%!TEX root = 162AB Lecture Notes v2.tex
\chapter{Frame Fields}
\begin{introduction}
\item The dot product revisited
\item The Frenet formulas
\end{introduction}



\section{Dot Product}
We have discussed inner product in Chapter~\ref{chp0}. The dot product is a special case of inner product. So we shall very quick go through it.

\begin{definition}{}{}
Let $\p=(p_1,p_2,p_3)$ and $\q=(q_1,q_2,q_3)$ in $\R^3$. The \red{dot product } is defined by
\[
\p\cdot\q=p_1q_1+p_2q_2+p_3q_3.
\]
The \red{norm } of a point $\p$ is defined by 
\[
\|\p\|=\sqrt{\p\cdot\p}.
\]
The \red{Euclidean distance } from $\p$ to $\q$ is the number 
\[
d(\p,\q)=\|\p-\q\|.
\]
Vectors are called \red{orthogonal }, if 
\[
\p\cdot\q=0.\]
More generally, the \red{angle } $\theta$ between vectors $\p,\q$ is defined by the equation
\[
\p\cdot\q=\|\p\|\cdot\|\q\|\cdot\cos\theta.
\]
A vector $\p$  is called a \red{unit } vector, if $\|\p\|=1$.
\end{definition}



\begin{definition}{}{}
A set $\{\ee_1,\ee_2,\ee_3\}$ of three mutually orthogonal unit vectors tangent to $\R^3$ at $\p$ is called a \red{frame}\footnote{it is also called an orthonormal basis.} at the point $\p$.
\end{definition}

Thus $\{\ee_1,\ee_2,\ee_3\}$ is a frame if and only if 
\begin{align*}
& \ee_1\cdot\ee_1=\ee_2\cdot\ee_2=\ee_3\cdot\ee_3=1;\\
&\ee_1\cdot\ee_2=\ee_1\cdot\ee_3=\ee_2\cdot\ee_3=0.
\end{align*}

Using the Einstein Convention, we have $\ee_i\cdot\ee_j=\delta_{ij}$.

\begin{definition}{}{}
Let $\ee_1,\ee_2,\ee_3$ be a frame at a point $\p$ of $\R^3$. The $3\times 3$ matrix
$A$ whose rows are the Euclidean coordinates of these three vectors is called
the \red{attitude matrix } of the frame.
\end{definition}

Explicitly, if 
\begin{align*}
& \ee_1=(a_{11},a_{12},a_{13})_\p,\\
& \ee_2=(a_{21},a_{22},a_{23})_\p,\\
& \ee_3=(a_{31},a_{32},a_{33})_\p,
\end{align*} then
\[
A=\begin{bmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{bmatrix}
\]

We shall prove that $A$ is an \emph{\red{orthogonal}} matrix: we have
\[
\delta_{ij}=\ee_i\cdot\ee_j=a_{ik}a_{jk}.
\]

The matrix expression of the above equation is 
\[
AA^T=I,
\]
where $A^T$ is the transpose of $A$. 

\begin{note} If $AA^T=I$, then $A^T=A^{-1}$. Therefore 
\[
A^TA=A^{-1}A=I.
\]
Note that this is not a trivial result. Take a $2\times 2$ matrix, for example, Let
\[
A=\begin{bmatrix}a&b\\c&d\end{bmatrix}.
\]
Then $AA^T=I$ is equivalent to 
\[
a^2+b^2=1,\qquad ac+bd=0,\qquad c^2+d^2=1,
\]
and
\[
A^TA=I
\]
is equivalent to 
\[
a^2+c^2=1,\qquad ab+cd=0,\qquad b^2+d^2=1.
\]
Can you find an elementary proof of the above fact?  Hint: we can prove the identity
\begin{align*}
&
(a^2+c^2-1)^2+(b^2+d^2-1)^2+2(ab+cd)^2\\
& =(a^2+b^2-1)^2+(c^2+d^2-1)^2+2(ac+bd)^2.
\end{align*}
\end{note}



\section{Curves}

Let $\alpha(t)=(\alpha_1(t),\alpha_2(t),\alpha_3(t))$ be a curve. It is well known that 
\[
\alpha'(t)=(\alpha_1'(t),\alpha_2'(t),\alpha_3'(t))
\]
is the velocity of the curve. The velocity is a vector field. Its norm, $\|\alpha'(t)\|$, is called the \emph{\red{speed}} of the curve. The speed of a curve is a function along the curve. 

In terms of the components, we can write
\[
\nu=\|\alpha'\|=\left(\left(\frac{d\alpha_1}{dt}\right)^2+\left(\frac{d\alpha_2}{dt}\right)^2+\left(\frac{d\alpha_3}{dt}\right)^2\right)^{1/2}.
\]
The length of the curve, from $t=a$ to $t=b$, can be expressed by
\[
\int_a^b\|\alpha'(t)\|\, dt.
\]

\begin{definition}{}{}
A curve $\alpha: I\to \R^3$ is called \red{regular}, if the velocity vector field $\alpha'(t)\neq 0$ for any $t\in I$. This is equivalently  to say that the speed function $\nu=\|\alpha'(t)\|$ is not zero at any point $t\in I$. If $\|\alpha'(t)\|=1$, then the curve is called a \red{unit speed } curve. 
\end{definition}
\begin{definition}{}{}
A \red{reparametrization } is a map $t:[c,d]\to[a,b]$ such that the function is one-to-one and onto, and that $t'(s)\neq 0$ for any $s\in[c,d]$.
\end{definition}

\begin{proposition}{}{}
Let $t(s)$ be a reparametrization. Then we have the following two cases:
\begin{enumerate}[label={\rm (\arabic*)}]
\item $t'(s)>0$ and $t(c)=a, t(d)=b$;
\item  $t'(s)<0$ and $t(c)=b, t(d)=a$.
\end{enumerate}
\end{proposition}

\begin{proof}
Since $t'(s)\neq0$, we must have either $t'(s)>0$ or $t'(s)<0$. In the first case, $t(s)$ is monotonically increasing. Thus $t(c)$ must be minimal and hence equal to $a$, and $t(d)$ must be maximum, hence equal to $b$. This proves the first case. 

\end{proof}


\begin{theorem}{}{}
The length of a curve is an \red{invariant}.
\end{theorem}
\begin{proof}
Let $t=t(s)$ be a reparametrization, that is $t: [c,d]\to [a,b]$ such that the function $t$ is one-to-one and onto and  we assume that $t'(s)\neq 0$. Let $s=s(t): [a,b]\to[c,d]$ be the inverse function. Let $\beta(s)=\alpha(t(s))$. Then the length of the curve $\beta$, from $c$ to $d$, is
\[
\int_c^d\|\beta'(s)\|ds.
\]
By the chain rule, $\beta'(s)=t'(s)\alpha'(t(s))$. We then have
\[
\int_c^d\|\beta'(s)\|ds=\int_c^d|t'(s)|\cdot\|\alpha'(t(s))\|ds=\int_a^b\|\alpha'(t)\|\, dt.
\]
\end{proof}




\begin{theorem}{}{}
If $\alpha$ is a regular curve, then there is a parametrization $\beta$ such that $\beta$ has unit speed. 
\end{theorem}


\begin{proof}
Let $t=t(s)$ be a reparametrization and let $\beta(s)=\alpha(t(s))$. The requirement is that 
\[
1=\|b'(s)\|=|t'(s)|\cdot\|\alpha'(t(s))\|
\]
for any $s$. 
Thus in order to find the unit speed parametrization, we need to solve the differential equation
\[
t'(s)=\frac{1}{\|\alpha'(t(s))\|}
\]
with the initial value $t(0)=a$. This is a separable equation. We write $s'(t)=ds/dt$. Let $s=s(t)$ be the inverse function of $t(s)$. Then
\[
s(t)=\int_a^t\|\alpha'(u)\| du
\]
defines the unit speed reparametrization. 

\end{proof}


\begin{note}
Note that $s(a)=0$, and $s(b)$ is the length of the curve from $a$ to $b$.
\end{note}


\begin{example}
Consider the helix
\[
\alpha(t)=(a\cos t,a\sin t, bt).
\]
The velocity of $\alpha$ is
\[
\alpha'(t)=(-a\sin t, a\cos t, b).
\]
The speed of $\alpha$ is
\[
\|\alpha'(t)\|=\sqrt{a^2\sin^2 t+a^2\cos t^2 +b^2}=\sqrt{a^2+b^2}.
\]
Therefore $\alpha$ has \emph{constant speed} $c=\sqrt{a^2+b^2}$. 

Let 
\[
s(t)=\int_0^tc \,du=ct.
\]
Then $t=s/c$. We thus have
\[
\beta(s)=\alpha(s/c)=\left(a\cos\frac sc,a\sin \frac sc,\frac{bs}{c}\right)
\]
is the unit speed reparametrization.
\end{example}

\begin{definition}{}{}
Let $Y=y_iU_i$ be a vector field along a curve $\alpha$. Thus each function $y_i$ can be expressed as a function of $t$ via $\alpha$. The \red{differentiation } of $Y$ is simply differentiation on its Euclidean coordinate functions. 
\end{definition}


\begin{example}
Let 
\[
Y(t)=t^2 U_1-tU_3.
\]
Then 
\[
Y'(t)=2t U_1-U_3.
\]
\end{example}

In particular, we can define the \emph{\red{acceleration}}
$\alpha''(t)$ of the curve $\alpha(t)$.

\begin{lemma}{}{}
\begin{enumerate}[label={\rm (\arabic*)}]
\item A curve $\alpha$ is constant if and only if its velocity is zero, $\alpha'=0$.
\item A non-constant curve $\alpha$ is a straight line if  and only if its acceleration is zero, $\alpha''=0$.
\item A vector field $Y$ on a curve is parallel if and only if its derivative is zero, $Y'=0$\footnote{This is the definition of the  parallelism rather than 
a statment.}.
\end{enumerate}
\end{lemma}


\section{The Frenet Formulas}
Let $\beta: I\to\R^3$ be a unit-speed curve. Let $T=\beta'(s)$ be the velocity vector field. Then we have $\|T\|=1$. We consider $T'=\beta''(s)$.
Since $\|T\|=1$, we have $T\cdot T=1$. Taking derivative\footnote{This technique will be used repeatedly throughout the  rest of the book.} on both sides, we have
$T'\cdot T+T\cdot T'=0$. Thus $T\cdot T'=0$, and $T'$ is always orthogonal to $T$.

\begin{definition}{}{}
The \red{curvature } $\kappa(s)=\|T'(s)\|=\|\beta''(s)\|$. 
\begin{center}
\includegraphics[height=4cm]{images/fig-2.7.png}
\end{center}
\end{definition}


By the above definition, we know that $k(s)\geq 0$. In order to introduce the Frenet Formulas, we further assume that $\kappa>0$.

When $\kappa>0$, we have $N=T'/\kappa$. By definition, we have $\|N\|=1$. By the similar argument as above, we have $N\cdot N'=0$. 



\begin{definition}{}{}
Assume that $\kappa(s)>0$. Define $N=T'/\kappa$, and $B=T\times N$. Then $(T,N,B)$ is an orthonormal basis of the tangent space at point $\beta(s)$. We call $(T,N,B)$ the \red{Frenet frame field}, or \red{TNB frame field } on $\beta$. The collection $\{T,N,B,\kappa,\tau\}$ is called the \red{Frenet Apparatus}.
\end{definition}

\begin{remark}
If $\kappa\equiv 0$, then the curve is a straight line. On the other hand, if $\kappa$ is nowhere zero, then we are able to define the Frenet frame field. It is beyond the scope of this book to discuss curves with vanishing curvature at isolated points. 
\end{remark}

By definition, we have $\|N\|=1$. By the similar argument as above, we have $N\cdot N'=0$. Thus 
\begin{definition}{}{}
We can define
\[
\tau=N'\cdot B,
\]
where $\tau$ is called the \red{torsion } of the curve. 
\end{definition}





\begin{theorem}{Frenet Formulas}{thm1}
If $\beta: I\to\R^3$ is a unit-speed curve with curvature $\kappa>0$ and torsion $\tau$, then we have the following system of ordinary differential equations:
\[
\begin{matrix}
T'=&&\kappa N\\
N'=&-\kappa T&&+\tau B\\
B'=&&-\tau N
\end{matrix}.
\]
In matrix notation, we have
\begin{equation}\label{7}
(T, N, B)'=(T, N, B)\begin{bmatrix}0&-\kappa&0\\\kappa&0&-\tau\\0&\tau&0\end{bmatrix}.
\end{equation}
\end{theorem}

\begin{proof}
The proof is an application of the orthogonality of the frame. The first equation follows by definition. Since $N\cdot N'=0$, we can write
\[
N'=aT+bB.
\]
Thus $a=(N'-\tau B)\cdot T=N'\cdot T$. Since $N\cdot T=0$, we have $N'\cdot T+N\cdot T'=0$. Thus
\[
N'\cdot T=-N\cdot T'=-N\cdot \kappa T=-\kappa.
\]
By definition, we have 
\[
b=N'\cdot B=\tau.
\]
This proves the second equation. 

In order to obtain the third equation, we used the similar method. Write
\[
B'=pT+qN+rB.
\]
We have
\[
p=B'\cdot T=-B\cdot T'=-B\cdot \kappa N=0,
\]
and
\[
q=B'\cdot N=-B\cdot N'=-B\cdot(-\kappa T+\tau B)=-\tau,
\]
and
\[
r=B'\cdot B=0.
\]
This proves the third formula. 

\end{proof}
\vspace{.2in}
\begin{example}
We compute the Frenet frame $T, N, B$ and the curvature
and torsion functions of the unit-speed helix
\[
\beta(s)=\left(a\cos\frac sc, a\sin\frac sc,\frac{bs}{c}\right),
\]
where $c=(a^2+b^2)^{1/2}$ and $a>0$. Now
\[
T(s)=\beta'(s)=\left(-\frac ac\sin\frac sc, \frac ac\cos\frac sc,\frac{b}{c}\right).
\]
Hence 
\[
T'(s)=\left(-\frac{a}{c^2}\cos\frac sc, -\frac{a}{c^2}\sin\frac sc,0\right).
\]
Thus
\[
\kappa(s)= \|T'(s)\|=\frac{a}{c^2}=\frac{a}{a^2+b^2}>0.
\]
Since $T'=\kappa N$, we get
\[
N(s) =\left(-\cos\frac sc, -\sin\frac sc,0\right).
\]
Therefore, we have
\[
B=T\times N=\left(\frac bc\sin\frac sc,-\frac bc\cos\frac sc,\frac ac\right).
\]
The torsion is given by
\[
\tau=N'\cdot B=\frac{b}{c^2}.
\]

\end{example}


In what follows, we use Frenet formulas to study the properties of curves. 

\begin{definition}{}{}
A plane curve in $\R^3$ is a curve that lies in a single plane of $\R^3$.
\end{definition}

\begin{theorem}{}{}
Let $\beta$ be a unit-speed curve in $\R^3$ with $\kappa>0$. Then $\beta$ is a
plane curve if and only if $\tau=0$.
\end{theorem}

\begin{proof}
 If a curve $\beta(s)=(\beta_1(s),\beta_2(s),\beta_3(s))$ lies in a plane, then there are constants $a,b,c,d$ such that 
\[
a\beta_1+b\beta_2+c\beta_3=d.
\]
Using the vector notation, if $\vv=(a,b,c)$, then $\vv\cdot \beta=d$. As a result, we have
\[
\vv\cdot\beta'(s)=\vv\cdot\beta''(s)=\vv\cdot\beta'''(s)=0.
\]
Since $\beta'\perp \beta''$, they are linearly independent. Thus we can write $\beta'''$ as a linear combination of $\beta'$ and $\beta''$.
\[
\beta'''=p\beta'+q\beta'',
\]
where $p,q$ are constants. Since
\[
B=T\times N=\kappa^{-1}\beta'\times\beta'',
\]
we have
\[
\tau=N'\cdot B=(\kappa^{-1} \beta'')'\cdot B=0.
\]

On the other hand, if $\tau\equiv 0$, then $B'=0$. Thus $B$ is a constant vector. Let
\[
f(s)=\beta(s)\cdot B-\beta(0)\cdot B.
\]
Then $f'=0$ and since $f(0)=0$, we know that $f\equiv 0$. Thus $\beta(s)$ is on the plane
\[
\xx\cdot B-\beta(0)\cdot B=0.
\]

\end{proof}

In the following, we shall use the 
Taylor's formula to study the curve $\beta$. We have
\[
\beta(s)=\beta(0)+\beta'(0)s+\frac 12\beta''(0)s^2+\frac 16\beta'''(0)s^3+o(s^3).
\]
Apparently, we have
\begin{align*}
&
\beta'(0)=T(0),\\
&\beta''(0)=\kappa(0)N(0),\\
&\beta'''(0)=(\kappa N)'=\kappa'(0)N+\kappa(0)(-\kappa(0) T(0)+\tau(0) B(0)).
\end{align*}

Therefore, we have
\begin{align}\label{23}
\begin{split}
&
\beta(s)-\beta(0)=sT(0)+\frac 12 s^2\kappa(0)N(0)\\
&+\frac 16s^3(\kappa'(0)N+\kappa(0)(-\kappa(0) T(0)+\tau(0) B(0)))+o(s^3).
\end{split}
\end{align}
From the above formula, we know that the Frenet Apparatus completely determined the curve, at least for the first three terms in the Taylor's expansion. 
\vspace{.2in}

\begin{remark}
We have emphasized all along the distinction between a
tangent vector and a point of $\R^3$. However, Euclidean space has, as we have
seen, the remarkable property that given a point $\p$, there is a natural one-to-one
correspondence between points $(v_1, v_2, v_3)$ and tangent vectors $(v_1, v_2, v_3)_\p$
at $\p$. Thus one can transform points into tangent vectors (and vice versa) by
means of this canonical isomorphism. In the next two sections particularly,
it will often be convenient to switch quietly from one to the other without
change of notation. Since corresponding objects have the same Euclidean coordinates,
this switching can have no effect on scalar multiplication, addition,
dot products, differentiation, or any other operation defined in terms of
Euclidean coordinates.

So in ~\eqref{23}, the left side $\beta(s)-\beta(0)$ is a vector on $\R^3$, and the right hand side, which is a linear combination of $T(0),N(0)$ and $B(0)$, is a vector in the tangent space of $\beta(0)$. We identify $T_{\beta(0)}\R^3$ with $\R^3$ so the  equality makes sense. 
\end{remark}



\section{Arbitrary-Speed Curves}

It is a simple matter to adapt the results of the previous section to the study
of a regular curve $\alpha: I\to\R^3$ that does not necessarily have unit speed. We
merely transfer to $\alpha$ the Frenet apparatus of a unit-speed reparametrization $\bar\alpha$ 
of $\alpha$. Explicitly, if $s$ is an arc length function for $\alpha$, then
\[
\alpha(t)=\bar\alpha(s(t))
\]
for all $t$. Or, we can write $\alpha(t)=\bar\alpha(s)$. 

Assume that $\bar T, \bar N, \bar B,\bar\kappa>0,\bar\tau$ are the Frenet Apparatus with respect to the unit speed curve $\bar\alpha(s)$. Then we 
define
\begin{itemize}
\item \emph{\red{curvature}} function: $\kappa=\bar\kappa(s)$,
\item \emph{\red{torsion}} function: $\tau=\bar\tau(s)$,
\item \emph{\red{unit tangent}} vector field: $T=\bar T(s)$,
\item \emph{\red{principal}} vector field: $N=\bar N(s)$,
\item\emph{\red{binomial}} vector field: $B=\bar B(s)$.
\end{itemize}
 \vspace{.1in}
 
The speed of the curve is defined by $\nu=\|\alpha'(t)\|$. We can regard $\nu$ as a function of $t$ but throught $t=t(s)$, it can be regarded as a function of $s$ as well. Since $\alpha(t)=\bar \alpha(s)$, we have $\alpha'(t)=\bar \alpha'(s)\frac{ds}{dt}$. Thus we have
\[
\left|\frac{ds}{dt}\right|=\|\alpha'(t)\|=\nu.
\]




\begin{lemma}{}{} Assume that $ds/dt>0$\footnote{We shall always assume this for the rest of the lecture notes. Thus we have $\nu=ds/dt$.}. 
If $\alpha$ is a regular curve in $\R^3$ with $\kappa>0$, then we have
\[
\begin{matrix}
T'=&&\kappa\nu N\\
N'=&-\kappa\nu T&&+\tau\nu B\\
B'=&&-\tau \nu N
\end{matrix}.
\]
\end{lemma}

\begin{proof}
We shall use the Frenet formulas for unit speed curves. Since $T(t)=\bar T(s), N(t)=\bar N(s)$ and $B(t)=\bar B(s)$, we have
$T'(t)=\bar T'(s)\nu, N'(t)=\bar N'(s)\nu$, and $B'(t)=\bar B(s)\nu$. The lemma then follows from Theorem~\ref{thm:thm1}.



\end{proof}

\begin{note}
There is a commonly used notation for the calculus that completely ignores
change of parametrization. For example, the same letter would designate
both  a curve $\alpha$ and its unit-speed parametrization $\bar\alpha$, and similarly with the
Frenet apparatus of these two curves. Differences in derivatives are handled
by writing, say, $dT/dt$ for $T'(t)$ and $dT/ds$ for $\bar T'(s)$. 
\end{note}




\begin{lemma}{}{23}
If $\alpha$ is a regular curve with speed function $\nu$, then the velocity
and acceleration of $\alpha$ are given by
\[
\alpha'=\nu T,\quad\alpha''=\frac{d\nu}{dt}\, T+\kappa\nu^2N.
\]
\end{lemma}

\begin{proof}
The proof is not difficult but the notations are confusing. 
By the context, $\alpha'=\alpha'(t)$ and $\alpha''=\alpha''(t)$. We thus have
\[
\alpha'=T\,\frac{ds}{dt}=T\,\nu,
\]
and
\[
\alpha''=\frac{d\nu}{dt}\,T+\nu\,T'=\frac{d\nu}{dt}\, T+\kappa\nu^2N.
\]

\end{proof}



\begin{remark}
The formula $\alpha'=\nu\, T$ is to be expected since $\alpha'$  and $T$ are each tangent
to the curve and $T$ has a unit length, while $\|\alpha'\| =\nu$. 
\begin{center}
\includegraphics[height=4cm]{images/fig-2.15.png}
\end{center}
The formula for
acceleration is more interesting. By definition, $\alpha''$ is the rate of change of the
and in general both the length and the direction of $\alpha'$ are changing.
The \emph{\red{tangential component}} $\frac{d\nu}{dt}T$ of $\alpha''$ measures the rate of change of
the length of $\alpha'$  (that is, of the speed of $\alpha$). The \emph{\red{normal component}} $\kappa\nu^2\,N$ measures
the rate of change of the direction of $\alpha'$. Newton’s laws of motion show
that these components may be experienced as forces. For example, in a car
that is speeding up or slowing down on a straight road, the only force one
feels is due to $\frac{d\nu}{dt}T$. If one takes an unbanked curve at speed $\nu$, the resulting
sideways force is due to $\kappa\nu^2 N$. Here $\kappa$ measures how sharply the \emph{\red{road}} turns;
the effect of speed is given by $\nu^2$, so 60 miles per hour is four times as unsettling
as 30.
\end{remark}

\begin{note}
Assume that we live in a $1$-dimensional space defined by the above curve $\alpha$. We then can only  measure the tangential component of the acceleration $\alpha''$. As a $1$-dimensional creature, it is not possible for the creature  to understand $\kappa$, the curvature of the curve. It would think it lives in a straight line. 
\end{note}


\begin{theorem}{}{43}
Let $\alpha$ be a regular curve in $\R^3$. Then
\begin{align*}
& T=\alpha'/\|\alpha'\|,
\\
&N=B\times T,&&\kappa=\|\alpha'\times\alpha''\|/\|\alpha'\|^3,
\\
&B=\alpha'\times\alpha''/\|\alpha'\times\alpha''\|,&&\tau=(\alpha'\times\alpha'')\cdot\alpha'''/\|\alpha'\times\alpha''\|^2.
\end{align*}
\end{theorem}

\begin{proof}
The proof is just a matter of applications of the chain rule and the Frenet formulas. But it contains several basic techniques in differential geometry. 

First, we have $T=\alpha'/\|\alpha'\|$. Next, using the above Lemma~\ref{lem:23}, we have
\begin{equation}\label{24}
\alpha'\times\alpha''=\|\alpha'\|\,T\times (\nu'T+\kappa\nu^2 N)=\kappa\nu^3\, B.
\end{equation}
Taking the norm of the above equation, we 
prove the  formula for $\kappa$ is proved. 


 That $N=B\times T$ follows from the definition. From Lemma~\ref{lem:23},  we have
 \[
 \alpha'''=(\nu'T)'+(\kappa\nu^2\,N)'.
 \]
From the above, we know that the $B$ components of $\alpha'''$ is $\kappa\tau\nu^3$. Thus we have
\[
(\alpha'\times\alpha'')\cdot\alpha'''=\kappa^2\tau\nu^6.
\]
Since $\|\alpha'\times\alpha''\|^2=\kappa^2\nu^6$, the formula for $\tau$ is proved. 

\end{proof}

\begin{elink}
As we know, the definition of the curvature is $\kappa=\nu^{-1}\|T'(t)\|$. Can we obtain the above formula by a straightforward computation. Yes, it is complicated, but such kind of  computation contains useful techniques in differential geometry. See \href{https://youtu.be/uVhg4ild8Z8}{here} for the method. 
\end{elink}

\begin{example}
We compute the Frenet apparatus of the 3-curve
\[
\alpha(t)=(3t-t^3, 3t^2, 3t+t^3).
\]

We have
\begin{align*}
& \alpha'(t)=3\,(1-t^2, 2t, 1+t^2),\\
&\alpha''(t)=6\,(-t,1,t),\\
&\alpha'''(t)=6\,(-1,0,1).
\end{align*}

First, we have
\[
\nu=\|\alpha'\|=\sqrt{\alpha'\cdot\alpha'}=3\sqrt{2}(1+t^2). 
\]
We also have 
\[
\alpha'\times\alpha''=18\,
\begin{vmatrix}
U_1&U_2&U_3\\1-t^2& 2t& 1+t^2\\-t&1&t
\end{vmatrix}=18\,(-1+t^2, -2t, 1+t^2).
\]
Hence 
\[
\|\alpha'\times\alpha''\|=18\sqrt 2(1+t^2).
\]
We compute
\[
(\alpha'\times\alpha'')\cdot\alpha'''=6\cdot 18\cdot 2.
\]
It remains only to substitute this data into the formulas in Theorem~\ref{thm:43} with
$N$ being computed by another cross product. The final results are
\begin{align*}
& T=\frac{(1-t^2, 2t, 1+t^2)}{\sqrt 2(1+t^2)},\\
& N=\frac{(-2t, 1-t^2,0)}{1+t^2},\\
& B=\frac{(-1+t^2, -2t, 1+t^2)}{\sqrt 2(1+t^2)},\\
&\kappa=\tau=\frac{1}{3(1+t^2)^2}.
\end{align*}

\end{example}

\vspace{.2in}


For the rest of the section, we shall 
do some applications of the above formulas.

\begin{definition}{}{}
The \red{spherical image } of a unit-speed curve $\beta(s)$ is the curve $\sigma(s)=T(s)=\beta'(s)$. 
\end{definition}

Let $\sigma$ be the spherical image of $\beta$. Then $\sigma'=\beta''=\kappa N$, where $\kappa$ is the curvature of $\beta$. That $\kappa>0$ ensures that $\sigma$ is a regular curve. In order to compute the curvature $\kappa_\sigma$ of $\sigma$, we compute\footnote{If $\kappa>0$, then $\sigma''\neq 0$. Thus the Frenet Apparatus always exists for $\sigma$.}
\begin{align*}
& \sigma'=\beta''=\kappa\, N,\\
&\sigma''=\beta'''=\kappa'\, N+\kappa\, N'=\kappa'\, N+\kappa(-\kappa\, T+\tau B).
\end{align*}
Thus
\[
\sigma'\times\sigma''=\kappa^2(\kappa B+\tau T). 
\]
Using the formula in Theorem~\ref{thm:43}, we have
\[
\kappa_\sigma=\frac{\|\sigma'\times\sigma''\|}{\nu^3}=(1+(\tau/\kappa)^2)^{1/2}\geq 1.
\]


\begin{definition}{}{}
A regular curve $\alpha$ in $\R^3$ is a \red{cylindrical helix } provided the
unit tangent vector field $T$ of $\alpha$ has constant angle $\theta$ with some fixed unit vector $\uu$; that is, $T(t)\cdot \uu=\cos\theta$ for all $t$.
\end{definition}

\begin{theorem}{}{}
A regular curve $\alpha$ with $\kappa>0$ is a cylindrical helix if and
only if the ratio $\tau/\kappa$ is constant.
\end{theorem}
\begin{proof}
First assume that a unit speed curve is a cylindrical helix curve. Then there is a constant vector $\uu$ such that 
$T\cdot \uu=c$, a constant. Taking derivative on both sides, we get $\kappa\, N\cdot \uu=0$. Since $\kappa>0$, we get $N\cdot \uu=0$.
Therefore, if we write $\uu$ as a linear combination of $T,N,B$, there would be no $N$ component.

By the assumption, we would get
\[
\uu=\cos\theta\, T+\sin\theta\, B.
\]
Taking derivative of the above equation again, we get
\[
0=\cos\theta\, \kappa\, N-\sin\theta\,\tau N.
\]
We thus have $\kappa\cos\theta-\tau\sin\theta=0$. Hence $\tau/\kappa=\cot\theta$ is a constant. \\

Conversely, if $\tau/\kappa$ is a constant, we write $\tau/\kappa=\cot\theta$. Let 
\[
\uu=\cos\theta\, T+\sin\theta\, B.
\]
Then
\[
\uu'=\cos\theta\, \kappa\, N-\sin\theta\, \tau\, N=0.
\]
Therefore $\uu$ is a constant vector field. Obviously, $\|\uu\|=1$. Then $T\cdot\uu=\cos\theta$ which means that the curve is a 
 cylindrical helix.




\end{proof}


Can we prove the above result by solving the Frenet differential equations? Yes, in the following, we give another proof of the fact that if $\tau/\kappa=c$ be a constant, then there must be a unit vector $\uu$  such that $T\cdot\uu$ is a constant. 

We use ~\eqref{7} to obtain
\[
(T, N, B)'=(T, N, B)\begin{bmatrix}0&-\kappa&0\\\kappa&0&-\tau\\0&\tau&0\end{bmatrix}
=\kappa\,(T, N, B)\begin{bmatrix}0&-1&0\\1&0&-c\\0&c&0\end{bmatrix}.
\]
Thus we have (See the remark below)
\[
(T, N, B)=(T(0), N(0), B(0))e^{A\int_0^t\kappa(u)\, du},
\]
where 
\[
A=\begin{bmatrix}0&-1&0\\1&0&-c\\0&c&0\end{bmatrix}.
\]
Since $A$ is a skew-symmetric $3\times 3$ matrix, it is singular\footnote{One can prove this fact by a straightforward computation. Alternatively, we observe that for skew-symmetric matrices, all eigenvalues must be purely imaginary unless they are zero. Since all purely imaginary eigenvalues must be in pairs (conjugate eigenvalues), there must be at least one zero eigenvalue for odd dimensional matrices.}. Let $\vv$ be a unit eigenvector  of 
$A$ with respect to the zero eigenvalue. Let $\uu$ be the vector such that 
\[
\uu\cdot (T(0), N(0), B(0))=\vv. 
\]
Then
\[
T\cdot \uu=\uu\cdot T=\uu\cdot (T(0), N(0), B(0))e^{A\int_0^t\kappa(u)\, du}\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}.
\]
Note that $\vv, A=0$, we thus have 
\[
\vv\,e^{A\int_0^t\kappa(u)\, du}=\vv\sum_{k=0}^\infty\frac{1}{k!}A^k\left(\int_0^t\kappa(u)\, du\right)^k=\vv.
\]
Thus 
\[
T\cdot \uu=\vv\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}
\]
is a constant. 

\vspace{.1in}

\begin{remark}
Let $A(t)$ be a matrix-valued function. We consider the system of differential equations
\[
A'(t)=A(t)B(t),\qquad A(0)=A_0,
\]
where $B(t)$ is a matrix-valued function. Then in general, we don't have a formula for the (unique) solution.  If for any $t, t'$, we have
\[
B(t)B(t')=B(t') B(t),
\]
then we have the solution
\[
A(t)=A_0\, e^{\int_0^t B(u)du}.
\] 
For further explanation, see \href{https://youtu.be/2tf4U9d-Pdk}{here}.
\end{remark}

\section{Covariant Derivatives}
In this section, we shall define probably one of the most important concepts in differential geometry, or may be even one of the most important concepts in mathematics: \emph{\red{covariant derivatives}} on vector fields. 



\begin{definition}{}{}
Let $W$ be a vector field on $\R^3$, and let $\vv$ be a tangent vector
to $\R^3$ at the point $\p$. Then the \red{covariant derivative } of $W$ with respect to
$\vv$ is the tangent vector
\[
\nabla_\vv W=W(\p+t\vv)'(0)
\]
at the point $\p$.
\end{definition}

On surface, the covariant derivative is just another version of directional derivative, especially we have already defined such kind of derivative along a curve before. But we shall systematically use  this kind of derivative under the frame of differential operators. 

\begin{example} Let $W=x^2 U_1+yz U_3$, and let 
\[
\vv=(-1,0,2),
\]
at $\p=(2,1,0)$. Then 
\[
\p+t\vv=(2-t, 1, 2t).
\]
So 
\[
W(\p+t\vv)=(2-t)^2 U_1+2t U_3. 
\]
Hence 
\[
\nabla_\vv W=W(\p+t\vv)'(0)=-4\, U_1(\p)+2\, U_3(\p).
\]

\end{example}

\begin{lemma}{}{}
If $W=w_i U_i$ is a vector field on $\R^3$, and $\vv$ is a tangent vector at $\p$, then 
\[
\nabla_\vv W=\vv[w_i] U_i(\p).
\]
\end{lemma}
\begin{proof}
We have\footnote{This  equation is from the book which is, of course, correct. But in all previous formulas,  we use $U_i$ instead of $U_i(\p+t\vv)$.}
\[
W(\p+t\vv)=w_i(\p+t\vv) U_i(\p+t\vv).
\]
Differentiation of the above at $t=0$, we obtain
\[
\nabla_\vv W=W(\p+t\vv)'(0)=\vv[w_i] U_i(\p).
\]
\end{proof}

In short, to apply $\nabla_\vv$  to a vector field, apply $\vv$ to its Euclidean coordinates.

\begin{theorem}{}{}
Let $\vv$ and $\ww$  be tangent vectors $\R^3$ at $\p$, and let $Y$ and $Z$ be vector fields on $\R^3$. Then for numbers $a,b$ and function $f$
\begin{enumerate}[label={\rm (\arabic*)}]
\item $\nabla_{a\vv+b\ww} Y=a\nabla_{\vv} Y+b\nabla_{\ww} Y$,
\item $\nabla_\vv(aY+bZ)=a\nabla_\vv Y+b\nabla_\vv Z$,
\item $\nabla_\vv(fY)=\vv[f]Y(\p)+f(\p)\nabla_\vv Y$,
\item $\vv[Y\cdot Z]=\nabla_\vv Y\cdot Z(\p)+Y(\p)\cdot\nabla_\vv Z$.
\end{enumerate}
\end{theorem}

\begin{proof}
The proof  is straightforward. For example, to prove (3), let $Y=y_iU_i$ and then
$fY=(fy_i)U_i$.
Thus
\[
\nabla_\vv(fY)=\vv[fy_i] U_i=\vv[f] y_iU_i+f\vv[y_i] U_i=\vv[f]Y(\p)+f(\p)\nabla_\vv Y.
\]


\end{proof}

If we regard the set of vector fields as a module over the algebra of smooth functions, then the operator $\nabla_\vv$ is linear with respect to the addition and scalar multiplication. It is a \emph{\red{derivative }} on the multiplication of smooth functions to vector fields. 



\begin{corollary}{}{}
Let $V,W,Y$ and $Z$ be vector fields on $\R^3$; $f,g$ be smooth functions. Then
\begin{enumerate}[label={\rm (\arabic*)}]
\item $\nabla_{fV+gW} Y=f\nabla_{V} Y+g\nabla_{W} Y$,
\item $\nabla_V(aY+bZ)=a\nabla_V Y+b\nabla_V Z$,
\item $\nabla_V(fY)=V[f]Y+f\nabla_V Y$,
\item $V[Y\cdot Z]=\nabla_V Y\cdot Z+Y\cdot\nabla_VZ$.
\end{enumerate}
\end{corollary}

\section{Frame Fields}
When the Frenet formulas were discovered (by Frenet in 1847, and independently
by Serret in 1851), the theory of surfaces in $\R^3$ was already a richly
developed branch of geometry. The success of the Frenet approach to curves
led Darboux (around 1880) to adapt this ``method of moving frames'' to the
study of surfaces. Then it was Cartan who brought
the method to full generality. His essential idea was very simple: To each point
of the object under study (a curve, a surface, Euclidean space itself, . . .)
assign a frame; then using orthonormal expansion express the rate of change
of the frame in terms of the frame itself. This, of course, is just what the
Frenet formulas do in the case of a curve.



In the next three sections we shall carry out this scheme for the Euclidean
space $\R^3$. We shall see that geometry of curves and surfaces in $\R^3$ is not merely
an analogue, but actually a corollary, of these basic results. 

\begin{definition}{}{}
Vector fields $E_1, E_2, E_3$ on $\R^3$ constitute a \red{frame field }  on $\R^3$ provided 
\[
E_i\cdot E_j=\delta_{ij}.
\]
\end{definition}


Apparently, $U_1, U_2, U_3$ is a frame field. We call such a frame field \emph{\red{Euclidean frame field}}. In the following, we shall introduce two more important frame fields. 

\begin{example} (\emph{\red{The cylindrical frame field}})  Let $r,\theta, z$ be the usual cylindrical coordinate functions on
$\R^3$. We shall pick a unit vector field
in the direction in which each coordinate increases (when the other two are
held constant). For $r$, this is evidently
\[
E_1=\cos\theta \, U_1+\sin\theta\, U_2.
\]
Then
\[
E_2=-\sin\theta\, U_1+\cos\theta\, U_2
\]
points in the direction of increasing $\theta$ as in Fig. 2.19.
Finally, the direction
of increase of $z$ is, of course, straight up, so
\[
E_3=U_3.
\]
\begin{center}
\includegraphics[height=5cm]{images/fig-2.19.png}
\end{center}
\end{example}

\begin{remark}
I don't think in the above example, deducing the cylindrical frame field from the cylindrical coordinates by the picture,  is mathematically rigid, or \emph{even correct}. In fact, as long as the above $\{E_1,E_2,E_3\}$ is an orthonormal basis at each point, it defines a frame field. We can \emph{\red{call}} it the cylindrical frame field. There is no proof needed here.

A better way to show the relationship between the cylindrical coordinates and the cylindrical frame field is to use 
the chain rule. We have $x=r\cos\theta, y=r\sin\theta, z=z$. Then by the chain rule, for any function $f$, we have 
\[
\frac{\pa f}{\pa r}=\frac{\pa x}{\pa r}\cdot\frac{\pa f}{\pa x}+\frac{\pa y}{\pa r}\cdot\frac{\pa f}{\pa y}+\frac{\pa z}{\pa r}\cdot\frac{\pa f}{\pa z}. 
\]
In terms of vector notation, this implies that 
\[
\frac{\pa }{\pa r}=\frac{\pa x}{\pa r}\, U_1+\frac{\pa y}{\pa r}\, U_2+\frac{\pa z}{\pa r}\, U_3=\cos\theta\, U_1+\sin\theta\, U_2. 
\]

Using the same method, we have
\[
\frac{\pa}{\pa z}=U_3.
\]

However, 
\[
\frac{\pa }{\pa \theta}=\frac{\pa x}{\pa \theta}\, U_1+\frac{\pa y}{\pa \theta}\, U_2+\frac{\pa z}{\pa \theta}\, U_3=-r\sin\theta \, U_1+r\cos\theta\, U_2
\]
which is not $E_2$ but $rE_2$. 

In general, if $\{x_1,x_2,x_3\}$ are another set of coordinate functions, then the vector fields 
\[
\left\{\frac{\pa}{\pa x_1},\frac{\pa}{\pa x_2},\frac{\pa}{\pa x_3}\right\}
\]
do not define a frame field (they may not be orthonormal). 

\end{remark}



\begin{example} (\emph{\red{The spherical frame field}}) Now we define the spherical frame field $\{F_1, F_2, F_3\}$. Let $\rho, \theta,\varphi$ be the spherical coordinates. We then have 
\begin{align*}
& x=\rho\cos\varphi\cos\theta,\\
& y=\rho\cos\varphi\sin\theta,\\
&z=\rho\sin\varphi.
\end{align*}
The spherical frame field is defined by
\begin{align*}
& F_1=\cos\varphi \,(\cos\theta \,U_1+\sin\theta \,U_2)+\sin\varphi \,U_3\\
& F_2=-\sin\theta\, U_1+\cos\theta\, U_2,
\\
& F_3=-\sin\varphi\,(\cos\theta\, U_1+\sin\theta\, U_2)+\cos\varphi\, U_3.
\end{align*}
\begin{center}
\includegraphics[height=5cm]{images/fig-2.20.png}
\end{center}
\end{example}

By a straightforward computation, we have 
\begin{align*}
& \frac{\pa}{\pa\rho}=E_1,\\
&  \frac{\pa}{\pa\theta} =-\rho\cos\varphi\sin\theta\, U_1+\rho\cos\varphi\cos\theta\, U_2=\rho\cos\varphi\, F_2,\\
& \frac{\pa}{\pa\varphi} = -\rho\sin\varphi\cos\theta\,U_1-\rho\sin\varphi\sin\theta\, U_2 +\rho\cos\varphi\, U_3=\rho\, F_3.
\end{align*}


\begin{lemma}{}{}
Let $\{E_1,E_2, E_3\}$ be a frame field on $\R^3$.

{\rm (1)}. If $V$ is a vector field on $\R^3$, then $V=f_iE_i$, where the functions $f_i=V\cdot E_i$ are called \red{coordinate functions } of $V$ with
respect to $\{E_1,E_2, E_3\}$. 

{\rm (2)}. If $V=f_iE_i$ and $W=g_iE_i$, then $V\cdot W=f_ig_i$. In particular, $\|V\|=(\sum f_i^2)^{1/2}$.
\end{lemma}

\section{Connection forms}
Let $\{E_1,E_2, E_3\}$ be a frame field, and let $\vv$ be a vector at $\p\in \R^3$. Let $\p\in\R^3$. Then we are able to write 
\begin{align*}
&\nabla_{\vv} E_1=c_{11} E_1(\p)+c_{12} E_2(\p)+c_{13} E_3(\p),\\
&\nabla_{\vv} E_2=c_{21} E_1(\p)+c_{22} E_2(\p)+c_{23} E_3(\p),\\
&\nabla_{\vv} E_3=c_{31} E_1(\p)+c_{32} E_2(\p)+c_{33} E_3(\p).
\end{align*}
Using the Einstein Convention, we have
\[
\nabla_\vv E_i=c_{ij} E_j(\p).
\]

Since $\{E_1,E_2, E_3\}$ is a frame, we then have\footnote{We know that $1\leq i,j\leq 3$. But this is implied by the Einstein convention so can be omitted.}
\[
c_{ij}=\nabla_\vv E_i\cdot E_j(\p).
\]
We then can define a one form $\omega_{ij}$ such that
\[
\omega_{ij}= c_{ij}=\nabla_\vv E_i\cdot E_j(\p).
\]





\begin{lemma}{}{}
Using the above notations, then $\omega_{ij}$ are $1$-forms satisfying
\[
\omega_{ij}=-\omega_{ji}.
\]
These $1$-forms are called \red{connection forms } of the frame field $\{E_1,E_2,E_3\}$. 
\end{lemma}

\begin{proof}
In order to prove $\omega_{ij}$ is a $1$-form, we just need to prove that 
\[
\omega_{ij} (a\vv+b\ww)=a\,\omega_{ij}(\vv)+b\,\omega_{ij}(\ww),
\]
where $\vv,\ww$ are vectors and $a,b$ are real numbers. 
But this follows from 
\[
\nabla_{a\vv+b\ww} E_i=a\,\nabla_{\vv}E_i+b\,\nabla_{\ww} E_i.
\]

To prove $\omega_{ij}=-\omega_{ji}$, we observe that since $E_i\cdot E_j=\delta_{ij}$. Then
\[
0=\vv[E_i\cdot E_j]=\nabla_\vv E_i\cdot E_j+E_i\cdot \nabla_\vv E_j=\omega_{ij}(\vv)+\omega_{ji}(\vv).
\]




\end{proof}



\begin{theorem}{}{}
Let $\omega_{ij}$ be the connection $1$-froms of a frame field $\{E_1,E_2,E_3\}$ on $\R^3$. Then for any vector field $V$ on $\R^3$, we have
\[
\nabla_V E_i=\omega_{ij}(V) E_j.
\]
\end{theorem}


By $\omega_{ij}=-\omega_{ji}$, we have $\omega_{ii}=0$. We write
\[
\omega=\begin{bmatrix} \omega_{11} & \omega_{12}& \omega_{13}\\
\omega_{21} & \omega_{22}& \omega_{23}\\
\omega_{31} & \omega_{32}& \omega_{33}
\end{bmatrix}
=\begin{bmatrix} 0& \omega_{12}& \omega_{13}\\
-\omega_{12} &0 & \omega_{23}\\
-\omega_{13} & -\omega_{23}&0
\end{bmatrix}.
\]

If we write 
\[
E=(E_1, E_2, E_3),
\]
then in matrix notations, we have
\[
\nabla_\vv E=E\cdot\omega^T.
\]
This can be used to compare with the Frenet formula
\[
(T, N, B)'=(T, N, B)\begin{bmatrix}0&-\kappa&0\\\kappa&0&-\tau\\0&\tau&0\end{bmatrix}.
\]

\vspace{.2in}



\newpage
%\setcounter{chapter}{+3}



