Differential Geometry

This is that phenomenally fascinating subject that answers many of the questions you might have had lingering from Multivariable Calculus and many more you never thought to ask.

$\gdef\pp{\mathbf{p}} \gdef\uu{\mathbf{u}} \gdef\vv{\mathbf{v}} \gdef\FF{\mathbf{F}} \gdef\GG{\mathbf{G}} \gdef\ww{\mathbf{w}} \gdef\RR{\mathbb{R}} \gdef\st{\; \mid \;} \gdef\cross{\times} \gdef\dot{\cdot} \gdef\grad{\nabla} \gdef\curl{\grad\cross} \gdef\div{\grad\dot} \gdef\tgrad{\text{grad}\;} \gdef\tcurl{\text{curl}\;} \gdef\tdiv{\text{div}\;} \gdef\pdiff#1#2{\frac{\partial #1}{\partial #2}} \gdef\vvec#1{\left<#1\right>}$

Basics

• Tangent vector: $\vv_\pp \in T_\pp \RR^3$: vector $\vv$ rooted at point $\pp$
• Vector basis at $\pp$: $\{U_1(\pp), U_2(\pp), U_3(\pp)\}$
• Directional derivative of $f:\RR^3\rightarrow \RR$: $df(\vv) = \vv[f] = \grad_\vv f = \grad f \cdot \vv$

Curves

Type Description
curve (trajectory) Differentiable function $\alpha : I \rightarrow \RR^3$, $I\subset \RR$ is open
Curve (set of points; implicit) $\{\pp \st f(\pp)=0\}$ in $\RR^2$

From Calculus:

• $\beta(s) = \alpha(h(s)) \quad\Rightarrow\quad \beta'(s) = \alpha'(h(s)) h'(s)$
• $\alpha'(t)[f] \quad=\quad df(\alpha'(t)) \quad=\quad \grad f(\alpha(t)) \cdot \alpha'(t) \quad=\quad \frac{d}{dt}\left(f(\alpha(t))\right)$

Differential 1-form

The total differential $df = f_x dx + f_y dy + f_z dz$ from multivariable calculus is actually a 1-form: a linear function that maps tangent vectors to real numbers so that

• $df(\vv_\pp) = (f_x dx + f_y dy + f_z dz)(v_1, v_2, v_3)_\pp = \grad f\cdot \vv_\pp = \vv_\pp[f]$ (directional derivative)

If a differential geometry vector is thought of as a linear algebra column vector, a 1-form is like a row vector at each point. This table summarizes the relationship in $\RR^3$ (it generalizes easily to $\RR^n$).

(Tangent) Vector Field1-form
Def$V: \RR^3 \rightarrow T_\pp \RR^3$
(assigns $\vv_\pp$ to each $\pp\in\RR^3$)
$\phi:T_\pp \RR^3 \rightarrow \RR$
(linear at each $\pp\in\RR^3$)
Basis$\{U_1(\pp), U_2(\pp), U_3(\pp)\}$$\{dx, dy, dz\}$ or $\{dx_1, dx_2, dx_3\}$
Basis interaction
$dx_i(U_j) = \left\{\begin{matrix}1 \text{ if } i=j \\0 \text{ if } i \ne j\end{matrix}\right. \qquad \Rightarrow \qquad \left[\begin{matrix}dx_1\\dx_2\\dx_3\end{matrix}\right] \left[\begin{matrix}U_1 & U_2 & U_3\end{matrix}\right] = I$
Like a dot product
$\underbrace{(3dx+4dy+dz)}_{\left[\begin{matrix}3&4&1\end{matrix}\right] \left[\begin{matrix}dx_1\\dx_2\\dx_3\end{matrix}\right]}\underbrace{(U_1+3U_2+U_3)}_{\left[\begin{matrix}U_1 & U_2 & U_3\end{matrix}\right]\left[\begin{matrix}1\\3\\1\end{matrix}\right]} = \left[\begin{matrix}3&4&1\end{matrix}\right] \left[\begin{matrix}1\\3\\1\end{matrix}\right] = 16$
Projecting to basis$\vv = \sum_i dx_i(\vv) U_i$$\phi = \sum_i \phi(U_i) dx_i$
Chain rule$\frac{d}{dt}(f(\alpha(t))) = \alpha'(t)[f] = df(\alpha'(t))$$d(h(f)) = h'(f) df$

Further insights into curves, 1-forms, and how to integrate 1-forms along a curve.

Differential $k$-forms

It is easier to understand $k$-forms in terms of $k$-vectors. In $\RR^3$, we have 0-forms, 1-forms, 2-forms, and 3-forms. These can be identified with scalars, vectors, oriented areas, and signed volumes, respectively. The specific case of the 1-form and its interpretation as a row vector in matrix multiplication is shown in the previous section. The wedge product ($\wedge$) combines lower forms to give higher ones, but it is orientation-sensitive like the cross-product. It is an alternating multilinear operator:

• $dx dy = dx\wedge dy = -dy \wedge dx = -dy dx$ ($2$-form)
• $dx dx = dx\wedge dx = 0$

$k$ Calculus 3 $k$-vectors (Clifford algebra) $k$-form $k$-form basis
0 1 (scalar) 1 (0-vector) $f$ $\{1\}$
1 $\vv$ (length vector) $\vv$ (vector) $\phi$ $\{dx, dy, dz\}$
2 $\vv\cross\ww$
(area vector)
$\vv\wedge \ww$
(bivector, oriented area)
$\phi\wedge\psi$ $\{dy dz, dz dx, dx dy\}$
3 $\vv\cross\ww\cdot\uu$
(volume scalar)
$\vv\wedge \ww\wedge \uu$
(trivector, signed volume)
$\phi\wedge\psi\wedge\theta$ $\{dx dy dz\}$

As one can imagine, based on the analogy with Calculus 3 objects above, the wedge ($\wedge$) operator computes a determinant ($k$-dimensional volume). Thus:

• $dx dy dz = dy dz dx=dz dx dy=-dy dx dz$

just like

• $\vv\cross\ww\cdot\uu=\ww\cross\uu\cdot\vv=\uu\cross\vv\cdot\ww=-\ww\cross\vv\cdot\uu$

Exterior Derivative

The exterior derivative operator $d$ is a linear differential operator that takes a $k$-form to a $(k+1)$-form and is defined via:

• $df(\vv_\pp)=\vv_\pp[f]$ (0-form to 1-form, base case)
• $d(f dx_i \wedge dx_j\wedge\ldots\wedge dx_k) = df \wedge dx_i \wedge dx_j\wedge\ldots\wedge dx_k$ ($k$-form to $(k+1)$-form)
• For example, $d(f dx_i) = df \wedge dx_i$ (1-form to 2-form)

Product rule: $d(\phi\wedge\psi) = d\phi\wedge\psi + (-1)^k \phi\wedge d\psi$ (where $\phi$ is a $k$-form)

Properties in $\RR^3$ (prove these!)
$k$-formexterior derivativeCalculus 3 analog
$f$$df = f_x dx + f_y dy + f_z dz$$\grad f = \vvec{f_x, f_y, f_z}$
$\phi = f dx + g dy + h dz$$d\phi = \left\|\begin{matrix}dydz&dzdx&dxdy\\\pdiff{}{x}&\pdiff{}{y}&\pdiff{}{z}\\f&g&h\end{matrix}\right\|$$\curl\vvec{f,g,h}=\left\|\begin{matrix}U_1&U_2&U_3\\\pdiff{}{x}&\pdiff{}{y}&\pdiff{}{z}\\f&g&h\end{matrix}\right\|$
$\phi = f dydz + g dzdx + h dxdy$$d\phi = (f_x+g_y+h_z)dz dy dz$$\div\vvec{f,g,h} = f_x+g_y+h_z$
$k \ge 3$$d\phi=0$

RecallAny dimension
$\curl\grad f=\mathbf{0}$$\div\curl\FF = 0$$dd\phi = 0$
Conservative vector field $\FF$: satisfies $\FF = \grad f$ for some $f$Curl-exact vector field $\GG$: satisfies $\GG = \curl \FF$ for some $\FF$Exact form $\phi$: satisfies $\phi = d\psi$ for some $\psi$
Curl-free: satisfies $\curl \FF = \mathbf{0}$Diverence-free: satisfies $\div \GG = 0$Closed form $\phi$: satisfies $d\phi = 0$
Conservative ⇒ curl-freeCurl-exact ⇒ divergence-freeExact ⇒ closed
In a simply-connected domain, the reverse (⇐) is also true.
• See also the last two tables in my Calculus 3 summary and observe that there is a generalized Stokes' theorem in differential geometry that works in any dimension: $\int_{\Omega} d\omega = \int_{\partial \Omega} \omega$.

• Question: How can one represent (Laplace operator) $\div\grad f = \grad^2 f$ in differential geometry? To answer this and also to get a big picture of how everything connects, look at my notes on the bigger picture.