Shapes of euclidean polyhedra and hyperbolic geometry Ivan - PowerPoint PPT Presentation

Shapes of euclidean polyhedra and hyperbolic geometry Ivan Izmestiev joint work with François Fillastre (University of Cergy-Pontoise), http://arxiv.org/abs/1310.1560 Second ERC Workshop Delaunay Geometry: Polytopes, Triangulations, and Spheres Berlin, October 9, 2013 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 1 / 25

Main result V ∶= a (positively spanning) configuration of n unit vectors in R d . M ( V ) ∶= the space of (homothety classes of) polytopes with outer facet normals V . Theorem The space M ( V ) is a polyhedral ball of dimension n − d − 1 . Each cell of M ( V ) carries a natural hyperbolic metric. Example 6 vectors in R 3 determine a 2-dimensional complex. Metrically this is a right-angled hyperbolic hexagon. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 2 / 25

Main ingredients Combinatorics (linear algebra): studying the combinatorial structure of the polyhedron P ( h ) ∶= { Vx ≤ h } ⊂ R d , V ∈ R n × d fixed , h ∈ R n variable Gale diagrams ↝ h ∈ C ( V ) a subfan of the secondary fan of V . Geometry (bilinear algebra): studying the second intrinsic volume (“quermassintegral”) ⎧ area ( P ( h )) for d = 2 ⎪ ⎪ ⎪ ⎪ ⎪ 2 area ( ∂ P ( h )) for d = 3 vol 2 ( h ) = ⎨ 1 ⎪ ⎪ ⎪ α F ( P ( h )) vol 2 ( F ) for d ≥ 4 ⎪ ∑ ⎪ ⎩ dim F = 2 Alexandrov-Fenchel inequalities for mixed volumes ↝ quadratic form of signature (+ , − ,..., −) . Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 3 / 25

Case d = 2 Bavard, Ghys’92: hyperbolic metric on the space of polygons with fixed edge directions ℓ 1 ,...,ℓ n ∈ C ( V ) ⇔ v i n ℓ i v i = 0 , ℓ i ≥ 0 ∀ i ∑ ℓ i i = 1 Thus C ( V ) = R n ≥ 0 ∩ ( n − 2 ) -subspace Polygons up to translation = pointed ( n − 2 ) -cone with ≤ n facets. Polygons up to similarity = ( n − 3 ) -polytope with n − 2, n − 1, or n facets. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 4 / 25

Facets of C ( V ) ℓ i = 0 ⇔ i -th edge contracts to a point n − 1 facets n − 2 facets n facets tr 2 ( ∆ n − 3 ) tr ( ∆ n − 3 ) ∆ n − 3 ℓ 3 = 0 v 2 v 1 ℓ 5 = 0 ℓ 1 = 0 C ( V ) ℓ 2 = 0 ℓ 4 = 0 v 3 v 5 v 4 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 5 / 25

The area is a quadratic form Consider the area of polygons with edge normals V : C ( V ) → R ℓ ↦ area ( P ( ℓ )) area ( P ( ℓ )) = q ( ℓ ) Theorem where q is a quadratic form of signature (+ , − ,..., −) . p 4 Why quadratic: Vertex coordinates are linear functions of ℓ p 3 p 5 area ( P ( ℓ )) = 1 2 ( det ( p 2 − p 1 , p 3 − p 1 ) + det ( p 3 − p 1 , p 4 − p 1 )+ det ( p 5 − p 1 , p 4 − p 1 )) p 1 p 2 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 6 / 25

The signature q ( ℓ ) = area ( S 1 ) − area ( S 2 ) − area ( S 3 ) ℓ 2 ℓ 1 area ( S 1 ) = ( a ℓ 2 + b ℓ 3 ) 2 = ( a ′ ℓ 1 + b ′ ℓ 5 ) 2 P ( ℓ ) area ( S 2 ) = ( c ℓ 3 ) 2 , area ( S 3 ) = ( d ℓ 5 ) 2 ℓ 3 ℓ 5 S 2 S 3 ℓ 4 After coordinate change x 0 = a ℓ 2 + b ℓ 3 , x 1 = c ℓ 3 , x 2 = d ℓ 5 obtain area ( P ) = x 2 0 − x 2 1 − x 2 2 a quadratic form of signature (+ , − , −) . Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 7 / 25

The hyperbolic space q h ( x ) = x 2 q s ( x ) = x 2 0 + x 2 1 + ... + x 2 0 − x 2 1 − ... − x 2 d S d ∶= { x ∣ q ( x ) = 1 } d H d ∶ = { x ∣ q ( x ) = 1 , x 0 > 1 } Metric of constant curvature − 1. Metric of constant curvature 1. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 8 / 25

Hyperbolic polyhedra angle of a polygon = dihedral angle of the cone = arccos q ( ν 1 ,ν 2 ) (in the hyperbolic as in the spherical case) q ( ν 1 ,ν 2 ) = 0 ⇔ sides form a right angle If q ( x ) = q ′ ( x 0 ,..., x d − 2 ) − x 2 d , then { x d − 1 = 0 } ⊥ { x d = 0 } . d − 1 − x 2 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 9 / 25

Space of polygons is a (truncated) orthoscheme q has signature (+ , − ,..., −) ⇒ M ( V ) ∶= C ( V ) ∩ { ℓ ∣ q ( ℓ ) = 1 } is a hyperbolic polyhedron Theorem (Bavard,Ghys’92) If ∣ i − j ∣ ≥ 2 , then F i ⊥ F j , where F i is the facet of M ( V ) corresponding to contraction of the i-th edge. Proof. q ( ℓ ) = x 2 0 − x 2 1 − x 2 ℓ 2 ℓ 1 2 F 3 = { x 1 = 0 } F 5 = { x 2 = 0 } P ( ℓ ) ℓ 3 ℓ 5 S 2 S 3 ℓ 4 Example The space of equiangular pentagons of area 1 is isometric to the regular right-angled hyperbolic pentagon. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 10 / 25

Hyperbolic orthoschemes and complex hyperbolic orbifolds Bavard, Ghys’92 Polygones du plan et polyédres hyperboliques ▸ Realization of hyperbolic orthoschemes from the Im Hof’s list. ▸ Complete list: Im Hof’90, Tumarkin’07, Kistler’11. This was motivied by Thurston’98 Shapes of polyhedra and triangulations of the sphere ▸ Complex hyperbolic structure on the space of Euclidean metrics on S 2 with cone angles α 1 ,...,α n . ▸ Realization of some non-arithmetic complex Coxeter orbifolds. We present a generalization of the Bavard-Ghys construction to higher dimensions. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 11 / 25

From d = 2 to d > 2 Combinatorics ▸ facet normals V no more determine the combinatorial type of the polytope P ▸ ⇒ C ( V ) no more a cone, but a fan (made of type cones) Geometry ▸ What to use instead of area ( P ) ? ▸ Is there a canonical quadratic form of signature ( + , − ,..., − ) ? ▸ Can the dihedral angles of M ( V ) be computed? Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 12 / 25

Type cones V = ( v 1 ,..., v n ) a positively spanning vector configuration in R d C ( V ) ∶ = translation classes of polytopes with facet normals V P , P ′ ∈ C ( V ) are normally equivalent if they have the same normal fan: N( P ) = N( P ′ ) C ( V ) = ⊔ T ( ∆ ) , ∆ where T ( ∆ ) = { P ∣ N( P ) = ∆ }/{ translations } . Theorem The closure of C ( V ) is a pointed fan with convex support. Example For V = {± e 1 ,..., ± e d } all P are normally equivalent. Monotypic polytopes: McMullen, Schneider, Shephard’74. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 13 / 25

Support numbers and Gale dual Given V ∈ R n × d with rows v i ∈ R d of norm 1, consider P ( h ) = { x ∈ R n ∣ Vx ≤ h } , h ∈ R n Support numbers ( h i ) n i = 1 . We have P ( h ) + p = P ( h + Vp ) Hence C ( V ) ⊂ R n / im V . h i p v i π ∶ R n → R n / im V , π ( e i ) =∶ ¯ v i Definition Vector configuration V = ( ¯ v n ) is called Gale dual to V. v 1 ,..., ¯ V ⊺ V = E n , rank V + rank V = n V positively spanning ⇔ V lies in an open half-space Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 14 / 25

C ( V ) lies in pos ( V ) Theorem P ( h ) ≠ ∅ ⇔ π ( h ) ∈ pos ( V ) ∶ = { ∑ v i ∣ λ i ≥ 0 } λ i ¯ i Proof. Note: 0 ∈ P ( h ) ⇔ h i ≥ 0 ∀ i . P ( h ) ≠ ∅ ⇔ ∃ p ∈ P ( h ) ⇔ 0 ∈ P ( h ) − p = P ( h − Vp ) ⇔ h − Vp ∈ R n ≥ 0 ⇔ π ( h ) ∈ pos ( ¯ v n ) v 1 ,..., ¯ Example V are normals of the triangular bipyramid ⇒ V span a hexagonal cone v 2 ¯ v 4 ¯ ¯ v 6 v 3 ¯ ¯ v 1 ¯ v 5 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 15 / 25

C ( V ) is the 2-core of V The k -core of a vector configuration W ⊂ R m is core k ( W ) = { x ∈ R m ∣ ∀ y s.t. ⟨ y , x ⟩ ≥ 0 ∃ w i 1 ,..., w i k s.t. ⟨ y , w i α ⟩ ≥ 0 } E.g. core 1 ( W ) = pos ( W ) core 2 ( W ) = ⋂ pos ( W ∖ { w }) w ∈ W Theorem The closure of C ( V ) is core 2 ( V ) . Proof. i -th facet non-empty ⇔ ∃ p ∈ R n such that ⟨ v i , p ⟩ = h i and ⟨ v j , p ⟩ < h j for all j ≠ i . Then use P ( h − Vp ) = P ( h ) − p etc. Example ¯ v 2 v 4 ¯ ¯ v 6 ¯ v 3 v 1 ¯ v 5 ¯ Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 16 / 25

Type cones are the chambers of the chamber fan The chamber fan Ch ( W ) of W is the coarsest common subdivision of all cones spanned by W . Theorem Closures of type cones form the chamber fan of V. Lemma pos ( V σ ) ∈ N ( P ( h )) ⇔ π ( h ) ∈ pos ( V ¯ σ ) , where ¯ σ = { 1 ,..., n } ∖ σ . Proof. Similar to the preceding two arguments. Example v 2 ¯ ¯ v 4 ¯ v 6 ¯ v 3 v 1 ¯ ¯ v 5 Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 17 / 25

References All the above arguments on type cones and their arrangement appeared in ▸ McMullen’73 Representations of polytopes and polyhedral sets ▸ Shephard’71 Spherical complexes and radial projections of polytopes From this, there is only one step to the secondary polyhedron. Ivan Izmestiev (FU Berlin) Shapes of euclidean polyhedra and hyperbolic geometry Berlin’2013 18 / 25