3d − 1 points.
Classical Enumerative Geometry and Quantum Cohomology – p.19 Pick two auxiliary lines l1 and l2 and set p3d−1 = l1 ∩ l2.
Pick four points of Ct, p = p1, q = p2, r a point of l1 and s a point of l2.
Observe that Ct passes through iff r = s.
An argument due to Kontsevich
2 Fix 3d − 2 points p1, p2, . . . , p3d−2 in P . Then we get a 1-dimensional family of rational curves Ct of degree d, which contain these points.
Classical Enumerative Geometry and Quantum Cohomology – p.20 Pick four points of Ct, p = p1, q = p2, r a point of l1 and s a point of l2.
Observe that Ct passes through iff r = s.
An argument due to Kontsevich
2 Fix 3d − 2 points p1, p2, . . . , p3d−2 in P . Then we get a 1-dimensional family of rational curves Ct of degree d, which contain these points. Pick two auxiliary lines l1 and l2 and set p3d−1 = l1 ∩ l2.
Classical Enumerative Geometry and Quantum Cohomology – p.20 Observe that Ct passes through iff r = s.
An argument due to Kontsevich
2 Fix 3d − 2 points p1, p2, . . . , p3d−2 in P . Then we get a 1-dimensional family of rational curves Ct of degree d, which contain these points. Pick two auxiliary lines l1 and l2 and set p3d−1 = l1 ∩ l2.
Pick four points of Ct, p = p1, q = p2, r a point of l1 and s a point of l2.
Classical Enumerative Geometry and Quantum Cohomology – p.20 An argument due to Kontsevich
2 Fix 3d − 2 points p1, p2, . . . , p3d−2 in P . Then we get a 1-dimensional family of rational curves Ct of degree d, which contain these points. Pick two auxiliary lines l1 and l2 and set p3d−1 = l1 ∩ l2.
Pick four points of Ct, p = p1, q = p2, r a point of l1 and s a point of l2.
Observe that Ct passes through iff r = s.
Classical Enumerative Geometry and Quantum Cohomology – p.20 Picture
Here is a picture of what is going on back :
q=p2
s
p=p1
r
Ct
Classical Enumerative Geometry and Quantum Cohomology – p.21 Note that the cross-ratio is infinity if r = s. By the principle of continuity, the number of times the cross-ratio is zero, is equal to the number of times the cross-ratio is infinity. So, when is the cross-ratio zero, and when is it infinity? cross-ratio It is zero when p = s or r = q and it is infinity if r = s or p = q.
Keeping track of the cross-ratio
Note that there is a map B −→ P1 which assigns to a point t ∈ B, the cross-ratio of the four points p, q, r and s.
Classical Enumerative Geometry and Quantum Cohomology – p.22 By the principle of continuity, the number of times the cross-ratio is zero, is equal to the number of times the cross-ratio is infinity. So, when is the cross-ratio zero, and when is it infinity? cross-ratio It is zero when p = s or r = q and it is infinity if r = s or p = q.
Keeping track of the cross-ratio
Note that there is a map B −→ P1 which assigns to a point t ∈ B, the cross-ratio of the four points p, q, r and s. Note that the cross-ratio is infinity if r = s.
Classical Enumerative Geometry and Quantum Cohomology – p.22 So, when is the cross-ratio zero, and when is it infinity? cross-ratio It is zero when p = s or r = q and it is infinity if r = s or p = q.
Keeping track of the cross-ratio
Note that there is a map B −→ P1 which assigns to a point t ∈ B, the cross-ratio of the four points p, q, r and s. Note that the cross-ratio is infinity if r = s. By the principle of continuity, the number of times the cross-ratio is zero, is equal to the number of times the cross-ratio is infinity.
Classical Enumerative Geometry and Quantum Cohomology – p.22 It is zero when p = s or r = q and it is infinity if r = s or p = q.
Keeping track of the cross-ratio
Note that there is a map B −→ P1 which assigns to a point t ∈ B, the cross-ratio of the four points p, q, r and s. Note that the cross-ratio is infinity if r = s. By the principle of continuity, the number of times the cross-ratio is zero, is equal to the number of times the cross-ratio is infinity. So, when is the cross-ratio zero, and when is it infinity? cross-ratio
Classical Enumerative Geometry and Quantum Cohomology – p.22 Keeping track of the cross-ratio
Note that there is a map B −→ P1 which assigns to a point t ∈ B, the cross-ratio of the four points p, q, r and s. Note that the cross-ratio is infinity if r = s. By the principle of continuity, the number of times the cross-ratio is zero, is equal to the number of times the cross-ratio is infinity. So, when is the cross-ratio zero, and when is it infinity? cross-ratio It is zero when p = s or r = q and it is infinity if r = s or p = q.
Classical Enumerative Geometry and Quantum Cohomology – p.22 Picture of family Ct r p q
B
Classical Enumerative Geometry and Quantum Cohomology – p.23 But looking at this picture, picture , in fact it would seem this cannot occur (and nor can p = q). In fact what is happening, is that a copy of P1 is bubbling off. Ct is forced to break into two curves, one of degree d1 and d2, where d = d1 + d2.
Bubbling off
From the previous picture, it would seem that p = r and q = r can occur.
Classical Enumerative Geometry and Quantum Cohomology – p.24 In fact what is happening, is that a copy of P1 is bubbling off. Ct is forced to break into two curves, one of degree d1 and d2, where d = d1 + d2.
Bubbling off
From the previous picture, it would seem that p = r and q = r can occur. But looking at this picture, picture , in fact it would seem this cannot occur (and nor can p = q).
Classical Enumerative Geometry and Quantum Cohomology – p.24 Bubbling off
From the previous picture, it would seem that p = r and q = r can occur. But looking at this picture, picture , in fact it would seem this cannot occur (and nor can p = q). In fact what is happening, is that a copy of P1 is bubbling off. Ct is forced to break into two curves, one of degree d1 and d2, where d = d1 + d2.
Classical Enumerative Geometry and Quantum Cohomology – p.24 Singular fibre
r p s
p q q
Classical Enumerative Geometry and Quantum Cohomology – p.25 C1 passes through 3d1 − 1 points. Choose these points 3d1 − 1 points, from amongst 3d − 1 − 3 points. Choose which nodes to smooth. Choose two curves C1 and C2 through given points. Choose the points r and s.
Choices
d1 and d2.
Classical Enumerative Geometry and Quantum Cohomology – p.26 Choose which nodes to smooth. Choose two curves C1 and C2 through given points. Choose the points r and s.
Choices
d1 and d2. C1 passes through 3d1 − 1 points. Choose these points 3d1 − 1 points, from amongst 3d − 1 − 3 points.
Classical Enumerative Geometry and Quantum Cohomology – p.26 Choose two curves C1 and C2 through given points. Choose the points r and s.
Choices
d1 and d2. C1 passes through 3d1 − 1 points. Choose these points 3d1 − 1 points, from amongst 3d − 1 − 3 points. Choose which nodes to smooth.
Classical Enumerative Geometry and Quantum Cohomology – p.26 Choose the points r and s.
Choices
d1 and d2. C1 passes through 3d1 − 1 points. Choose these points 3d1 − 1 points, from amongst 3d − 1 − 3 points. Choose which nodes to smooth. Choose two curves C1 and C2 through given points.
Classical Enumerative Geometry and Quantum Cohomology – p.26 Choices
d1 and d2. C1 passes through 3d1 − 1 points. Choose these points 3d1 − 1 points, from amongst 3d − 1 − 3 points. Choose which nodes to smooth. Choose two curves C1 and C2 through given points. Choose the points r and s.
Classical Enumerative Geometry and Quantum Cohomology – p.26 Recursive Formula
Putting all this together we get
2 2 3d − 4 3 3d − 4 Nd = Nd1 Nd2 d1d2 − d1d2 , 3d1 − 1 3d1 − 2 d1X+d2=d
where N1 = 1 and N2 = 1. In fact N3 = 12, . . . .
Classical Enumerative Geometry and Quantum Cohomology – p.27 We have a new product, a deformation of the old product. Instead of counting number of intersection points, it counts the number of rational curves meeting given cycles (ie Gromov-Witten invariants).
Recursion formula corresponds to associativity of quantum product.
Quantum Cohomology
Set up a ring to count these objects, Z[x] QH∗(P2) = . hx3 = qi This is the (small) quantum cohomology ring.
Classical Enumerative Geometry and Quantum Cohomology – p.28 Recursion formula corresponds to associativity of quantum product.
Quantum Cohomology
Set up a ring to count these objects, Z[x] QH∗(P2) = . hx3 = qi This is the (small) quantum cohomology ring. We have a new product, a deformation of the old product. Instead of counting number of intersection points, it counts the number of rational curves meeting given cycles (ie Gromov-Witten invariants).
Classical Enumerative Geometry and Quantum Cohomology – p.28 Quantum Cohomology
Set up a ring to count these objects, Z[x] QH∗(P2) = . hx3 = qi This is the (small) quantum cohomology ring. We have a new product, a deformation of the old product. Instead of counting number of intersection points, it counts the number of rational curves meeting given cycles (ie Gromov-Witten invariants). Recursion formula corresponds to associativity of quantum product.
Classical Enumerative Geometry and Quantum Cohomology – p.28 Formula . In particular, we derive N2(0, 5, 0) = 3264, the correct answer to the question, how many conics tangent to five given lines?
Tangency condition
For example, set Nd(a, b, c) to be the number of curves of degree d through a general points, tangent to b lines, and tangent to c lines, at specified general points.
Classical Enumerative Geometry and Quantum Cohomology – p.29 In particular, we derive N2(0, 5, 0) = 3264, the correct answer to the question, how many conics tangent to five given lines?
Tangency condition
For example, set Nd(a, b, c) to be the number of curves of degree d through a general points, tangent to b lines, and tangent to c lines, at specified general points. Formula .
Classical Enumerative Geometry and Quantum Cohomology – p.29 Tangency condition
For example, set Nd(a, b, c) to be the number of curves of degree d through a general points, tangent to b lines, and tangent to c lines, at specified general points. Formula . In particular, we derive N2(0, 5, 0) = 3264, the correct answer to the question, how many conics tangent to five given lines?
Classical Enumerative Geometry and Quantum Cohomology – p.29 Theorem. (Xi Chen) Let S be a general K3 surface in n P , such that OS(1) is not a multiple of another line bundle. Then every rational curve which is a hyperplane section is nodal.
Further Work
Theorem. (Beauville, Yau-Zaslow) Let S be a general K3 surface in Pg. Then the number n(g) of rational curves on S which are hyperplane sections is equal to
∞ q n(g)qg = . q ∞ (1 − qn)24 Xg=1 n=1 Q provided every such curve is nodal.
Classical Enumerative Geometry and Quantum Cohomology – p.30 Further Work
Theorem. (Beauville, Yau-Zaslow) Let S be a general K3 surface in Pg. Then the number n(g) of rational curves on S which are hyperplane sections is equal to
∞ q n(g)qg = . q ∞ (1 − qn)24 Xg=1 n=1 Q provided every such curve is nodal. Theorem. (Xi Chen) Let S be a general K3 surface in n P , such that OS(1) is not a multiple of another line bundle. Then every rational curve which is a hyperplane section is nodal. Classical Enumerative Geometry and Quantum Cohomology – p.30