\part{Fundamentals}
\chapter{Differentiable curves}

Differential geometry is basically the study of smooth geometric objects, such as curves, surfaces, and more generally, generalizing the geometric ideas from these structures (like curvature) to differentiable manifolds (particularly with the idea of doing calculus on these spaces).

To distingush, differential topology tends to concern itself more with topological properties of the differentiable manifolds. That said, ideas from differential topology are often used for terminology and to rigorously build the theory of differential geometry.


This means that differential geometry draws extensively from vector analysis, multilinear algebra, general topology, and differential topology; a background in all 4 fields is necessary to study this field.


\begin{definition}[Differentiable curve]
A \emph{differentiable curve} is a $C^{\infty}$ function $\gamma : I \to \mathbb{R}^n $ mapping a real interval to a Euclidean space.
Sometimes the image of this function is also reffered to as the differentiable curve.
\end{definition}

We also have the notion of closed differentiable curves and so forth.
We should note that differentiable curves may have cusps and loops or 'back tracking' (going backwards along its previous path).

Folium of Descartes
Line
Circle
Helix

One important geometric property of curves is their \emph{arclength}; how long is a curve?
\begin{definition}
The \emph{length of a curve}
\[ \ell(\gamma) = \int^{t_1}_{t_0} \| \gamma' (t) \|dt\]
The \emph{arc-length of a curve}
\[ s_{\gamma}(t) = \int^{t}_{t_0} \| \gamma' (u) \|du\]
\end{definition}

It's important to note that there are mutliple ways to parametrize a curve.

\begin{definition}
A \emph{reparametrization of a curve}
\end{definition}

differentiable curves have the same length

Note that Curves may also have other pathologial, like retracing previous paths. 

Although differentiable curves have infinite reparametrizations, . We can do this by using the arc-length function as a diffeomorphism. Unfortunately, it is not always a diffeomorphism; we require a 'nicer' class of differentiable curves called regular differentiable curves.

\begin{definition}
	A \emph{regular curve} is a differentiable curve $\gamma$ where $\gamma'(t) \neq \mathbf{0}$
\end{definition}

It would be nice if we could just reparametrize a curve based on its arc-length.

\begin{definition}
An \emph{arc-length reparametrization of a differentiable curve} is 
	\[\gamma \circ s_{\gamma} \]
\end{definition}


It turns out that we need regular curves we can make an arc-length reparametrization of the curve; otherwise the arc length function will only be in $C^1$ rather than $C^{\infty}$.We'll now prove that being regular is the sufficient and necessary condition for such a reparametrization.

\begin{theorem}
A differentiable curve is \emph{arc-length reparametrizable} iff it is regular.
\end{theorem}

There is something interesting to note about the derivative of an arc-length reparametrization.

\begin{proposition}
A differentiable curve is in its arc-length reparametrization iff the norm of its derivative is always 1.
\[ \| \gamma'(s) \| = 1  \]
\end{proposition}
This requires nothing more than the chain rule and inverse function theorem


\begin{definition}
The \emph{winding number}
\end{definition}



\section{Frenet-Serret frame}

Given a arc-length parametrized curve, we can define a frame related to this differentiable curve.

The following result will be useful in understanding the nature of arc-length parametrized curves as well as constructing our frame.

\begin{proposition}
\[\mathbf{r}(s) \cdot \mathbf{r}'(s)=0\]
\[ \| \mathbf{r}'(s) \| =0\]
\end{proposition}


\begin{definition}
The \emph{tangent vector}
\[\mathbf{T}(s) = \mathbf{r}'(s)\]
\end{definition}

\begin{definition}
The \emph{normal vector}
\[\mathbf{N}(s) = \frac{\mathbf{T}'(s)}{\| \mathbf{T}'(s) \|}\]
\end{definition}

\begin{definition}
The \emph{binormal vector}
	\[\mathbf{B}(s) = \mathbf{T}(s) \times \mathbf{N}(s)\]
\end{definition}

Note that $\times$ is the cross product. Some authors prefer to use wedge notation $\land$ since it is the notation for the wedge product (the generalization of the cross product). Not going to lie, I like that idea but I stick to the cross notation since we're dealing with a Euclidean space.

\begin{definition}
The \emph{Frenet-Serret frame} is an orthonormal frame spanning $\mathbb{R}^3$ derived from a differentiable curve $\mathbf{r}(s)$ parametrized by arclength.
\[ \mathbf{T}(s) = \mathbf{r}'(s) \]
\[ \mathbf{N}(s) =  \frac{\mathbf{T}'(s)}{ \| \mathbf{T}'(s) \|} \]
\[ \mathbf{B}(s) = \mathbf{T}(s) \times \mathbf{N}(s) \]
The chain rule can be used to represent these basis functions with respect to time.
\end{definition}



\begin{definition}[Curvature]
The \emph{curvature} is the measure of directional change on the tangential plane
\[ \kappa = \| \mathbf{T}' \| \]
\[ \kappa = \mathbf{T}' \cdot \mathbf{N} \]
\end{definition}

\begin{definition}[Torsion]
The \emph{torsion} is the measure of directional change out of the tangential plane.
\[ \tau = \| \mathbf{B}' \| \]
\[ \tau = - \mathbf{B}' \cdot \mathbf{N} \]
\end{definition}

\begin{theorem}[Frenet-Serret equations]
	\[ \mathbf{T}'(s) = \kappa(s) \mathbf{N}(s)\]
	\[ \mathbf{N}'(s) = -\kappa(s) \mathbf{T}(s) - \tau(s) \mathbf{B}(s)\]
	\[ \mathbf{B}'(s) = \tau(s) \mathbf{N}(s)\]
\end{theorem}




\begin{proposition}
\[ \kappa = \frac{\| \mathbf{r}' \times \mathbf{r}'' \|}{\| \mathbf{r}'\|^{3}} \]
\end{proposition}


\begin{proposition}
\[ \tau= \frac{ \mathbf{r}' \cdot ( \mathbf{r}'' \times \mathbf{r}''' ) }{\| \mathbf{r}' \times \mathbf{r}''\|^{2}} \]
\end{proposition}








One may wonder, can regular differentiable curves be categorized exactly by their Frenet-Serret equations?

\begin{theorem}[Fundamental theorem of curves]
Given scalar functions $\kappa_0,\tau_0$ where $\kappa_0 >0$, there exists an arc-length parametrized curve in $\mathbb{R}^3$ with curvature $\kappa_0$ and torsion $\tau_0$. Moreover, this curve is unique up to an isometry.
\end{theorem}
This is proven by using the Frenet-Serret equations as an ODE, proving that any solution forms an orthonormal frame.


It is worth noting that the idea of torsion being the 'twist' factor out of the tangent-normal plane gives a good criteria for planar curves.

\begin{proposition}
A differentiable curve is a planar curve iff its torsion is always 0.
\end{proposition}

\begin{proposition}
A differentiable curve is a piece of a line iff its curvature is always 0.
\end{proposition}


We'll now tackle an ancient problem; which curve with length $\ell$ has the largest area?

\begin{proposition}[Isoperimetric inequality]
Let $\gamma$ be a closed differentiable curve, then trhe following holds.
\[(\ell (\gamma) )^2 \geq 4\pi \mathrm{A}\]
\end{proposition}



\section{Local canonical form}


\[\gamma(s) = \sum^{3}_{n=1} \frac{s^n}{n!} \gamma^{(n)} (0) + R(s)\]

By the definitions of the Frenet-Serret frame and the Frenet-Serret equations, we have the following.

\[\gamma(s)-\gamma(0) = (s- \frac{\kappa^2(s) s^3}{3!}) \mathbf{T} + (\frac{s^2 \kappa (s)}{2}+ \frac{s^3 \kappa'(s)}{3!})\mathbf{N} - \frac{\kappa(s)\tau(s)s^3}{3!} \mathbf{B} +R(s)\]


\section{Total curvature}


\begin{comment}


			<h3 class=cyan>Frenet-Serret frame</h3>
			<p>Orthonormal basis spanning \(\mathbb{R}^3\) derived from a point in a path function \(\mathbf{r}(t)\) with arclength \(s\).</p>
			<ul>
				<li><b>Tangent</b>, unit vector with direction tangent to the path at the point</li>
				<li><b>Normal</b>, unit vector with direction normal to the path at the point (differentiation of tangent with respect to arclength)</li>
				<li><b>Binormal</b>, unit vectot with direction normal to the two other vectors</li>
			</ul>
			<ul>
				<li>\(\mathbf{T}(s) = \mathbf{r}'(s)\)</li>
				<li>\(\mathbf{N}(s) =  \frac{\mathbf{T}'(s)}{ \| \mathbf{T}'(s) \|}\)</li>
				<li>\(\mathbf{B}(s) = \mathbf{T}(s) \times \mathbf{N}(s)\)</li>
			</ul>
			<p>The chain rule can be used to represent these basis functions with respect to time</p>

			<h5 class=cyan>Proof</h5>
			<p>The unit vector in the direction of the path is \(\hat{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{|\frac{d\mathbf{r}}{dt}|}\) </p>
			<p>\(\hat{T} \cdot \hat{T}=1 \)</p>
			<p>\(\implies \frac{d}{dt}(\hat{T} \cdot \hat{T}) =0\)</p>
			<p>\(\implies 2\hat{T} \cdot \frac{d\hat{T}}{dt} = 0 \)</p>
			<p>\(\implies \hat{T} \perp \frac{d\hat{T}}{dt} \)</p>
			<p>Therefore after reducing to a unit vector, \(\hat{N}(t) =  \frac{\frac{d\hat{T}}{dt}}{| \frac{d\hat{T}}{dt} |}\) is added to the orthonormal basis.</p>
			<p>The cross product \(\hat{B}(t) = \hat{T} \times \hat{N}\) can be used to find one of the two unit vectors that can complete the orthonormal basis </p>-->





			<h3 class=cyan>Frenet-Serret formulae</h3>
			<ul>
				<li>\(\mathbf{T}' = \kappa \mathbf{N}\)</li>
				<li>\(\mathbf{N}' = -\kappa \mathbf{T} +\tau \mathbf{B}\)</li>
				<li>\(\mathbf{B}' = -\tau \mathbf{N}\)</li>
			</ul>

			<h3 class=cyan>Fundamental planes</h3>
			<ul>
				<li>Osculating plane \(\text{span}\{ \mathbf{T} , \mathbf{N} \} \)</li>
				<li>Rectifying plane \(\text{span}\{ \mathbf{T} , \mathbf{B} \} \)</li>
				<li>Normal plane \(\text{span}\{ \mathbf{N} , \mathbf{B} \} \)</li>
			</ul>

\end{comment}




