Planar maps and continued fractions
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1 Batz 2010 p. 1/4 Planar maps and contnued fractons n collaboraton wth Jéréme Boutter
2 maps and dstances: generaltes Batz 2010 p. 2/4
3 Batz 2010 p. 3/4
4 degree of a face = number of ncdent edge sdes (= 3 here) Batz 2010 p. 4/4
5 enumerate maps wth a control of the face degrees Batz 2010 p. 4/4
6 Batz 2010 p. 4/4 compute the generatng functon of maps wth a weght g k per face of degree k g 1 g 2 g 3 g 4 g 10 n practce, consder bounded degrees (fnte number of non-zero g k s)
7 Batz 2010 p. 5/4 dstance n maps dstance d between two vertces on the maps
8 Batz 2010 p. 5/4 dstance n maps dstance d between two vertces on the maps = mnmal number of edges to connect them (d = 3 here)
9 Batz 2010 p. 6/4 d dstance-dependent two-pont functon = g.f. of maps wth two marked vertces at dstance d
10 Batz 2010 p. 6/4 d d dstance-dependent two-pont functon = g.f. of maps wth a marked vertex and a marked orented edge at dstance d.e: of type (d,d): t d ({g k }) (d 0)
11 Batz 2010 p. 6/4 d 1 d dstance-dependent two-pont functon = g.f. of maps wth a marked vertex and a marked orented edge at dstance d.e: of type (d 1,d): r d ({g k }) (d 1)
12 Batz 2010 p. 7/4 more convenent to demand that the dstance be less than a gven value,.e. consder the g.f. and so that S = R = d=1 ( d=0 r d ( 1) t d ) 1/2 r d = R d R d 1, ( 0) t d = S 2 d S2 d 1
13 Batz 2010 p. 8/4 the BDG bjecton start wth a planar map wth a marked vertex
14 Batz 2010 p. 8/4 the BDG bjecton start wth a planar map wth a marked vertex
15 Batz 2010 p. 8/4 the BDG bjecton start wth a planar map wth a marked vertex
16 Batz 2010 p. 8/4 the BDG bjecton start wth a planar map wth a marked vertex
17 Batz 2010 p. 8/4 the BDG bjecton start wth a planar map wth a marked vertex end up wth a moble tree wth nteger labels
18 Batz 2010 p. 9/4 R and S may be nterpreted as moble g.f. R = and S = recurson relatons for R and S
19 Batz 2010 p. 10/4 examples trangulatons g k = g 3 δ k,3
20 Batz 2010 p. 10/4 examples trangulatons g k = g 3 δ k,3 R = 1 + g 3 R (S + S 1 ), S = g 3 (S 2 + R + R +1 ) (R 0 = 0)
21 Batz 2010 p. 10/4 examples trangulatons g k = g 3 δ k,3 R = 1 + g 3 R (S + S 1 ), S = g 3 (S 2 + R + R +1 ) (R 0 = 0) quadrangulatons g k = g 4 δ k,4 R = 1 + g 4 R (R 1 + R + R +1 ), S = 0 (R 0 = 0)
22 Batz 2010 p. 10/4 examples trangulatons g k = g 3 δ k,3 R = 1 + g 3 R (S + S 1 ), S = g 3 (S 2 + R + R +1 ) (R 0 = 0) R = R (1 y )(1 y +2 ) (1 y +1 ) 2, S = S g 3 R 2 y (1 y)(1 y 2 ) (1 y +1 )(1 y +2 ) R = 1 + 2g 3 RS, S = g 3 (S 2 + 2R) y + 1 y + 2 = 1 g 2 3 R3 quadrangulatons g k = g 4 δ k,4 R = 1 + g 4 R (R 1 + R + R +1 ), S = 0 (R 0 = 0)
23 Batz 2010 p. 10/4 examples trangulatons g k = g 3 δ k,3 R = 1 + g 3 R (S + S 1 ), S = g 3 (S 2 + R + R +1 ) (R 0 = 0) R = R (1 y )(1 y +2 ) (1 y +1 ) 2, S = S g 3 R 2 y (1 y)(1 y 2 ) (1 y +1 )(1 y +2 ) R = 1 + 2g 3 RS, S = g 3 (S 2 + 2R) y + 1 y + 2 = 1 g 2 3 R3 quadrangulatons g k = g 4 δ k,4 R = 1 + g 4 R (R 1 + R + R +1 ), S = 0 (R 0 = 0) R = R (1 y )(1 y +3 ) (1 y +1 )(1 y +2 ), R = 1+3g 4R 2, y+ 1 y +1 = 1 g 4 R 2
24 Batz 2010 p. 11/4 so far the soluton was guessed for - trangulatons - bpartte maps.e. wth faces of even degrees only (g 2k+1 = 0 k)
25 Batz 2010 p. 11/4 so far the soluton was proved for - trangulatons - bpartte maps wth small degrees 4,6,8
26 Batz 2010 p. 11/4 so far the soluton was proved for - trangulatons - bpartte maps wth small degrees 4,6,8 can we get a general expresson for R and S for arbtrary maps? can we get t n a constructve way (not a smple guess)? can we explan the partcular form observed so far (nvolvng (b-)ratos)?
27 Batz 2010 p. 12/4 an unexpected connecton wth a much smpler problem
28 rooted map wth a marked orented edge Batz 2010 p. 13/4
29 Batz 2010 p. 13/4 control the degree n of the root face (on the rght of the edge) n =10
30 Batz 2010 p. 13/4 choose the root face as external face n the plane maps wth a boundary" of length n n =10
31 Batz 2010 p. 14/4 F n ({g k }) n =10 by conventon, no weght for the external face and F 0 = 1 - global quantty (no dstance nvolved) - satsfes Tutte s equatons ( matrx model loop equatons) - well understood snce Tutte s work (60 s)
32 Batz 2010 p. 15/4 consder the (well-understood) resolvent" F(z) n 0 F n z n
33 Batz 2010 p. 15/4 consder the (well-understood) resolvent" F(z) n 0 F n z n we have the remarkable property F(z) = 1 S 0 z 1 R 1 z 2 1 S 1 z R 2z 2 1.e. the dstance-dependent two-pont generatng functons S and R are gven by the contnued fracton expanson of the resolvent
34 Batz 2010 p. 16/4 Σ n F n z n F(z) =
35 Batz 2010 p. 16/4 F(z) = 1 1 zs0 z 2 R1 1 zs1 z 2 R2 1 zs... 2
36 Batz 2010 p. 17/4 contnued fractons and Motzkn paths t s a classcal result of combnatorcs (Flajolet) that 1 S 0 z 1 R 1 z 2 1 S 1 z R 2z 2 1 = n 0 Z + 0,0 (n)zn where Z 0,0 + (n) s the generatng functon for weghted Motzkn paths of length n:
37 Batz 2010 p. 17/4 contnued fractons and Motzkn paths t s a classcal result of combnatorcs (Flajolet) that 1 S 0 z 1 R 1 z 2 1 S 1 z R 2z 2 1 = n 0 Z + 0,0 (n)zn + Z 0,0 Z ( n) = + 0,0 ( n): Z Σ paths n + 0, ( n) Π R 1 Π S 0 0 n
38 Batz 2010 p. 18/4 we have the remarkable property F n = Z + 0,0(n) where Z 0,0 + (n) enumerates Motzkn paths weghted by the dstance-dependent two-pont functons R and S
39 Batz 2010 p. 19/4 proof 0 d closest vertex at dstance d maps wth a boundary of length n, a marked vertex and a marked orented edge on the boundary whose orgn s at dstance d and s a closest vertex" ( closest to the marked vertex among those vertces lyng on the boundary) f n;d
40 Batz 2010 p. 19/4 proof 0 closest vertex at dstance 0 we have F n = f n;0
41 Batz 2010 p. 19/4 proof 0 d closest vertex at dstance d d d n
42 Batz 2010 p. 19/4 proof 0 1 draw the leftmost geodesc path from each boundary vertex to the marked vertex
43 Batz 2010 p. 19/4 proof 0 ~ S 1 1 ~ R decomposton nto two types of slces assocated resp. to down- ( 1) and level-steps ( )
44 Batz 2010 p. 20/4 ~ + Z ( n) = d,d Σ paths n + Z ( n) d,d Π ~ R 1 Π ~ S + Z ( n): d,d d d n
45 Batz 2010 p. 21/4 ~ S ~ R
46 Batz 2010 p. 21/4 d d d 1 Z+ 0 d,d (n) = and n partcular d =0 Z+ 0,0 (n) = F n f n;d
47 Batz 2010 p. 22/4 t remans to show that R = R and S = S 1 d 1 d ~ R ~ S = Σ d= 1 ~ = Σs d= 0 ~ ~ r r d d d ~ s d d 0 0 d
48 Batz 2010 p. 23/4 showng R = R showng r d = r d ~ r d 1 d d 0 d 1 0 d leftmost geodesc r d
49 Batz 2010 p. 23/4 showng S 2 = S2 = d=0 t d showng d,d 0 max(d,d )=d s d s d = t d ~ s d s ~ d 0 d d d d d= max( d, d ) leftmost geodesc d d 0 leftmost geodesc 0 t d
50 Batz 2010 p. 24/4 consequence classcal result about contnued fractons: the coeffcents S and R can be n expressed n terms of the coeffcents F n of the assocated seres expanson va Hankel determnants, namely: R = H H 2 H 2 1, H = det 0 m,n F m+n S = H H H 1 H 1, H = det 0 m,n F m+n+δ n, wth the conventon H 1 = 1 and H 1 = 0
51 Batz 2010 p. 25/4 our strategy: 1 take a nce expresson for F n 2 compute the Hankel determnants H and H 3 deduce explct expressons for R and S (and eventually for r d and t d )
52 1 a nce expresson for F n Batz 2010 p. 26/4
53 Batz 2010 p. 27/4 consder R = lm R, S = lm S
54 Batz 2010 p. 27/4 consder R = lm R, S = lm S and ntroduce the unformly weghted path g.f. ζ + 0 ( n) = Z Σ paths n + 0 ( n) Π R Π S Z + 0 ( n): 0 0 n
55 consder R = lm R, S = lm S and ntroduce the unformly weghted path g.f. ζ 0 Σ ( n) = paths n Z ( n) 0 Π R Π S Z 0 ( n): 0 0 n Batz 2010 p. 27/4
56 consder R = lm R, S = lm S and ntroduce the unformly weghted path g.f. ζ l Σ ( n) = paths n Z ( n) l Π R Π S l Z ( n): 0 l n Batz 2010 p. 27/4
57 Batz 2010 p. 28/4 R and S are determned by the relatons R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1)
58 Batz 2010 p. 28/4 R and S are determned by the relatons R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1) whle F n has the nce expresson: A q = R F n = A q ζ 0 + (n + q) q 0 δ q,0 g k ζ 0 (k q 2) k=q+2.e. F n s a lnear combnaton of g.f. for Motzkn paths of ncreasng lengths n,n+1,,n+q max wth q max = k max 2
59 Batz 2010 p. 28/4 R and S are determned by the relatons R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1) whle F n has the nce expresson: F n = q 0 A q ζ + 0 (n + q) example: trangulatons R = 1 + 2g 3 R S = g 3 (S 2 + 2R) F n = ζ + 0 (n)(1 + g 3RS) ζ + 0 (n + 1)g 3R
60 Batz 2010 p. 29/4 2 after smple row and column manpulatons H = R (+1) 2 det 0 l,l (B l l B l+l +2), B l = R l 2 q=0 A q ζ l (q) = B l
61 Batz 2010 p. 29/4 2 after smple row and column manpulatons H = R (+1) 2 det 0 l,l (B l l B l+l +2), B l = R l 2 q=0 A q ζ l (q) = B l F m+n = Σ q A q ζ + 0 ( m+n+q ) ζ l l l+l +2 ( q) ζ ( q ) ζ + 0 ( ) : m+n+q l l m q n
62 Batz 2010 p. 29/4 2 after smple row and column manpulatons H = R (+1) 2 det 0 l,l (B l l B l+l +2), B l = R l 2 q=0 A q ζ l (q) = B l H = ( + 1)SH + R det 0 l,l (B l l δ l, B l+l +δ l, +2)
63 Batz 2010 p. 29/4 2 after smple row and column manpulatons H = R (+1) 2 det 0 l,l (B l l B l+l +2), B l = R l 2 q=0 A q ζ l (q) = B l settng p = k max 2, we have B n = 0 for n > p denote x = (x 1, 1/x 1,x 2, 1/x 2,...,x p, 1/x p ) the solutons of p B n x n = 0 n= p
64 Batz 2010 p. 29/4 2 after smple row and column manpulatons H = R (+1) 2 det 0 l,l (B l l B l+l +2), B l = R l 2 q=0 A q ζ l (q) = B l settng p = k max 2, we have B n = 0 for n > p denote x = (x 1, 1/x 1,x 2, 1/x 2,...,x p, 1/x p ) the solutons of p B n x n = 0 n= p det 0 l,l (e p+l l (x) e p+l+l +2(x)) n terms of elementary symmetrc functons e j (x)
65 Batz 2010 p. 30/4 then H and H are dentfed as symplectc Schur functons H =( 1) p(+1) B +1 p R (+1) 2 sp 2p (λ p,+1,x) H ( + 1)S H =( 1) p(+1)+1 B +1 p for the parttons λ p,+1 : +1 R sp 2p ( λ p,+1,x) p +1 λ p,+1 : p
66 Batz 2010 p. 30/4 then H and H are dentfed as symplectc Schur functons H =( 1) p(+1) B +1 p R (+1) 2 sp 2p (λ p,+1,x) H ( + 1)S H =( 1) p(+1)+1 B +1 p R sp 2p ( λ p,+1,x) 3 R = R sp 2p(λ p,+1,x) sp 2p (λ p, 1,x) sp 2p (λ p,,x) 2 S = S R ( sp2p ( λ p,+1,x) sp 2p (λ p,+1,x) sp ) 2p( λ p,,x) sp 2p (λ p,,x)
67 Batz 2010 p. 31/4 fnal formula the symplectc Schur functons may be expressed as determnants of fxed sze p R = R det 1 m,n p (x+1+n m ( x 1 n m ) det det 1 m,n p (x+n m 1 m,n p (x 1+n m ) 2 x n m ) x +1 n m ) S = S R det 1 m,n p (x+1+n δ n,1 m ) m x 1 n+δ n,1 det 1 m,n p (x+1+n m x 1 n m ) det 1 m,n p (x+n δ n,1 m x n+δ n,1 det 1 m,n p (x+n m m ) x n m )
68 Batz 2010 p. 32/4 trangulatons and hard dmers F n = ζ + 0 (n)(1 + g 3RS) ζ + 0 (n + 1)g 3R A 0 = 1 + g 3 RS, A 1 = g 3 R B 0 = A 0 + SA 1 = 1, B 1 = R 1 2 A1 = g 3 R 3 2 x + 1 x = 1 g 3 R 3 2 settng y = x 2 y + 1 y + 2 = 1 g 2 3 R3 R = R (1 y )(1 y +2 ) (1 y +1 ) 2, S = S g 3 R 2 y (1 y)(1 y 2 ) (1 y +1 )(1 y +2 )
69 Batz 2010 p. 32/4 trangulatons and hard dmers F n = ζ + 0 (n)(1 + g 3RS) ζ + 0 (n + 1)g 3R A 0 = 1 + g 3 RS, A 1 = g 3 R B 0 = A 0 + SA 1 = 1, B 1 = R 1 2 A1 = g 3 R 3 2 F m+n g 3 R 3/2 g 3 R 3/2 1 S 1/2 R R 1/2 m n+1
70 Batz 2010 p. 33/4 from the LGV lemma: det confgs of avodng paths g 2 R 3 3 ( +1) 2 R n confgs of hard dmers on a segment H Zhard dmers on [0,] = snh( + 2)θ (2 cosh θ) +1 snh θ g 2 3R 3 = 1 4 cosh 2 θ
71 Batz 2010 p. 33/4 from the LGV lemma: det confgs of avodng paths g 2 R 3 3 ( +1) 2 R n confgs of hard dmers on a segment H Zhard dmers on [0,] = x +2 x 2 (x + x 1 ) +1 (x x 1 ) g 2 3R 3 = 1 (x + x 1 ) 2
72 Batz 2010 p. 34/4 concluson the dstance-dependent two-pont functons R and S are coded n the contnued fracton expanson of the resolvent F(z) = 1 1 zs0 z 2 R1 1 zs1 z 2 R2 1 zs... 2
73 Batz 2010 p. 34/4 concluson the dstance-dependent two-pont functons R and S are coded n the contnued fracton expanson of the resolvent ths leads to explct expressons of R and S as (b-)ratos of symplectc Schur functons R = R sp 2p(λ p,+1,x) sp 2p (λ p, 1,x) sp 2p (λ p,,x) 2 S = S R ( sp2p ( λ p,+1,x) sp 2p (λ p,+1,x) sp ) 2p( λ p,,x) sp 2p (λ p,,x)
74 Batz 2010 p. 34/4 concluson the dstance-dependent two-pont functons R and S are coded n the contnued fracton expanson of the resolvent ths leads to explct expressons of R and S as (b-)ratos of symplectc Schur functons what about unbounded degrees?
75 Batz 2010 p. 34/4 concluson the dstance-dependent two-pont functons R and S are coded n the contnued fracton expanson of the resolvent ths leads to explct expressons of R and S as (b-)ratos of symplectc Schur functons what about unbounded degrees? what about hgher genus maps?
76 Batz 2010 p. 35/4 + Z d,d+j ( n) = Z Σ paths n + d,d+j ( n) Π R 1 Π S + Z ( n): d,d+j d d+j n
77 Batz 2010 p. 35/4 Z d,d+j Σ ( n) = paths n Z ( n ) d,d+j Π R 1 Π S Z ( n): d,d+j d d+j n
78 Batz 2010 p. 36/4 send d : Z + d,d+j Σ ( n) ζ ( n) = Π R Π j + Z paths n + j ( n) S Z + j ( n): j n
79 Batz 2010 p. 36/4 send d : Z d,d+j ( n) ζ j ( n) = Z j Σ paths n ( n) Π R Π S Z j ( n): j n
80 Batz 2010 p. 37/4 recurson relatons d d degree d k d S : 0 d < 0 d < k 1 S = k 1 g k Z, (k 1)
81 Batz 2010 p. 37/4 recurson relatons d d degree d k d S : 0 d < 0 d < k 1 S = k 1 g k Z, (k 1) S = k 1 g k ζ 0 (k 1)
82 Batz 2010 p. 38/4 recurson relatons d d 1 degree 1 d k d R : 0 d < 0 d < k 1 1 R = 1 + k 2 g k Z, 1 (k 1)
83 Batz 2010 p. 38/4 recurson relatons d d 1 degree 1 d k d R : 0 d < 0 d < k 1 1 R = 1 + k 2 g k Z, 1 (k 1) R = 1 + k 2 g k ζ 1 (k 1)
84 Batz 2010 p. 39/4 expresson for F n F n = f n;0 = d d =0 f n;d d d =1 f n;d
85 Batz 2010 p. 39/4 expresson for F n F n = f n;0 = d d =0 f n;d d d =1 f n;d 0 d d d d <d n Z + d,d (n)
86 Batz 2010 p. 39/4 expresson for F n F n = f n;0 = d d =0 f n;d d d =1 f n;d d 0 g k d d d 1 1 1
87 Batz 2010 p. 39/4 expresson for F n F n = f n;0 = d d =0 f n;d d d =1 f n;d d +j d 1 d d <d k 3 0 g k j 1 d d+j d 1 n k 1 Z + d,d+j (n)z d+j,d 1(k 1)
88 Batz 2010 p. 40/4 F n = Z + d,d (n) k 3 g k j 1 Z + d,d+j (n)z d+j,d 1(k 1) vald for any d 0 ( conserved quantty)
89 Batz 2010 p. 40/4 F n = Z + d,d (n) k 3 send d : g k j 1 Z + d,d+j (n)z d+j,d 1(k 1) F n = ζ + 0 (n) k 3 g k j 1 ζ + j (n)ζ j 1(k 1) n terms of paths weghted by R and S only
90 Batz 2010 p. 41/4 R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1) F n = ζ + 0 (n) k 3 g k j 1 ζ + j (n)ζ j 1(k 1)
91 Batz 2010 p. 41/4 R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1) F n = ζ + 0 (n) k 3 g k j 1 ζ + j (n)ζ j 1(k 1) example: trangulatons R = 1 + 2g 3 R S = g 3 (S 2 + 2R) F n = ζ + 0 (n) g 3R 2 ζ + 1 (n) usng ζ + 0 (n + 1) = S ζ+ 0 (n) + R ζ+ 1 (n), we end up wth F n = ζ + 0 (n)(1 + g 3RS) ζ + 0 (n + 1)g 3R
92 Batz 2010 p. 41/4 R = 1 + k 2 g k ζ 1 (k 1) S = k 1 g k ζ 0 (k 1) F n = ζ + 0 (n) k 3 g k j 1 ζ + j (n)ζ j 1(k 1) n all generalty A q = R F n = A q ζ 0 + (n + q) q 0 δ q,0 g k ζ 0 (k q 2) k=q+2.e. F n s a lnear combnaton of g.f. for Motzkn paths of ncreasng lengths n,n+1,,n+q max wth q max = k max 2
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