Intersection theory on moduli spaces of curves via hyperbolic geometry

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Intersection theory on moduli spaces of curves via hyperbolic geometry The University of Melbourne In the past few decades, moduli spaces of curves have become the centre of a rich confluence of rather disparate areas such as geometry, combinatorics, integrable systems and theoretical physics. Starting from baby principles, I will describe exactly what a moduli space is and motivate the study of its intersection theory. The talk will include a discussion of recent results from my PhD thesis, including a new proof of a formula of Kontsevich. 8 April 2008

What is a moduli space? A moduli space parametrises a family of geometric objects. Different points in a moduli space represent different geometric objects and nearby points represent objects with similar structure. Toy example: The moduli space of triangles Consider a triangle with side lengths a, b and c. M = { (a, b, c) R 3 + a + b > c, b + c > a, and c + a > b } y c x x b z a = y + z x = b+c a 2 b = z + x y = c+a b 2 c = x + y z = a+b c 2 y a z M = { (x, y, z) R 3 } +

What is a moduli space good for? Baby enumerative geometry question How many triangles with vertices labelled A, B and C are isosceles; have at least one side of length 5; and have at least one side of length 7? Define X iso M, the locus of isosceles triangles. Define X 5 M, the locus of triangles with a side of length 5. Define X 7 M, the locus of triangles with a side of length 7. Same question What is X iso X 5 X 7?

Intuitive intersection theory (a.k.a. cohomology) An (N d)-dimensional subset of an N-dimensional space is said to have codimension d. A generic intersection between subsets with codimension d 1 and codimension d 2 has codimension d 1 + d 2. A generic intersection between m subsets of an N-dimensional space with codimensions d 1 + d 2 + + d m = N is a set of points. The number of these points is called an intersection number. In order to obtain a well-defined intersection number, it is necessary to jiggle the picture, live in a compact space and count with signs. We will use the following (non-standard) notation for intersection numbers. { X1 X X 1 X 2 X m = 2... X m if finite 0 otherwise

The geometry of surfaces TOPOLOGY Topologists classified (compact, connected, orientable) surfaces by genus. genus = 0 genus = 1 genus = 2 GEOMETRY What does a geometer do with a surface? It depends... algebraic geometry complex analysis hyperbolic geometry algebraic structure complex structure hyperbolic metric up to isomorphism up to biholomorphism up to isometry algebraic curve Riemann surface hyperbolic surface... but actually it doesn t, since these objects are all the same!

Moduli spaces of curves M g,n can be defined as the moduli space of genus g smooth algebraic curves with n marked points; genus g Riemann surfaces with n punctures; or genus g hyperbolic surfaces with n cusps. The marked points or punctures or cusps are labelled from 1 up to n. Three technical problems Problem: M g,n does not always exist Solution: Do not allow (g, n) = (0, 0), (0, 1), (0, 2) or (1, 0) Problem: Solution: Problem: Solution: M g,n is not compact Use the Deligne Mumford compactification M g,n Points in M g,n correspond to stable algebraic curves M g,n is not a manifold Treat M g,n like an orbifold Intersection numbers may be rational

Why do we care about moduli spaces of curves? Algebraic Geometry Combinatorics Hyperbolic Geometry Topology MODULI SPACES String Theory Matrix Models Integrable Systems

Why do we care about moduli spaces of curves? Algebraic Geometry Combinatorics Hyperbolic Geometry Topology MODULI SPACES String Theory Matrix Models Integrable Systems BECAUSE THEY ARE INTERESTING AND FUN!

Psi-classes Important fact The dimension of M g,n is 6g 6 + 2n. The psi-classes ψ 1, ψ 2,..., ψ n are codimension 2 subsets of M g,n. In the more technical language of cohomology, ψ 1, ψ 2,..., ψ n H 2 (M g,n, Q). Choose non-negative integers a 1 + a 2 + + a n = 3g 3 + n and consider the intersection number of psi-classes ψ a1 1 ψa2 2 ψan n Q. There are various other important subsets of M g,n such as κ 1. Examples of psi-class intersection numbers On M 0,5, the intersection number ψ 1 ψ 2 is 2. On M 1,1, the intersection number ψ 1 is 1 24.

Constructing the psi-classes M g,n+1 M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 π M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 π M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 dim D k = 6g 6 + 2n codim D k = 2 in M g,n+1 π M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 dim D k = 6g 6 + 2n codim D k = 2 in M g,n+1 dim E k = 6g 6 + 2n codim E k = 2 in M g,n+1 π M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 π dim D k = 6g 6 + 2n codim D k = 2 in M g,n+1 dim E k = 6g 6 + 2n codim E k = 2 in M g,n+1 codim D k E k = 4 in M g,n+1 codim D k E k = 2 in D k M g,n

Constructing the psi-classes π : M g,n+1 M g,n is known as the forgetful map M g,n+1 π dim D k = 6g 6 + 2n codim D k = 2 in M g,n+1 dim E k = 6g 6 + 2n codim E k = 2 in M g,n+1 codim D k E k = 4 in M g,n+1 codim D k E k = 2 in D k codim ψ k = 2 in M g,n ψ k M g,n

A brief history of Witten s conjecture Big question How do you calculate intersection numbers of psi-classes? Witten (1991): I conjecture that if we put all of the intersection numbers of psi-classes into a generating function F, then F satisfies the infinite sequence of partial differential equations known as the KdV hierarchy. Kontsevich (1992): Witten is right! I have a formula which relates intersection numbers with ribbon graphs. Okounkov and Pandharipande (2001): Witten is right! We have a formula which relates intersection numbers with Hurwitz numbers. Mirzakhani (2004): Witten is right! I have a formula which relates intersection numbers with volumes of moduli spaces.

Volumes of moduli spaces Let M g,n (L 1, L 2,..., L n ) be the moduli space of genus g hyperbolic surfaces with n boundary components of lengths L 1, L 2,..., L n. The moduli space M g,n (L 1, L 2,..., L n ) has a natural symplectic structure so one can measure its volume. Let V g,n (L 1, L 2,..., L n ) be the volume of M g,n (L 1, L 2,..., L n ). Mirzakhani s recursion The volumes V g,n (L 1, L 2,..., L n ) satisfy the following recursive formula. 2 Z Z L 1V g,n(l 1,..., L n) = xy H(x + y, L 1) V g 1,n+1(x, y, L 2,..., L n) dx dy L 1 0 0 + X Z Z xy H(x + y, L 1) V g1, I 1 +1(x, L I1 ) V g2, I 2 +1(y, L I2 ) dx dy g 1 +g 2 =g I 1 I 2 =[2,n] 0 0 nx Z + x[h(x, L 1 + L k ) + H(x, L 1 L k )] V g,n 1(x, L 2,..., b L k,..., L n) dx k=2 0 One corollary of this formula is that V g,n (L 1, L 2,..., L n ) is a polynomial.

From volumes to intersection numbers Mirzakhani s theorem The volume V g,n (L 1, L 2,..., L n ) is given by the following formula. a +m=3g 3+n (2π 2 ) m ψ a1 1 ψa2 2 ψan n κ m 1 2 a a 1!a 2!... a n!m! L 2a1 1 L2a2 2... L 2an n. Mirzakhani s recursion lets you calculate V g,n (L 1, L 2,..., L n ). Mirzakhani s theorem says that V g,n (L 1, L 2,..., L n ) is a polynomial whose coefficients store intersection numbers on M g,n. The psi-class intersection numbers are stored in the top degree. Mirzakhani s recursion+mirzakhani s theorem = Witten s conjecture Philosophy: Any meaningful statement about V g,n gives a meaningful statement about intersection numbers on M g,n.

Examples of volume polynomials V 0,3 = 1 V 0,4 = 1 2 (L2 1 + L2 2 + L2 3 + L2 4 + 4π2 ) V 0,5 = 1 8 L 4 i + 1 2 L 2 i L 2 j + 3π 2 L 2 i + 10π 4 V 1,1 = 1 48 (L2 1 + 4π2 ) V 1,2 = 1 192 L4 1 + π2 12 L2 1 + π4 4 + π2 12 L2 2 + 1 192 L4 2 + 1 96 L2 1 L2 2 V 2,1 = 139π4 23040 L4 1 + 169π6 2880 L2 1 + 29π8 192 + 29π2 138240 L6 1 + 1 442368 L8 1

New volume polynomial relations Generalised string equation, generalised dilaton equation and more The volume polynomials V g,n+1 and V g,n satisfy the following relations. V g,n+1 (L 1,..., L n, 2πi) = V g,n+1 n L k V g,n dl k k=1 (GSE) (L 1,..., L n, 2πi) = 2πi(2g 2 + n)v g,n (GDE) L n+1 2 V g,n+1 n V g,n L 2 (L 1,..., L n, 2πi) = L k (4g 4 + n)v g,n n+1 L k. k=1 k V g,n+1 L k (L 1,..., L n, 2πi) = [??? ] V g,n n+1

Three proofs Algebraic geometry Mirzakhani s theorem translates these results into relations between intersection numbers on M g,n+1 and M g,n. Such relations emerge from analysing the forgetful map π : M g,n+1 M g,n. Mirzakhani s recursion Mirzakhani s recursion determines all volumes V g,n (L 1, L 2,..., L n ), so it should encode these relations in some sense. Interestingly, these proofs use identities among the Bernoulli numbers. Hyperbolic geometry? A purely imaginary length usually corresponds to an angle. So these results suggest a connection between intersection numbers on M g,n and the geometry of hyperbolic surfaces with cone points.

What is a ribbon graph? A ribbon graph of type (g, n) is a graph with a cyclic ordering of the edges at every vertex which can be thickened to give a surface of genus g and n boundary components labelled from 1 up to n. Trivalent ribbon graph of type (1, 1) 1 1 Trivalent ribbon graphs of type (0, 3) 2 3 1 3 1 2 1 2 3 1 2 3

Kontsevich s combinatorial formula explained Kontsevich s combinatorial formula X ny ψ a 1 1 ψa 2 2 ψan n a =3g 3+n k=1 (2a k 1)!! s 2a k +1 k = X Y 22g 2+n Aut(Γ) Γ TRG g,n e Γ 1 s l(e) + s r(e) LHS: polynomial in 1 s 1, 1 s 2,..., 1 s n coefficients store all intersection numbers of psi-classes on M g,n RHS: rational function in s 1, s 2,..., s n strange enumeration over trivalent ribbon graphs of type (g, n) Kontsevich s combinatorial formula is incredible!

Kontsevich s combinatorial formula for g = 0 and n = 3 The LHS is easy. LHS = ψ 0 1 ψ 0 2 ψ 0 3 1 s 1 s 2 s 3 The RHS has one term for each trivalent ribbon graph of type (0, 3). RHS = 2 2s 1 (s 1 + s 2 )(s 1 + s 3 ) + 2 2s 2 (s 2 + s 3 )(s 2 + s 1 ) 2 + 2s 3 (s 3 + s 1 )(s 3 + s 2 ) + 2 (s 1 + s 2 )(s 2 + s 3 )(s 3 + s 1 ) = s 2s 3 (s 2 + s 3 ) + s 3 s 1 (s 3 + s 1 ) + s 1 s 2 (s 1 + s 2 ) + 2s 1 s 2 s 3 s 1 s 2 s 3 (s 1 + s 2 )(s 2 + s 3 )(s 3 + s 1 ) (s 1 + s 2 )(s 2 + s 3 )(s 3 + s 1 ) = s 1 s 2 s 3 (s 1 + s 2 )(s 2 + s 3 )(s 3 + s 1 ) 1 = s 1 s 2 s 3 Conclusion: ψ 0 1 ψ0 2 ψ0 3 = 1.

A new approach to Kontsevich s combinatorial formula Kontsevich s combinatorial formula simplified INTERSECTION NUMBERS OF PSI-CLASSES = RIBBON GRAPHS Sketch proof: Step 1 of 3 INTERSECTION NUMBERS OF PSI-CLASSES = ASYMPTOTICS OF VOLUMES OF MODULI SPACES Recall that the intersection numbers of psi-classes are stored in the top degree coefficients of V g,n (L 1, L 2,..., L n ). You can access the top degree coefficients of a polynomial using asymptotics. Sketch proof: Step 2 of 3 ASYMPTOTICS OF VOLUMES OF MODULI SPACES = HYPERBOLIC SURFACES WITH VERY LONG BOUNDARIES To understand V g,n(nl 1, Nl 2,..., Nl n) for large N, one must understand hyperbolic surfaces with boundaries of lengths Nl 1, Nl 2,..., Nl n for large N.

A new approach to Kontsevich s combinatorial formula Sketch proof: Step 3 of 3 HYPERBOLIC SURFACES WITH VERY LONG BOUNDARIES = RIBBON GRAPHS Fact: The Gauss Bonnet theorem implies that all hyperbolic surfaces of genus g with n boundary components have the same surface area. Crucial geometric reasoning: If you take a hyperbolic surface and stretch its boundary lengths to infinity, then you will obtain a ribbon graph after rescaling.

A problem about ribbon graphs and determinants Ribbon graphs and determinants Consider a trivalent ribbon graph of type (g, n). Colour n of the edges blue and the remaining edges red. Let A be the matrix formed from the adjacency between the blue edges and the boundaries. Let B be the matrix formed from the oriented adjacency between the red edges. Then det B = 2 2g 2 (det A) 2.

A problem about ribbon graphs and determinants Ribbon graphs and determinants Consider a trivalent ribbon graph of type (g, n). Colour n of the edges blue and the remaining edges red. Let A be the matrix formed from the adjacency between the blue edges and the boundaries. Let B be the matrix formed from the oriented adjacency between the red edges. Then det B = 2 2g 2 (det A) 2. The following is a ribbon graph with g = 0 and n = 26.

A problem about ribbon graphs and determinants Ribbon graphs and determinants Consider a trivalent ribbon graph of type (g, n). Colour n of the edges blue and the remaining edges red. Let A be the matrix formed from the adjacency between the blue edges and the boundaries. Let B be the matrix formed from the oriented adjacency between the red edges. Then det B = 2 2g 2 (det A) 2. The following is a ribbon graph with g = 0 and n = 26. det B = 2 2g 2 (det A) 2 256 = 1 4 32 2

End matter Slides http://www.ms.unimelb.edu.au/ norm Article A tourist s guide to intersection theory on moduli spaces of curves To appear in the Australian Mathematical Society Gazette Volume 35, No. 2 (May 2008) or No. 3 (July 2008) Seminar Geometry and Topology seminar on 6 May 2008 Thesis Coming soon! Thanks to my supervisors: Paul Norbury, Craig Hodgson and Iain Aitchison.