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.
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
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{$df(\vv_\pp) = (f_x dx + f_y dy + f_z dz)(v_1, v_2, v_3)_\pp = \grad f\cdot \vv_\pp = \vv_\pp[f]$} (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 Field | 1-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!) | ||
|---|---|---|
| {$k$}-form | exterior derivative | Calculus 3 analog |
| {$f$} | {$df = f_x dx + f_y dy + f_z dz$} | {$\grad f = \vvec{f_x, f_y, f_z}$} |
| {$\phi = f dx + g dy + h dz$} | {$d\phi = \left\|\begin{matrix}dydz&dzdx&dxdy\\\pdiff{}{x}&\pdiff{}{y}&\pdiff{}{z}\\f&g&h\end{matrix}\right\|$} | {$\curl\vvec{f,g,h}=\left\|\begin{matrix}U_1&U_2&U_3\\\pdiff{}{x}&\pdiff{}{y}&\pdiff{}{z}\\f&g&h\end{matrix}\right\|$} |
| {$\phi = f dydz + g dzdx + h dxdy$} | {$d\phi = (f_x+g_y+h_z)dz dy dz$} | {$\div\vvec{f,g,h} = f_x+g_y+h_z$} |
| {$k \ge 3$} | {$d\phi=0$} | |
| Recall | Any 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-free | Curl-exact ⇒ divergence-free | Exact ⇒ closed |
| In a simply-connected domain, the reverse (⇐) is also true. | ||
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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: {$\int_{\Omega} d\omega = \int_{\partial \Omega} \omega$}.
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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.