Discrete Differential Geometry 6

DDG Lecture 6

Exterior Derivative

two big ideas in calculus: differentiation vs. integration. linked by fundamental theorem of calculus.

Exterior calculus:

  • differentiation of k-forms (exterior derivative)
  • integration of k-forms (measure volume)
  • linked by Stoke's Theorem

Goal: integrate over meshes to get discrete exterior calculus.

Motivation for exterior calculus.

Why generalize vector calculus?

  • Hard to measure change in volume
  • Duality clarifies distinction between different concepts/quantities.
  • Topology: notion of differentiation doesn't require metric (e.g. cohomology)
  • Geometry: clear language for calculus on curved surfaces.
  • Physics: clear distinction between physical quantities (e.g. velocity vs. momentum)
  • CS: leads directly to discretization

Exterior Derivative

Derivative as limit, slope, tangent plane, "best linear approximation".

Vector derivatives: gradient RRnR \rightarrow R^n, divergence RnRR^n \rightarrow R, curl RnRnR^n \rightarrow R^n

In coordinates:

ϕ:RnR\phi : \mathbb{R}^n \rightarrow \mathbb{R}

X=ux+vy+wzX = u \frac{\partial}{\partial x} + v \frac{\partial}{\partial y} + w \frac{\partial}{\partial z} where u,v,w:RnRu,v,w : \mathbb{R^n} \rightarrow \mathbb{R} are coordinate functions and x,y,z\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} are the basis vector fields.



ϕ=ϕxx+ϕyy+ϕzz\nabla \phi = \frac{\partial \phi}{\partial x}\frac{\partial}{\partial x} + \frac{\partial \phi}{\partial y} \frac{\partial}{\partial y} + \frac{\partial \phi}{\partial z} \frac{\partial}{\partial z}


X=ux+vy+wz\nabla \cdot X = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}


×X=(wyvz)x+(uzwx)y+(vxuy)z\begin{aligned} \nabla \times & X = \\ & \left( \frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} \right) \frac{\partial}{\partial x} + \\ & \left( \frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} \right) \frac{\partial}{\partial y} + \\ & \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \frac{\partial}{\partial z} \end{aligned}

The Exterior Derivative

Let Ωk\Omega^k denote the space of all differential k-forms.

Unique linear map d:ΩkΩk+1d: \Omega^k \rightarrow \Omega^{k+1} such that

  • differential : for k=0, dϕ(X)=DXϕd \phi(X) = D_X \phi
    • take the "directional derivative of ϕ\phi along XX
  • product rule : d(αβ)=dαβ+(1)kαdβd(\alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^k \alpha \wedge d \beta
  • exactness : dd=0d \circ d = 0

Exterior Derivative -- Differential

review: Directional Derivative

defn The directional derivative of a scalar function ϕ\phi at a point pp with respect to a vector XX is the rate at which that function increases as we walk away from pp with velocity XX.

DXϕp:=limϵ0ϕ(p+ϵX)ϕ(p)ϵD_X \phi |_p := \lim_{\epsilon \rightarrow 0} \frac{\phi(p+\epsilon X) - \phi(p)}{\epsilon}

This is trivially extended to XX vector field instead of at point pp by applying previous definition pointwise. The result is a scalar function.


<ϕ,X>=DXϕ X\left< \nabla \phi, X \right> = D_X \phi ~ \forall X

i.e the gradient is the unique vector field ϕ\nabla \phi whose inner product with any vector field X yields the directional derivative along X. (Note: depends on defn of inner product.)

Differential of a function

0-forms are scalar functions. Change in a scalar function can be measured via the differential. Two ways to define differential:

  1. As unique 1-form such that applying any vector field gives directional derivative along those directions.
- $d\phi(X) = D_X \phi$
  1. In coordinates:
- $d \phi := \frac{\partial \phi}{\partial x^1} dx^1 + \cdots + \frac{\partial \phi}{\partial x^n} dx^n$

Conceptually similar to the gradient.

<ϕ>=DXϕ\left< \nabla \phi \right> = D_X \phidϕ(X)=DXϕd \phi(X) = D_X\phi

what's the difference?

  • type: vector field vs. differential 1-form
  • dependence on inner product.
    • dϕ()=<ϕ,>d\phi(\cdot) = \left<\nabla \phi, \cdot \right>
    • we have no dependence on geometry for differential.
(dϕ)=ϕ    dϕ()=<ϕ,>    (ϕ)=dϕ(d\phi)^\sharp = \nabla \phi \iff d\phi(\cdot) = \left<\nabla \phi, \cdot \right> \iff (\nabla \phi)^\flat = d\phi

Exterior Derivative -- Product Rule

Q: why is it true that ab = ba for a,b in R rectangle area implies commutativity.

Preduct rule of differentiation of real functions.

(We neglect the O(n^2) term as differentiation is linear approximation on crack.)

let α\alpha be a k-form and β\beta be an l-form. Then

d(αβ)=(dα)β+(1)kα(dβ)d(\alpha \wedge \beta) = (d \alpha) \wedge \beta + (-1)^k \alpha \wedge (d\beta)

Recursive evaluation of differential form

Ex. Let α:=udx, β:=vdy, γ:=wdz\alpha := u dx, ~ \beta := v dy, ~ \gamma := w dz be differential 1-forms on Rn\mathbb{R}^n.

where u,v,w:RnRu,v,w : \mathbb{R}^n \rightarrow \mathbb{R} are 0-forms. Also, let ω:=αβ\omega := \alpha \wedge \beta. Then,

d(ωγ)=(dω)γ+(1)2ω(dγ)=[(dα)βα(dβ)]γ+(1)2ω(dγ)=(dα)βγα(dβ)γ+αβ(dγ)\begin{aligned} d(\omega \wedge \gamma) &= (d\omega) \wedge \gamma + (-1)^2 \omega \wedge (d\gamma) \\ &= \left[ (d\alpha) \wedge \beta - \alpha \wedge (d\beta) \right] \wedge \gamma + (-1)^2 \omega \wedge (d\gamma) \\ &= (d\alpha) \wedge \beta \wedge \gamma - \alpha \wedge (d \beta) \wedge \gamma + \alpha \wedge \beta \wedge (d \gamma) \end{aligned}


dα=(du)dx+u(ddx)d\alpha = (du) \wedge dx + u (ddx)


d(ωγ)=(dα)βγα(dβ)γ+αβ(dγ)=(dudx)βγα(dvdy)γ+αβ(dwdz)\begin{aligned} d(\omega \wedge \gamma) &= (d\alpha) \wedge \beta \wedge \gamma - \alpha \wedge (d \beta) \wedge \gamma + \alpha \wedge \beta \wedge (d \gamma) \\ &= (du \wedge dx) \wedge \beta \wedge \gamma - \alpha \wedge (dv \wedge dy) \wedge \gamma + \alpha \wedge \beta \wedge (dw \wedge dz) \end{aligned}

Ex. Let ϕ(x,y):=12e(x2+y2)\phi(x,y) := \frac{1}{2} e^{-(x^2+y^2)}. Then

dϕ=ϕxdx+ϕydy=2ϕ(xdx+ydy)d\phi = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy \\ = -2 \phi(xdx + ydy)

Ex. Let α(x,y)=xdx+ydy\alpha(x,y) = xdx + ydy . Then.

dα=(xxdx+xydy)dx+(yx+yy)dy=dxdx+dydy=0\begin{aligned} d\alpha &= \left( \frac{\partial x}{\partial x}dx + \frac{\partial x}{\partial y}dy \right) \wedge dx + \left( \frac{\partial y}{\partial x} + \frac{\partial y}{\partial y} \right) dy \\ &= dx \wedge dx + dy \wedge dy \\ &= 0 \end{aligned}

Ex. Let α\alpha as above, then dαd \star \alpha

dα=d((xdx+ydy))=d(xdyydx)=dxdydydx=2(dxdy)\begin{aligned} d \star \alpha &= d(\star(xdx + ydy)) \\ &= d(xdy - ydx) \\ &= dx \wedge dy - dy \wedge dx \\ &= 2 (dx \wedge dy) \end{aligned}

Exterior Derivative -- Exactness

dd=0d \circ d = 0

Q: let α=udx+vdy+wdz\alpha = udx + vdy + wdz where u,v,w:R3Ru,v,w : \mathbb{R}^3 \rightarrow \mathbb{R}. What is dαd\alpha? A:

dα=d(udx+vdy+wdz)=dudx+ud(dx)+dvdy+vd(dv)+dwdz+wd(dz)=dudx+dvdy+dwdz=uy(dydx)+uz(dzdx)+vx(dxdy)+vz(dzdy)+wx(dxdz)+wy(dydz)=(vxuy)dxdy+(wyvz)dydz+(uzwx)dzdx\begin{aligned} d\alpha &= d(udx + vdy + wdz) = du \wedge dx + u \wedge d(dx) + dv \wedge dy + v \wedge d(dv) + dw \wedge dz + w \wedge d(dz) \\ &= du \wedge dx + dv \wedge dy + dw \wedge dz \\ &= \frac{\partial u}{\partial y}(dy \wedge dx) + \frac{\partial u}{\partial z}(dz \wedge dx) + \frac{\partial v}{\partial x}(dx \wedge dy) + \frac{\partial v}{\partial z}(dz \wedge dy) + \frac{\partial w}{\partial x}(dx \wedge dz) + \frac{\partial w}{\partial y}(dy \wedge dz) \\ &= \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dx \wedge dy + \left(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} \right) dy \wedge dz + \left(\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} \right) dz \wedge dx \end{aligned}

Vaguely reminiscent of the cross product. Thus reminiscent of curl.


Let X:=ux+vy+wzX := u \frac{\partial}{\partial x} + v \frac{\partial}{\partial y} + w \frac{\partial}{\partial z}.


×Xdα=(vxuy)dxdy+(wyvz)dydz+(uzwx)dzdx\begin{aligned} \nabla \times X \cong d \alpha = \\ & \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dx \wedge dy \\ & + \left(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} \right) dy \wedge dz \\ & + \left(\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} \right) dz \wedge dx \end{aligned}

In other words, ×X=(dX)\nabla \times X = (\star dX^\flat)^\sharp


Ex. Let ambient space be R3\mathbb{R}^3. Let α\alpha as before. What is dαd \star \alpha?

dα=d((udx+vdy+wdz))=d(udydz+vdzdx+wdxdy)=dudydz+dvdzdx+dwdxdy=uxdxdydz+vydydzdx+wzdzdxdy=(ux+vy+wz)dxdydz\begin{aligned} d \star \alpha &= d (\star(u dx + v dy + wdz)) \\ &= d(u dy\wedge dz + v dz \wedge dx + w dx \wedge dy) \\ &= du \wedge dy \wedge dz + dv \wedge dz \wedge dx + dw \wedge dx \wedge dy \\ &= \frac{\partial u}{\partial x} dx \wedge dy \wedge dz + \frac{\partial v}{\partial y} dy \wedge dz \wedge dx + \frac{\partial w}{\partial z} dz \wedge dx \wedge dy \\ &= \left( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} \right) dx \wedge dy \wedge dz \end{aligned}

Vaguely reminiscent to the dot product or the divergence.

i.e X=(dX)\nabla \cdot X = \star (d \star X^\flat)

Define the codifferential δ:=d\delta := \star d \star.

Exterior vs. Vector Derivatives

gradϕ\phidiv XXcurl YY
(dϕ)(d\phi)^\sharpd(X)\star d (\star X^\flat)((dY))(\star(d Y^\flat))^\sharp


Δ=\Delta = \nabla \cdot \nabla


Δ=dd\Delta = \star d \star d

We can generalize to k-forms:

Δ:=dd+dd\Delta := \star d \star d + d \star d \star