siegel s theorem edge coloring and a holant dichotomy
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Siegels Theorem, Edge Coloring, and a Holant Dichotomy Tyson - PowerPoint PPT Presentation

Sep 28, 2022 •11 likes •1.07k views

Siegels Theorem, Edge Coloring, and a Holant Dichotomy Tyson Williams (University of Wisconsin-Madison) Joint with: Jin-Yi Cai and Heng Guo (University of Wisconsin-Madison) 1 / 41 2 / 41 Finiteness Theorems Theorem (Siegels Theorem)

  1. Tutte Polynomial Definition The Tutte polynomial of an undirected graph G is E ( G ) = ∅ ,  1    xT ( G \ e ; x , y ) e ∈ E ( G ) is a bridge,  T ( G ; x , y ) = yT ( G \ e ; x , y ) e ∈ E ( G ) is a loop,     T ( G \ e ; x , y ) + T ( G / e ; x , y ) otherwise, where G \ e is the graph obtained by deleting e and G / e is the graph obtained by contracting e . 17 / 41

  2. Tutte Polynomial Definition The Tutte polynomial of an undirected graph G is E ( G ) = ∅ ,  1    xT ( G \ e ; x , y ) e ∈ E ( G ) is a bridge,  T ( G ; x , y ) = yT ( G \ e ; x , y ) e ∈ E ( G ) is a loop,     T ( G \ e ; x , y ) + T ( G / e ; x , y ) otherwise, where G \ e is the graph obtained by deleting e and G / e is the graph obtained by contracting e . The chromatic polynomial is χ ( G ; λ ) = ( − 1) | V |− 1 λ T( G ; 1 − λ, 0) . 17 / 41

  3. Tutte Polynomial Dichotomy Theorem (Vertigan) For any x , y ∈ C , the problem of evaluating the Tutte polynomial at ( x , y ) over planar graphs is #P-hard unless ( x − 1)( y − 1) ∈ { 1 , 2 } or ( x , y ) ∈ { (1 , 1) , ( − 1 , − 1) , ( ω, ω 2 ) , ( ω 2 , ω ) } , where ω = e 2 π i / 3 . In each of these exceptional cases, the computation can be done in polynomial time. y 4 3 2 1 x � 2 � 1 1 2 3 4 � 1 � 2 18 / 41

  4. Medial Graph Definition (a) (b) (c) A plane graph (a), its medial graph (c), and the two graphs superimposed (b). 19 / 41

  5. Directed Medial Graph Definition (a) (b) (c) A plane graph (a), its directed medial graph (c), and the two graphs superimposed (b). 20 / 41

  6. Eulerian Graphs and Eulerian Partitions Definition A graph is Eulerian if every vertex has an even degree. 1 21 / 41

  7. Eulerian Graphs and Eulerian Partitions Definition A graph is Eulerian if every vertex has an even degree. 1 A digraph is Eulerian if “in degree” = “out degree” at every vertex. 2 21 / 41

  8. Eulerian Graphs and Eulerian Partitions Definition A graph is Eulerian if every vertex has an even degree. 1 A digraph is Eulerian if “in degree” = “out degree” at every vertex. 2 An Eulerian partition of an Eulerian digraph � G is a partition of the 3 edges of � G such that each part induces an Eulerian digraph. Let π κ ( � G ) be the set of Eulerian partitions of � G into at most κ parts. 21 / 41

  9. Eulerian Graphs and Eulerian Partitions Definition A graph is Eulerian if every vertex has an even degree. 1 A digraph is Eulerian if “in degree” = “out degree” at every vertex. 2 An Eulerian partition of an Eulerian digraph � G is a partition of the 3 edges of � G such that each part induces an Eulerian digraph. Let π κ ( � G ) be the set of Eulerian partitions of � G into at most κ parts. Example with two monochromatic vertices (of degree 4). 21 / 41

  10. Crucial Identity Theorem (Ellis-Monaghan) For a plane graph G, � 2 µ ( c ) , κ T( G ; κ + 1 , κ + 1) = c ∈ π κ ( � G m ) where µ ( c ) is the number of monochromatic vertices in c. 22 / 41

  11. Connection to Holant Then 2 µ ( c ) = Holant( G m ; E ) , � c ∈ π κ ( � G m ) where E  2 if w = x = y = z    1 if w = x � = y = z     E ( w z x y ) = 0 if w = y � = x = z  1 if w = z � = x = y      0 otherwise.  Denote E by � 2 , 1 , 0 , 1 , 0 � . 23 / 41

  12. Connection to Holant Then 2 µ ( c ) = Holant( G m ; E ) , � c ∈ π κ ( � G m ) where E  2 if w = x = y = z    1 if w = x � = y = z     E ( w z x y ) = 0 if w = y � = x = z  1 if w = z � = x = y      0 otherwise.  Denote E by � 2 , 1 , 0 , 1 , 0 � . 23 / 41

  13. Connection to Holant Then 2 µ ( c ) = Holant( G m ; E ) , � c ∈ π κ ( � G m ) where E  2 if w = x = y = z    1 if w = x � = y = z     E ( w z x y ) = 0 if w = y � = x = z  1 if w = z � = x = y      0 otherwise.  Denote E by � 2 , 1 , 0 , 1 , 0 � . 23 / 41

  14. Connection to Holant Then 2 µ ( c ) = Holant( G m ; E ) , � c ∈ π κ ( � G m ) where E  2 if w = x = y = z    1 if w = x � = y = z     E ( w z x y ) = 0 if w = y � = x = z  1 if w = z � = x = y      0 otherwise.  Denote E by � 2 , 1 , 0 , 1 , 0 � . 23 / 41

  15. Connection to Holant Then 2 µ ( c ) = Holant( G m ; E ) , � c ∈ π κ ( � G m ) where E  2 if w = x = y = z    1 if w = x � = y = z     E ( w z x y ) = 0 if w = y � = x = z  1 if w = z � = x = y      0 otherwise.  Denote E by � 2 , 1 , 0 , 1 , 0 � . 23 / 41

  16. #P-hardness of # κ -EdgeColoring Theorem # κ - EdgeColoring is #P-hard over planar κ -regular graphs for κ ≥ 3 . 24 / 41

  17. #P-hardness of # κ -EdgeColoring Theorem # κ - EdgeColoring is #P-hard over planar κ -regular graphs for κ ≥ 3 . Proof for κ = 3 . Reduce from Holant( · ; � 2 , 1 , 0 , 1 , 0 � ) to Holant( · ; AD 3 ) in two steps: polynomial Holant( · ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant( · ; � 0 , 1 , 1 , 0 , 0 � ) ( interpolation ) ≤ T Holant( · ; AD 3 ) (gadget construction) 24 / 41

  18. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  19. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  20. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  21. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  22. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  23. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  24. Gadget Construction Step Holant( G ; � 0 , 1 , 1 , 0 , 0 � ) ≤ T Holant( G ′ ; AD 3 ) AD 3 w z y AD 3 x  0 if w = x = y = z    if w = x � = y = z 1     f ( w z x y ) = � 0 , 1 , 1 , 0 , 0 � = 1 if w = y � = x = z  if w = z � = x = y 0      0 otherwise.  25 / 41

  25. Polynomial Interpolation Step: Recursive Construction Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) N s N 1 N 2 N s +1 Vertices are assigned � 0 , 1 , 1 , 0 , 0 � . 26 / 41

  26. Polynomial Interpolation Step: Recursive Construction Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) N s N 1 N 2 N s +1 Vertices are assigned � 0 , 1 , 1 , 0 , 0 � . Let f s be the function corresponding to N s . Then f s = M s f 0 , where  0 2 0 0 0   1  1 1 0 0 0 0         M = 0 0 0 1 0 and f 0 = 0 .         0 0 1 0 0 1     0 0 0 0 1 0 Obviously f 1 = � 0 , 1 , 1 , 0 , 0 � . 26 / 41

  27. Polynomial Interpolation Step: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 27 / 41

  28. Polynomial Interpolation Step: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1     x 0 0 0 0 + 1 3 x − 1 0 1 0 0 0    3  f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 27 / 41

  29. Polynomial Interpolation Step: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1     x 0 0 0 0 + 1 3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 27 / 41

  30. Polynomial Interpolation Step: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1     x 0 0 0 0 + 1 3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 Note f (4) = E = � 2 , 1 , 0 , 1 , 0 � . 27 / 41

  31. Polynomial Interpolation Step: Eigenvectors and Eigenvalues Spectral decomposition M = P Λ P − 1 , where  1 − 2 0 0 0   2 0 0 0 0  1 1 0 0 0 0 − 1 0 0 0         P = 0 0 1 1 0 and Λ = 0 0 1 0 0 .         0 0 1 − 1 0 0 0 0 − 1 0     0 0 0 0 1 0 0 0 0 1 Let x = 2 2 s . Then x − 1     x 0 0 0 0 + 1 3 x − 1 0 1 0 0 0    3  f ( x ) = f 2 s = P Λ 2 s P − 1 f 0 = P   P − 1 f 0 =   0 0 1 0 0 0 .         0 0 0 1 0 1     0 0 0 0 1 0 Note f (4) = E = � 2 , 1 , 0 , 1 , 0 � . (Side note: picking s = 1 so that x = 4 only works when κ = 3.) 27 / 41

  32. Polynomial Interpolation Step: The Interpolation Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) 28 / 41

  33. Polynomial Interpolation Step: The Interpolation Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant( G ; f (4)) ≤ T Holant( G ; f ( x )) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) 28 / 41

  34. Polynomial Interpolation Step: The Interpolation Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant( G ; f (4)) ≤ T Holant( G ; f ( x )) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( x ) = Holant( G ; f ( x )) ∈ Z [ x ] has degree n . 28 / 41

  35. Polynomial Interpolation Step: The Interpolation Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant( G ; f (4)) ≤ T Holant( G ; f ( x )) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( x ) = Holant( G ; f ( x )) ∈ Z [ x ] has degree n . Let G s be the graph obtained by replacing every vertex in G with N s . Then Holant( G 2 s ; � 0 , 1 , 1 , 0 , 0 � ) = p (2 2 s ). 28 / 41

  36. Polynomial Interpolation Step: The Interpolation Holant( G ; � 2 , 1 , 0 , 1 , 0 � ) = T Holant( G ; f (4)) ≤ T Holant( G ; f ( x )) ≤ T Holant( G s ; � 0 , 1 , 1 , 0 , 0 � ) If G has n vertices, then p ( x ) = Holant( G ; f ( x )) ∈ Z [ x ] has degree n . Let G s be the graph obtained by replacing every vertex in G with N s . Then Holant( G 2 s ; � 0 , 1 , 1 , 0 , 0 � ) = p (2 2 s ). Using oracle for Holant( · ; � 0 , 1 , 1 , 0 , 0 � ), evaluate p ( x ) at n + 1 distinct points x = 2 2 s for 0 ≤ s ≤ n . By polynomial interpolation, efficiently compute the coefficients of p ( x ). QED. 28 / 41

  37. Proof Outline for Dichotomy of Holant( · ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( · ; � a , b , c � ) is in P or #P-hard. 29 / 41

  38. Proof Outline for Dichotomy of Holant( · ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( · ; � a , b , c � ) is in P or #P-hard. Attempt to construct a special arity 1 local constraint using � a , b , c � . 1 Attempt to interpolate all arity 2 local constraints of a certain form, 2 assuming we have the special arity 1 local constraint. Construct a ternary local constraint that we show is #P-hard, 3 assuming we have these arity 2 local constraints. 29 / 41

  39. Proof Outline for Dichotomy of Holant( · ; � a , b , c � ) For all a , b , c ∈ C , want to show that Holant( · ; � a , b , c � ) is in P or #P-hard. Attempt to construct a special arity 1 local constraint using � a , b , c � . 1 Attempt to interpolate all arity 2 local constraints of a certain form, 2 assuming we have the special arity 1 local constraint. Construct a ternary local constraint that we show is #P-hard, 3 assuming we have these arity 2 local constraints. For some a , b , c ∈ C , our attempts fail. In those cases, we either show the problem is in P or 1 prove #P-hardness without the help of additional signatures. 2 29 / 41

  40. edge coloring k=r Eulerian partition directed medial graph parity condition major interpolate results ternary and quaternary resutls Tutte diagonal as state sum tal_color: f(P_0) = 0 check orthogonality condition lattice condition (LC) Puiseux series planar T utte dichotomy only 5 solutions in Z for p(x,y) state sum as Holant problem any arity interpolation LC characterization for cubic polys LC satisfied by Sn or An Galois Gps 3R & 2C roots in x for p(x,y) condition for Sn Galois gp Dedkind's Theorem condition from same norm roots edge coloring k>r hard generic binary interpolation Bobby Fischer gadget planar Eulerian partition hard (tau_color) reducible p(x,y) satisfies LC for y>3 irreducible p(x,y) satisfies LC for y>3 p(x,3) satisfies LC local holographic transformation interpolate all binary edge coloring k=r hard eigenvalue shifted triple (EST) special binary interpolation binary interpolation eigenvalues planar pairing def chomatic in Tutte obtain =_4 planar Eulerian partition hard (tau_4) p(x,y) satisfies LC for y=>3 Triangle gadget obtain any a+(k-3)b-(k-2)c=0 construct unary 2nd distinct norms 1st distinct norms extra special cases EST distinct norms interpolate all binaries find planar pairing reduction to vertex coloring construct <1> in two cases generalized edge coloring hard <3(k-1),k-3,-3> hard for k>3 <6,0,-3> hard obtain <a',b',b'> assuming a+(k-3)b-(k-2)c!=0 obtain <3(k-1),k-3,-3> generic generalized anti-gadget interpolation 4th special case arity reduction local holographic transformation <(k-1)(k-2),2-k,2> hard ternary construction summary 2nd special case typical case 3rd special case 1st special case 5th special case edge coloring k>r hard a+(k-3)b-(k-2)c=0 dichotomy binary interpolation summary <a,b,c> dichotomy 30 / 41

  41. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . , x d are distinct 31 / 41

  42. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct 31 / 41

  43. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct 31 / 41

  44. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct � x is not a root of unity 31 / 41

  45. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct � x is not a root of unity Let q d ( X , Y ) ∈ Z [ X , Y ] be a homogeneous polynomial of degree d . ∀ d ∈ N , Can interpolate q d ( X , Y ) from q d ( x 0 , y 0 ) , q d ( x 1 , y 1 ) , . . . , q d ( x d , y d ) � ? 31 / 41

  46. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct � x is not a root of unity Let q d ( X , Y ) ∈ Z [ X , Y ] be a homogeneous polynomial of degree d . ∀ d ∈ N , Can interpolate q d ( X , Y ) from q d ( x 0 , y 0 ) , q d ( x 1 , y 1 ) , . . . , q d ( x d , y d ) � 31 / 41

  47. Interpolating Multivariate Polynomials Let p d ( X ) ∈ Z [ X ] be a polynomial of degree d . ∀ d ∈ N , Can interpolate p d ( X ) from p d ( x 0 ) , p d ( x 1 ) , . . . , p d ( x d ) � x 0 , x 1 , . . . are distinct � x is not a root of unity Let q d ( X , Y ) ∈ Z [ X , Y ] be a homogeneous polynomial of degree d . ∀ d ∈ N , Can interpolate q d ( X , Y ) from q d ( x 0 , y 0 ) , q d ( x 1 , y 1 ) , . . . , q d ( x d , y d ) � lattice condition 31 / 41

  48. Lattice Condition Definition We say that λ 1 , λ 2 , . . . , λ ℓ ∈ C − { 0 } satisfy the lattice condition if ℓ ∀ x ∈ Z ℓ − { 0 } � with x i = 0 , i =1 we have ℓ � λ x i i � = 1 . i =1 32 / 41

  49. Lattice Condition Definition We say that λ 1 , λ 2 , . . . , λ ℓ ∈ C − { 0 } satisfy the lattice condition if ℓ ∀ x ∈ Z ℓ − { 0 } � with x i = 0 , i =1 we have ℓ � λ x i i � = 1 . i =1 Lemma Let p ( x ) ∈ Q [ x ] be a polynomial of degree n ≥ 2 . If the Galois group of p over Q is S n or A n and 1 the roots of p do not all have the same complex norm, 2 then the roots of p satisfy the lattice condition. 32 / 41

  50. Interpolation Lemma Lemma If there exists an infinite sequence of F -gates defined by an initial signature s ∈ C n × 1 and a recurrence matrix M ∈ C n × n satisfying the following conditions, M is diagonalizable with n linearly independent eigenvectors; 1 s is not orthogonal to exactly ℓ of these linearly independent row 2 eigenvectors of M with eigenvalues λ 1 , . . . , λ ℓ ; λ 1 , . . . , λ ℓ satisfy the lattice condition; 3 then Holant( · ; F ∪ { f } ) ≤ T Holant( · ; F ) for any signature f that is orthogonal to the n − ℓ of these linearly independent eigenvectors of M to which s is also orthogonal. 33 / 41

  51. Interpolation Lemma Lemma If there exists an infinite sequence of F -gates defined by an initial signature s ∈ C n × 1 and a recurrence matrix M ∈ C n × n satisfying the following conditions, M is diagonalizable with n linearly independent eigenvectors; 1 s is not orthogonal to exactly ℓ of these linearly independent row 2 eigenvectors of M with eigenvalues λ 1 , . . . , λ ℓ ; λ 1 , . . . , λ ℓ satisfy the lattice condition; 3 then Holant( · ; F ∪ { f } ) ≤ T Holant( · ; F ) for any signature f that is orthogonal to the n − ℓ of these linearly independent eigenvectors of M to which s is also orthogonal. Our proof applies this with n = 9 and ℓ = 5. 33 / 41

  52. The Recurrence Matrix ( κ − 1) ( κ 2 +9 κ − 9 )  12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 1)(2 κ − 3)(4 κ − 3) 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)  3( κ − 3)( κ − 1) 3 κ 3 − 28 κ 2 +60 κ − 36 − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 2) ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3)   (2 κ − 3)(4 κ − 3) 12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) 9 κ 3 − 26 κ 2 +27 κ − 9 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)     2 ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 3) ( κ 3 − 12 κ 2 +22 κ − 12 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)   ( κ − 3) 2 κ 3 +6 κ 2 − 30 κ +36 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2) − 4( κ − 3)(2 κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 − 4( κ − 3)(2 κ − 3) 3( κ − 3) κ 3 +6 κ 2 − 30 κ +36 3( κ − 3) 6( κ − 3)( κ − 2) ( κ − 3) 2 ( κ − 1) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3) 2 ( κ − 2)  ( κ − 3) 2 ( κ − 3) 2 ( κ − 1) (2 κ − 3) ( 2 κ 2 − 9 κ +18 ) − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3) 34 / 41

  53. The Recurrence Matrix ( κ − 1) ( κ 2 +9 κ − 9 )  12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 1)(2 κ − 3)(4 κ − 3) 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)  3( κ − 3)( κ − 1) 3 κ 3 − 28 κ 2 +60 κ − 36 − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 2) ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3)   (2 κ − 3)(4 κ − 3) 12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) 9 κ 3 − 26 κ 2 +27 κ − 9 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)     2 ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 3) ( κ 3 − 12 κ 2 +22 κ − 12 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)   ( κ − 3) 2 κ 3 +6 κ 2 − 30 κ +36 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2) − 4( κ − 3)(2 κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 − 4( κ − 3)(2 κ − 3) 3( κ − 3) κ 3 +6 κ 2 − 30 κ +36 3( κ − 3) 6( κ − 3)( κ − 2) ( κ − 3) 2 ( κ − 1) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3) 2 ( κ − 2)  ( κ − 3) 2 ( κ − 3) 2 ( κ − 1) (2 κ − 3) ( 2 κ 2 − 9 κ +18 ) − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3) Characteristic polynomial is ( x − κ 3 ) 4 f ( x , κ ), where f ( x , κ ) = x 5 − κ 6 (2 κ − 1) x 3 − κ 9 ( κ 2 − 2 κ + 3) x 2 + ( κ − 2)( κ − 1) κ 12 x + ( κ − 1) 3 κ 15 . 34 / 41

  54. The Recurrence Matrix ( κ − 1) ( κ 2 +9 κ − 9 )  12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 1)(2 κ − 3)(4 κ − 3) 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)  3( κ − 3)( κ − 1) 3 κ 3 − 28 κ 2 +60 κ − 36 − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 2) ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3)   (2 κ − 3)(4 κ − 3) 12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) 9 κ 3 − 26 κ 2 +27 κ − 9 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)     2 ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 3) ( κ 3 − 12 κ 2 +22 κ − 12 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)   ( κ − 3) 2 κ 3 +6 κ 2 − 30 κ +36 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2) − 4( κ − 3)(2 κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 − 4( κ − 3)(2 κ − 3) 3( κ − 3) κ 3 +6 κ 2 − 30 κ +36 3( κ − 3) 6( κ − 3)( κ − 2) ( κ − 3) 2 ( κ − 1) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3) 2 ( κ − 2)  ( κ − 3) 2 ( κ − 3) 2 ( κ − 1) (2 κ − 3) ( 2 κ 2 − 9 κ +18 ) − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3) Characteristic polynomial is ( x − κ 3 ) 4 f ( x , κ ), where f ( x , κ ) = x 5 − κ 6 (2 κ − 1) x 3 − κ 9 ( κ 2 − 2 κ + 3) x 2 + ( κ − 2)( κ − 1) κ 12 x + ( κ − 1) 3 κ 15 . 1 κ 15 f ( κ 3 x , κ ), we get After replacing κ by y + 1 in p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . 34 / 41

  55. The Recurrence Matrix ( κ − 1) ( κ 2 +9 κ − 9 )  12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 1)(2 κ − 3)(4 κ − 3) 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)  3( κ − 3)( κ − 1) 3 κ 3 − 28 κ 2 +60 κ − 36 − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 2) ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3)   (2 κ − 3)(4 κ − 3) 12( κ − 3)( κ − 1) 2 ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) ( κ − 3) 2 ( κ − 1) 2( κ − 3) 2 ( κ − 2)( κ − 1) 9 κ 3 − 26 κ 2 +27 κ − 9 6( κ − 3)( κ − 2)( κ − 1) 2 ( κ − 3) 3 ( κ − 2)( κ − 1)     2 ( κ 3 − 14 κ 2 +30 κ − 18 ) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) − ( κ − 3)(2 κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1) 2 ( κ − 3) ( κ 3 − 12 κ 2 +22 κ − 12 ) − ( κ − 3) 2 ( κ − 2)(2 κ − 3) 3( κ − 3)( κ − 1)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)   ( κ − 3) 2 κ 3 +6 κ 2 − 30 κ +36 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2)  − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 κ 3 +3 κ − 9 ( κ − 3) 2 ( κ − 1) 3( κ − 3) 2 ( κ − 2) − 4( κ − 3)(2 κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3)    ( κ − 3) 2 − 4( κ − 3)(2 κ − 3) 3( κ − 3) κ 3 +6 κ 2 − 30 κ +36 3( κ − 3) 6( κ − 3)( κ − 2) ( κ − 3) 2 ( κ − 1) − 2( κ − 3)( κ − 2)(2 κ − 3) 3( κ − 3) 2 ( κ − 2)  ( κ − 3) 2 ( κ − 3) 2 ( κ − 1) (2 κ − 3) ( 2 κ 2 − 9 κ +18 ) − 4( κ − 3)(2 κ − 3) 3( κ − 3) 6( κ − 3)( κ − 2) 3( κ − 3) 6( κ − 3)( κ − 2) − 2( κ − 3)( κ − 2)(2 κ − 3) Characteristic polynomial is ( x − κ 3 ) 4 f ( x , κ ), where f ( x , κ ) = x 5 − κ 6 (2 κ − 1) x 3 − κ 9 ( κ 2 − 2 κ + 3) x 2 + ( κ − 2)( κ − 1) κ 12 x + ( κ − 1) 3 κ 15 . 1 κ 15 f ( κ 3 x , κ ), we get After replacing κ by y + 1 in p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . Want to prove: for all integers y ≥ 4, the roots of p ( x , y ) satisfy the lattice condition. 34 / 41

  56. Irreducible over Q ? We suspect that for any integer y ≥ 4, p ( x , y ) is irreducible in Q [ x ]. 35 / 41

  57. Irreducible over Q ? We suspect that for any integer y ≥ 4, p ( x , y ) is irreducible in Q [ x ]. Don’t know how to prove this. 35 / 41

  58. Irreducible Quintic Lemma For any integer y ≥ 1 , the polynomial p ( x , y ) in x has three distinct real roots and two nonreal complex conjugate roots. Proof. Discriminant. 36 / 41

  59. Irreducible Quintic Lemma For any integer y ≥ 1 , the polynomial p ( x , y ) in x has three distinct real roots and two nonreal complex conjugate roots. Proof. Discriminant. Lemma For any integer y ≥ 4 , if p ( x , y ) is irreducible in Q [ x ] , then the roots of p ( x , y ) satisfy the lattice condition. Proof. Three distinct real roots do not have the same norm. An irreducible polynomial of prime degree n with exactly two nonreal roots has S n as its Galois group over Q . Hence the roots satisfy the lattice condition. 36 / 41

  60. What if Reducible? 37 / 41

  61. What if Reducible? We know five integer solutions to p ( x , y ) = 0. For these solutions, p ( x , y ) is reducible: ( x − 1)( x 4 + x 3 + 2 x 2 − x + 1)  y = − 1   x 2 ( x 3 − x − 2)  y = 0     ( x + 1)( x 4 − x 3 − 2 x 2 − x + 1) p ( x , y ) = y = 1 ( x − 1)( x 2 − x − 4)( x 2 + 2 x + 2)  y = 2     ( x − 3)( x 4 + 3 x 3 + 2 x 2 − 5 x − 9)  y = 3 .  37 / 41

  62. What if Reducible? We know five integer solutions to p ( x , y ) = 0. For these solutions, p ( x , y ) is reducible: ( x − 1)( x 4 + x 3 + 2 x 2 − x + 1)  y = − 1   x 2 ( x 3 − x − 2)  y = 0     ( x + 1)( x 4 − x 3 − 2 x 2 − x + 1) p ( x , y ) = y = 1 ( x − 1)( x 2 − x − 4)( x 2 + 2 x + 2)  y = 2     ( x − 3)( x 4 + 3 x 3 + 2 x 2 − 5 x − 9)  y = 3 .  Lemma Only integer solutions to p ( x , y ) = 0 are (1 , − 1) , (0 , 0) , ( − 1 , 1) , (1 , 2) , (3 , 3) . 37 / 41

  63. If Reducible Lemma For any integer y ≥ 4 , if p ( x , y ) is reducible in Q [ x ] , then the roots of p ( x , y ) satisfy the lattice condition. Proof. By previous lemma, no linear factor over Z . By Gauss’ Lemma, no linear factor over Q . Then more Galois theory if p ( x , y ) factors as a product of two irreducible polynomials of degrees 2 and 3. 38 / 41

  64. Effective Siegel’s Theorem p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . 39 / 41

  65. Effective Siegel’s Theorem p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . Let ( a , b ) be an integer solution to p ( x , y ) = 0 with a � = 0. 39 / 41

  66. Effective Siegel’s Theorem p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . Let ( a , b ) be an integer solution to p ( x , y ) = 0 with a � = 0. One can show that a | b 2 . 39 / 41

  67. Effective Siegel’s Theorem p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . Let ( a , b ) be an integer solution to p ( x , y ) = 0 with a � = 0. One can show that a | b 2 . Consider g 2 ( x , y ) = y 2 x + y − x 2 + 1 . g 1 ( x , y ) = y − x 2 and (This particular choice is due to Aaron Levin.) Then g 1 ( a , b ) and g 2 ( a , b ) are integers. 39 / 41

  68. Effective Siegel’s Theorem p ( x , y ) = x 5 − (2 y + 1) x 3 − ( y 2 + 2) x 2 + ( y − 1) yx + y 3 . Let ( a , b ) be an integer solution to p ( x , y ) = 0 with a � = 0. One can show that a | b 2 . Consider g 2 ( x , y ) = y 2 x + y − x 2 + 1 . g 1 ( x , y ) = y − x 2 and (This particular choice is due to Aaron Levin.) Then g 1 ( a , b ) and g 2 ( a , b ) are integers. However, if | a | > 16, then either g 1 ( a , b ) or g 2 ( a , b ) is not an integer. 39 / 41

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