COUNTING CURVES ON SURFACES Norm Do (joint with Musashi Koyama and - PowerPoint PPT Presentation

COUNTING CURVES ON SURFACES Norm Do (joint with Musashi Koyama and Dan Mathews) Monash University Discrete Mathematics Seminar 29 February 2016 Choose an even number of points on the boundary of a surface. How many ways are there to pair up these points with disjoint arcs on the surface? The most basic instance of this problem produces the Catalan numbers while the problem in general exhibits a surprisingly rich structure. For example, we will show that this enumeration obeys an effective recursion and exhibits polynomial behaviour. Moreover, there are unexpected connections to algebraic geometry and mathematical physics.

Counting curves creates Catalan How many ways are there to pair up b points on the boundary of a disk with disjoint arcs? For b even, we get the sequence 1 , 1 , 2 , 5 , 14 , 42 , . . . of Catalan numbers. For b odd, we get zero.

From disks to surfaces Surfaces are classified by their genus ( g ) and number of boundaries ( n ). g = 1 b 1 = 4 1 2 b 2 = 6 n = 2 Question Label the boundaries 1 , 2 , . . . , n and choose b i points on boundary i . How many ways are there to pair up these points with disjoint arcs? Denote the answer by G g , n ( b 1 , . . . , b n ). Example G g , n (0 , . . . , 0) = 1 G g , n ( b 1 , . . . , b n ) = 0 if � b i odd � 2 m � 1 G 0 , 1 (2 m ) = Cat m = m +1 m G 1 , 1 (2) = 3

Why isn’t the answer infinite? Two arc diagrams are equivalent if there is a continuous bijection (preserving orientation and boundary points) that takes one to the other. Equivalently, two arc diagrams are equivalent if they can be related by Dehn twists. (One creates a Dehn twist by cutting along a simple closed curve and gluing back the surface with a 360 ◦ twist.) So equivalent arc diagrams might look different “on the surface”!

Annuli — Insular arc diagrams Write G 0 , 2 ( b 1 , b 2 ) = T ( b 1 , b 2 ) + I ( b 1 , b 2 ) . � �� � � �� � traversing insular Fact � 2 m � For m a non-negative integer, I (2 m , 0) = . m Proof. � 2 m � There are ways to draw m arrows inwards and m arrows outwards. m One can draw “anticlockwise” arcs following the arrows uniquely.

Annuli — Insular arc diagrams (continued) Corollary � 2 m � From I (2 m , 0) = , we obtain m � 2 m 1 �� 2 m 2 � I (2 m 1 , 2 m 2 ) = , m 1 m 2 I (2 m 1 + 1 , 2 m 2 + 1) = 0 , � 2 m � 1 G 0 , 1 (2 m ) = . m +1 m Proof. Glue two annuli together to get I (2 m 1 , 2 m 2 ) = I (2 m 1 , 0) I (2 m 2 , 0). You can’t pair up the 2 m 1 + 1 points on boundary 1 with arcs. [Przytycki, 1999] For each arc diagram enumerated by G 0 , 1 (2 m ), there are m + 1 regions. Punching a hole in one of these regions � 2 m � yields one of the arc diagrams enumerated by I (2 m , 0). m

Annuli — Traversing arc diagrams Consider the case ( b 1 , b 2 ) = (2 m 1 , 2 m 2 ). � 2 m 1 �� 2 m 2 � There are ways to draw m i in/out-arrows on boundary i . m 1 m 2 There are m 1 m 2 ways to connect an in-arrow on boundary 1 to an out-arrow on boundary 2 by an arc γ . Cut along γ to obtain a disk with m 1 + m 2 − 1 in/out-arrows. Punch a hole in the disk to make an annulus. As above, draw “anticlockwise” arcs following the arrows. Remove the hole and mark the region it used to be in. Glue along γ to obtain an annulus divided into m 1 + m 2 regions. ∗ ∗

Annuli — Traversing arc diagrams (continued) We obtain a unique traversing arc diagram with a marked region, so � 2 m 1 �� 2 m 2 � m 1 m 2 T (2 m 1 , 2 m 2 ) = . m 1 + m 2 m 1 m 2 A similar argument leads to � 2 m 1 �� 2 m 2 � T (2 m 1 + 1 , 2 m 2 + 1) = (2 m 1 + 1)(2 m 2 + 1) . m 1 + m 2 + 1 m 1 m 2 Theorem Putting the insular and traversing arc diagrams together yields � 2 m 1 �� 2 m 2 � G 0 , 2 (2 m 1 , 2 m 2 ) = m 1 m 2 + m 1 + m 2 m 1 + m 2 m 1 m 2 � 2 m 1 �� 2 m 2 � G 0 , 2 (2 m 1 + 1 , 2 m 2 + 1) = (2 m 1 + 1)(2 m 2 + 1) . m 1 + m 2 + 1 m 1 m 2

Structure theorem Theorem (Polynomiality) � 2 m � Let C (2 m ) = C (2 m + 1) = . For ( g , n ) � = (0 , 1) or (0 , 2) , m G g , n ( b 1 , . . . , b n ) = C ( b 1 ) · · · C ( b n ) × � G g , n ( b 1 , . . . , b n ) , where � G g , n is a symmetric quasi-polynomial of degree 3 g − 3 + 2 n. Example � parity G g , n ( b 1 , . . . , b n ) g n 0 1 (0) 1 / ( m 1 + 1) 0 2 (0 , 0) ( m 1 m 2 + m 1 + m 2 ) / ( m 1 + m 2 ) 0 2 (1 , 1) (2 m 1 + 1)(2 m 2 + 1) / ( m 1 + m 2 + 1) 0 3 (0 , 0 , 0) ( m 1 + 1)( m 2 + 1)( m 3 + 1) 0 3 (1 , 1 , 0) (2 m 1 + 1)(2 m 2 + 1)( m 3 + 1) 12 ( m 2 + 5 m + 12) 1 1 1 (0)

Topological recursion Theorem For S = { 2 , 3 , . . . , n } and b 1 > 0 , � G g , n ( b 1 , b S ) = b k G g , n − 1 ( b 1 + b k − 2 , b S \{ k } ) k ∈ S � � � � + G g − 1 , n +1 ( i , j , b S ) + G g 1 , | I | +1 ( i , b I ) G g 2 , | J | +1 ( j , b J ) . i + j = b 1 − 2 g 1 + g 2 = g I ⊔ J = S Any G g , n ( b ) can be computed from the initial conditions G g , n ( 0 ) = 1 . Remark The ( g , n ) case depends on ( g , n − 1), ( g − 1 , n + 1), and � g 1 + g 2 = g , ( g 1 , n 1 ) × ( g 2 , n 2 ) for n 1 + n 2 = n + 1 .

Topological recursion — Sketch proof What happens when we cut along an arc that meets boundary 1? The arc has endpoints on distinct boundaries. ( g , n ) � ( g , n − 1) The arc has endpoints on boundary 1 and is “non-separating”. ( g , n ) � ( g − 1 , n + 1) The arc has endpoints on boundary 1 and is “separating”. ( g , n ) � ( g 1 , n 1 ) × ( g 2 , n 2 ) 1 2

Clean arc diagrams Call an arc diagram clean if there is no arc “parallel to the boundary” (so that cutting along it produces a disk). Definition Let N g , n ( b 1 , . . . , b n ) be the number of clean arc diagrams on a genus g surface with n labelled boundaries and b i points chosen on boundary i . Example N g , n (0 , . . . , 0) = 1 N 0 , 1 (2 m ) = δ m , 0 N 1 , 1 (2) = 1

Clean structure theorem Theorem Let ¯ b = b + δ b , 0 . For ( g , n ) � = (0 , 1) or (0 , 2) and b � = 0 , N g , n ( b 1 , . . . , b n ) = ¯ b 1 · · · ¯ b n × � N g , n ( b 2 1 , . . . , b 2 n ) , where � N g , n is a symmetric quasi-polynomial of degree 3 g − 3 + n. Example � N g , n ( b 2 1 , . . . , b 2 parity n ) g n 0 3 (0 , 0 , 0) 1 0 3 (1 , 1 , 0) 1 1 4 ( b 2 1 + b 2 2 + b 2 3 + b 2 0 4 (0 , 0 , 0 , 0) 4 ) + 2 1 4 ( b 2 1 + b 2 2 + b 2 3 + b 2 4 ) + 1 0 4 (1 , 1 , 0 , 0) 2 1 4 ( b 2 1 + b 2 2 + b 2 3 + b 2 0 4 (1 , 1 , 1 , 1) 4 ) + 2 48 ( b 2 1 1 1 (0) 1 + 20)

Clean topological recursion Theorem For S = { 2 , 3 , . . . , n } and ( g , n ) � = (0 , 1) , (0 , 2) , (0 , 3) and b 1 > 0 , � � ¯ � � � i m b 1 � � N g , n ( b 1 , b S ) = + N g , n − 1 ( i , b S \{ k } ) 2 k ∈ S i + m = b 1 + b k i + m = b 1 − b k � � stable � i ¯ ¯ � j m � N g 1 , | I | +1 ( i , b I ) � � + N g − 1 , n +1 ( i , j , b S ) + N g 2 , | J | +1 ( j , b J ) 2 g 1 + g 2 = g i + j + m = b 1 I ⊔ J = S Stable means we exclude terms with ( g , n ) = (0 , 1) or (0 , 2) . Proof. Similar in flavour to the “unclean” topological recursion.

Prototypical proof of polynomiality Consider ( g , n ) = (0 , 4) and let b 1 be largest. Recall that � N 0 , 3 ( b 1 , b 2 , b 3 ) = 1 if b 1 + b 2 + b 3 is even. � N 0 , 4 ( b 1 , b 2 , b 3 , b 4 ) = 1 We have b 1 � A 0 ( b 1 + b k ) + A 0 ( b 1 − b k ), 2 k =2 , 3 , 4 � 12 ( b 3 + 8 b ) � 1 b even , ¯ where A 0 ( b ) = i m = 12 ( b 3 − b ) 1 b odd . i + m = b m even If b 1 , b 2 , b 3 , b 4 are even, then � � 1 ( b 1 + b k ) 3 + 8( b 1 + b k ) � N 0 , 4 ( b 1 , b 2 , b 3 , b 4 ) = 24 b 1 k =2 , 3 , 4 � + ( b 1 − b k ) 3 + 8( b 1 − b k ) = 1 4( b 2 1 + b 2 2 + b 2 3 + b 2 4 ) + 2 .

Prototypical proof of polynomiality (continued) More generally, we need the following result. Theorem (Brion–Vergne, 1997) Let P be a convex lattice polytope in R n with non-empty interior I. Let f : R n → R be a degree d homogeneous polynomial. Then � � N P ( f , k ) = f ( x ) and N I ( f , k ) = f ( x ) x ∈ Z n ∩ kP x ∈ Z n ∩ kI are polynomials of degree n + d, with N I ( f , k ) = ( − 1) n + d N P ( f , − k ) . Corollary The functions � � p p 2 m q q p 2 m q 2 n r A m ( b ) = ¯ B m , n ( b ) = p ¯ ¯ and p + q = b p + q + r = b q even r even are odd quasi-polynomials of degree 2 m +3 and 2 m +2 n +5 , respectively.

Clean vs. unclean You can clean arc diagrams by removing arcs parallel to the boundary. Conversely, any arc diagram can be created by gluing cuffs to a clean one. Fact � b � There are ¯ a cuffs with b points on the outer boundary and a points b − a 2 on the inner boundary.

Clean vs. unclean (continued) Corollary It follows from the previous fact that � b i � n � � G g , n ( b 1 , . . . , b n ) = × N g , n ( a 1 , . . . , a n ) . b i − a i 2 a i ≡ b i i =1 Polynomiality for N g , n implies � b i � � � n finite � � n a i ( a i ) 2 d i G g , n ( b 1 , . . . , b n ) = × C ( d 1 , . . . , d n ) ¯ b i − a i 2 a i ≡ b i i =1 d 1 ,..., d n =0 i =1 � b i � finite n � � � a i ( a i ) 2 d i = C ( d 1 , . . . , d n ) ¯ . b i − a i 2 d 1 ,..., d n =0 i =1 a i ≡ b i � �� � ( 2 mi mi ) × quasi-polynomial in b i Thus, we obtain polynomiality for G g , n .