Differential Geometry: Discrete Exterior Calculus [Discrete Exterior Calculus (Thesis). Hirani, 2003] [Discrete Differential Forms for Computational Modeling. Desbrun et al., 2005] [Build Your Own DEC at Home. Elcott et al., 2006] Simplices Definition: A k-simplex σ is the non-degenerate convex hull of k+1 (ordered) geometrically distinct points. 0-simplex 1-simplex 2-simplex 3-simplex Simplices Definition: A k-simplex σ is the non-degenerate convex hull of k+1 (ordered) geometrically distinct points. 0-simplex 1-simplex 2-simplex 3-simplex A (k-1)-face of a k-simplex σ is a (k-1)-simplex spanned by k of the k+1 vertices. Simplices Definition: The boundary ∂σ of a k-simplex σ is the signed union of all (k-1)-faces. ∂ ∂ ∂ ∂ +1 -1 Simplices Definition: The boundary ∂σ of a k-simplex σ is the signed union of all (k-1)-faces. ∂ ∂ ∂ ∂ +1 -1 Formally, the boundary of σ={v0,…,vk} is: k j ∂{v0 ,...,vk }= ∑(−1) {v0 ,...,vˆ j ,...,vk } j=0 Simplices Definition: The boundary ∂σ of a k-simplex σ is the signed union of all (k-1)-faces. ∂ ∂ ∂ ∂ +1 Note: -1 The boundary of a boundary is empty since each (k-2) appears twice, with opposite signs: ∂∂ = 0/ Simplicial Complexes Definition: A simplicial complex Κ is a collection of simplices satisfying the following conditions: – Every face of each simplex σ∈Κ is also in Κ. – For any σ1, σ2∈Κ, the intersection σ1∩σ2 is either empty or a common face. Simplicial Complexes Not a Simplicial Complex Discrete Manifolds Definition: An n-dimensional discrete manifold M is an n- dimensional simplicial complex with the property that: – For each simplex σ∈M, the union of all incident n- simplices forms an n-dimensional (half-)ball. A Discrete Manifold Not a Discrete Manifold Chains Definition: A k-chain of a simplicial complex Κ is linear combination of the k-simplices in Κ: c = ∑c(σ )⋅σ σ∈Κ k where c is a real-valued function. The space of k-chains is denoted Ck(Κ). Chains Definition: A k-chain of a simplicial complex Κ is linear combination of the k-simplices in Κ: c = ∑c(σ )⋅σ σ∈Κ k where c is a real-valued function. The space of k-chains is denoted Ck(Κ). Informally, a k-chain is a weighted combination of k-dimensional volume elements. Boundaries of Chains Since k-chains are linear combinations of k- simplices and since we know how to compute the boundary of a k-simplex, we can extend the notion of the boundary operator to chains: ∂c = ∂ ∑c(σ )⋅σ = ∑c(σ )⋅∂σ σ∈Κ k σ∈Κ k Boundaries of Chains Since k-chains are linear combinations of k- simplices and since we know how to compute the boundary of a k-simplex, we can extend the notion of the boundary operator to chains: ∂c = ∂ ∑c(σ )⋅σ = ∑c(σ )⋅∂σ σ∈Κ k σ∈Κ k That is, the boundary operator ∂:Ck(Κ)→Ck-1(Κ), takes weights assigned to k-simplices in Κ and transforms them into weights on their (k-1)- faces. Cochains Definition: The dual of a k-chain is a k-cochain which is a linear map taking a k-chain to a real value. Cochains Definition: The dual of a k-chain is a k-cochain which is a linear map taking a k-chain to a real value. The space of k-cochains is denoted Ωk (Κ). Cochains Informally, we can think of chains and cochains as being related through integration: – Cochain: What we would like to integrate. – Chain: The domain over which we perform the integration. Cochains Informally, we can think of chains and cochains as being related through integration: – Cochain: What we would like to integrate. – Chain: The domain over which we perform the integration. Given a k-chain and a k-cochain, we evaluate the cochain on the chain by integrating. We denote the evaluation of ω∈Ωk(Κ) on c∈Ck(Κ) by: ω = ω = ω ∫ ∫ i ∑ci ∫ i i σ c ∑ciσ i i i Discrete Exterior Derivative Definition: We define the exterior derivative d:Ωk→ Ωk+1 as the adjoint (complement) of the boundary operator. Discrete Exterior Derivative Definition: We define the exterior derivative d:Ωk→ Ωk+1 as the adjoint (complement) of the boundary operator. Given a ω∈Ωk(Κ) and c∈Ck+1(Κ), we can evaluate ω on the boundary ∂c. Discrete Exterior Derivative Definition: We define the exterior derivative d:Ωk→ Ωk+1 as the adjoint (complement) of the boundary operator. Given a ω∈Ωk(Κ) and c∈Ck+1(Κ), we can evaluate ω on the boundary ∂c. The derivative dω is the (k+1)-cochain whose value on c is equal to the value of ω on ∂c: ∫ dω = ∫ω c ∂c Discrete Exterior Derivative Example 1: v3 -1 -2 Given the 1-form ω which, when v4 0.5 v evaluated on the edges of the 2 complex, gives: 0 0.5 2 1 v v1 ∫ω =1 ∫ω = 2 ∫ω = −2 ∫ω = −1 0 {v0 ,v1} {v1 ,v2 } {v2 ,v3} {v3 ,v4 } ∫ω = 0 ∫ω = .5 ∫ω = .5 {v4 ,v0 } {v1 ,v4 } {v2 ,v4 } we need to compute the evaluations of the 2- form dω with the property: ∫ dω = ∫ω c ∂c Discrete Exterior Derivative Example 1: v3 -1 -2 v Consider the evaluation of dω on 4 0.5 v2 the 2-simples {v0,v1,v4}. In this case we must have: 0 0.5 2 dω = ω 1 v1 ∫ ∫ v0 {v0 ,v1 ,v4 } ∂{v0 ,v1 ,v4 } = ∫ω {v0 ,v1}+{v1 ,v4 }+{v4 ,v0 } = ∫ω + ∫ω + ∫ω {v0 ,v1} {v1 ,v4 } {v4 ,v0 } =1+ 0.5 + 0 Discrete Exterior Derivative Example 1: v3 -1 -2 v Consider the evaluation of dω on 4 0.5 v2 the 2-simples {v0,v1,v4}. In this case we must have: 0 0.5 2 dω = ω 1 v1 ∫ ∫ v0 {v0 ,v1 ,v4 } ∂{v0 ,v1 ,v4 } = ∫ω {v0 ,v1}+{v1 ,v4 }+{v4 ,v0 } = ∫ω + ∫ω + ∫ω {v0 ,v1} {v1 ,v4 } {v4 ,v0 } =1+ 0.5 + 0 Discrete Exterior Derivative Example 1: v3 -1 -2 v Consider the evaluation of dω on 4 0.5 v2 the 2-simples {v0,v1,v4}. In this case we must have: 0 0.5 2 dω = ω 1 v1 ∫ ∫ v0 {v0 ,v1 ,v4 } ∂{v0 ,v1 ,v4 } = ∫ω {v0 ,v1}+{v1 ,v4 }+{v4 ,v0 } = ∫ω + ∫ω + ∫ω {v0 ,v1} {v1 ,v4 } {v4 ,v0 } =1+ 0.5 + 0 Discrete Exterior Derivative Example 1: v3 -1 -2 v Consider the evaluation of dω on 4 0.5 v2 the 2-simples {v0,v1,v4}. In this case 0 0.5 2 we must have: 1.5 dω = ω 1 v1 ∫ ∫ v0 {v0 ,v1 ,v4 } ∂{v0 ,v1 ,v4 } = ∫ω {v0 ,v1}+{v1 ,v4 }+{v4 ,v0 } = ∫ω + ∫ω + ∫ω {v0 ,v1} {v1 ,v4 } {v4 ,v0 } =1+ 0.5 + 0 Discrete Exterior Derivative Example 1: v3 -1 -2 v Similarly, on the 2-simplex {v1,v2,v4} 4 0.5 v2 the evaluation of dω becomes: 2 0 0.5 2 ∫ dω = ∫ω + ∫ω + ∫ω 1.5 {v1 ,v2 ,v4 } {v1 ,v2 } {v2 ,v4 } {v4 ,v1} 1 v v1 = 2 + 0.5 − 0.5 0 Discrete Exterior Derivative Example 1: v3 -1 -2 v -3.5 And on the 2-simplex {v2,v3,v4} the 4 0.5 v2 evaluation of dω becomes: 2 0 0.5 2 ∫ dω = −2 −1− 0.5 1.5 {v2 ,v3 ,v4 } 1 v1 v0 Discrete Exterior Derivative Example 2: -2 v3 Given the 0-form ω which, when v4 2 -1 v evaluated on the vertices of the 2 complex, gives: v 1 v1 ∫ω = 0 ∫ω =1 ∫ω = 2 ∫ω = −2 ∫ω = −2 0 0 {v0 } {v1} {v2 } {v3} {v4} we can compute the evaluation of dω on the edges in an analogous manner. Discrete Exterior Derivative Example 2: -2 v3 1 -4 v Given the 0-form ω which, when 4 -3 2 -1 v evaluated on the vertices of the 2 -2 complex, gives: 1 1 1 v 1 v1 ∫ω = 0 ∫ω =1 ∫ω = 2 ∫ω = −2 ∫ω = −2 0 0 {v0 } {v1} {v2 } {v3} {v4} we can compute the evaluation of dω on the edges in an analogous manner: ∫ dω = ∫ω − ∫ω ∫ dω = ∫ω − ∫ω ... {v0 ,v1} {v1} {v0 } {v1 ,v2 } {v2 } {v1} Discrete Exterior Derivative Example 2: v3 1 -4 v Using the evaluation of dω on the 4 -3 v edges, we can compute the 2 -2 1 1 evaluation of d(dω) on the triangles: 0 1 v v1 ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 0 {v0 ,v1 ,v4 } {v0 ,v1} {v1 ,v4 } {v4 ,v0 } Discrete Exterior Derivative Example 2: v3 1 -4 v Using the evaluation of dω on the 4 -3 v2 edges, we can compute the 0 -2 1 1 evaluation of d(dω) on the triangles: 0 1 v v1 ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 0 {v0 ,v1 ,v4 } {v0 ,v1} {v1 ,v4 } {v4 ,v0 } ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 {v1 ,v2 ,v4 } {v1 ,v2 } {v2 ,v4 } {v4 ,v1} Discrete Exterior Derivative Example 2: v3 1 -4 v 0 Using the evaluation of dω on the 4 -3 v2 edges, we can compute the 0 -2 1 1 evaluation of d(dω) on the triangles: 0 1 v v1 ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 0 {v0 ,v1 ,v4 } {v0 ,v1} {v1 ,v4 } {v4 ,v0 } ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 {v1 ,v2 ,v4 } {v1 ,v2 } {v2 ,v4 } {v4 ,v1} ∫ d(dω) = ∫ dω + ∫ dω + ∫ dω = 0 {v2 ,v3 ,v4 } {v2 ,v3} {v3 ,v4 } {v4 ,v2 } Discrete Exterior Derivative Note: v3 1 -4 k v 0 By definition, for any ω∈Ω (Κ) and 4 -3 v k+2 2 any c∈C (Κ), we must have: 0 -2 1 1 ∫ d(dω) = ∫ dω = ∫ω 0 c ∂c ∂∂c 1 v1 v0 Discrete Exterior Derivative Note: v3 1 -4 k v 0 By definition, for any ω∈Ω (Κ) and 4 -3 v k+2 2 any c∈C (Κ), we must have: 0 -2 1 1 ∫ d(dω) = ∫ dω = ∫ω 0 c ∂c ∂∂c 1 v v1 But since the boundary of a boundary 0 is always empty, this implies that: ∫ d(dω) = 0 c Discrete Exterior Derivative Note: v3 1 -4 k v 0 By definition, for any ω∈Ω (Κ) and 4 -3 v k+2 2 any c∈C (Κ), we must have: 0 -2 1 1 ∫ d(dω) = ∫ dω = ∫ω 0 c ∂c ∂∂c 1 v v1 But since the boundary of a boundary 0 is always empty, this implies that: ∫ d(dω) = 0 c Since this is true for all (k+2)-chains, we have: d(dω) = 0.