Differential Geometry: Discrete Exterior Calculus

Transcription

1 Differential Geometry: Discrete Exterior Calculus [Discrete Exterior Calculus (Thesis). Hirani, 23] [Discrete Differential Forms for Computational Modeling. Desbrun et al., 25] [Build Your Own DEC at Home. Elcott et al., 26]

2 Chains Recall: A k-chain of a simplicial complex Κ is linear combination of the k-simplices in Κ: c = c(σ ) σ k σ Κ where c is a real-valued function. The dual of a k-chain is a k-cochain which is a linear map taking a k-chain to a real value. Chains and cochains related through evaluation: Cochains: What we integrate Chains: What we integrate over

3 Boundary Operator Recall: The boundary σ of a k-simplex σ is the signed union of all (k-)-faces. + - The boundary operator extends linearly to act on chains: c = c( σ ) σ = c( σ ) σ σ Κ k σ Κ k

4 Discrete Exterior Derivative Recall: The exterior derivative d:ω k Ω k+ is the operator on cochains that is the complement of the boundary operator, defined to satisfy Stokes Theorem Since the boundary of the boundary is empty: = dd =

5 The Laplacian Definition: The Laplacian of a function f is a function defined as the divergence of the gradient of f: f = div( f ) = f

6 The Laplacian Definition: The Laplacian of a function f is a function defined as the divergence of the gradient of f: f = div( f ) = f Q. What do the gradient and divergence operators act on?

7 The Gradient Definition: Given a function f:r n R, the gradient of f at a point p is defined as: f f f ( p) =,..., x x n

8 Definition: Given a function f:r n R, the gradient of f at a point p is defined as: Given a direction vector v, we have: The Gradient ( ) ) (, ) (... ) ( p f v p f x f v x f v p f v p f + = = x n f x f p f,..., ) (

9 The Gradient Definition: Given a function f:r n R, the gradient of f at a point p is defined as: f f f ( p) =,..., x Given a direction vector v, we have: f ( p + v) f ( p) + v, f ( p) The gradient is an operator (not really a vector) telling us how moving in direction v will affect the value of the function. x n

10 The Gradient Stokes Theorem: Given an oriented curve c R n with endpoints c and c, integrating the gradient of a function over the curve, we get the difference between the values of the function at the endpoints: f ( p) = f c ) f ( ) c ( c f c c

11 The Gradient Stokes Theorem: Given an oriented curve c R n with endpoints c and c, integrating the gradient of a function over the curve, we get the difference between the values of the function at the endpoints: f ( p) = f c ) f ( ) c ( c The gradient satisfies Stokes Theorem and takes something we evaluate at points to something we evaluate over curves. f c c

12 The Gradient Stokes Theorem: Given an oriented curve c R n with endpoints c and c, integrating the gradient of a function over the curve, we get the difference between the values of the function at the endpoints: f ( p) = f c ) f ( ) c ( c The gradient satisfies Stokes Theorem and takes something we evaluate at points to something we evaluate n n over curves. = d : Ω ( R ) Ω ( R ) The gradient operator is the exterior derivative taking -forms to -forms: f c c

13 Divergence Definition: Given a vector-field X:R n R n, representing a flow, the flux of the vector field through a patch of hyper-surface S is the amount of flow that passes through the surface: Flux = S where n is the surface normal. S ( f ) X, n ds X S

14 Divergence Stokes Theorem: Given a volume V R n with boundary V=S, integrating the divergence of a function over the volume, we get the flux across the boundary: = V X X, n S ds V V=S X

15 Divergence Stokes Theorem: Given a volume V R n with boundary V=S, integrating the divergence of a function over the volume, we get the flux across the boundary: = V X X, n The divergence satisfies Stokes Theorem and takes something we evaluate over hyper-surfaces to something we evaluate over volumes. S ds V V=S X

16 Divergence Stokes Theorem: Given a volume V R n with boundary V=S, integrating the divergence of a function over the volume, we get the flux across the boundary: = V X X, n The divergence satisfies Stokes Theorem and takes something we evaluate over hyper-surfaces The divergence to something operator is we the exterior evaluate over volumes. S ds derivative taking (n-)-forms to n-forms: = d : Ω ( R ) Ω ( R n n n n ) V V=S X

17 The Laplacian f = div ( f ) = f Q. What do the gradient and divergence operators act on? A. The gradient and divergence operators are exterior derivatives that act on -forms and (n-) forms respectively: Gradient :Ω (R n ) Ω (R n ) Divergence :Ω n- (R n ) Ω n (R n )

18 The Laplacian f = div ( f ) = f Gradient :Ω (R n ) Ω (R n ) Divergence :Ω n- (R n ) Ω n (R n ) Problem: The Laplacian is supposed to take functions (-forms) to functions, but the divergence doesn t take -forms as input and it doesn t return -forms as output!!!

19 Poincaré Duality Observation: In an n-dimensional inner-product space, specifying a k-dimensional subspace is the same as specifying an (n-k)-dimensional subspace.

20 Poincaré Duality Observation: In an n-dimensional inner-product space, specifying a k-dimensional subspace is the same as specifying an (n-k)-dimensional subspace. Moreover, if the space is oriented, specifying a k-dimensional oriented subspace is the same as specifying an (n-k)-dimensional oriented subspace.

21 Poincaré Duality If we are given a k-dimensional subspace spanned by vectors {v,,v k }, we can always find vectors {v k+,,v n } such that the vectors {v,,v n } are a basis for the whole space and: v i, v j = for all i k and all k+ j n.

22 Poincaré Duality If we are given a k-dimensional subspace spanned by vectors {v,,v k }, we can always find vectors {v k+,,v n } such that the vectors {v,,v n } are a basis for the whole space and: v i, v j = for all i k and all k+ j n. If the space has an orientation, we can order the {v k+,,v n } so that the vectors {v,,v n } have positive orientation.

23 Poincaré Duality Example: Consider R 2, with the standard inner product and orientation x y:

24 Poincaré Duality Example: Consider R 2, with the standard inner product and orientation x y: R R 2 : x y R R : x y, y -x R 2 R : x y

25 Poincaré Duality Example: Consider R 3, with the standard inner product and orientation x y z:

26 Poincaré Duality Example: Consider R 3, with the standard inner product and orientation x y z: R R 2 : x y z R R 2 : x y z, y -x z, z x y R 2 R : x y z, x z -y, y z x R 3 R : x y z

27 The Laplacian f = div ( f ) = f Gradient :Ω (R n ) Ω (R n ) Divergence :Ω n- (R n ) Ω n (R n ) Problem: The Laplacian is supposed to take functions (-forms) to functions, but the divergence doesn t take -forms as input and it doesn t return -forms as output!!!

28 The Laplacian f = div ( f ) = f Gradient :Ω (R n ) Ω (R n ) Divergence :Ω n- (R n ) Ω n (R n ) Problem: The Laplacian is supposed to take functions (-forms) to functions, but the divergence doesn t take -forms as input and it doesn t return -forms as output!!! Solution: Use the duality between k-dimensional and (n-k)-dimensional subspaces to turn -forms into (n-)-forms and n-forms into -forms.

29 Hodge Star Locally, a k-form take k-dimensional, oriented, surface elements and return a value. Q: The corresponding dual (n-k)-form should act on the (oriented) orthogonal complement, but what value should it return?

30 Hodge Star Locally, a k-form take k-dimensional, oriented, surface elements and return a value. Q: The corresponding dual (n-k)-form should act on the (oriented) orthogonal complement, but what value should it return? A: Informally, we think of a k-form as assigning a density per unit k-volume (the average ). We define the corresponding (n-k)-form with the same density per unit (n-k)-volume.

31 Hodge Star Definition: The map taking a k-form to the corresponding (n-k)-forms is called the Hodge Star operator: : Ω k ( R n ) Ω n k ( R n )

32 Hodge Star For R 2 : *:Ω (R 2 ) Ω 2 (R 2 ): x y *:Ω (R 2 ) Ω (R 2 ): x y, y -x *:Ω 2 (R 2 ) Ω (R 2 ): x y For R 3 : *:Ω (R 3 ) Ω 3 (R 3 ): x y z *:Ω (R 3 ) Ω 2 (R 3 ): x y z, y -x z, z x y *:Ω 2 (R 3 ) Ω (R 3 ): x y z, x z -y, y z x *:Ω 3 (R 3 ) Ω (R 3 ): x y z

33 Hodge Star For R 2 : *:Ω (R 2 ) Ω 2 (R 2 ): x y *:Ω (R 2 ) Ω (R 2 ): x y, y -x *:Ω 2 (R 2 ) Ω (R 2 ): x y For R 3 : *:Ω (R 3 ) Ω 3 (R 3 ): x y z *:Ω (R 3 ) Ω 2 (R 3 ): x y z, y -x z, z x y *:Ω 2 (R 3 ) Ω (R 3 ): x y z, x z -y, y z x Note that *:Ω 3 the (R 3 Hodge ) Ω Star (R 3 operator from k-forms to (n-k)-forms and the operator from (n-k)-forms ): x y z to k-forms are almost inverses: k n k = ( ) k ( n k )

34 Dual Complexes In order to be able to define a discrete Laplacian, we need to have a discrete version of the Hodge Star.

35 Dual Complexes In order to be able to define a discrete Laplacian, we need to have a discrete version of the Hodge Star. In the same way as we defined dual/orthogonal subspaces in the continuous case, we would like to define dual/orthogonal simplexes in the discrete case.

36 Dual Complexes Definition: Given a discrete manifold, we define the Voronoi dual complex *K to be the (non-simplicial) dual complex obtained by associating simplices with their Voronoi duals. K *K

37 Dual Complexes Definition: Given a discrete manifold, we define the Voronoi dual complex *K to be the (non-simplicial) dual complex obtained by associating simplices with their Voronoi duals. K *K

38 Dual Complexes Definition: Given a discrete manifold, we define the Voronoi dual complex *K to be the (non-simplicial) dual complex obtained by associating simplices with their Voronoi duals. K *K

39 Dual Complexes Definition: Given a discrete manifold, we define the Voronoi dual complex *K to be the (non-simplicial) dual complex obtained by associating simplices with their Voronoi duals. K *K

40 Dual Complexes Definition: Given a discrete manifold, we define the Voronoi dual complex *K to be the (non-simplicial) dual complex obtained by associating simplices with their Voronoi duals.

41 Dual Complexes Using the dual complex, we can associate each (primal) k-simplex with a (dual) (n-k)-cell.

42 Dual Complexes Using the dual complex, we can associate each (primal) k-simplex with a (dual) (n-k)-cell. Furthermore, because we are using the Voronoi dual, the dual (n-k)-cells are orthogonal to the primal k-simplices.

43 Dual Complexes Definition: We denote by * the map that takes primal simplices to the corresponding (n-k)-cells: : Κ ( k ) * Κ ( n k )

44 Dual Complexes Definition: Using the association between primal simplices and dual cells, we define the discrete Hodge Star operator that takes k-cochains on the primal complex to (n-k)-cochains on the dual complex: : Ω k ( ) n k Κ Ω (* Κ)

45 Dual Complexes Definition: If ω is a k-cochain with value ω(σ) on σ, we treat ω as having a density of ω(σ)/ σ per unit k- volume on σ (the average value of ω on σ).

46 Dual Complexes Definition: If ω is a k-cochain with value ω(σ) on σ, we treat ω as having a density of ω(σ)/ σ per unit k- volume on σ (the average value of ω on σ). Thus, we define *ω so it has the same density ω(σ)/ σ per unit (n-k)-volume on *σ.

47 Dual Complexes Definition: If ω is a k-cochain with value ω(σ) on σ, we treat ω as having a density of ω(σ)/ σ per unit k- volume on σ (the average value of ω on σ). Thus, we define *ω so it has the same density ω(σ)/ σ per unit (n-k)-volume on *σ. Or in other words: ω σ ( σ ) = ω( σ ) σ

48 Putting it All Together In 3D: We have both primal and dual complexes. We have a boundary operator, and hence exterior derivative on both. We have k-cochains over both complexes. We have the Hodge Star operator to transition between the two.

49 Putting it All Together Using this structure, we define the Laplacian:. Compute the gradient using the ext. derivative: primal -cochain primal -cochain

50 Putting it All Together Using this structure, we define the Laplacian:. Compute the gradient using the ext. derivative: primal -cochain primal -cochain 2. Apply the Hodge star operator: primal -cochain dual (n-)-cochain

51 Putting it All Together Using this structure, we define the Laplacian:. Compute the gradient using the ext. derivative: primal -cochain primal -cochain 2. Apply the Hodge star operator: primal -cochain dual (n-)-cochain 3. Compute the divergence using the ext. derivative: dual (n-)-cochain dual n-cochain

52 Putting it All Together Using this structure, we define the Laplacian:. Compute the gradient using the ext. derivative: primal -cochain primal -cochain 2. Apply the Hodge star operator: primal -cochain dual (n-)-cochain 3. Compute the divergence using the ext. derivative: dual (n-)-cochain dual n-cochain 4. Apply the Hodge star operator: dual n-cochain primal -cochain : = ( ) n( k ) + * d * d

53 Putting it All Together In Practice: To work with DEC in practice, we only need to define two types of matrices:

54 In Practice: To work with DEC in practice, we only need to define two types of matrices:. The boundary operators : Specify the faces and orientations of each simplex/cell in the primal/dual complex. Putting it All Together v v v 2 v 3 v 4 e e e 2 e 3 e 6 e 5 e 4 t t t 2 = 2 =

55 In Practice: To work with DEC in practice, we only need to define two types of matrices:. The boundary operators : Specify the faces and orientations of each simplex/cell in the primal/dual complex. Putting it All Together v v v 2 v 3 v 4 e e e 2 e 3 e 6 e 5 e 4 t t t 2 = 2 = The exterior derivative is just the transpose matrix.

56 Putting it All Together In Practice: To work with DEC in practice, we only need to define two types of matrices: 2. The discrete Hodge Star operators: Specify the ratio of between volumes of dual (nk)-simplices and volumes of primal k-simplices. v 3 e 3 t e 2 v 2 4 e 6 v 2 e 4 e 5 t t e v e v *v 3 *e 3 *e 2 *t *v 2 4 *e 6 *e *t 5 *e *e 4 *t *v *v *e

57 Putting it All Together In Practice: To work with DEC in practice, we only need to define two types of matrices: 2. The discrete Hodge Star operators: Specify the ratio of between volumes of dual (nk)-simplices and volumes of primal k-simplices. v 3 e 3 t e 2 v 2 4 e 6 v 2 *v 4 *v 3 *e 3 *e 2 *e*t 6 2 This is just a diagonal t e *t e 5 matrix whose *e 4 e *e 5 entries are the *e ratios of the volumes t 4 of the dual (n-k)-cells *t *v *v *e and the v e volumes of v the corresponding primal k-simplices.

58 Putting it All Together Example (2D Laplacian): Starting with a -cochain (function) ω, we evaluate the cochain to get a value ω i at each vertex. v 3 ω 3 ω 2 v 2 ω ω 4 v v v 4 ω ω 5 v 5

59 Putting it All Together Example (2D Laplacian): Applying the exterior derivative, we get a - cochain that evaluates to: λ ω on primal edges from v to v i. i = i ω v 3 ω 3 ω 2 v 2 ω ω 4 λ v 4 λ v v 4 λ 3 λ 2 ω λ 5 ω 5 v 5

60 Putting it All Together Example (2D Laplacian): Applying the Hodge Star, we get a -cochain that evaluates to µ i on corresponding dual edges: µ ( ) i = λ i l * i l i where l i is the length of the primal edge from v to v i, and l * i is the length of its dual. v 3 λ 3 µ 3 µ λ 2 2 µ 4 v 2 v 4 λ 4 v λ µ v µ 5 λ 5 v 5

61 Putting it All Together Example (2D Laplacian): Applying the exterior derivative again, we get a 2-cochain that evaluates to L on the 2-cell dual to v : L = µ i i v 3 v 2 µ 3 L µ 2 µ 4 µ v 4 v v µ 5 v 5

62 Putting it All Together Example (2D Laplacian): Finally, applying the Hodge Star again, we get a -chain whose value at vertex v is the Laplacian: ( ) A ω ( v ) = L where A is the area of the 2-cell dual to v. v 3 v 2 ω L v 4 v v v 5

63 Putting it All Together Example (2D Laplacian): ( l * l ) µ i = λ i i i If we denote by α i and β i the two angles adjacent to the primal edge from v to v i v 3 β 3 v 2 α 3 β 2 α 2 v 4 α 4 β 4 v β v α α 5 β 5 v 5

64 Putting it All Together Example (2D Laplacian): ( l * l ) µ i = λ i i i If we denote by α i and β i the two angles adjacent to the primal edge from v to v i, the angles between the edges from v to the v dual vertices and the dual edges 3 β 3 β β 2 3 α themselves become α i and β i. 3 v 4 α 4 β 4 α 5 β 5 v 2 α 3 α 2 β 3 α 4 v β v α 5β4 β 5 v 5 α β α 2 α

65 Putting it All Together Example (2D Laplacian): ( l * l ) µ i = λ i i i Thus, for the edge from v to v i, we have: l i = 2 sin = * βi 2sinα l = cos + i+ i βi cosαi+ v 3 β 3 v 2 α 3 β 2 α 2 β 3 α 3 v 4 α 4 β 4 β 3 α 4 v β v α 5β4 β 5 α β α 2 α α 5 β 5 v 5

66 Putting it All Together Example (2D Laplacian): ( l * l ) µ i = λ i i i Thus, for the edge from v to v i, we have: and the ratio of dual to primal edge-lengths is: l l * i i l i = 2 sin = * βi 2sinα = cos + i+ i i cos i+ cos β cosα ( cot β cotα ) i i+ = = + sin sin i i βi αi+ 2 l v 4 β α 4 β 4 v 3 β 3 α 5 β 5 v 2 α 3 β 2 α 2 β 3 α 4 β 3 α 3 v β v α 5β4 β 5 α α β α 2 α Thanks Barry. v 5