Discrete Differential Geometry 4

DDG Week 4


Week 3: Exterior algebra (k-vectors) Week 4: Exterior calculus (k-forms)

Exterior Calculus, "how do lengths, areas, volumes, change over curved surfaces?

Measurement devices have the same dimensionality as the thing we're measuring.

Study exterior calculus in flat spaces (Rn\mathbb{R}^n) May seem redundant, but will help when space starts curving.

Vectors & Covectors

covectors do measurements vectors get measured

Wedging covectors get k-forms, which are duals to k-vectors.

Similar to row/column vectors. A row vector is a linear map which maps the column vector to a real value.

let α\alpha be a unit covector, uu be a vector of any magnitude. α(u)\alpha(u) tells us how long uu is in the direction α\alpha


Let VV be a ceal vector space. Its dual space VV^* is the collection of all linear functions α:VR\alpha : V \mapsto \mathbb{R} together with the operations of addition.

(α+β)(u):=α(u)+β(u)(\alpha + \beta)(u) := \alpha(u) + \beta(u)

and scalar multiplication

(cα)(u):=c(α(u))(c \alpha)(u) := c (\alpha(u))

α,βV,uV,cR\forall \alpha,\beta \in V^*, u \in V, c \in \mathbb{R}


An element of a dual vector space is called a dual vector or covector

(unrelated to the Hodge Star/Dual)

e.g V=C0V = C^0 and δ(f):=f(0)V\delta(f) := f(0) \in V^* (Dirac delta)

Let [x,y,z][r,g,b][x,y,z] [r,g,b] be column vectors, then the transpose converts between row/column. Likewise, let u,vVu,v \in V be vectors, then u,vVu^\flat, v^\flat \in V^*. let α,βV\alpha, \beta \in V^* be covectors, then α,βV\alpha^\sharp, \beta^\sharp \in V.

let the inner product be aTMba^T M b instead of the normal inner product. u(v)=uTMvu^\flat(v) = u^T M v and α(β)=αM1βT\alpha(\beta^\sharp) = \alpha M^{-1} \beta^T Equivalently, u()=<u,>u^\flat(\cdot) = <u, \cdot> and α()=<α,>\alpha(\cdot) = <\alpha^\sharp, \cdot>

k-forms will be multilinear maps from a k-vector.

Measurement of 2-vectors

The "shadow" of a parallelogram onto another parallelogram. Given (u,v)(u,v), pick orthonormal basis for plane α,β\alpha, \beta.

  1. project onto plane
  • u(α(u),β(u))u \mapsto (\alpha(u), \beta(u))
  • v(α(v),β(v))v \mapsto (\alpha(v), \beta(v))
  1. calculate area via cross product
  • α(u)β(v)α(v)β(u)\alpha(u)\beta(v) - \alpha(v)\beta(u)

Can apply same expression when α,β\alpha, \beta not orthonormal.

(αβ)(u,v):=α(u)β(v)α(v)β(u)(\alpha \wedge \beta)(u,v) := \alpha(u)\beta(v) - \alpha(v)\beta(u)

This is the definiton of the application of a 2-form to two vectors. Note: result scales with "area" of αβ\alpha \wedge \beta

Note that the application of the 2-form is antisymettric

(αβ)(u,v)=(αβ)(v,u)(\alpha \wedge \beta)(u,v) = - (\alpha \wedge \beta)(v,u)

Geometrically, this shows that the orientation of the face is different.

The arguments to the wedge are also antisymmetric.

(αβ)(u,v)=(βα)(u,v)(\alpha \wedge \beta)(u,v) = -(\beta \wedge \alpha)(u,v)

this has the same meaning, as the 2-form's orientation is flipped.

Measurement of 3-vectors

in R3\mathbb{R}^3, all 3-vectors have the same "direction". Can only measure magnitude.

Let u,v,wu,v,w be the edges of a parallelopiped. Let (α,β,γ)(\alpha, \beta, \gamma) be any orthonormal basis.

  1. project u,v,wu,v,w onto the basis
  • u(α(u),β(u),γ(u))u \mapsto (\alpha(u), \beta(u), \gamma(u))
  • v(α(v),β(v),γ(v))v \mapsto (\alpha(v), \beta(v), \gamma(v))
  • w(α(w),β(w),γ(w))w \mapsto (\alpha(w), \beta(w), \gamma(w))
  1. apply standard formula for volume (det)
α(u)β(v)γ(w)+α(v)β(w)γ(u)+α(w)β(u)γ(v)α(u)β(w)γ(v)α(w)β(v)γ(u)α(v)β(u)γ(w)\alpha(u)\beta(v)\gamma(w) + \alpha(v)\beta(w)\gamma(u) + \alpha(w)\beta(u)\gamma(v) \\ - \alpha(u)\beta(w)\gamma(v) - \alpha(w)\beta(v)\gamma(u) - \alpha(v)\beta(u)\gamma(w)

Generally, a k-form is a fully antisymmetric, multilinear measurement of a vector.

Typically, think of a k-form as a map from a k-vector to a scalar.

α:V×VR\alpha : V \times \cdots V \rightarrow \mathbb{R}
(α1αk)(u1,,uk):=det((α1(u1),...,αk(uk)),,(αk(u1),...,αk(uk)))(\alpha_1 \wedge \cdots \wedge \alpha_k)(u_1, \dots, u_k) := \det((\alpha_1(u_1), ..., \alpha_k(u_k)), \dots, (\alpha_k(u_1), ..., \alpha_k(u_k)))

For historical reasons, we do not write

(αβ)(uv)(\alpha \wedge \beta) (u \wedge v)

and instead use the notation we have been using above.

note: a 0-form is a scalar

k-forms in coordinates

For vectors in a basis e1,,ene_1, \dots, e_n, write

v=v1e1++vnenv = v^1 e_1 + \cdots + v^n e_n

The scalar values viv^i are the coordinates of v.

For covectors in a so-called dual basis e1,,ene^1, \dots, e^n, write

α=α1e1++αnen\alpha = \alpha_1 e^1 + \cdots + \alpha_n e^n


ei(ej)=δije^i(e_j) = \delta_{ij}

where δij\delta_{ij} is the Kronecker delta.


v=2e1+2e2v = 2 e_1 + 2 e_2 α=2e1+3e2\alpha = -2 e^1 + 3 e^2

α(v)=(2e1+3e2)(2e1+2e2)=2e1(2e1+2e2)+3e2(2e1+2e2)=4e1(e1)4e1(e2)+6e2(e1)+6e2(e2)=4+6=2\alpha(v) = (-2e^1 + 3e^2)(2e_1 + 2e_2) \\ \quad = -2e^1(2e_1 + 2e_2) + 3e^2(2e_1 + 2e_2) \\ \quad = -4 e^1(e_1) - 4e^1(e_2) + 6e^2(e_1) + 6e^2(e_2) \\ \quad = -4 + 6 = 2

ex. 2-form

u=2e1+2e2α=e1+3e2v=2e1+2e2β=2e1+e2u = 2e_1 + 2e_2 \quad \alpha = e^1 + 3e^2 \\ v = -2e_1 + 2e_2 \quad \beta = 2e^1 + e^2


(αβ)(u,v)=α(u)β(v)α(v)β(u)α(u)=2+6=8α(v)=2+6=4β(u)=4+2=6β(v)=4+2=2(\alpha \wedge \beta)(u,v) = \alpha(u)\beta(v) - \alpha(v)\beta(u) \\ \alpha(u) = 2 + 6 = 8\\ \alpha(v) = -2 + 6 = 4\\ \beta(u) = 4 + 2 = 6\\ \beta(v) = -4 + 2 = -2


(αβ)(u,v)=8(2)(4)6=40(\alpha \wedge \beta)(u,v) = 8\cdot(-2) - (4)\cdot6 = -40

Recall  and \sharp \text{ and } \flat If α=u  u=α\alpha^\sharp = u ~\land~ u^\flat = \alpha then

α=a1e1++anen    u=u1e1++unen\alpha = a_1 e^1 + \cdots + a_ne^n \stackrel{\sharp}{\implies} u = u^1e_1 + \cdots + u^ne_n α=a1e1++anen    u=u1e1++unen\alpha = a_1 e^1 + \cdots + a_ne^n \stackrel{\flat}{\impliedby} u = u^1e_1 + \cdots + u^ne_n

This is sometimes called the musical isomorphism