Discrete Euclidean Curvature Flows
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1 Discrete Euclidean Curvature Flows 1 1 Department of Computer Science SUNY at Stony Brook Tsinghua University 2010
2 Isothermal Coordinates Relation between conformal structure and Riemannian metric Isothermal Coordinates A surface Σ with a Riemannian metric g, a local coordinate system (u,v) is an isothermal coordinate system, if g=e 2λ(u,v) (du 2 + dv 2 ).
3 Gaussian Curvature Gaussian Curvature Suppose ḡ=e 2λ g is a conformal metric on the surface, then the Gaussian curvature on interior points are K = Δ g λ = 1 e 2λΔλ, where Δ= 2 u v 2
4 Conformal Metric Deformation Definition Suppose Σ is a surface with a Riemannian metric, ( ) g11 g g= 12 g 21 g 22 Suppose λ :Σ R is a function defined on the surface, then e 2λ g is also a Riemannian metric on Σ and called a conformal metric. λ is called the conformal factor. g e 2λ g Angles are invariant measured by conformal metrics. Conformal metric deformation.
5 Curvature and Metric Relations Yamabi Equation Suppose ḡ=e 2λ g is a conformal metric on the surface, then the Gaussian curvature on interior points are K = e 2λ ( Δ g λ + K), geodesic curvature on the boundary k g = e λ ( n λ + k g ).
6 Uniformization Theorem (Poincaré Uniformization Theorem) Let (Σ, g) be a compact 2-dimensional Riemannian manifold. Then there is a metric g=e 2λ g conformal to g which has constant Gauss curvature. Spherical Euclidean Hyperbolic
7 Surface Ricci Flow Key Idea K = Δ g λ, Roughly speaking, Let dλ dt = K, Heat equation! dk dt dk dt =Δ g dλ dt =Δ g K + K 2
8 Surface Ricci Flow Definition (Hamilton s Surface Ricci Flow) A closed surface with a Riemannian metric g, the Ricci flow on it is defined as dg ij = 2Kg ij. dt If the total area of the surface is preserved during the flow, the Ricci flow will converge to a metric such that the Gaussian curvature is constant every where.
9 Ricci Flow Theorem (Hamilton 1982) For a closed surface of non-positive Euler characteristic, if the total area of the surface is preserved during the flow, the Ricci flow will converge to a metric such that the Gaussian curvature is constant (equals to K ) every where. Theorem (Bennett Chow) For a closed surface of positive Euler characteristic, if the total area of the surface is preserved during the flow, the Ricci flow will converge to a metric such that the Gaussian curvature is constant (equals to K ) every where.
10 Summary Surface Ricci Flow Conformal metric deformation g e 2u g Curvature Change - heat diffusion dk dt =Δ g K + K 2 Ricci flow du dt = K K.
11 Generic Surface Model - Triangular Mesh Surfaces are represented as polyhedron triangular meshes. Isometric gluing of triangles in E 2. Isometric gluing of triangles in H 2,S 2.
12 Generic Surface Model - Triangular Mesh Surfaces are represented as polyhedron triangular meshes. Isometric gluing of triangles in E 2. Isometric gluing of triangles in H 2,S 2.
13 Generic Surface Model - Triangular Mesh Surfaces are represented as polyhedron triangular meshes. Isometric gluing of triangles in E 2. Isometric gluing of triangles in H 2,S 2.
14 Discrete Generalization Concepts 1 Discrete Riemannian Metric 2 Discrete Curvature 3 Discrete Conformal Metric Deformation
15 Discrete Metrics Definition (Discrete Metric) A Discrete Metric on a triangular mesh is a function defined on the vertices, l : E ={all edges} R +, satisfies triangular inequality. A mesh has infinite metrics.
16 Discrete Curvature Definition (Discrete Curvature) Discrete curvature: K : V ={vertices} R 1. K(v)=2π i α i,v M;K(v)=π α i,v M i Theorem (Discrete Gauss-Bonnet theorem) K(v)+ K(v)=2πχ(M). v M v M v α 1 α 2 α 3 α 1 α2 v
17 Discrete Metrics Determines the Curvatures l j θ i v i cosine laws v k v k θ k l i lj θ l j k li θ θ i θ j j v i l k v l j v k i v j θ i θ k θ j R 2 H2 S 2 l k v k l i v j cosl i = cosθ i + cosθ j cosθ k sinθ j sinθ k (1) coshl i = coshθ i+ coshθ j coshθ k sinhθ j sinhθ k (2) 1 = cosθ i + cosθ j cosθ k sinθ j sinθ k (3)
18 Discrete Conformal Metric Deformation - Geometric Approach Conformal maps Properties transform infinitesimal circles to infinitesimal circles. preserve the intersection angles among circles. Idea - Approximate conformal metric deformation Replace infinitesimal circles by circles with finite radii.
19 Discrete Conformal Metric Deformation vs CP
20 Circle Packing Metric CP Metric We associate each vertex v i with a circle with radius γ i. On edge e ij, the two circles intersect at the angle of Φ ij. The edge lengths are φ12 φ r 1 v 31 1 e 12 e 31 e23 v 2 v 3 r 2 φ 23 r 3 l 2 ij = γ 2 i + γ 2 j + 2γ i γ j cosφ ij CP Metric (Σ,Γ,Φ), Σ triangulation, Γ={γ i v i },Φ={φ ij e ij }
21 Discrete Conformal Factor Conformal Factor Defined on each vertex u:v R, logγ i R 2 u i = logtanh γ i 2 H 2 logtan γ i 2 S 2 Properties Symmetry Discrete Laplace Equation K i u j = K j u i dk=δdu, Δ is a discrete Lapalce-Beltrami operator.
22 Unified Framework of Discrete Curvature Flow Analogy Curvature flow Energy E(u)= Hessian of E denoted as Δ, du dt = K K, ( K i K i )du i, i dk=δdu.
23 Criteria for Discretization Key Points Convexity of the energy E(u) Convexity of the metric space (u-space) Admissible curvature space (K-space) Preserving or reflecting richer structures Conformality
24 Cosine law v k l j θ k l i 2l j l k cosθ i = lj 2 + lk 2 l2 i θ i θ j dθ 2l j l k sinθ i i dl i = 2l i v i l k A=l j l k sinθ i v j dθ i dl i = l i A
25 Cosine law v k 2l j l k cosθ i = lj 2 + lk 2 l2 i l j θ k l i 2l j = 2l k cosθ i 2l j l k sin θ i θ i θ j v i v l j k l j = l i cosθ k + l k cosθ i dθ i dl j = l k cosθ i l j A = l i cosθ k A = dθ i dl i
26 Cosine law τ ki v i l j θ i θ k v k l k τ ij l i θ j τ jk l 2 k = r 2 i + r 2 j + 2cosτ ij r i r j v j 2l i dl i dr j l 2 i = r 2 j + r 2 k + 2r jr k cosτ jk = 2r j + 2r k cosτ jk dl i = 2r j 2 + 2r j r k cosτ jk dr j 2l i r j = r j 2 + rk 2+ 2r jr k cosτ jk + rj 2 2l i r j = l2 i + r 2 j r 2 k 2l i r j
27 Cosine law Let u i = logr i, then dθ 1 dθ 2 dθ 3 0 = 1 A l l l 3 l1 2+r 2 2 r 3 2 l1 2+r 3 2 r 2 2 2l 1 r 2 2l 1 r 3 l2 2+r 1 2 r 3 2 2l 2 r 1 0 l3 2+r 1 2 r 2 2 l3 2+r 2 2 r 1 2 2l 3 r 1 2l 3 r 2 0 l 2 2 +r 2 3 r 2 1 2l 2 r 3 1 cosθ 3 cosθ 2 cosθ 3 1 cosθ 1 cosθ 2 cosθ 1 1 r r r 3 du 1 du 2 du 3
28 Cosine law τ ki v i l j θ i v k θ k w k l k τ jk l i θ φ j ji d ji v j 2l k dl k dr j = 2r j + 2r i cosτ ij dl r k j = 2r j 2 + 2r i r j cosτ ij dr j 2l k = l2 k + r j 2 ri 2 2l k In triangle [v i,v j,w k ], τ ij l 2 k = r 2 i + r 2 j + 2cosτ ij r i r j dl k = 2 l kr j cosφ ji = r j cosφ ji = d ji du j 2l k
29 Cosine law τ ki v i l j θ i h j There is a unique circle orthogonal to three circles (v i,r i ), the center is o, the distance from o to edge [v i,v j ] is h k. θ k v k τ ij l k h i o h k l i τ jk θ j v j Theorem dθ i du j dθ j du k dθ k du i = dθ j du i = h k l k = dθ k du j = h i l i = dθ i du k = h j l j
30 Cosine law Proof. τ ki v i l j θ i v k θ k h i h k θ j l k o τ jk l i d d jk θ j d ji v j θ i u j = θ i l i l i u j + θ i l k l k u j = θ i l i ( l i u j l k u j cosθ j ) = l i A (d jk d ji cosθ j ) dl = i l i l k sinθ j τ ij dθ i du j = h k l k = h k sinθ j l k sin θ j = h k l k
31 Cosine law v j v i,v j v i u j = 2 v j u j,v j v i τ ki v i l j θ i v k θ k h k l k h i o τ jk l i d jk θ j d ji v j l 2 k u j l k u j d ji = 2 v j u j,v j v i = v j u j, v j v i l k = v j u j, v j v i l k τ ij Similarly v j u j = v j o So v j u j = v j 0. d jk = v j u j, v j v k l i
32 Metric Space τki vk τjk Lemma For any three obtuse angles τ ij,τ jk,τ ki [0, π 2 and any three positive numbers r 1,r 2 and r 3, there is a configuration of 3 circles in Euclidean geometry, unique upto isometry, having radii r i and meeting in angles τ ij. lj hj θk hi o li Proof. θi hk θj vi lk vj max{r 2 i,r 2 j }<r 2 i +r 2 j +2r i r j cosτ ij (r i +r j ) 2 τij so max{r 2 i,r 2 j }<l k r i + r j l k r i + r j < l i + l j.
33 Discrete Ricci Energy Lemma ω is closed 1-form in Ω:={(u 1,u 2,u 3 ) R 3 }. τki vi vk θk hi lj hj o hk θi lk τjk li θj vj Because θ i u j dω = θ j u i, so = ( θ i u j θ j u i )du j du i + τij ( θ j u k θ k u j )du k du j + ω = θ i du i +θ j du j +θ k du j ( θ k θ i )du i du k u i u k = 0.
34 Discrete Ricci Energy τki vi lj θi hj θk vk lk hi o hk li τjk θj vj Lemma The Ricci energy E(u 1,u 2,u 3 ) is well defined. Because Ω=R 3 is convex, closed 1-form τij is exact, therefore E(u 1,u 2,u 3 ) is well defined. (u1,u 2,u 3 ) E(u 1,u 2,u 3 )= ω (0,0,0)
35 Discrete Ricci Energy τki vi lj θi hj θk vk τij (u1,u 2,u 3 ) E(u 1,u 2,u 3 )= ω (0,0,0) lk hi o hk li τjk θj vj Lemma The Ricci energy E(u 1,u 2,u 3 ) is strictly concave on the subspace u 1 + u 2 + u 3 = 0. The gradient E =(θ 1,θ 2,θ 3 ), the Hessian matrix is H = θ 1 u 1 θ 1 u 2 θ 1 u 3 θ 2 u 1 θ 2 u 2 θ 2 u 3 θ 3 u 1 θ 3 u 2 θ 3 u 3 because of θ 1 + θ 2 + θ 3 = π, θ i u i = θ i u j θ i u k = θ j u i θ k u i
36 Ricci energy Proof. H = h 3 l3 + h 2 l2 h 3 l3 h 2 l2 h 3 h 3 l3 l3 + h 1 l1 h 1 l1 h 2 l2 h 1 h 2 l1 l2 + h 1 l1 H is diagonal dominant, it has null space (1, 1, 1), on the subspace u 1 + u 2 + u 3 = 0, it is strictly negative definite. Therefore the discrete Ricci energy E(u 1,u 2,u 3 ) is strictly concave.
37 Discrete Ricci Energy Lemma The Ricci energy E(u) is strictly convex on the subspace vi M u i = 0. M The gradient E =(K 1,K 2,,K n ). The Ricci energy M E(u)=2π v i M u i [v i,v j,v k ] M E ijk (u i,u j,u k ) u ω = K i du i,e(u)= v i M 0 where E ijk is the ricci energy defined on the face [v i,v j,v k ]. The linear term won t affect the convexity of the energy. The ω. null space of the Hessian is (1,1,,1). In the subspace u i = 0, the energy is strictly convex.
38 Uniqueness Lemma Suppose Ω R n is a convex domain, f :Ω R is a strictly convex function, then the map is one-to-one. Proof. x f(x) Suppose x 1 = x 2, f(x 1 )= f(x 2 ). Because Ω is convex, the line segment (1 t)x 1 + tx 2 is contained in Ω. construct a convex function g(t)=f((1 t)x 1 + tx 2 ), then g (t) is monotonous. But g (0)= f(x 1 ),x 2 x 1 = f(x 2 ),x 2 x 1 =g (1), contradiction.
39 Uniqueness x f(x) f(x) f(x) x 2 x 1 Ω Lemma Suppose Ω R n is a convex domain, f :Ω R is a strictly convex function, then the map x f(x) is one-to-one.
40 Uniqueness Theorem Suppose M is a mesh, with circle packing metric, all edge intersection angles are non-obtuse. Given the target curvature (K 1,K 2,,K n ), i K i = 2πχ(M). If the solution (u 1,u 2,,u n ) Ω(M), i u i = 0 exists, then it is unique. Proof. The discrete Ricci energy E on Ω { i u i = 0} is convex, E(u 1,u 2,,u n )=(K 1,K 2, K n ). Use previous lemma.
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