CSM Workshop 1: Zeros of Graph Polynomials Enumeration of Spanning Subgraphs with Degree Constraints Dave Wagner University of Waterloo I. The Set-Up graph notation G=(V,E) a (big) finite graph graph notation G=(V,E) H E (G ) a (big) finite graph a set of edges, i.e. a spanning subgraph graph notation G=(V,E) H E (G ) a (big) finite graph a set of edges, i.e. a spanning subgraph deg( H ) : V N the degree function of H graph notation G=(V,E) H E (G ) a (big) finite graph a set of edges, i.e. a spanning subgraph deg( H ) : V N the degree function of H the set of vertices of degree k in H is Vk ( H ) v V : deg( H , v) k energy of a subgraph J the energy of a single edge energy of a subgraph J the energy of a single edge k the “chemical potential” of a vertex of degree k energy of a subgraph J the energy of a single edge k the “chemical potential” of a vertex of degree k the energy of a (spanning) subgraph H is U ( H ) J # H k #Vk ( H ) k partition function T the absolute temperature partition function T the absolute temperature 1 k BT the inverse temperature partition function T the absolute temperature 1 k BT the inverse temperature the Boltzmann weight of a subgraph H is e U ( H ) partition function T the absolute temperature 1 k BT the inverse temperature e the Boltzmann weight of a subgraph H is the partition function is ZG e U ( H ) U ( H ) H E(G) polynomial expression let ye J and uk e k polynomial expression let ye J and for a subgraph H let uk e k u deg(H ) udeg(H ,v ) vV polynomial expression let ye J and for a subgraph H let uk e k u deg(H ) udeg(H ,v ) vV the partition function is Z G (u, y ) u H E(G) deg( H ) y #H multivariate version let x xv : v V and x deg( H ) x deg( H ,v ) v vV multivariate version let x xv : v V and x deg( H ) x deg( H ,v ) v vV the multivariate partition function is ~ ZG (u, x) u H E(G) deg( H ) x deg( H ) multivariate version let x xv : v V and x deg( H ) x deg( H ,v ) v vV the multivariate partition function is ~ ZG (u, x) then u H E(G) deg( H ) ~ 1/ 2 ZG (u, y) ZG (u, y ) x deg( H ) example let u0 u1 1 and uk 0 for all k>=2 example let u0 u1 1 and u deg(H ) udeg(H ,v ) vV uk 0 for all k>=2 1 if H is a matching otherwise 0 example let u0 u1 1 and u deg(H ) udeg(H ,v ) vV ~ ZG (u, x) uk 0 for all k>=2 1 if H is a matching otherwise 0 and Z G (u, y ) are, respectively, the multivariate and univariate matching polynomials of G vertex-dependent activities the chemical potentials can vary from vertex to vertex: vertex-dependent activities the chemical potentials can vary from vertex to vertex: let u (v) u , u ,..., u (v) 0 (v) 1 (v) d where d deg( G, v) vertex-dependent activities the chemical potentials can vary from vertex to vertex: let u (v) u , u ,..., u and redefine (v) 0 (v) 1 (v) d u deg(H ) u vV where d deg( G, v) (v) deg( H ,v ) vertex-dependent activities the chemical potentials can vary from vertex to vertex: let u (v) u , u ,..., u and redefine (v) 0 (v) 1 (v) d u deg(H ) u where d deg( G, v) (v) deg( H ,v ) vV the multivariate partition function is still ~ ZG (u, x) u H E(G) deg( H ) x deg( H ) II. The Results the key polynomials for each vertex v of G form the key polynomial d (v) k K v ( z ) uk z k 0 k d in which d deg( G, v) the key polynomials for each vertex v of G form the key polynomial d (v) k K v ( z ) uk z k 0 k d Since u (v) k e k( v ) in which d deg( G, v) this polynomial depends on T the key polynomials for each vertex v of G form the key polynomial d (v) k K v ( z ) uk z k 0 k d Since u (v) k e except when all k( v ) (v ) k in which d deg( G, v) this polynomial depends on T 0, the key polynomials for each vertex v of G form the key polynomial d (v) k K v ( z ) uk z k 0 k d Since u (v) k e except when all that is, when all k( v ) in which d deg( G, v) this polynomial depends on T (v ) k 0, (v) k 0,1 u first theorem Assume that all zeros of all the keys are within an angle of the negative real axis 0 / 2. Then… first theorem Assume that all zeros of all the keys are within an angle of the negative real axis 0 / 2. Then… 1. If arg( xv ) 2 ~ for all v then ZG (u, x) 0. first theorem Assume that all zeros of all the keys are within an angle of the negative real axis 0 / 2. Then… 1. If arg( xv ) 2. If 2 ~ for all v then ZG (u, x) 0. arg( y) 2 then Z G (u, y ) 0. first theorem Assume that all zeros of all the keys are within an angle of the negative real axis 0 / 2. Then… 1. If arg( xv ) 2. If 2 ~ for all v then ZG (u, x) 0. arg( y) 2 then Z G (u, y ) 0. This statement is independent of the size of the graph…. first theorem Assume that all zeros of all the keys are within an angle of the negative real axis 0 / 2. Then… 1. If arg( xv ) 2. If 2 ~ for all v then ZG (u, x) 0. arg( y) 2 then Z G (u, y ) 0. This statement is independent of the size of the graph…. so it can be used for thermodynamic limits. first theorem first theorem first theorem Consider the case 0: first theorem Consider the case 0: Assume that all zeros of all the keys are nonpositive real numbers. Then… first theorem Consider the case 0: Assume that all zeros of all the keys are nonpositive real numbers. Then… 1. If arg( xv ) 2 ~ for all v then ZG (u, x) 0. first theorem Consider the case 0: Assume that all zeros of all the keys are nonpositive real numbers. Then… 1. If arg( xv ) 2 ~ for all v then ZG (u, x) 0. (the half-plane property) first theorem Consider the case 0: Assume that all zeros of all the keys are nonpositive real numbers. Then… 1. If arg( xv ) 2 ~ for all v then ZG (u, x) 0. (the half-plane property) 2. All zeros of Z G (u, y ) are nonpositive real numbers. the Heilmann-Lieb (1972) theorem let u0 u1 1 and uk 0 for all k>=2 the Heilmann-Lieb (1972) theorem let u0 u1 1 for each vertex v, and uk 0 for all k>=2 K v ( z ) 1 dz has only real nonpositive zeros…. d deg( G, v) the Heilmann-Lieb (1972) theorem let u0 u1 1 for each vertex v, and uk 0 for all k>=2 K v ( z ) 1 dz d deg( G, v) has only real nonpositive zeros.… 1. The multivariate matching polynomial has the half-plane property. ~ ZG (u, x) the Heilmann-Lieb (1972) theorem let u0 u1 1 for each vertex v, and uk 0 for all k>=2 d deg( G, v) K v ( z ) 1 dz has only real nonpositive zeros…. 1. The multivariate matching polynomial has the half-plane property. 2. The univariate matching polynomial has only real nonpositive zeros. ~ ZG (u, x) Z G (u, y ) a generalization fix functions f , g : V (G ) N such that f g f 1 (at every vertex) a generalization fix functions f , g : V (G ) N such that f g f 1 (at every vertex) choose vertex chemical potentials so that u (v) k 1 if f (v) k g (v) otherwise 0 a generalization fix functions f , g : V (G ) N such that f g f 1 (at every vertex) choose vertex chemical potentials so that u (v) k 1 if f (v) k g (v) otherwise 0 Then every key has only real nonpositive zeros, so that 1. ~ ZG (u, x) has the half-plane property (new) 2. Z G (u, y ) has only real nonpositive zeros (W. 1996) a theorem of Ruelle (1999) let u0 u1 u2 1 and uk 0 for all k>=3 a theorem of Ruelle (1999) let u0 u1 u2 1 for each vertex v, and uk 0 for all k>=3 d 2 K v ( z ) 1 dz z 2 has all its zeros within /4 of the negative real axis a theorem of Ruelle (1999) let u0 u1 u2 1 uk 0 for all k>=3 d 2 K v ( z ) 1 dz z 2 for each vertex v, has all its zeros within 1. If arg( xv ) and 4 /4 of the negative real axis ~ for all v then ZG (u, x) 0. (new) a theorem of Ruelle (1999) let u0 u1 u2 1 has all its zeros within 2. If arg( y ) uk 0 for all k>=3 d 2 K v ( z ) 1 dz z 2 for each vertex v, 1. If arg( xv ) and 4 2 /4 of the negative real axis ~ for all v then ZG (u, x) 0. (new) then Z G (u, y ) 0. a theorem of Ruelle (1999) let u0 u1 u2 1 has all its zeros within 2. If arg( y ) uk 0 for all k>=3 d 2 K v ( z ) 1 dz z 2 for each vertex v, 1. If arg( xv ) and 4 2 /4 of the negative real axis ~ for all v then ZG (u, x) 0. (new) then Z G (u, y ) 0. 2 (Ruelle proves that for 2. it suffices that Re( y ) ( 1) 2 for a graph with maximum degree .) second theorem Assume that all zeros of all the keys have modulus at least . Then… ~ 1. If xv for all v then ZG (u, x) 0. 2. If y 2 then Z G (u, y ) 0. third theorem Assume that all zeros of all the keys have modulus at most , and that the degree of each key equals the degree of the corresponding vertex. Then… ~ 1. If xv for all v then ZG (u, x) 0. 2. If y 2 then Z G (u, y ) 0. corollary If all zeros of all keys are on the unit circle, and all keys have the same degree as the corresponding vertex, then every zero of Z G (u, y ) is on the unit circle. corollary If all zeros of all keys are on the unit circle, and all keys have the same degree as the corresponding vertex, then every zero of Z G (u, y ) is on the unit circle. For any graph G, every zero of y H E (G ) is on the unit circle. #H deg( G, v) vV deg( H , v) 1 application consider a sequence of graphs G whose union is an infinite graph application consider a sequence of graphs G whose union is an infinite graph assume that each graph G is d-regular application consider a sequence of graphs G whose union is an infinite graph assume that each graph G is d-regular that all keys are the same K ( , z ) application consider a sequence of graphs G whose union is an infinite graph assume that each graph G is d-regular that all keys are the same K ( , z ) and that the thermodynamic limit free energy exists: 1 f ( , J , μ) lim log Z G (u, y ) G # V (G ) application consider a sequence of graphs G whose union is an infinite graph assume that each graph G is d-regular that all keys are the same K ( , z ) and that the thermodynamic limit free energy exists: 1 f ( , J , μ) lim log Z G (u, y ) G # V (G ) If the free energy is non-analytic at a nonnegative real * then K (* , z ) has a zero not at the origin with nonnegative real part. example 1. let u0 1 and uk 0 for all k>=3 example 1. let u0 1 and uk 0 for all k>=3 d 2 the key is K ( z ) 1 du1 z u2 z 2 example 1. let u0 1 and uk 0 for all k>=3 d 2 the key is K ( z ) 1 du1 z u2 z 2 if u1 e 1 0 then the zeros of K(z) have negative real part…. No phase transitions for any physical (J,T) example 1. let u0 1 and uk 0 for all k>=3 d 2 the key is K ( z ) 1 du1 z u2 z 2 if u1 e 1 0 then the zeros of K(z) have negative real part…. No phase transitions for any physical (J,T) from the second theorem it follows that when there is no phase transition for J 2 k BT log d2 u1 0 example 2. fix functions f , g : V (G ) N such that (at every vertex) f g f 3 example 2. fix functions f , g : V (G ) N such that (at every vertex) choose vertex chemical potentials so that u (v) k 1 if f (v) k g (v) otherwise 0 f g f 3 example 2. fix functions f , g : V (G ) N such that (at every vertex) f g f 3 choose vertex chemical potentials so that u (v) k 1 if f (v) k g (v) otherwise 0 When the thermodynamic limit f ( , J , μ) exists it is analytic for all physical values of (J,T). (no phase transitions) example 3. in a 2d-regular graph, consider the key 2d d K ( z ) 1 z u z 2 d d example 3. in a 2d-regular graph, consider the key 2d d K ( z ) 1 z u z 2 d d for a thermodynamic limit a phase transition with can only happen at f ( , J , μ) of these k BT 2d 2 log log 4 d 2 2 d 1 (dJ 2 ) log e 2 d III. Summary summary * very general set-up, but it records no global structure summary * very general set-up, but it records no global structure * unifies a number of previously considered things summary * very general set-up, but it records no global structure * unifies a number of previously considered things * very mild hypotheses, but similarly weak conclusions about absence of phase transitions: summary * very general set-up, but it records no global structure * unifies a number of previously considered things * very mild hypotheses, but similarly weak conclusions about absence of phase transitions: * many general “soft” results summary * very general set-up, but it records no global structure * unifies a number of previously considered things * very mild hypotheses, but similarly weak conclusions about absence of phase transitions: * many general “soft” results * some quantitative “hard” versions of qualitatively intuitive results summary * very general set-up, but it records no global structure * unifies a number of previously considered things * very mild hypotheses, but similarly weak conclusions about absence of phase transitions: * many general “soft” results * some quantitative “hard” versions of qualitatively intuitive results * proofs are short and easy: (half-plane property/polarize & Grace-Walsh-Szego/ “monkey business”/diagonalize)
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