Geometric Modeling - New York Universitypanozzo/gp18/CotanLaplacian.pdf• If M had boundary curves, we’d need to specify boundary conditions. u = b in M CSCI-GA.3033-018 - Geometric

CSCI-GA.3033-018 - Geometric Modeling - Daniele Panozzo

Geometric ModelingCotangent Laplacian Derivation

Acknowledgements: Olga Diamanti, Julian Panetta


Cotan Laplacian Motivation

Normal ComputationMean Curvature

RANDOM EQUATIONS FOR SLIDES 3

Y =

2

66664

�f(v1)�f(v2)�f(v3)

· · ·�f(vN )

3

77775

L =

2

6664

w11 w12 · · · w1Nw21 w22 · · · w2N...

... · · ·...

wN1 wN2 · · · wNN

3

7775= {wij}

wij =

8<

:

0 i 6= j, @ edge (i, j)1 i 6= j, 9 edge (i, j)�|N(vi)| i = j

wij =

8>><

>>:

0 i 6= j, @ edge (i, j)cot↵j+cot�j

2A(vi)i 6= j, 9 edge (i, j)

�P

vk2N(vi)wik i = j

F =

2

66664

f(v1)f(v2)f(v3)· · ·

f(vN )

3

77775

(I � �dtL)vn+1 = vn

vn+1 = vn + �dtLvn+1

vn+1 � vn = �dtLvn

vn+1 = (I + �dtL)vn

@v

@t= �L(v)

@f

@t= ��f

n = �[Lv]i2Hi

Hi =1

2

�� [Lv]i

��

Smoothing

• Discrete Laplacian enables many useful operations on triangle meshes:

• Cotangent weights give less mesh-dependent results, ensuring proper convergence under mesh refinement

…


Where We’re Headed

• “Cotan Laplacian” refers to weight on each edge contribution.

• We’ll consider two different ways to derive this formula:

• “Traditional graphics approach:” Finite Volume-style

• Finite Element Approach (more general)

�cotanM f(v) =1

2A(v)

X

vi2N(v)

(cot(↵i) + cot(�i))

✓f(vi)� f(v)

◆

Discrete Laplace-Beltrami

Second Approach: Cotangent Formula

CGL slideset 10

1

2(cot(↵i) + cot(�i))


Goal: Mesh Independence• We want discrete Laplacian to  

converge to the Laplace Beltrami operator of smooth surface S as mesh M is refined to better approximate S.

• F is a vector of discrete values (per vertex),  f(x) is a smooth function.

• Idea: reinterpret F as a function interpolating the smooth “f,” and directly compute its Laplacian. Converge as F —> f

RANDOM EQUATIONS FOR SLIDES 3

Y =

2

66664

�f(v1)�f(v2)�f(v3)

· · ·�f(vN )

3

77775

L =

2

6664

w11 w12 · · · w1Nw21 w22 · · · w2N...

... · · ·...

wN1 wN2 · · · wNN

3

7775= {wij}

wij =

8<

:

�|N(vi)| i = j0 i 6= j, @ edge (i, j)1 i 6= j, 9 edge (i, j)

F =

2

66664

f(v1)f(v2)f(v3)· · ·

f(vN )

3

77775

@f

@t= ��f

�cotanM F �! �Sf


Piecewise-Linear Interpolation• Since F holds per-vertex values, it naturally corresponds to a  

linear function on each triangle

• (Corner values uniquely determine linear function over triangle)

xy

f(x, y)


“Hat Function” Basis• Piecewise linear functions are expressed most simply using the  

“hat functions” (linear Lagrange polynomials)

• One basis function per vertex,

• Unique linear function (on triangles) having:

• Interpolation of F on mesh M is just:

�i(vj) =

(1 if i = j

0 otherwise

�i(x) vi

fM (x) =X

i

f(vi)�i(x)


Gradients• Recall, the Laplacian is the divergence of the gradient:

• So we first need to compute our interpolant’s gradient:

• In other words, the gradient is just a linear combination of the basis function gradients weighted by the per-vertex f values.

rfM (x) = r X

i

f(vi)�i(x)

!=X

i

f(vi)r�i(x)

r�i(x)

�f = r · (rf)


Hat Function Gradients• Basis functions are linear over each triangle, so these gradients are

constant on each triangle

• We can determine gradient with simple geometric arguments:�0(v0) = 1

�0(v1) = 0 �0(v2) = 0

Where does the gradient point? What about its magnitude?


Gradient Derivation�0(v0) = 1

�0(v1) = 0 �0(v2) = 0e0

e?0

r�0 · e0 = 0 =) r�0 = ↵e0?

h=) r�0 =

1

h

e?0��e?0��

r�0 · h

e?0��e?0��

!= 1

=) r�0 =e?02A

(area is 1/2 * base * height)


Final Gradient Expression

• Note, this is the gradient evaluated on one triangle T incident vertex i. (Different gradient on others).

• Works on curved triangle meshes too! Just need triangle area and inward-pointing perpendicular edge vectors:

• Full gradient is then the linear combination

rfM (x) =X

i

f(vi)r�i(x)

r�i|T =e?i2AT

e?i = n⇥ ei


Now the Laplacian!• Now we want to compute

• But the gradient is piecewise constant!  What does divergence even mean?

�fM = r ·rfM (x) = r · X

i

f(vi)r�i(x)!


Too Many Derivatives!• Problem: function is piecewise linear (only C0 continuous) and we’re

trying to take second derivatives

• But we can make sense of things by using the Divergence Theorem over some region “R”:

• (This allows us to avoid explicitly computing divergence; just need dot products with the curve normals on R’s boundary.)

n

nn

n

n

n

R@R

Z

Rr · (r�) dx =

Z

@Rn ·r� ds


Approach 1: (Traditional in Graphics)• Idea: define Laplacian on a vertex to be the  

average over some region surrounding the vertex.

• To get cotan Laplacian, you need to average over the vertices’ Voronoi regions (region for which v is the closest vtx).

• This region seems somewhat arbitrary; approach 2 won’t need this choice.

R(vi)vi

@R(vi)

[LF]i :=1

|R(vi)|

Z

R(vi)r ·rfM (x) dx

=1

|R(vi)|

Z

@R(vi)n ·rfM (x) ds


Approach 1 Continued• Break into sum over triangles (so gradient is constant):

n1

n2

vi

`1`2

[LF]i =1

|R(vi)|

Z

@R(vi)n ·rfM (x) ds

=1

|R(vi)|X

T3vi

Z

T\@R(vi)

n ·rfM (x) ds

=1

|R(vi)|X

T3vi

rfT ·Z

T\@R(vi)

n ds

=1

|R(vi)|X

T3vi

rfT · (n1`1 + n2`2)


Voronoi Diagram Properties

• Voronoi diagram edges bisect vi’s incident edges

n1

n2

vi

`1`2

n1

n2

vi

vj

vkn3

vi + vj2

vi + vk2 n1`1 + n2`2 + n3`3 = 0

(Tri edge vectors sum to zero; so do their perpendiculars.)

ei

[LF]i =1

|R(vi)|X

T3vi

rfT · (n1`1 + n2`2)

=) [LF]i =1

|R(vi)|X

T3vi

rfT ·e?i2

=) n1`1 + n2`2 = �n3`3 =✓vk + vi

2� vj + vi

2

◆?

=1

2(vk � vj)? =

e?i2


Plug in per-Tri Gradient

e?i + e?j + e

?k = 0

edges sum to zero:( )

vi

vj

vk

ei

ej

ek

[LF]i =1

|R(vi)|X

T3vi

rfT ·e?i2

=1

|R(vi)|X

T3vi

(fir�i + fjr�j + fkr�k) ·e?i2

=1

|R(vi)|X

T3vi

"fi

e?i2AT

+ fje?j2AT

+ fke?k2AT

#· e

?i

2

=1

|R(vi)|X

T3vi

"�fi

e?j2AT

+e?k2AT

!+ fj

e?j2AT

+ fke?k2AT

#· e

?i

2

=1

|R(vi)|X

T3vi

"(fj � fi)

e?j2AT

+ (fk � fi)e?k2AT

#· e

?i

2


Final Steps

e?i · e?k4AT

=keikkekk cos(�)2keikkekk sin(�)

=1

2cot(�)

e?i · e?j4AT

=keikkejk cos(↵)2keikkejk sin(↵)

=1

2cot(↵)

Note:

vi

vj

vk

ei

ej

ek

↵

�

[LF]i =1

|R(vi)|X

T3vi

"(fj � fi)

e?j2AT

+ (fk � fi)e?k2AT

#· e

?i

2

=1

|R(vi)|X

T3vi

(fj � fi)e?j · e?i4AT

+ (fk � fi)e?k · e?i4AT

[LF]i =1

|R(vi)|X

T3vi

(fj � fi) cot(↵) + (fk � fi) cot(�)


• We have the cotan weights now, but expressed as a sum over triangles:

• To get the standard formula, we can regroup these terms as a sum over edges:

• |R(vi)| is the area of the ith vertex’s Voronoi region (the averaging area)

Regrouping Terms

[LF]i =1

2|R(vi)|X

T3vi

(fj � fi) cot↵+ (fk � fi) cot�

[LF]i =1

2|R(vi)|X

vj2N(vi)

(cot(↵ij) + cot(�ij))

✓f(vj)� f(vi)

◆


Approach 2: Finite Element Method• Instead of trying to discretize the Laplacian itself (using averaging

regions), we discretize the whole PDE containing it. For instance:

• Here “b” is a forcing term (e.g. injecting/sinking heat at each point), and we assume M is a closed surface mesh without a boundary.

• If M had boundary curves, we’d need to specify boundary conditions.

��u = b in M


Weak Form of the Poisson Equation• We can’t expect a piecewise linear function “u” to solve this PDE exactly

(why does its Laplacian mean, anyway?)

• But we can ask it to be a “weak solution:”

• In other words, we ask for zero residual “along” all test functions.

• Linear FEM: choose piecewise linear functions for both and

• Physical interpretation: minimizing a potential energy over the space of piecewise linear functions

�Z

M �u dx =

Z

M b dx 8 “test functions”

u


Expand Functions in Basis�Z

M �u dx =

Z

M b dx

u(x) :=X

j

uj�j(x) (x) :=X

i

ai�i(x) b(x) :=X

j

bj�j(x)

�Z

M

X

i

ai�i(x)

!�

0

@X

j

uj�j(x)

1

A dx =Z

M

X

i

ai�i(x)

!0

@X

j

bj�j(x)

1

A dx

=) �X

i,j

aiuj

✓Z

M�i��j dx

◆=X

i,j

aibj

✓Z

M�i�j dx

◆


Integration by Parts for• To avoid second derivatives, apply divergence theorem in the form

of an integration by parts (move derivative onto )

• Note the “product rule”: r · (�ir�j) = �i��j +r�i ·r�j =)�i��j = r · (�ir�j)�r�i ·r�j

�Z

M�i��j dx = �

Z

Mr · (�ir�j)�r�i ·r�j dx

= �Z

@Mn · (�ir�j) ds+

Z

Mr�i ·r�j dx

=

Z

Mr�i ·r�j dx

Apply divergence theorem

There is no boundary!

Z

M�i��j dx

�i


Laplacian and Mass Matrices• Now our weak PDE looks like:    

• Expressed in matrix form:   

• Plugging in gradient expressions:

X

i,j

aiuj

✓Z

Mr�ir�j dx

◆=

X

i,j

aibj

✓Z

M�i�j dx

◆

aTLu = aTMb 8a

Lij =

Z

Mr�i ·r�j dx =

X

T3vi,vj

e?T,i2AT

·e?T,j2AT

AT =X

T3vi,vj

e?T,i · e?T,j4AT

Lij Mij

=) Lu = Mb

Weak form must hold for all test functions!


Laplacian Matrix Cotangent Weights

Lij =X

T3vi,vj

e?T,i · e?T,j4AT

Lij =

8>>>><

>>>>:

1

2(cot(↵ij) + cot(�ij)) if i 6= j

X

T3vi

e?T,i ·⇣�e?T,j � e?T,k

⌘

4AT= �1

2

X

vk2N(vi)

(cot(↵ik) + cot(�ik)) if i = j

vi

vj↵ij

�ij


FEM vs “Graphics Approach”• Get exactly the same cotan weights (up to division by region area)

• FEM doesn’t require defining an averaging region

• FEM computes Laplacian integrated against test function instead of an averaged (point) quantity: 

• FEM uses exact mass matrix to integrate RHS against test functions; graphics approach uses diagonal “Lumped” approximation Mdiag based on region areas (e.g. Voronoi)—still converges, but higher error. (Lumping effectively treats the RHS function “b” as piecewise constant over the averaging region, while FEM+exact mass matrix treats it as piecewise linear).

• FEM generalizes to higher degree; just use more basis functions!

LFEM = MdiagLAvg Mdiag = diag([R(v0), R(v1), · · · ])

Documents

Geometric Modeling - New York Universitypanozzo/gp18/CotanLaplacian.pdf• If M had boundary curves, we’d need to specify boundary conditions. u = b in M CSCI-GA.3033-018 - Geometric