# Differential Geometry – Integrals

## Curves

A smooth, oriented curve (trajectory) {$\ct$} is a map from an interval {$ [a,b]$} into {$\RR^n$} such that {$\ct'(t)$} exists and {$|\ct'(t)| \ne 0$}. The image {$\ct([a,b])$} (i.e. the collection of corresponding points in {$\RR^n$} is called a Curve {$C$}.

**Parametrization independent quantities:** Those that depend only on {$C$} and
possibly the orientation, but not on the specific trajectory {$\ct$}.

**unit tangent:**{$\TT = \frac{\ct'(t)}{|\ct'(t)|}$}**unit normal (2D):**{$\NN = \TT^\perp$}**total displacement**,**total arc length**

Two different curves {$\ct(t)$} and {$\cs(u)$} are reparametrizations of each other if their images are equal. These two curves correspond to the traversal of the same Curve with different speeds. E.g.

are different parametrizations/traversals of the Curve
`{$\{(x,x^2,x^3) \st x \in [1,4]\}$}`

. Observe that these two
parametrizations have opposite orientation.

Our goal is to compute useful quantities with curves that are **independent of parametrization**, but may be dependent on **orientation**. To do this, we need
the following concepts.

### Velocity and speed

Below, assume that {$C$} is the Curve, equipped with an orientation, corresponding to the curve {$\ct$}.

velocity | {$\frac{d\ct}{dt} = \ct'(t) \quad \Rightarrow \quad d\ct = \ct'(t) dt$} | displacement function | {$\ct(t)-\ct(a) = \int_a^t \ct'(u) du = \int_C d\ct$} |
---|---|---|---|

speed | {$\frac{ds}{dt} = \|\ct'(t)\| \quad \Rightarrow \quad ds = \|\ct'(t)\| dt$} | arc length | {$s(t) = \int_a^t \|\ct'(u)\| du = \int_C ds$} |

As mentioned before, the **total displacement** and **total arc length** are
**parametrization-independent** (up to orientation).

**Parametrization-independent integration**

Since {$s(t)$} is strictly increasing, its inverse {$t(s)$} exists so that the unit-speed parametrization {$\cs(s) = \ct(t(s))$} with the same orientation as {$\ct$} is well-defined.

**Verify:**

- {$\cs'(s) = \TT$} and thus also {$|\cs'(s)| = 1$} (parametrization independent!)
- {$\TT ds = d\ct$} and {$\TT \cdot d\ct = ds$}

Can now integrate functions along oriented curve {$C$} in a parametrization-independent way. These are called **line integrals.**

scalar value |
{$\int_C f ds = \int_C f |\ct(t)'| dt$} | arc length when {$f=1$} |

vector field along curve |
{$\int_C \FF \cdot \TT ds \quad = \quad \int_C \FF \cdot d\alpha$} | how much {$\FF$} is flowing along {$C$} |

vector flux through curve (2D) |
{$\int_C \FF \cdot \NN ds$} | how much {$\FF$} is flowing through {$C$} |

### Differential geometry perspective

Recall from differential geometry that we had the following properties between the standard basis {$U_1, U_2, U_3$} for vector fields and their dual 1-forms {$dx_1, dx_2, dx_3$}. Specifically, {$dx_i$} is defined so that
```
$$
\begin{aligned}
dx_i(U_j) = \delta_{ij} = \left\{\begin{matrix}1 & \text{ if } i = j\\0 & \text{ if } i\ne j\end{matrix}\right. \qquad \Rightarrow \qquad
\left[\begin{matrix}
dx_1\\dx_2\\dx_3
\end{matrix}\right]
\left[\begin{matrix}
U_1 & U_2 & U_3
\end{matrix}\right]
= I
\end{aligned}
$$
```

The following equivalent interpretations for the dual 1-forms {$dx_i$} offer some insights that will help in subsequent discussion.

- {$dx_i(\vv)$} gives the coefficient of {$\vv$} in the basis
`{$\{U_1, U_2, U_3\}$}`

. - {$dx_1,dx_2,dx_3$} act like row vectors of the inverse of the basis matrix {$A = [U_1 \; U_2 \; U_3]$}. In this special case, since the basis is orthonormal, the inverse is simply the transpose {$A^T$}. However, for a more general basis as we will soon see, this is not the case.

### Curve differentials as 1-forms

**Goal:** The description of line integrals seems to use two differentials {$dt$} and {$ds$}, as well as a vector {$d\ct$} of differentials. Understanding the 1-form interpretation of these differentials allows for a formalization of 1-form integration.

Recall that differential 1-forms, by definition, are *linear scalar-valued functions on tangent vectors* at each point and thus have the *same dimension as the tangent space*. Since at each point on a curve, the tangent space is a 1-dimensional, so is the space of 1-forms. So, we need to define just a single basis 1-form of the tangent space of a curve.

**Definition of the {$dt$} 1-form:** Follow the pattern above from differential geometry, define {$dt$} to be the unique 1-form along the curve such that {$dt(\pdiff{\ct}{t}) = dt(\ct'(t)) = 1$} and {$dt(\vv) = 0$} for all {$\vv \perp \ct'(t)$}. In other words, **{$dt$} is the dual 1-form of the vector field {$\pdiff{\ct}{t}$}** which spans the tangent space along the curve.

**Exercises**

**Definition implications:**With {$\ct(t) = \vvec{x(t),y(t),z(t)}$}, write {$dt$} in the standard 1-form basis; i.e. figure out the coefficients in {$dt = a\,dx + b\,dy + c\,dz$} using the second definition above. What is the magnitude/norm of the 1-form {$dt$}? How is this consistent with the magnitude of {$\ct'(t)$}?{$dt = \frac{x'(t) dx+y'(t) dy+z'(t) dz}{\|\ct'(t)\|^2}$}, so that {$dt(\vv) = \frac{\ct'(t) \cdot \vv}{\ct'(t) \cdot \ct'(t)}$}. The magnitude {$\frac{1}{\|\ct'(t)\|}$} is the reciprocal of {$\|\ct'(t)\|$} as expected from the definition.**Arc-length parametrization:**Show that in the special case of a unit-speed parametrization {$\cs(s) = \vvec{\tilde{x}(t),\tilde{y}(t),\tilde{z}(t)}$}, write {$ds$} in the 1-form basis`{$\{dx,dy,dz\}$}`

as above, leading to the conclusion that {$ds(\vv) = \TT \cdot \vv$}, or equivalently, {$ds(\TT) = 1$}. Verify the conversion formula between the two 1-forms {$ds = |\ct'(t)|dt$}.{$ds = \tilde{x}'(s) dx + \tilde{y}'(s) dy + \tilde{z}'(s) dz = \frac{x'(t) dx+y'(t) dy+z'(t) dz}{\|\ct'(t)\|} = \|\ct'(t)\|dt$}. Note that {$\vvec{\tilde{x}'(s), \tilde{y}'(s), \tilde{z}'(s)} = \TT$} and the magnitude is 1 as expected.**Note:**Try these problems before going into the next section as the solutions would be referred to below.

### Integration of an arbitrary 1-form along a curve

Observing that {$dt(\ct'(t)) = 1$} and that {$ds(\ct'(t)) = |\ct'(t)|$}, one can define the integration of a 1-form along the curve via Thus,

- {$\int g\, dt = \int g\, dt(\ct'(t))\,dt = \int g\, (1)\, dt = \int g\, dt$} as expected
- {$\int f\, ds = \int f\, ds(\ct'(t))\, dt = \int f\, \TT \cdot \ct'(t)\, dt = \int f\, |\ct'(t)| dt$} as expected But this definition applies to arbitrary 1-forms in {$\RR^3$} leading to the following observation.

**Line integrals of vector fields:**
One can represent a vector field in {$\RR^3$} directly as {$\FF = \vvec{P,Q,R}$} (where each coordinate is a function), or alternatively, in its 1-form representation in the ambient space: {$\phi = P dx + Q dy + R dz$}. Noting that {$d\ct$} ({$=\TT ds$}) is a vector {$\vvec{dx,dy,dz}$} of differentials, here are equivalent ways of integrating the **tangential** component {$\FF\cdot\TT$} of vector field.