The map asymptotics constant t g

The map asymptotics constant t g Edward A. Bender Department of Mathematics University of California, San Diego La Jolla, CA 92093-02 ebender@ucsd.edu Zhicheng Gao School of Mathematics and Statistics Carleton University Ottawa, Ontario KS5B6 Canada zgao@math.carleton.ca L. Bruce Richmond Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario N2L 3G Canada Submitted: Jan 28, 2008; Accepted: Mar 22, 2008; Published: Mar 27, 2008 Mathematics Subject Classification: 05C30 Research supported by NSERC Research supported by NSERC the electronic journal of combinatorics 5 (2008), #R5

Abstract The constant t g appears in the asymptotic formulas for a variety of rooted maps on the orientable surface of genus g. Heretofore, studying this constant has been difficult. A new recursion derived by Goulden and Jackson for rooted cubic maps provides a much simpler recursion for t g that lea to estimates for its asymptotics. Introduction Let Σ g be the orientable surface of genus g. A map on Σ g is a graph G embedded on Σ g such that all components of Σ g G are simply connected regions. These components are called faces of the map. A map is rooted by distinguishing an edge, an end vertex of the edge and a side of the edge. With M n,g the number of rooted maps on Σ g with n edges, Bender and Canfield [] showed that M n,g t g n 5(g )/2 2 n as n, () where the t g are positive constants which can be calculated recursively using a complicated recursion involving, in addition to g, many other parameters. The first three values are t 0 = 2 π, t = 24 and t 2 = 7 4320 π. Gao [3] showed that many other interesting families of maps also satisfy asymptotic formulas of the form αt g (βn) 5(g )/2 γ n (2) and presented a table of α, β and γ for eleven families. Richmond and Wormald [5] showed that many families of unrooted maps have asymptotics that differ from the rooted asymptotics by a factor of four times the number of edges. See Goulden and Jackson [4] for a discussion of connections with mathematical physics. Although α, β and γ in (2) seem relatively easy to compute, the common factor t g has been difficult to study. A recursion for rooted cubic maps derived by Goulden and Jackson [4] lea to a much simpler recursion for t g than that in []. We will use it to derive the following recursion and asymptotic estimate for t g. Theorem Define u g by u = /0 and u g = u g + g R (g, h)r 2 (g, h) u hu g h for g 2, (3) where R (g, h) = [/5] g [/5] h [/5] g h, R 2 (g, h) = [4/5] g [4/5] h [4/5] g h the electronic journal of combinatorics 5 (2008), #R5 2

and [x] k is the rising factorial x(x + ) (x + k ). Then t g = 8 [/5] ( ) g[4/5] g 25 g Γ ( ) u 5g g 96 2 ( ) 40 sin(π/5)k 440g g/2 as g, (4) 2π e where u g K. = 0.04869 is a constant. 2 Cubic Maps A map is called cubic if all its vertices have degree 3. The dual of cubic maps are called triangular maps whose faces all have degree 3. Let T n,g be the number of triangular maps on Σ g with n vertices and let C n,g be the number of cubic maps on Σ g with 2n vertices. It was shown in [2] that T n,g 3 ( 3 7 2 9) (g )/2 tg n 5(g )/2 (2 3) n as n. (5) Since a triangular map on Σ g with v vertices has exactly 2(v + 2g 2) faces, Define C n,g = T n 2g+2,g 3 6 (g )/2 t g n 5(g )/2 (2 3) n as n. (6) H n,g = (3n + 2)C n,g for n, (7) H,0 = /2, H 0,0 = 2 and H,g = H 0,g = 0 for g 0. Goulden and Jackson [4] derived the following recursion for (n, g) (, 0): H n,g = ( n 4(3n + 2) g n(3n 2)H n 2,g + H i,h H n 2 i,g h ). (8) n + i= h=0 This is significantly simpler than the recursion derived in [2]. We will use it to derive information about t g. 3 Generating Functions Define the generating functions T g (x) = T n,g x n, C g (x) = C n,g x n, H g (x) = H n,g x n and F g (x) = x 2 H g (x). It was shown in [2] that T g (x) is algebraic for each g 0, and T 0 (x) = 2 t3 ( t)( 4t + 2t 2 ) with x = t( t)( 2t), (9) 2 the electronic journal of combinatorics 5 (2008), #R5 3

where t = t(x) is a power series in x with non-negative coefficients. It follows from (6) and (7) that We also have C g (x) = x 2g 2 T g (x) for g 0, (0) F g (x) = 3x 3 C g(x) + 2x 2 C g (x) for g. () F 0 (x) = H 0,0 x 2 + (3n + 2)C n,0 x n+2 n = 2x 2 + 3x 3 C 0(x) + 2x 2 C 0 (x) = 2x 2 + 3xT 0(x) 4T 0 (x) = 2 t2 ( t), (2) where we have used (9). Hence C g (x) and F g (x) are both algebraic for all g 0. In the following we assume g. From the recursion (8), we have n + 4 3n + 2 H n,gx n = n(3n 2)H n 2,g x n n + 2 H,0 H n,g x n + x 2 g H h (x)h g h (x). h=0 Using (7) with a bit manipulation, we can rewrite the above equation as 4 (n + )C n,g x n = 3x 2 F g (x) + xf g (x) + xh,g With δ i,j the Kronecker delta, this becomes g + x F g (x) + x 2 F h (x)f g h (x). h=0 x 3 C g(x) + x 2 C g (x) = 2x 4 F g (x) + 4x 3 F g (x) + 2x 3 δ g, It follows from () that + 4xF g (x) + 8F 0 (x)f g (x) + 4 g F h (x)f g h (x). ( 2x 24F 0 (x)) F g (x) = 36x 4 F g (x) + 2x 3 F g (x) + 6x 3 δ g, + 2 g F h (x)f g h (x) x 2 C g (x). (3) the electronic journal of combinatorics 5 (2008), #R5 4

Substituting (2) and (9) into (3), we obtain F g (x) = ( 36x 4 F 6t + 6t g (x) + 2x 3 F g (x) + 6x 3 δ 2 g, g ) + 2 F h (x)f g h (x) x 2 C g (x). (4) We now show that this equation can be used to calculate C g (x) more easily than the method in [2]. For this purpose we set s = 6t + 6t 2 and show inductively that C g (x) is a polynomial in s divided by s a for some integer a = a(g) > 0. (It can be shown that a = 5g 3 is the smallest such a, but we do not do so.) The method for calculating C g (x) follows from the proof. Then we have x 2 = 432 (s )2 (2s + ) and dx = 44x s(s ). (5) Thus x d dx = x dx d = (s )(2s + ) 3s d, d 2 dx 2 = ( ) 2 d 2 dx 2 + d(/dx) dx d = 48(2s + ) d 2 48(s + ) s 2 2 s 3 d. From the above and () F g (x) + x 2 ( C g = x 2 3x dc g 6t + 6t 2 dx + (2s + )C ) g s With some algebra, (4) can be rewritten as = x2 (2s + ) s d((s )C g ). d((s )C g ) = 4(s )2 (2s + ) d 2 F g 4(s ) df g + s 2 2 s 3 g 584 + F (s ) 2 (2s + ) 2 h F g h for g 2. (6) In what follows P (s) stan for a polynomial in s and a a positive integer, both different at each occurrence. It was shown in [2] that C (x) = T (x) = s 2s 2. By (), (5) and the induction hypothesis, the right hand side of (6) has the form P (s)/s a. Integrating, (s )C g = P (s)/s a + K log s. Since we know C g (x) is algebraic, so is (s )C g and hence K = 0. Since s = correspon to x = 0, C g is defined there. It follows that P (s) in (s )C g = P (s)/s a is divisible by s, completing the proof. the electronic journal of combinatorics 5 (2008), #R5 5

Using Maple, we obtained C 2 = C 3 = C 4 = C 5 = where (2s + )(7s 2 + 60s + 28)( s) 3, 2 6 3 4 s 7 (5052s 4 747s 3 33960s 2 35620s 9800)(2s + ) 2 (s ) 5, 2 9 3 8 s 2 P 4 (s)(2s + ) 3 (s ) 7, 2 4 3 s 7 P 5 (s)(2s + ) 4 ( s) 9, 2 7 3 4 s 22 P 4 (s) = 2458544 63378560s 03689240s 2 4286406s 3 + 3477893s 4 + 20750256s 5 + 47636s 6, P 5 (s) = 7703740800 + 50294009360s + 778660480s 2 + 0038608272s 3 6827627792s 4 67700509763s 5 2455389524s 6 + 4783020s 7 + 394857272s 8. 4 Generating Function Asymptotics Suppose A(x) is an algebraic function and has the following asymptotic expansion around its dominant singularity /r: A(x) = k a j ( rx) j/2 + O ( ( rx) (k+)/2), j=l where a j are not all zero. Then we write A(x) k a j ( rx) j/2. j=l The following lemma is proved in [2]. Lemma For g 0, T g (x) is algebraic, 3 T 0 (x) 72 5 26 + 54 6 ( 2 3x) 3/2, T g (x) 3 ( 3 7 2 9) ( ) (g )/2 5g 3 tg Γ ( 2 3x) (5g 3)/2 for g. 2 the electronic journal of combinatorics 5 (2008), #R5 6

Let ( ) 5g f g = 24 3/2 6 g/2 Γ t g. (7) 2 Using Lemma, (0) and (), we obtain C g (x) 288 (5g 3) f g( 2 3x) (5g 3)/2 for g, F g (x) f g ( 2 3x) (5g )/2 for g. As noted in [2], the function t(x) of (9) has the following asymptotic expansion around its dominant singularity x = 2 : 3 t 3 3 2 6 6 ( 2 3x) /2. Using this and (2), we obtain F 0 (x) 3 3 + f 0 ( 2 3x) /2, 72 6 6t + 6t 2 2 ( 2 3x) /2. Comparing the coefficients of ( 2 3x) (5g )/2 on both sides of (4), we obtain 6 f g = 96 (5g 4)(5g 6)f g + 6 g 6 f h f g h. (8) Letting ( ) g 25 6 6 6 u g = f g. 96 [/5] g [4/5] g and using (7), the recursion (8) becomes (3). 5 Asymptotics of t g It follows immediately from (3) that u g u g for all g 2. To show that u g approaches a limit K as g, it suffices to show that u g is bounded above. The value of K is then calculated using (3). We use induction to prove u g for all g. Since u = and u 0 2 = u +, we 480 can assume g 3 for the induction step. From now on g 3. Note that R (g, )R 2 (g, ) = 5(g 4 5 )(g 6 5 ) > 5(g 4 5 )(g 9 5 ) R (g, 2)R 2 (g, 2) = 25 24 (g 6 5 )(g 5 )( 5(g 4 5 )(g 9 5 )) > 25 24 (g 6 5 + 4 5 )(g 5 4 5 )( 5(g 4 5 )(g 9 5 )) 2(g 3) ( 5(g 4 5 )(g 9 5 )). the electronic journal of combinatorics 5 (2008), #R5 7

Note that R i (g, h) = R i (g, g h) and, for h < g/2, R i(g,h+) R i (g,h) observations and the induction hypothesis with (3) we have Hence u g = u g + < u g + < u g + g u h u g h R (g, h)r 2 (g, h) g 2 2u u g 5(g 4)(g 9) + R 5 5 (g, 2)R 2 (g, 2) h=2 /5 5(g 4)(g 9) + /2 5(g 4)(g 9) 5 5 5 5 < u g + 5g 9 5g 4. g ( u g < u 2 + k=3 5k 9 ) 5k 4 < u 2 + The asymptotic expression for t g in (4) is obtained by using [x] k = Γ(x + k), Γ(/5)Γ(4/5) = Γ(x). Combining all these 5 3 9 <. π sin(π/5), and Stirling s formula Γ(ag + b) ( ) ag ag 2π(ag) b /2 e as g, for constants a > 0 and b. 6 Open Questions We list some open questions. From (8), we can show that f(z) = g f g z g satisfies the following differential equation f(z) = 6 ( ) 6 6 6(f(z)) 2 + 96 z 25z 2 f (z) + 25zf (z) f(z) +. 72 The asymptotic expression of f g implies that f(z) cannot be algebraic. Can one show that f(z) is not D-finite, that is, f(z) does not satisfy a linear differential equation? There is a constant p g that plays a role for maps on non-orientable like t g plays for maps on orientable surfaces [3]. Is there a recursion for maps on non-orientable surfaces that can be used to derive a theorem akin to Theorem for p g? Find simple recursions akin to (8) for other classes of rooted maps that lead to simple recursive calculations of their generating functions as in (6). the electronic journal of combinatorics 5 (2008), #R5 8

References [] E.A. Bender and E.R. Canfield, The asymptotic number of maps on a surface, J. Combin. Theory, Ser. A, 43 (986), 244 257. [2] Z.C. Gao, The Number of Rooted Triangular Maps on a Surface, J. Combin. Theory, Ser. B, 52 (99), 236 249. [3] Z.C. Gao, A Pattern for the Asymptotic Number of Rooted Maps on Surfaces, J. Combin. Theory, Ser. A, 64 (993), 246 264. [4] I. Goulden and D.M. Jackson, The KP hierarchy, branched covers and triangulations, preprint (2008). [5] L.B. Richmond and N.C. Wormald, Almost all maps are asymmetric, J. Combin. Theory, Ser. B, 63 (995), 7. the electronic journal of combinatorics 5 (2008), #R5 9