The Euclidean Volume of the Moduli Space. Kevin M. Chapman

Transcription

1 The Euclidean Volume of the Moduli Space Kevin M Chapman Abstract A topological recursion formula is developed for the Euclidean volume of the Moduli Space of curves with n marked points, proved by the computation of the inverse Laplace Transform of another formula This is the only known proof for the volume recursion formula without reference to the symplectic volume recursion formula In particular, there are no straightforward geometric arguments that lead to a proof, another indication of the effectiveness of the method developed in the current paper Contents Introduction Moduli Space and Ribbon Graphs 3 Euclidean Volume and Lattice Point Count Functions 4 4 The Laplace Transform of the Lattice Point Count Functions 7 5 The Laplace Transform of the Euclidean Volume 6 Examples 5 7 Acknowledgements 6 References 6 Introduction The goal of this paper is to derive a topological recursion formula for the Euclidean volume of the Moduli space of curves with n marked points A topological recursion formula is used for computing a function f g,n+ p,, p n+ ) from previous functions f g,n, with g + n + ) > g + n > This has effectively been done already, because the symplectic and Euclidean volumes satisfy vg,np) S 5g 5+n vg,np) E from, and there is a topological recursion formula for the symplectic volume from ; substituting would give the desired recursion formula for the Euclidean volume This paper will derive the recursion formula for the Euclidean volume without appeal to the symplectic recursion formula Why do the Euclidean and symplectic volume of the Moduli space of curves matter? The fact that the ratio of the two is a constant ignoring g and n) is used in Kontsevich s proof of the Witten conjecture, a statement that two different models of two dimensional quantum gravity give the same results The methods in and provide an elementary way to compute the ratio of these two functions, so it is worth understanding For more motivation, including the relation with Eynard-Orantin theory, see The strategy to achieve the goal uses the relation between lattice point counting quasipolynomials and Euclidean volume derived in This relation is between the Laplace transform of the lattice point count functions and the Laplace transform of the Euclidean volume, and it gives rise to a topological recursion formula for the Laplace transform M Chapman, Originally published in Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) The Regents of the University of California

2 the Euclidean volume From here, the inverse Laplace transform is taken to obtain the topological recursion formula for the Euclidean volume Moduli Space and Ribbon Graphs The Moduli space of Riemann surfaces is defined, as a set, as all complex structures on Σ g, where Σ g is a Riemann surface of genus g, modulo biholomorphic equivalence Consider a slightly different object, the Moduli space with n marked points, M g,n, It is defined as the set of all complex structures, or complex coordinate systems, on a Riemann surface of genus g, Σ g, with n marked points, modulo an equivalance relation Two complex structures on Σ g are equivalent if there is a biholomorphic mapping h h and h are holomorphic) between the two coordinate systems As an example, consider M,3 This is the set of all complex structures on the Riemann sphere with three fixed points Any complex structure on the Riemann sphere must be diffeomorphic to C { }, so assume that the complex structure is of the form C { } First, pick three distinct points on the sphere, a, a, a, considered as points in C { } Suppose a i for each i Define h by hx) a a a a x a x a Then the function h maps a to, a to, and a to If one of the a i, then without loss of generality, let a, and define g by gx) x a a a Then g maps a to, a to, and a to Since h and g are biholomorphic, any complex structure on the sphere with three distinct marked points can be brought to another complex structure with the marked points {,, } Therefore, M,3 {one single point}, since any complex structure on the Riemann sphere with three arbitrary marked points can be brought to another complex structure with {,, } as the marked points Consider now M,4 and start with four points, a, a, a, a 3 Map three of them to {,, }, using the map h After mapping by h, there is one point left, b The question now becomes, is there some biholomorphic map, fixing {,, } pointwise, that could map any arbitrary b C \ {, } to some fixed point p C \ {, }? The answer is no; the only biholomorphic function mapping the Riemann sphere to itself and fixing,, and is the identity map To show this, singularities and Laurent series are considered Let f be a biholomorphic mapping of C { } into itself, fixing,, and pointwise Then f restricted to C is entire, so f has a power series expansion fx) a i x i, i valid for all x C Define gx) fx ) for x, Then g has Laurent series expansion gx) a i x i, valid for all x, Since Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) i lim gx) lim x x fx ) lim fx), x

3 g has an isolated singularity, in this instance a pole, at x A result from complex analysis states that if g has an isolated pole at x and g is holomorphic in a punctured neighborhood of, then the principal part of its Laurent series expansion around has only finite terms; in other words, there is a N such that n > N a n Therefore, the original function f was a polynomial Since f is biholomorphic, and since only f), it must be that fx) x n, because the only root of f is at x and because f) But this function is only biholomorphic when n, since an nth root of unity ξ n for n gives fξ n ) Therefore, f is the identity Thus, M,4 is, as a set, C \ {, } General g and n cases are difficult to deal with What kinds of spaces are they? To help our understanding, there is a theorem by 3, 5, 7 which relates the moduli space with n marked points to metric ribbon graphs The isomorphism is an orbifold isomorphism For more information, refer to 4 Theorem M g,n R n + RG g,n Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) RG g,n is a space of all metric ribbon graphs of type g, n), defined as Γ ribbon graph of type g,n) R eγ) + AutΓ), where eγ) is the number of edges of a ribbon graph Γ, and AutΓ) is the automorphism group of Γ, which is finite for g + n > The automorphism group of Γ can be considered as the set of all homeomorphisms ψ: Σ g Σ g which fix the set of vertices and edges V and E of Γ setwise, modulo the equivalence relation ψ φ if ψ and φ agree on V and E A ribbon graph of type g, n), Γ, is the corresponding graph to the -skeleton of a cell decomposition of a genus g surface with n -cells, satisfying the following equation: v e+n g, where v and e are the number of vertices and edges of Γ, respectively The equality arises because the ribbon graphs are cell decompositions of the Riemann surface, and so it is possible to compute the Euler characteristic of the surface in two different ways Every vertex of the graph also is required to have valency greater than or equal to three As an example, a figure eight is a ribbon graph of type, 3), because it has two edges and one vertex, so ) + n, and so n 3 The dumbbell shape is also a ribbon graph of type, 3), because it has three edges and two vertices, and by the previous computation, this means n 3 What is the real dimension of RG g,n? It is determined by the maximum number of edges a ribbon graph can have This is the case when every vertex has valency three the number of half-edges coming into the vertex is three) This is because if there is a vertex of valency four or more, an edge can be added to create two new vertices, both of which have less valency As an example, consider the figure eight; it has a vertex of valency four Adding an edge creates the dumbbell shape, which makes two vertices of valency three But no more edges can be added to the dumbbell, since then some vertex would have valency less than three Since the maximum number of edges occur in a trivalent ribbon graph, it is enough to know eγ) for Γ trivalent In this case, every vertex has valency three, and every edge touches two possibly nondistinct) vertices, so 3v e Plugging into the equation v e+n g gives e 3e + 3n 6 6g, or that e 6g 6 + 3n and v 4g 4 + n For the case, 3), e 3 and v, which is consistent with previous information For the general case, this gives that the real dimension of RG g,n is 6g 6 + 3n Due to theorem, this implies that 3 3

4 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) 4 Figure The dumbbell, the figure eight, and a double bubble shape, all ribbon graphs of type, 3) These are the only ribbon graphs of type, 3) the Moduli space with n marked points has real dimension 6g 6 + n although this is known through other methods) For the case, 3), the Moduli space is zero dimensional, which is consistent with what was shown earlier 3 Euclidean Volume and Lattice Point Count Functions A metric ribbon graph is a ribbon graph which has a positive real number assigned to every edge By theorem, M g,n R n + RG g,n There is a map π : RG g,n R n +, which is a projection map This map takes a metric ribbon graph of type g, n) and gives an n dimensional real vector, with entries given by the boundary lengths of the faces of the ribbon graph As an example, consider the double bubble shaped ribbon graph of type, 3) in figure 3 b 3 b 5 b 7 Figure 3 The edges have length, 5, and 7, and each face has a label, b, b, and b 3 Each face has boundary length 35, 85, and 9, respectively Therefore, π would take this metric ribbon graph to the vector 35, 85, 9) 4

5 A natural question to ask is, what is π 35, 85, 9)? That is, what are the metric ribbon graphs of type, 3) that are taken to 35, 85, 9) when π is applied? The double bubble described above is taken to 35, 85, 9), but what about other graphs? For the figure eight, label the interior faces b and b and the external face b 3, with edge lengths e and e ; then the length of b is e, the length of b is e, and the length of b e is e + e Consider the system of equations e 35, e 85 and e + e 9, which has no solution Thus, no metric ribbon graph in the shape of a figure eight is taken to 35, 85, 9) If the metric ribbon graph is of the form of a dumbbell, the following equations are appropriate: e 35, e 85, and e + e + e 3 9 But all the edge lengths are positive, so again there is no solution Therefore, there is only one metric ribbon graph that is taken to 35, 85, 9) Hence M,3 π 35, 85, 9), because M,3 is just one point This kind of result is true in general; π p) M g,n for p R n + For a specific ribbon graph Γ of type g, n), label all of the edges and faces Let A Γ denote the edge-face incidence matrix of the matrix That is, A Γ is a n eγ) dimensional matrix, with entries a ij which count how many times an edge j appears in a face i Since an edge can appear no times, one time, or twice in a face, it follows that a ij is zero, one, or two As an example, consider the double bubble ribbon graph as shown in figure Let e, e, e 3 denote the edges of length, 5, and 7, respectively Then the incidence matrix is A Γ The incidence matrix will map the vector with the edge lengths to a vector with the lengths of the boundaries of the faces For Γ of type g, n), call P Γ p) A Γ p), p Rn +, where A Γ is considered as a map from R eγ) + to R n + The Euclidean volume of the polytope P Γ p), denoted by volp Γ p)), is defined by the pushforward measure of A Γ as A Γ) µ) ψ, where µ is the Lebesgue p measure on R eγ) and ψ is the Lebesgue measure on R n The pushforward measure is denoted by A Γ ) µ) and defined by A Γ ) µ)b) µa Γ B)) for all measurable B Rn This definition is equivalent to imposing that, for all measurable B, 3) volp Γ p))dψ, B Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) dµ A Γ B) which is the method used to compute the first few volumes of polytopes For the double bubble ribbon graph, A Γ was computed In that case, ψ µ, and integration over A Γ B) is simply µb), since A Γ has determinant two Thus the volume function is simply, since the relation must hold for all open sets B Note that in the double bubble case, there is an implicit assumption that, for p p, p, p 3 ), he inequalities p +p p 3, p +p 3 p, p 3 +p p do not hold, so that the volume function is really when those conditions are not satisfied, and otherwise In the dumbbell case, the volume function is when one of p + p < p 3, p + p 3 < p, or p 3 + p < p hold, and in the figure eight case, one of p + p p 3, p + p 3 p, or p 3 + p p holds Note that for any measurable B R 3 satisfying p + p p 3 for all p B is two-dimensional, so that volp Γ p)) is not well-defined by equation 3 in this case In general, this problem of being unable to compute certain volumes only occurs when eγ) < n Since the smallest number of vertices possible is, e + n g, or that the smallest eγ) can be is g + n Thus, this problem only occurs when g To solve 5 5

6 the problem, the volume when the integration is over the n dimensional space is just defined to be With the Euclidean volumes of the polytopes, it is possible to define the Euclidean volume of the Moduli space to be, for g + n >, 3) v E g,np) Γ trivalent ribbon graph of type g,n) volp Γ p)) AutΓ) vg,n E is a polynomial, a fact which will be proven later By the previous computations, v,3 E p) Also, ve, p) 96 p To show this, first note that there is only one trivalent ribbon graph of type, ) with automorphism group Z 6 Since there is only one face and three edges, A Γ Let b and consider now A Γ b) This is the set of all x, y, z) such that x + y + z) b, with x, y, z Now, A Γ, q) is the union of all A Γ b) with b, q This union is a tetrahedron with vertices,, ), q,, ),, q, ),,, q ) If V is the volume of this tetrahedron, then q volp Γp))dp V But the volume of the tetrahedron is q )3 3! Differentiating both sides shows that volp Γ q)) q 6 This gives that ve p, p) 96 Consider now metric ribbon graphs with integer edge lengths It is natural to consider the discrete analogue of the volume, the lattice point count functions They are defined by, for p Z n + and g + n >, 33) N g,n p) Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) Γ of type g,n) #{x Z eγ) + A Γx p} AutΓ) These functions were first considered by Norbury in 6 They have the property that, for p p,, p n ), if p + + p n mod, then N g,n p) This is because every edge appears uniquely in two faces sometimes in the same face two times), and therefore for e,, e m the edge lengths in a metric ribbon graph, e + + e m ) p + + p n It is easy to check that N,3 p) if p + p + p 3 is even and is otherwise and that N, p) 48 p when p is even The lattice point count functions satisfy the following topological recursion relation, which was first proven in : Theorem 3 The lattice point count functions satisfy the following recursion formula: 6 34) p N g,n p n ) n p +p j q qp + p j q)n g,n q, p n\{,j} ) p p j + Hp p j ) q q qp p j q)n g,n q, p n\{,j} ) p j p + Hp j p ) qp j p q)n g,n q, p n\{,j} ) + q q p q q ) N g,n+ q, q, p n\{} )+ q +q p 6

7 stable g +g g I Jn\{} N g, I +q, p I )N g, J +q, p J ) where n {,,, n}, p n p, p,, p n ), the stability condition is g + I > and g + J > and where { x > Hx) x is the Heaviside step function The equation is proven by consideration of edge removal of metric ribbon graphs with cilium, the details of which can be found in There is a relation between the lattice point count and Euclidean volume, satisfied for all bounded and closed B R n + and for continuous functions f from R n into R, 35) lim k p B k Zn + N g,n kp) fp) k3g +n) B v E g,np)fp)dp dp dp n This relation holds because of the definition of the Euclidean volume in terms of the pushforward measure Because of this relation between the lattice point count functions and the Euclidean volume, it is natural to expect a recursion formula for the Euclidean volume functions If one naively replaces the lattice point count functions with the Euclidean volume and substitutes appropriate summations with integrations, one obtains a false recursion formula The naive formula has, instead of factors of 4, factors of Theorem 3 The Euclidean volume satisfies the following recursion relation: p vg,np E n ) n p +p j 36) qp + p i q)v E 4 g,n q, p n\{,i} )dq i Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) p p j + Hp p j ) qp p j q)vg,n q, E p n\{,i} )dq ) pj p Hp j p ) qp j p q)vg,n q, E p n\{,i} )dq + q q p q q ) v 4 g,n+q E, q, p n\{} ) q +q p + stable g +g g I Jn\{} v E g,n q, p I )v E g,n q, p J ), ) 7 dq dq The initial conditions v,3 E p) and ve, p) 96 p completely determine vg,n E for g + n > It is obvious that vg,n E is a polynomial, because the integration of a polynomial over each region is itself a polynomial 4 The Laplace Transform of the Lattice Point Count Functions Because N g,n is defined only over Z n +, while the Euclidean volume is defined over R n +, it is natural to consider the Laplace transform of N g,n and v E g,n, so as to extend the functions 7

8 to the same domain for comparison For any complex w n C n, with Rew j ) > for each j n, define the Laplace transform of N g,n by 4) L g,n w n ) N g,n p)e p,w n, p Z n + where p, w n p w + + p n w n, and define V E 4) V E g,nt n ) where the variable substitution t t n R n + g,n by 43) e w j t j + t j is used Define v E g,np)e p,w n dp n ), 44) L g,n t n ) t n t n L g,n wt n )), Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) where wt n ) w t ),, w n t n )) The variable substitution turns L g,n into a Laurent polynomial in the t j variables and V E g,n into a polynomial in t j, as will be shown From here, the Laplace transform of the lattice point count recursion formula is taken to obtain a new recursion formula Theorem 4 The Laplace transform of N g,n satisfies the following recursion formula: 8 45) L g,n t n ) n t j 6 t j t t j 3 t )3 t t )3 t L g,n+t, t, t n\{} ) + L g,n t n\{j} ) t j ) )3 t L g,n t n\{} ) j stable g +g g I Jn\{} L g, I +t, t I )L g, J +t, t J ) Proof : Multiply each side of equation 34) by p n\{} and take the Laplace transform The left hand side of the equation becomes L g,n w w n ) L g,n w n w w n ), which n will now be denoted by L g,n w n ) Consider for now the first line, which becomes n p +p j p Z n q p j p + p j q)qp p j p n )N g,n q, p n\{,j} )e p,w n q, p n\{,j} ), where p j indicates the absence of p j Summation over nonnegative p is allowed and changes nothing, since multiplication by p i for each i was performed This can be rewritten as n qp p j p n N g,n q, p n\{,j} ) e p n\{,j},w n\{,j} e qw q p n\{,j} Z n 8

9 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) 9 l le lw where p + p j q l The first summation q+l p j q+l p j p j e p jw w j ), p j e p jw w j ) is just e qw le lw e wj e wj e +q+l)w wj) w e w j e w, and is equal to e qw le lw e w e w j ) e w +w j + q + l)e w e q+l)w w j ) + q + l)e w +w j e q+l)w w j ), which is equal to l + q + l)e w +w j e qw j e lw j e w e w j ) e w +w j e qw le lw + lq + l)e w e qw j e lw j Summation over l and q is performed, which gives n e w +w j e w e w j ) Lg,n w n\{j} ) e w e w ) ) L g,n w n\{} ) e w j e w j) Consider now the second line of equation 34): ew e w e w j Lg,n w n\{} ) w j e w j e w j) n p Z n q p p j Hp p j ) e qw q p p j q p j qp p j p n N g,n q, p n\{,j} ) e p,w p n\{,j} Z n n l le lw p j e p jw +w j ) p j where p p j q l is used For the third line, qp p j p n N g,n q, p n\{,j} ) e p n\{,j},w n\{,j} n e w +w j Lg,n w n\{j} ) e w +w j ) e w e w ), 9

10 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) n p Z n p j p Hp j p ) n p n\{,j} Z n n q q l p p j p q p j qp p j p n N g,n q, p n\{,j} ) e p,w p + q + l)le p w +w j ) e lw j e qw j qp p j p n N g,n q, p n\{,j} ) e p n\{,j},w n\{,j} e w +w j Lg,n w n\{} ) e w +w j ) e w j e w j) + where the substitution p j p q l is used Summing the first three lines gives n e w +w j e w e w j ) n Lg,n w n\{j} ) e w e w ) e w e w e w j + n e w +w j Lg,n w n\{} ) e w +w j ) n w j e w n ) L g,n w n\{} ) e w j e w j) e w e w e w j Lg,n w n\{} ) w j e w j e w j) e w +w j Lg,n w n\{j} ) e w +w j ) e w e w ) e w j e w j) + e w e w j ew+wj e w +w j For the fourth line, first note that, for any f, p n q +q p q q p q q )e pw fq, q ) where p q q l, and q q l fw, w ) e w e w e w j ) Lg,n w n\{j} ) e w e w ) le lw e q +q )w q q fq, q ) q q The Laplace transform of the fourth line, then, is Lg,n w n\{} ) w j e w j e w j), Lg,n w n\{} ) w j e w j e w j) q q fq, q )e q w +q w ) ) L g,n w n\{} ) e w j e w j) e w e w ) fw, w ),

11 e w e w ) L g,n+ w, w, w n\{} ) + Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) stable g +g g I Jn\{} L g, I +w, w I ) L g, J +w, w J ) The relations e w j e w j) t j ) e w 6 t, j e w e w j ew+wj e w +w j t jt ) t, t j and L g,n w t),, w n t) ) ) n n L g,n t,, t n )t ) t n ) hold, when changing from w n to t n using formula 43 By replacing each expression in w by one in t, formula 45 is obtained Now, notice that L g,n t n ) is a Laurent polynomial in the t j variables This can be proven by induction over g + n, using the base case g, n), 3) or, ) The initial values are L,3 t 3 ) ) and 6 t L, t) t ) 3 t t 3 8 t 4 The proof that it is a Laurent polynomial is as follows: check that each term being summed over in formula 43 is a Laurent polynomial The only potential difficulty arises in checking for the term t j t j t t )3 t j t L g,n t n\{j} ) t j )3 t L g,n t n\{} ) j First, denote L g,n t n\{j} ) by ft ) and L g,n t n\{} ) by ft j ) Then the expression being considered is t j t j t t )3 t j t ft ) t j )3 t ft j ) j By induction hypothesis, ft ) and ft j ) are Laurent polynomials in the t i variables Therefore, ft ) ft j ) is a Laurent polynomial in the t i variables As t t j, the expression tends to zero Thus, it is divisible by t t j, and the quotient is a Laurent polynomial in the t i variables For P a Laurent polynomial in the t i variables, t j t j P t, t j ) is still a Laurent polynomial in the t i variables Thus L g,n t n ) is a Laurent polynomial in the t i variables Since L g,n t n ) is a Laurent polynomial, the degree is well-defined; denote the top degree terms by L g,n t n ) It is not difficult to show again by induction) that the following formula holds Theorem 4 The top degree terms of the Laplace transform of N g,n satisfies the following recursion formula: 46) L g,n t n ) 6 n t j t j t t j 3 t4 L g,n+t, t, t n\{} ) + t 4 L g,n t n\{j} ) t 4 jl g,n t n\{} ) ) stable g +g g I Jn\{} L g, I +t, t I )L g, J +t, t J ) From here, it is easy to see that the degree of L g,n t n ) is 3g 3 + n)

12 5 The Laplace Transform of the Euclidean Volume Due to the relation of the Euclidean volume and the lattice point count functions, given by equation 35, L g,n t n ) Vg,nt E n ) Proof : From 35), vg,np)e E w,p dp dp n lim N g,n kp)e w,p k k 3g +n) 5) R n + lim k p Z n + p k Zn + N g,n p)e k w,p k 3g +n) lim L w g,n k k,, w n k The coordinate transformation, given by 43), has the expansion near w Since t tw) w w 6 + w3 36 w5 5 +, w wt) t 3t 3 5t 5 n L g,n t n ) L g,n wt ),, wt n ) ) t t n is a Laurent polynomial of degree 3g 3 + n), and since w w/k makes ) t k t + O k for a fixed t, n t t n R n + This result then gives v E g,np)e wt),p dp dp n lim k lim k L g,n n wt ) L g,n t t n k,, wt ) n) k ) )) kt + O,, kt n + O k k Theorem 5 V E g,n satisfies the following recursion formula: Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) k 3g +n) ) k 3g +n) k n k 3g +n) V E g,nt n ) 5) V E g,nt n ) 6 n t j t j t t j 3 t4 V g,n+t E, t, t n\{} ) + t 4 V E g,n t n\{j} ) t 4 jv E g,n t n\{} ) ) stable g +g g I Jn\{} g, I + t, t I )L g, J +t, t J ) V E

13 At this point, the Laplace transform of equation 36) can be computed after multiplying both sides by p p n ), and since every step is invertible, the previous recursion formula proves that formula 36) holds To compute the Laplace transform of 36), first consider the symmetric function V g,nw E n ) defined by the Laplace transform V E g,nw,, w n ) It satisfies the topological recursion E 53) V g,n w n ) w w j w w j w w j V E g,n+w, w, w n\{} ) + Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) w n R n + v E g,np)e w n,p dp dp n V E g,n w n\{j} ) w g +g g, I Jn\{} V g,n E w ) n\{}) wj V E g, I + w, w I ) V E g, J + w, w J ) 3 Proof : Since V E g,nw n ) ) n R n + p p n v S g,np)e w n,p dp dp n, to obtain the formula, multiply both sides of 36) by ) n p p n and take its Laplace transform The left-hand side gives V g,nw E n ) For a continuous function fq), by putting p + p j q l, 54) p +p j dp dp j dq p j qp + p j q)fq)e p w +p j w j ) w w j ) dq w w j ) q+l dq dl dp j qlfq)e qw e lw p j e p jw w j ) dl qlfq) e q+l)w e q+l)w j + q + l)w w j )e q+l)w j fw ) w fw ) ) j ) fwj ) wj w w j w j wj fw ) w j w w j w fw ) j ) wj, where fw) qfq)e qw dq Set p p j q l Then 55) dp dp j Hp p j ) dq dl p p j dq p j qp p j q)fq)e p w +p j w j ) dp j qlfq)e qw e lw p j e p jw +w j ) fw ) w + w j ), w 3

14 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) 4 and similarly, 56) dp dp j Hp j p ) dq dq Adding 55) and 56) gives dp dp j Hp p j ) pj p dq p j qp j p q)fq)e p w +p j w j ) dl dp qlfq)e qw j e lw j p + q + l)e p w +w j ) dl qlfq)e qw j e lw j w + w j ) + q + l w + w j ) fwj ) fwj ) w + w j ) + w + w j w j p p j dp dp j Hp j p ) fw ) w + w j ) w w j dq p j qp p j q)fq)e p w +p j w j ) pj p fw j ) w j dq p j qp j p q)fq)e p w +p j w j ) ) ) fwj ) + w + w j w j wj fw ) w j w + w j The sum of the right-hand sides of 54)-56) thus becomes ) fw ) w j w w j w + w j w fw ) j ) wj Therefore, the first three lines of 36) give w j V E g,n w n\{j} ) V g,n E w j w w j w w ) n\{}) wj For a continuous function f, dp q +q p q q p q q )fq, q )e p w dq dq w w j fw ) j ) wj dq dq dl q q lfq, q )e lw e q +q )w fw, w ), where fw, w ) R p p fp, p )e p w +p w ) dp dp Thus the last two lines of 36) + give w V E g,n+w, w, w n\{} ) + Hence 53) is proven g +g g, I Jn\{} V E g, I + w, w I ) V E w g, J + w, w J ) 4

15 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) 5 Now, change coordinates by a new coordinate change, w j t j Then dw j t j dt j, and t j w j t j hold Then V E g,nt n )dt n V E g,nw n )dw n, or V E g,nt n ) n V E g,nw n ) t t n holds So, multiply both sides of 53) by n t t n to get, for the first term of the first line, n t t n t j w j t t j t t j w j w w j V E g,n w n\{j}) w t 4 V E g,n t n\{j} ) 6 t j t j t t j t 4 V E g,n t n\{j} ) Similarly, the second term of the first line becomes n w j V g,n E w n\{}) t t n w j w w j wj t t j t j t j V g,n E t n\{}) t j t t j 4 t 6 t j t j t t j t 4 j V E g,n t n\{} ) The second line of 53) is changed similarly Since V E g,nt n ) is known to be continuous, every step in the Laplace transform of 36) is reversible, and so the formula for the Euclidean volume holds 6 Examples By using the recursion formulas, it is possible to compute v E g,n and N g,n Here are some of the results for small g, n), including the initial data The results for N g,n will only hold when the sum of its arguments is even; otherwise it will be zero These values except the initial conditions, of course) were computed using Mathematica 5

16 Explorations: The UC Davis Undergraduate Research Journal, Vol 4 ) 6 Lattice Point Count and Euclidean Volumes v E,3x, y, z) v E,4x, y, z, w) 8 x + y + z + w ) v E,5x, y, z, w, r) 64 x4 + y 4 + z 4 + w 4 + r 4 + 4x y + 4x z + 4x w + 4x r + 4y z + 4y w + 4y r + 4z w + 4z r + 4w r ) v,x) E 96 x v,x, E y) 768 x4 + x y + y 4 ) v,3x, E y, z) 96 x6 + y 6 + z 6 + 6x 4 y + 6x 4 z + 6y 4 z + 6x y 4 + 6x z 4 + 6y z 4 + x y z ) v,x) E x8 N,3 x, y, z) N,4 x, y, z, w) 4 x + y + z + w 4) N,5 x, y, z, w, r) 3 x4 + y 4 + z 4 + w 4 + r 4 + 4x y + 4x z + 4x w + 4x r + 4y z + 4y w + 4y r + 4z w + 4z r + 4w r x y z w r + 64) N, x) 48 x N, x, y) 384 x4 + y 4 + x y x y + 3) N, x) 343 3x8 84x x x + 638) 7 Acknowledgements I would like to acknowledge the people who supported me in preparation of this paper The first is Motohico Mulase This paper would have been impossible without his guidance and insight, so I thank him for all of his time and patience spent on me Next, I would like to thank my family for their support of me in all of my endeavors Minji Kim and Ian McConachie deserve thanks as well; as my peers in the research group, they greatly aided my understanding of certain concepts, including ribbon graphs and the notion of the volume of the Moduli space References J Bennett, D Cochran, B Safnuk, K Woskoff, Topological recursion for symplectic volumes of moduli spaces of curves, arxiv:747v3 ) K Chapman, M Mulase, B Safnuk, Topological Recursion and the Kontsevich Constants for the Volume of the Moduli of Curves, arxiv:955v ) 3 J L Harer, The cohomology of the moduli space of curves, in Theory of Moduli, Montecatini Terme, 985 Edoardo Sernesi, ed), Springer-Verlag 988, pp 38-4 M Mulase and M Penkava, Ribbon graphs, quadratic differentials on Riemann surfaces, and algebraic curves defined over Q, The Asian Journal of Mathematics 4), ) 5 D Mumford, Towards an enumerative geometry of moduli space of curves 983), in Selected Papers of David Mumford, ) 6 P Norbury, String and dilaton equations for counting lattice points in the moduli space of curves, arxiv:9544 9) 7 D D Sleator, R E Tarjan, and W P Thurston, Rotation distance, triangulations, and hyperbolic geometry, Journal of the American Mathematical Society, ) Department of Mathematics, University of California, Los Angeles, CA , USA address: kchapman@mathuclaedu 6