-Gaussian Ensembles and the Non-orientability of Polygonal Glueings - PowerPoint PPT Presentation

β -Gaussian Ensembles and the Non-orientability of Polygonal Glueings Michael La Croix Massachusetts Institute of Technology April 6, 2013

Gaussian Ensembles For β ∈ { 1 , 2 , 4 } an element of the β -Gaussian ensemble is constructed as A = G + G ∗ where G is n × n with i.i.d. Guassian entries selected from { R , C , H } . Motivating Question What is the value of E ( f ( A )) , when f is a symmetric function of the eigenvalues of its argument? Example 5 n + 5 n 2 + 2 n 3  β = 1  E (tr( A 4 )) = n + 2 n 3 β = 2 2 n 2 + 3 5 4 n − 5 4 n 3 β = 4  Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 1 / 15

General β via Eigenvalue Density The eigenvalues of A are all real with joint density proportional to � n � λ 2 − β | λ i − λ j | β exp � � i 2 2 i =1 1 ≤ i<j ≤ n Theorem For every θ , E ( p θ ( λ )) β is a polynomial in the variables n and b = 2 β − 1 . Example E ( p 4 ( λ )) β = (1 + b + 3 b 2 ) n + 5 bn 2 + 2 n 3 Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 2 / 15

A Recurrence behind the theorem 1 2 Set Ω := e − 2(1+ b ) p 2 ( x ) | V ( x ) | 1+ b , so that � f � = E ( f ( x )) = c b,n � R n f Ω d x . Integrate ∂ ∂ 1+ b e − p 2( x ) 2 x j +1 x j +1 p θ ( x )Ω = p θ ( x ) | V ( x ) | 2(1+ b ) 1 1 ∂x 1 ∂x 1 N x j +1 im i ( θ ) x i + j p θ ( x ) 1+ b x j +2 = ( j + 1) x j � 2 � 1 1 p θ ( x )Ω + p θ \ i ( x )Ω + 1 Ω − p θ ( x )Ω 1 1 1+ b x 1 − x i i ∈ θ i =2 to get An Algebraic recurrence Example j � � � � � p j +2 p θ � = b ( j + 1) � p j p θ � + α im i ( θ ) p i + j p θ \ i + � p l p j − l p θ � . i ∈ θ l =0 Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 3 / 15

What’s being counted? Example (for β ∈ { 1 , 2 , 4 } ) A ij i j tr( A 4 ) = � A li A jk A ij A jk A kl A li i,j,k,l l k A kl � n � � n � tr( A 4 ) � � E = 1! E( A 11 A 11 A 11 A 11 ) + 2! E(4 A 11 A 11 A 12 A 21 ) 1 2 � n � � n � + 2! E( A 12 A 21 A 12 A 21 ) + 3! E(2 A 12 A 21 A 13 A 31 ) 2 3 � n � � n � + 4! E( A 12 A 23 A 34 A 41 ) + 3! E(4 A 11 A 12 A 23 A 31 ) 4 3 � n � + 2! E(2 A 11 A 12 A 22 A 21 ) 2 Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 4 / 15

What’s being counted? Example (for β ∈ { 1 , 2 , 4 } ) 1 1 1 1 A ij i j 2 1 1 1 tr( A 4 ) = � A li A jk A ij A jk A kl A li 1 2 i,j,k,l 1 1 l k A kl 1 1 1 2 � n � � n � tr( A 4 ) � � E = 1! E( A 11 A 11 A 11 A 11 ) + 2! E(4 A 11 A 11 A 12 A 21 ) 1 1 1 2 2 1 � n � � n � + 2! E( A 12 A 21 A 12 A 21 ) + 3! E(2 A 12 A 21 A 13 A 31 ) 1 2 2 3 2 1 � n � � n � 2 1 + 4! E( A 12 A 23 A 34 A 41 ) + 3! E(4 A 11 A 12 A 23 A 31 ) 4 3 1 3 � n � 1 2 + 2! E(2 A 11 A 12 A 22 A 21 ) 2 3 1 Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 4 / 15

Expectations as Sums Since the entries of A are independent Guassians, � � E ( A i 1 j 1 A i 2 j 2 . . . A i k j k ) = E ( A u A v ) m ( u,v ) ∈ m summed over perfect matchings of the multiset { i 1 j 1 , i 2 j 2 , . . . , i k j k } � # { pairings consistent with p } p a painting � = # { paintings consistent with m } m a matching Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 5 / 15

Count the polygon glueings in 2 different ways ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 ( n ) 1 12 n 0 0 ( n ) 2 0 0 0 0 0 ( n ) 2 0 0 0 2 n ( n- 1) ( n ) 2 ( n ) 2 0 0 0 0 0 0 0 0 0 0 2 n ( n- 1) 0 0 ( n ) 2 0 0 ( n ) 2 0 0 0 0 0 0 2 n ( n- 1) 0 ( n ) 2 0 0 ( n ) 2 0 0 0 0 0 0 0 2 n ( n- 1) 0 ( n ) 2 ( n ) 2 0 0 0 0 0 0 ( n ) 2 0 0 3 n ( n- 1) 0 0 ( n ) 3 0 0 0 0 0 0 0 0 0 n ( n- 1)( n- 2) ( n ) 3 0 0 0 0 0 0 0 0 0 0 0 n ( n- 1)( n- 2) n 3 n 3 n 2 n 2 n 2 n 2 n 2 n n n n n Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 6 / 15

Polygon Glueings = Maps Identifying the edges of a polygon creates a surface. = Its boundary is a graph embedded in the surface. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 7 / 15

Polygon Glueings = Maps = = Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 7 / 15

Polygon Glueings = Maps Extra polygons give extra faces (and possibles extra components) Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 7 / 15

Graphs, Surfaces, and Maps Definition Example A surface is a compact 2 -manifold without boundary. (Non-orientable surfaces are permitted.) Definition A graph is a finite set of vertices together with a finite set of edges , such that each edge is associated with either one or two vertices. (It may have loops / parallel edges.) Definition A map is a 2-cell embedding of a graph in a surface. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 8 / 15

Equivalence of Maps Two maps are equivalent if the embeddings are homeomorphic. Homeomorphisms are more complicated than we might think Dehn Twists Y-Homeomorphisms Not Present for Photo Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 9 / 15

Equivalence of Maps Two maps are equivalent if the embeddings are homeomorphic. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 9 / 15

Ribbon Graphs, Flags, and Rooted Maps Example Definition The neighbourhood of the graph determines a ribbon graph , and the boundaries of ribbons determine flags. Definition Automorphisms permute flags, and a rooted map is a map together with a distinguished orbit of flags. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 10 / 15

Ribbon Graphs, Flags, and Rooted Maps Example Definition The neighbourhood of the graph determines a ribbon graph, and the boundaries of ribbons determine flags . Definition Automorphisms permute flags, and a rooted map is a map together with a distinguished orbit of flags. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 10 / 15

Ribbon Graphs, Flags, and Rooted Maps Example Definition The neighbourhood of the graph determines a ribbon graph, and the boundaries of ribbons determine flags. Definition Automorphisms permute flags, and a rooted map is a map together with a distinguished orbit of flags. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 10 / 15

Twisted Ribbons allow Non-Orientable Maps Example (A map on the Klein Bottle) Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 11 / 15

Twisted Ribbons allow Non-Orientable Maps Example (A map on the Torus) Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 11 / 15

Root Edge Deletion A rooted map with k edges can be thought of as a sequence of k maps. 2 n 2 n 2 2 2 α 2 b b bn n 2 2 bn bn n n Consecutive submaps differ in genus by 0, 1, or 2, and these steps are marked by 1 , b , and a = α 2 to assign a weight to a glueing. Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 12 / 15

Algebraic and Combinatorial Recurrences agree An Algebraic Recurrence Example j � � � � � p j +2 p θ � = b ( j + 1) � p j p θ � + α im i ( θ ) + � p l p j − l p θ � . p i + j p θ \ i i ∈ θ l =0 A Combinatorial Recurrence It corresponds to a combinatorial recurrence for counting polygon glueings. j-l i i+j j j+2 j+2 j+2 l Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 13 / 15

3 2 2 2 n n bn bn b 3 2 2 2 n n bn bn b 2 n 2 n b bn bn Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 14 / 15

The Future The combinatorial interpretation has a two-parameter refinement. Is there a corresponding matrix question? At b = 0 , we obtain glueings in 2 g : 1 correspondence with orientable glueings. Can this correspondence be made to preserve vertex degrees as well as face degrees? A similar recurrence describes moments of the β -Laguerre hypermaps . distribution, with maps replaced by For β ∈ { 1 , 2 , 4 } we can refine the combinatorial model and compute moments of XA . Is there a model for the β -Ensembles where this interpretation makes sense. For β = 1 and β = 2 , there is a natural duality between vertices and faces. What operation replaces it for b -weighted glueings? Jack symmetric functions that needs to be explored. The connection with Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 15 / 15

The End Thank You Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 15 / 15

Example � x 3 2 x 2 � � z 6 � is enumerated by ( y 3 y 4 y 5 ) . 3 2 ν = [2 3 , 3 2 ] ǫ = [2 6 ] φ = [3 , 4 , 5] Return Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 16 / 15

Explicit Formulae The hypermap series can be computed explicitly when H consists of orientable hypermaps or all hypermaps. Theorem (Jackson and Visentin - 1990) When H is the set of orientable hypermaps, �� �� = t ∂ � � � t | θ | H θ s θ ( x ) s θ ( y ) s θ ( z ) H O p ( x ) , p ( y ) , p ( z ); 0 ∂t ln � � � θ ∈ P t =1 . Theorem (Goulden and Jackson - 1996) When H is the set of all hypermaps (orientable and non-orientable), �� �� = 2 t ∂ t | θ | 1 � � � p ( x ) , p ( y ) , p ( z ); 1 ∂t ln Z θ ( x ) Z θ ( y ) Z θ ( z ) H A � H 2 θ � � θ ∈ P t =1 . Michael La Croix (MIT) β -Polygonal Glueings April 6, 2013 17 / 15