As usual we use indexes modulo $n$, in particular $p_{n+1}=\nobreak p_{1}$.
Assume $p_{i}=\gamma(t_{i})$. Set

$\displaystyle\text{\sc{w}}_{i}$ $\displaystyle=p_{i+1}-p_{i},$
$\displaystyle\text{\sc{v}}_{i}$ $\displaystyle=\gamma^{\prime}(t_{i}),$
$\displaystyle\alpha_{i}$
$\displaystyle=\measuredangle(\text{\sc{v}}_{i},\text{\sc{w}}_{i}),$
$\displaystyle\beta_{i}$
$\displaystyle=\measuredangle(\text{\sc{w}}_{i-1},\text{\sc{v}}_{i}),$

where $\alpha_{i},\beta_{i}\in(-\pi,\pi]$ are signed angles — $\alpha_{i}$ is
positive if $\text{\sc{w}}_{i}$ points to the left from $\text{\sc{v}}_{i}$.

By ➋ ‣ 5C, the value

$\Psi({\gamma_{i}})-\alpha_{i}-\beta_{i+1}$ ➍

is a multiple of $2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. Since
$\Phi({\gamma_{i}})<\pi$, by the chord lemma (3F), we also have that
$|\alpha_{i}|+\nobreak|\beta_{i}|<\pi$. By ➌ ‣ 5C, we have that
$|\Psi({\gamma_{i}})|\leqslant\Phi({\gamma_{i}})$; therefore the value in ➍ ‣
5C vanishes. In other word,for each $i$ we have

$\Psi({\gamma_{i}})=\alpha_{i}+\beta_{i+1}.$

Note that

$\delta_{i}=\pi-\alpha_{i}-\beta_{i}$ ➎

is the internal angle of $\beta$ at $p_{i}$; $\delta_{i}\in(0,2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi)$ for each $i$. Recall that the sum of the
internal angles of an $n$-gon is $(n-2){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi$ (see 0B); that is,

$\delta_{1}+\dots+\delta_{n}=(n-2){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

Therefore

$\displaystyle\Psi({\gamma})$
$\displaystyle=\Psi({\gamma})_{1}+\dots+\Psi({\gamma})_{n}=$ ➏
$\displaystyle=(\alpha_{1}+\beta_{2})+\dots+(\alpha_{n}+\beta_{1})=$
$\displaystyle=(\beta_{1}+\alpha_{1})+\dots+(\beta_{n}+\alpha_{n})=$
$\displaystyle=(\pi-\delta_{1})+\dots+(\pi-\delta_{n})=$
$\displaystyle=n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi-(n-2){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi=$ $\displaystyle=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

The case of piecewise smooth and regular curves is done the same way; we need
to subdivide the arcs in the cyclic concatenation further to meet the
requirement above and instead of equation ➎ ‣ 5C we have

$\delta_{i}=\pi-\alpha_{i}-\beta_{i}-\theta_{i},$

where $\theta_{i}$ is the signed external angle of $\gamma$ at $p_{i}$; it
vanishes if the curve $\gamma$ is smooth at $p_{i}$. Therefore instead of
equation ➏ ‣ 5C, we have

$\displaystyle\Psi({\gamma})$
$\displaystyle=\Psi({\gamma_{1}})+\dots+\Psi({\gamma_{n}})+\theta_{1}+\dots+\theta_{n}=$
$\displaystyle=(\alpha_{1}+\beta_{2})+\dots+(\alpha_{n}+\beta_{1})=$
$\displaystyle=(\beta_{1}+\alpha_{1}+\theta_{1})+\dots+(\beta_{n}+\alpha_{n}+\theta_{n})=$
$\displaystyle=(\pi-\delta_{1})+\dots+(\pi-\delta_{n})=$
$\displaystyle=n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi-(n-2){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi=$ $\displaystyle=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

∎

5.6. Exercise. Draw a smooth regular closed plane curve $\gamma$ such that

  * (a)

$\Psi({\gamma})=0$;

  * (b)

$\Psi({\gamma})=\Phi({\gamma})=10{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$;

  * (c)

$\Psi({\gamma})=2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$ and
$\Phi({\gamma})=4{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$.

5.7. Exercise. Let $\gamma\colon[a,b]\to\mathbb{R}$ be a smooth regular plane
curve with Frenet frame $\text{\sc{t}},\text{\sc{n}}$. Given a real parameter
$\ell$, consider the curve $\gamma_{\ell}(t)=\nobreak\gamma(t)+\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}(t)$; it is called a _parallel
curve of $\gamma$ at signed distance $\ell$_.

  * (a)

Show that $\gamma_{\ell}$ is a regular curve if $\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}k(t)\neq 1$ for all $t$, where $k(t)$ denotes
the signed curvature of $\gamma$.

  * (b)

Set $L(\ell)=\mathop{\rm length}\nolimits\gamma_{\ell}$. Show that

$L(\ell)=L(0)-\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\Psi({\gamma})$ ➐

for all $\ell$ sufficiently close to $0$.

  * (c)

Describe an example showing that formula ➐ ‣ (b) does not hold for all $\ell$.

###  D Osculating circline

5.8. Proposition. Given a point $p$, a unit vector t and a real number $k$,
there is a unique smooth unit-speed curve
$\sigma\colon\mathbb{R}\to\mathbb{R}^{2}$ that starts at $p$ in the direction
of t and has constant signed curvature $k$.

Moreover, if $k=0$, then it is a line $\sigma(s)=p+s{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{t}}$; if $k\neq 0$, then $\sigma$
runs around a circle of radius $\tfrac{1}{|k|}$ with center at
$p+\tfrac{1}{k}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}$, where
$\text{\sc{t}},\text{\sc{n}}$ is an oriented orthonormal frame.

Further we will use the term _circline_ for _a circle or a line_ ; these are
the only plane curves with constant signed curvature.

Proof. The proof is done by a calculation based on 5B and 5B.

Suppose $s_{0}=0$, choose a coordinate system such that $p$ is its origin and
t points in the direction of the $x$-axis. Therefore n points in the direction
of the $y$-axis. Then

$\displaystyle\theta(s)$ $\displaystyle=\int_{0}^{s}k{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dt=$ $\displaystyle=k{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s.$

Therefore

$\sigma^{\prime}(s)=(\cos[k{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}s],\sin[k{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}s]).$

It remains to integrate the last identity. If $k=0$, we get

$\sigma(s)=(s,0)$

which describes the line $\sigma(s)=p+s{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{t}}$.

If $k\neq 0$, we get

$\sigma(s)=(\tfrac{1}{k}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin[k{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s],\tfrac{1}{k}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(1-\cos[k{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}s])).$

which is the circle of radius $r=\tfrac{1}{|k|}$ centered at
$(0,\tfrac{1}{k})=p+\tfrac{1}{k}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}$. ∎

5.9. Definition. Let $\gamma$ be a smooth unit-speed plane curve; denote by
$k(s)$ the signed curvature of $\gamma$ at $s$.

The unit-speed curve $\sigma$ of constant signed curvature $k(s)$ that starts
at $\gamma(s)$ in the direction $\gamma^{\prime}(s)$ is called the _osculating
circline_ of $\gamma$ at $s$.

The center and radius of the osculating circle at a given point are called the
_center of curvature_ and _radius of curvature_ of the curve at that point.

The osculating circle $\sigma_{s}$ can be also defined as the unique circline
that has _second order of contact_ with $\gamma$ at $s$; that is,
$\rho(\ell)=\nobreak o(\ell^{2})$, where $\rho(\ell)$ denotes the distance
from $\gamma(s+\ell)$ to $\sigma_{s}$.

The following exercise is recommended to the reader familiar with the notion
of _inversion_.

5.10. Advanced exercise. Suppose $\gamma$ is a smooth regular plane curve that
does not pass thru the origin. Let $\hat{\gamma}$ be the inversion of $\gamma$
with respect to the unit circle centered at the origin. Show that the
osculating circline of $\hat{\gamma}$ at $s$ is the inversion of the
osculating circline of $\gamma$ at $s$.

###  E Spiral lemma

The following lemma was proved by Peter Tait [76] and later rediscovered by
Adolf Kneser [46].

5.11. Lemma. Assume that $\gamma$ is a smooth regular plane curve with
strictly decreasing positive signed curvature. Then the osculating circles of
$\gamma$ are nested; that is, if $\sigma_{s}$ denoted the osculating circle of
$\gamma$ at $s$, then $\sigma_{s_{0}}$ lies in the open disc bounded by
$\sigma_{s_{1}}$ for any $s_{0}<s_{1}$.

It turns out that the osculating circles of the curve $\gamma$ give a peculiar
foliation of an annulus by circles; it has the following property: if a smooth
function is constant on each osculating circle, then it must be constant in
the annulus [[, see]Lecture 10]fuchs-tabachnikov. Also note that the curve
$\gamma$ is tangent to a circle of the foliation at each of its points.
However, it does not run along any of those circles.

Proof. Let $\text{\sc{t}}(s),\text{\sc{n}}(s)$ be the Frenet frame,
$\omega(s)$, $r(s)$ the center and radius of curvature of $\gamma$. By 5D,

$\omega(s)=\gamma(s)+r(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}(s).$

Since $k>0$, we have that $r(s){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}k(s)=1$.
Therefore applying Frenet formula ➋ ‣ 5A, we get that

$\displaystyle\omega^{\prime}(s)$
$\displaystyle=\gamma^{\prime}(s)+r^{\prime}(s){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}(s)+r(s){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}^{\prime}(s)=$
$\displaystyle=\text{\sc{t}}(s)+r^{\prime}(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}(s)-r(s){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}k(s){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{t}}(s)=$
$\displaystyle=r^{\prime}(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}(s).$

Since $k(s)$ is decreasing, $r(s)$ is increasing; therefore
$r^{\prime}\geqslant 0$. It follows that $|\omega^{\prime}(s)|=r^{\prime}(s)$
and $\omega^{\prime}(s)$ points in the direction of $\text{\sc{n}}(s)$.

Since $\text{\sc{n}}^{\prime}(s)=-k(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{t}}(s)$, the direction of $\omega^{\prime}(s)$ cannot have
constant direction on a nontrivial interval; that is, the curve
$s\mapsto\omega(s)$ contains no line segments. Therefore

$\displaystyle|\omega(s_{1})-\omega(s_{0})|$ $\displaystyle<\mathop{\rm
length}\nolimits(\omega|_{[s_{0},s_{1}]})=$
$\displaystyle=\int_{s_{0}}^{s_{1}}|\omega^{\prime}(s)|{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}ds=$
$\displaystyle=\int_{s_{0}}^{s_{1}}r^{\prime}(s){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}ds=$ $\displaystyle=r(s_{1})-r(s_{0}).$

In other words, the distance between the centers of $\sigma_{s_{1}}$ and
$\sigma_{s_{0}}$ is strictly less than the difference between their radii.
Therefore the osculating circle at $s_{0}$ lies inside the osculating circle
at $s_{1}$ without touching it. ∎

The curve $s\mapsto\omega(s)$ is called the _evolute_ of $\gamma$; it traces
the centers of curvature of the curve. The evolute of $\gamma$ can be written
as

$\omega(t)=\gamma(t)+\tfrac{1}{k(t)}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}(t)$

and in the proof we showed that $(\tfrac{1}{k})^{\prime}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}$ is its velocity vector.

5.12. Exercise. Show that the stretched astroid

$\omega(t)=(\tfrac{a^{2}-b^{2}}{a}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\cos^{3}t,\tfrac{b^{2}-a^{2}}{b}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\sin^{3}t)$

is an evolute of the ellipse defined by

$\gamma(t)=(a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\cos t,b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\sin t).$

The following theorem states formally that _if you drive on the plane and turn
the steering wheel to the left all the time, then you will not be able to come
back to the place you started._

5.13. Theorem. Assume $\gamma$ is a smooth regular plane curve with positive
and strictly monotonic signed curvature. Then $\gamma$ is simple.

The same statement holds true without assuming positivity of curvature; the
proof requires only minor modifications.

Proof of 5E. Note that $\gamma(s)$ lies on the osculating circle $\sigma_{s}$
of $\gamma$ at $s$. If $s_{1}\neq s_{0}$, then by lemma 5E, $\sigma_{s_{0}}$
does not intersect $\sigma_{s_{1}}$. Therefore
$\gamma(s_{1})\neq\gamma(s_{0})$, hence the result.∎

5.14. Exercise. Show that a 3-dimensional analog of the theorem does not hold.
That is, there are self-intersecting smooth regular space curves with strictly
monotonic curvature.

5.15. Exercise. Assume that $\gamma$ is a smooth regular plane curve with
positive strictly monotonic signed curvature.

  * (a)

Show that no line can be tangent to $\gamma$ at two distinct points.

  * (b)

Show that no circle can be tangent to $\gamma$ at three distinct points.

Note that part 5E does not hold if we allow the curvature to be negative; an
example is shown on the diagram.

##  Chapter 6 Supporting curves

###  A Cooriented tangent curves

Suppose $\gamma_{1}$ and $\gamma_{2}$ are smooth regular plane curves. Recall
that the curves $\gamma_{1}$ and $\gamma_{2}$ are tangent at the time
parameters $t_{1}$ and $t_{2}$ if $\gamma_{1}(t_{1})=\gamma_{2}(t_{2})$ and
they share the tangent line at these time parameters. In this case the point
$p=\gamma_{1}(t_{1})=\gamma_{2}(t_{2})$ is called a _point of tangency_ of the
curves. If both curves are simple, then without ambiguity we may say that
$\gamma_{1}$ and $\gamma_{2}$ are tangent at the point $p$.

Note that if $\gamma_{1}$ and $\gamma_{2}$ are tangent at the time parameters
$t_{1}$ and $t_{2}$, then the velocity vectors $\gamma_{1}^{\prime}(t_{1})$
and $\gamma_{2}^{\prime}(t_{2})$ are parallel.

If $\gamma_{1}^{\prime}(t_{1})$ and $\gamma_{2}^{\prime}(t_{2})$ have the same
direction, we say that the curves are _cooriented_ , if their directions are
opposite, the curves are called _counteroriented_.

Note that reverting the parametrization of one of the curves, cooriented
curves become counteroriented and vice versa; so we can always assume the
curves are cooriented at any given point of tangency.

###  B Supporting curves

Let $\gamma_{1}$ and $\gamma_{2}$ be two smooth regular plane curves that
share a point

$p=\gamma_{1}(t_{1})=\gamma_{2}(t_{2})$

which is not an endpoint of any of the curves. Suppose that there is
$\varepsilon>0$ such that the arc
$\gamma_{2}|_{[t_{2}-\varepsilon,t_{2}+\varepsilon]}$ lies in a closed plane
region $R$ with the arc $\gamma_{1}|_{[t_{1}-\varepsilon,t_{1}+\varepsilon]}$
in its boundary, then we say that $\gamma_{1}$ _locally supports_ $\gamma_{2}$
at the time parameters $t_{1}$ and $t_{2}$. If both curves are simple, then
without ambiguity we also could say that $\gamma_{1}$ _locally supports_
$\gamma_{2}$ at the point $p$.

If the curves on the diagram are oriented according to the arrows, then
$\gamma_{1}$ supports $\gamma_{2}$ from the right at $p$ (as well as
$\gamma_{2}$ supports $\gamma_{1}$ from the left at $p$).

Suppose $\gamma_{1}$ is a simple proper curve, so it divides the plane into
two closed regions that lie on the left and right from $\gamma_{1}$. Then we
say that $\gamma_{1}$ _globally supports_ $\gamma_{2}$ at $t_{2}$ if
$\gamma_{2}$ runs in one of these closed regions and $\gamma_{2}(t_{2})$ lies
on $\gamma$.

Further, suppose $\gamma_{2}$ is closed, so it divides the plane into two
regions. We say that a point lies _inside_ (respectively, _outside_) of
$\gamma_{2}$ if it lies in the bounded region (respectively, unbounded)
region. In this case we say that $\gamma_{1}$ supports $\gamma_{2}$ _from
inside_ (_from outside_) if $\gamma_{1}$ supports $\gamma_{2}$ and lies in the
region inside it (respectively outside it).

Note that if $\gamma_{1}$ and $\gamma_{2}$ share a point
$p=\gamma_{1}(t_{1})=\gamma_{2}(t_{2})$ and are not tangent at $t_{1}$ and
$t_{2}$, then at time $t_{2}$ the curve $\gamma_{2}$ crosses $\gamma_{1}$
moving from one of its sides to the other. It follows that $\gamma_{1}$ cannot
locally support $\gamma_{2}$ at the time parameters $t_{1}$ and $t_{2}$.
Whence we get the following:

6.1. Definition-Observation. Let $\gamma_{1}$ and $\gamma_{2}$ be two smooth
regular plane curves. Suppose $\gamma_{1}$ locally supports $\gamma_{2}$ at
time parameters $t_{1}$ and $t_{2}$. Then $\gamma_{1}$ is tangent to
$\gamma_{2}$ at $t_{1}$ and $t_{2}$.

In such a case, if the curves are cooriented and the region $R$ in the
definition of supporting curves lies on the right (left) from the arc of
$\gamma_{1}$, then we say that $\gamma_{1}$ supports $\gamma_{2}$ from the
left (respectively right).

We say that a smooth regular plane curve $\gamma$ has a _vertex_ at $s$ if the
signed curvature function is critical at $s$; that is, if
$k^{\prime}(s)_{\gamma}=0$. If in addition to that, $\gamma$ is simple, we
could say that the point $p=\gamma(s)$ is a vertex of $\gamma$.

6.2. Exercise. Assume that the osculating circle $\sigma_{s}$ of a smooth
regular simple plane curve $\gamma$ locally supports $\gamma$ at
$p=\gamma(s)$. Show that $p$ is a vertex of $\gamma$.

###  C Supporting test

The following proposition resembles the second derivative test.

6.3. Proposition. Let $\gamma_{1}$ and $\gamma_{2}$ be two smooth regular
plane curves.

Suppose $\gamma_{1}$ locally supports $\gamma_{2}$ from the left (right) at
the time parameters $t_{1}$ and $t_{2}$. Then

$k_{1}(t_{1})\leqslant k_{2}(t_{2})\quad(\text{respectively}\quad
k_{1}(t_{1})\geqslant k_{2}(t_{2})).$

where $k_{1}$ and $k_{2}$ denote the signed curvature of $\gamma_{1}$ and
$\gamma_{2}$ respectively.

A partial converse also holds. Namely, if $\gamma_{1}$ and $\gamma_{2}$
tangent and cooriented at the time parameters $t_{1}$ and $t_{2}$ then
$\gamma_{1}$ locally supports $\gamma_{2}$ from the left (right) at the time
parameters $t_{1}$ and $t_{2}$ if

$k_{1}(t_{1})<k_{2}(t_{2})\quad(\text{respectively}\quad
k_{1}(t_{1})>k_{2}(t_{2})).$

Proof. Without loss of generality, we can assume that $t_{1}=t_{2}=0$, the
shared point $\gamma_{1}(0)=\gamma_{2}(0)$ is the origin and the velocity
vectors $\gamma^{\prime}_{1}(0)$, $\gamma^{\prime}_{2}(0)$ point in the
direction of $x$-axis.

Note that small arcs of $\gamma_{1}|_{[-\varepsilon,+\varepsilon]}$ and
$\gamma_{2}|_{[-\varepsilon,+\varepsilon]}$ can be described as a graph
$y=f_{1}(x)$ and $y=f_{2}(x)$ for smooth functions $f_{1}$ and $f_{2}$ such
that $f_{i}(0)=0$ and $f_{i}^{\prime}(0)=0$. Note that
$f_{1}^{\prime\prime}(0)=k_{1}(0)$ and $f_{2}^{\prime\prime}(0)=k_{2}(0)$ (see
3C)

Clearly, $\gamma_{1}$ supports $\gamma_{2}$ from the left (right) if

$f_{1}(x)\leqslant f_{2}(x)\quad(\text{respectively}\quad f_{1}(x)\geqslant
f_{2}(x))$

for all sufficiently small values of $x$. Applying the second derivative test,
we get the result. ∎

6.4. Advanced exercise. Let $\gamma_{0}$ and $\gamma_{1}$ be two smooth unit-
speed simple plane curves that are tangent and cooriented at the point
$p=\nobreak\gamma_{0}(0)=\nobreak\gamma_{1}(0)$. Assume $k_{0}(s)\leqslant
k_{1}(s)$ for any $s$. Show that $\gamma_{0}$ locally supports $\gamma_{1}$
from the right at $p$.

Give an example of two proper curves $\gamma_{0}$ and $\gamma_{1}$ satisfying
the above condition such that $\gamma_{0}$ does not support $\gamma_{1}$ at
$p$ globally.

Note that according to 3E for any closed smooth regular curve that runs in a
unit disc, the average of its absolute curvature is at least 1; in particular,
there is a point where the absolute curvature is at least 1. The following
exercise says that this statement also holds for loops.

6.5. Exercise. Assume a closed smooth regular plane loop $\gamma$ runs in a
unit disc. Show that there is a point on $\gamma$ with absolute curvature at
least 1.

6.6. Exercise. Assume a closed smooth regular plane curve $\gamma$ runs
between two parallel lines at distance 2 from each other. Show that there is a
point on $\gamma$ with absolute curvature at least 1.

Try to prove the same for a smooth regular plane loop.

6.7. Exercise. Assume a closed smooth regular plane curve $\gamma$ runs inside
of a triangle $\triangle$ with inradius 1; that is, the circle tangent to all
three sides of $\triangle$ has radius 1. Show that there is a point on
$\gamma$ with absolute curvature at least $1$.

The three exercises above are baby cases of 6E; try to find a direct solution,
without using 6E.

6.8. Exercise. Let $F$ be a plane figure bounded by two circle arcs
$\sigma_{1}$ and $\sigma_{2}$ of signed curvature 1 that run from $x$ to $y$.
Suppose $\sigma_{1}$ is shorter than $\sigma_{2}$. Assume a smooth regular arc
$\gamma$ runs in $F$ and has both endpoints on $\sigma_{1}$. Show that the
absolute curvature of $\gamma$ is at least 1 at some parameter value.

###  D Convex curves

Recall that a plane curve is convex if it bounds a convex region.

6.9. Proposition. Suppose that a closed simple plane curve $\gamma$ bounds a
figure $F$. Then $F$ is convex if and only if the signed curvature of $\gamma$
does not change sign.

6.10. Lens lemma. Let $\gamma$ be a smooth regular simple plane curve that
runs from $x$ to $y$. Assume that $\gamma$ runs on the right side (left side)
of the oriented line $xy$ and only its endpoints $x$ and $y$ lie on the line.
Then $\gamma$ has a point with positive (respectively negative) signed
curvature.

Note that the lemma fails for curves with self-intersections. For example, the
curve $\gamma$ on the diagram always turns right, so it has negative curvature
everywhere, but it lies on the right side of the line $xy$.

Proof. Choose points $p$ and $q$ on the line $xy$ so that the points $p,x,y,q$
appear in that order. We can assume that $p$ and $q$ lie sufficiently far from
$x$ and $y$, so that the half-disc with diameter $pq$ contains $\gamma$.

Consider the smallest disc segment with chord $[p,q]$ that contains $\gamma$.
Note that its arc $\sigma$ supports $\gamma$ at some point $w=\gamma(t_{0})$.

Let us parametrize $\sigma$ from $p$ to $q$. Note that the $\gamma$ and
$\sigma$ are tangent and cooriented at $w$. If not, then the arc of $\gamma$
from $w$ to $y$ would be trapped in the curvilinear triangle $xwp$ bounded by
the line segment $[p,x]$ and the arcs of $\sigma$, $\gamma$. But this is
impossible since $y$ does not belong to this triangle.

It follows that $\sigma$ supports $\gamma$ at $t_{0}$ from the right. By 6C,

$k(w)_{\gamma}\geqslant k_{\sigma}>0.$

∎

Remark. Instead of taking the minimal disc segment, one can take a point $w$
on $\gamma$ that maximizes the distance to the line $xy$. The same argument
shows that the curvature at $w$ is nonnegative, which is slightly weaker than
the required positive curvature.

Proof of 6D; ‘‘only-if’’ part. If $F$ is convex, then every tangent line of
$\gamma$ supports $\gamma$. If a point moves along $\gamma$, the figure $F$
has to stay on one side from its tangent line; that is, we can assume that
each tangent line supports $\gamma$ on one side, say on the right. Since a
line has vanishing curvature, the supporting test (6C) implies that
$k\geqslant 0$ at each point.

‘‘If’’ part. Denote by $K$ the convex hull of $F$. If $F$ is not convex, then
$F$ is a proper subset of $K$. Therefore $\partial K$ contains a line segment
that is not a part of $\partial F$. In other words, there is a line that
supports $\gamma$ at two points, say $x$ and $y$ that divide $\gamma$ in two
arcs $\gamma_{1}$ and $\gamma_{2}$, both distinct from the line segment
$[x,y]$.

Note that one of the arcs $\gamma_{1}$ or $\gamma_{2}$ is parametrized from
$x$ to $y$ and the other from $y$ to $x$. Passing to a smaller arc if
necessary we can ensure that only its endpoints lie on the line. Applying the
lens lemma, we get that the arcs $\gamma_{1}$ and $\gamma_{2}$ contain points
with signed curvatures of opposite signs. ∎

6.11. Exercise. Suppose $\gamma$ is a smooth regular simple closed convex
plane curve of diameter larger than 2. Show that $\gamma$ has a point with
absolute curvature less than 1.

6.12. Exercise. Suppose $\gamma$ is a simple smooth regular plane curve with
positive signed curvature. Assume $\gamma$ crosses a line $\ell$ at the points
$p_{1},p_{2},\dots\nobreak,p_{n}$ and these points appear on $\gamma$ in the
same order.

  * (a)

Show that $p_{2}$ cannot lie between $p_{1}$ and $p_{3}$ on $\ell$.

  * (b)

Show that if $p_{3}$ lies between $p_{1}$ and $p_{2}$ on $\ell$, then the
points appear on $\ell$ in the following order:

$p_{1},p_{3},\dots,p_{4},p_{2}.$

  * (c)

Describe all possible orders of the points $p_{i}$ on $\ell$.

###  E Moon in a puddle

The following theorem is a slight generalization of the theorem proved by
Vladimir Ionin and German Pestov in [43]. For convex curves, this result was
known earlier [11, §24].

6.13. Theorem. Assume $\gamma$ is a simple closed smooth regular plane loop
with absolute curvature bounded by 1. Then it surrounds a unit disc.

This theorem gives a simple but nontrivial example of the so-called _local to
global theorems_ — based on some local data (in this case the curvature of a
curve) we conclude a global property (in this case existence of a large disc
surrounded by the curve).

A straightforward approach would be to start with some disc in the region
bounded by the curve and blow it up to maximize its radius. However, as one
may see from the spinner-like example on the diagram it does not always lead
to a solution — a closed plane curve of curvature at most 1 may surround a
disc of radius smaller than 1 that cannot be enlarged continuously.

6.14. Key lemma. Assume $\gamma$ is a simple closed smooth regular plane loop.
Then at one point of $\gamma$ (distinct from its base) its osculating circle
$\sigma$ globally supports $\gamma$ from the inside.

First let us show that the theorem follows from the lemma.

Proof of 6E modulo 6E. Since $\gamma$ has absolute curvature at most 1, each
osculating circle has radius at least 1. According to the key lemma one of the
osculating circles $\sigma$ globally supports $\gamma$ from inside. In
particular $\sigma$ lies inside of $\gamma$, whence the result. ∎

Proof of 6E. Denote by $F$ the closed region surrounded by $\gamma$. We can
assume that $F$ lies on the left from $\gamma$. Arguing by contradiction,
assume that the osculating circle at each point $p\in\gamma$ does not lie in
$F$.

Given a point $p\in\gamma$ let us consider the maximal circle $\sigma$ that
lies completely in $F$ and tangent to $\gamma$ at $p$. The circle $\sigma$
will be called the _incircle_ of $F$ at $p$.

Note that the curvature $k_{\sigma}$ of the incircle $\sigma$ has to be larger
than $k(p)_{\gamma}$. Indeed, since $\sigma$ supports $\gamma$ from the left,
by 6C we have $k_{\sigma}\geqslant k(p)_{\gamma}$; in the case of equality,
$\sigma$ is the osculating circle at $p$. The latter is impossible by our
assumption.

It follows that $\sigma$ has to touch $\gamma$ at another point. Otherwise we
can increase $\sigma$ slightly while keeping it inside $F$.

Indeed, since $k_{\sigma}>k(p)_{\gamma}$, by 6C we can choose a neighborhood
$U$ of $p$ such that after a slight increase of $\sigma$, the intersection
$U\cap\sigma$ is still in $F$. On the other hand, if $\sigma$ does not touch
$\gamma$ at another point, then after some (maybe smaller) increase of
$\sigma$ the complement $\sigma\backslash U$ is still in $F$. That is, a
slightly increased $\sigma$ is still in $F$ — a contradiction.

Choose a point $p_{1}$ on $\gamma$ that is distinct from its base point. Let
$\sigma_{1}$ be the incircle at $p_{1}$. Denote by $\gamma_{1}$ an arc of
$\gamma$ from $p_{1}$ to a first point $q_{1}$ on $\sigma_{1}$. Denote by
$\hat{\sigma}_{1}$ and $\check{\sigma}_{1}$ two arcs of $\sigma_{1}$ from
$p_{1}$ to $q_{1}$ such that the cyclic concatenation of $\hat{\sigma}_{1}$
and $\gamma_{1}$ surrounds $\check{\sigma}_{1}$.

Two ovals pretend to be circles.

Let $p_{2}$ be the midpoint of $\gamma_{1}$. Denote by $\sigma_{2}$ the
incircle at $p_{2}$.

Note that $\sigma_{2}$ cannot intersect $\hat{\sigma}_{1}$. Otherwise, if
$\sigma_{2}$ intersects $\hat{\sigma}_{1}$ at some point $s$, then
$\sigma_{2}$ has to have two more common points with $\check{\sigma}_{1}$, say
$x$ and $y$ — one for each arc of $\sigma_{2}$ from $p_{2}$ to $s$. Therefore
$\sigma_{1}=\nobreak\sigma_{2}$ since these two circles have three common
points: $s$, $x$, and $y$. On the other hand, by construction,
$p_{2}\in\sigma_{2}$ and $p_{2}\notin\sigma_{1}$ — a contradiction.

Recall that $\sigma_{2}$ has to touch $\gamma$ at another point. From above it
follows that it can only touch $\gamma_{1}$ and therefore we can choose an arc
$\gamma_{2}\subset\gamma_{1}$ that runs from $p_{2}$ to a first point $q_{2}$
on $\sigma_{2}$. Since $p_{2}$ is the midpoint of $\gamma_{1}$, we have that

$\mathop{\rm length}\nolimits\gamma_{2}<\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm length}\nolimits\gamma_{1}.$ ➊

Repeating this construction recursively, we get an infinite sequence of arcs
$\gamma_{1}\supset\gamma_{2}\supset\dots$; by ➊ ‣ 6E, we also get that

$\mathop{\rm length}\nolimits\gamma_{n}\to 0\quad\text{as}\quad n\to\infty.$

Therefore the intersection

$\bigcap_{n}\gamma_{n}$

contains a single point; denote it by $p_{\infty}$.

Let $\sigma_{\infty}$ be the incircle at $p_{\infty}$; it has to touch
$\gamma$ at another point, say $q_{\infty}$. The same argument as above shows
that $q_{\infty}\in\gamma_{n}$ for any $n$. It follows that
$q_{\infty}=p_{\infty}$ — a contradiction. ∎

6.15. Exercise. Assume that a closed smooth regular curve $\gamma$ lies in a
figure $F$ bounded by a closed simple plane curve. Suppose that $R$ is the
maximal radius of discs that lie in $F$. Show that absolute curvature of
$\gamma$ is at least $\tfrac{1}{R}$ at some parameter value.

###  F Four-vertex theorem

Recall that a vertex of a smooth regular curve is defined as a critical point
of its signed curvature; in particular, any local minimum (or maximum) of the
signed curvature is a vertex. For example, every point of a circle is a
vertex.

6.16. Four-vertex theorem. Any smooth regular simple plane curve has at least
four vertices.

Evidently any closed smooth regular curve has at least two vertices — where
the minimum and the maximum of the curvature are attained. On the diagram the
vertices are marked; the first curve has one self-intersection and exactly two
vertices; the second curve has exactly four vertices and no self-
intersections.

The four-vertex theorem was first proved by Syamadas Mukhopadhyaya [61] for
convex curves. Many different proofs and generalizations are known. One of our
favorite proofs was given by Robert Osserman [63]. We give a proof of the
following stronger statement based on the key lemma in the previous section.

6.17. Theorem. Any smooth regular simple plane curve is globally supported by
its osculating circle at least at 4 distinct points; two from inside and two
from outside.

Proof of 6F modulo 6F. First note that if an osculating circline $\sigma$ at a
point $p$ supports $\gamma$ locally, then $p$ is a vertex. Indeed, if not,
then a small arc around $p$ has monotonic curvature.

Applying the spiral lemma (5E) we get that the osculating circles at this arc
are nested. In particular the curve $\gamma$ crosses $\sigma$ at $p$ and
therefore $\sigma$ does not locally support $\gamma$ at $p$. ∎

Proof of 6F. According to the key lemma (6E), there is a point $p\in\gamma$
such that its osculating circle supports $\gamma$ from inside. The curve
$\gamma$ can be considered as a loop with the base at $p$. Therefore the key
lemma implies the existence of another point $q\in\gamma$ with the same
property.

It shows the existence of two osculating circles that support $\gamma$ from
inside; it remains to show existence of two osculating circles that support
$\gamma$ from outside.

In order to get the osculating circles supporting $\gamma$ from outside, one
can repeat the proof of key lemma taking instead of incircle the circline of
maximal signed curvature that supports the curve from outside, assuming that
$\gamma$ is oriented so that the region on the left from it is bounded.

(Alternatively, if one applies to $\gamma$ an inversion with respect to a
circle whose center lies inside $\gamma$, then the obtained curve $\gamma_{1}$
also has two osculating circles that support $\gamma_{1}$ from inside.
According to 5D, these osculating circlines are inverses of the osculating
circlines of $\gamma$. Note that the region lying inside of $\gamma$ is mapped
to the region outside of $\gamma_{1}$ and the other way around. Therefore
these two circlines correspond to the osculating circlines supporting $\gamma$
from outside.) ∎

6.18. Advanced exercise. Suppose $\gamma$ is a closed simple smooth regular
plane curve and $\sigma$ is a circle. Assume $\gamma$ crosses $\sigma$ at the
points $p_{1},\dots,p_{2{{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}}n}$ and these
points appear in the same cycle order on $\gamma$ and on $\sigma$. Show that
$\gamma$ has at least $2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}n$ vertices.

Construct an example of a closed simple smooth regular plane curve $\gamma$
with only 4 vertices that crosses a given circle at arbitrarily many points.

##  Chapter 7 Definitions

###  A Topological surfaces

We will be mostly interested in smooth regular surfaces defined in the
following section. However we will sometimes use the following general
definition.

A connected subset $\Sigma$ in the Euclidean space $\mathbb{R}^{3}$ is called
a _topological surface_ (more precisely an _embedded surface without
boundary_) if any point of $p\in\Sigma$ admits a neighborhood $W$ in $\Sigma$
that can be parametrized by an open subset in the Euclidean plane; that is,
there is an injective continuous map $U\to W$ from an open set
$U\subset\mathbb{R}^{2}$ such that its inverse $W\to U$ is also continuous.

###  B Smooth surfaces

Recall that a function $f$ of two variables $x$ and $y$ is called _smooth_ if
all its partial derivatives $\frac{\partial^{m+n}}{\partial x^{m}\partial
y^{n}}f$ are defined and are continuous in the domain of definition of $f$.

A connected set $\Sigma\subset\mathbb{R}^{3}$ is called a _smooth surface_ (we
use it as a shortcut for the more precise term _smooth regular embedded
surface_) if it can be described locally as a graph of a smooth function in an
appropriate coordinate system.

More precisely, for any point $p\in\Sigma$ one can choose a coordinate system
$(x,y,z)$ and a neighborhood $U\ni p$ such that the intersection
$W=U\cap\Sigma$ is a graph $z=f(x,y)$ of a smooth function $f$ defined in an
open domain of the $(x,y)$-plane.

Examples. The simplest example of a smooth surface is the $(x,y)$-plane

$\Pi=\left\\{\,{(x,y,z)\in\mathbb{R}^{3}}\,:\,{z=0}\,\right\\}.$

The plane $\Pi$ is a surface since it can be described as the graph of the
function $f(x,y)=0$.

All other planes are smooth surfaces as well since one can choose a coordinate
system so that it becomes the $(x,y)$-plane. We may also present a plane as a
graph of a linear function $f(x,y)=a{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x+b{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+c$ for some constants $a$,
$b$ and $c$ (assuming the plane is not perpendicular to the $(x,y)$-plane, in
which case a different coordinate system is required to write the plane as the
graph of a function).

A more interesting example is the unit sphere

$\mathbb{S}^{2}=\left\\{\,{(x,y,z)\in\mathbb{R}^{3}}\,:\,{x^{2}+y^{2}+z^{2}=1}\,\right\\}.$

This set is not the graph of any function, but $\mathbb{S}^{2}$ is locally a
graph; it can be covered by the following 6 graphs:

$\displaystyle z$ $\displaystyle=f_{\pm}(x,y)=\pm\sqrt{1-x^{2}-y^{2}},$
$\displaystyle y$ $\displaystyle=g_{\pm}(x,z)=\pm\sqrt{1-x^{2}-z^{2}},$
$\displaystyle x$ $\displaystyle=h_{\pm}(y,z)=\pm\sqrt{1-y^{2}-z^{2}},$

where each function $f_{\pm},g_{\pm},h_{\pm}$ is defined in an open unit disc.
Any point $p\in\mathbb{S}^{2}$ lies in one of these graphs therefore
$\mathbb{S}^{2}$ is a surface. Since each function is smooth, so is the
surface $\mathbb{S}^{2}$.

###  C Surfaces with boundary

A connected subset in a surface that is bounded by one or more curves is
called a _surface with boundary_ ; such curves form the _boundary line_ of the
surface.

When we say _surface_ we usually mean a _smooth regular surface without
boundary_ ; we may use the term _surface without boundary_ if we need to
emphasize it; otherwise we may use the term _surface with possibly nonempty
boundary_.

###  D Proper, closed and open surfaces

If the surface $\Sigma$ is formed by a closed set, then it is called _proper_.
For example, for any smooth function $f$ defined on the whole plane, its graph
$z=f(x,y)$ is a proper surface. The sphere $\mathbb{S}^{2}$ gives another
example of proper surface.

On the other hand, the open disc

$\left\\{\,{(x,y,z)\in\mathbb{R}^{3}}\,:\,{x^{2}+y^{2}<1,\ z=0}\,\right\\}$

is not proper; this set is neither open nor closed.

A compact surface without boundary is called _closed_ (this term is closely
related to _closed curve_ but has nothing to do with _closed set_).

A proper noncompact surface without boundary is called _open_ (again the term
_open set_ is not relevant).

For example, the paraboloid $z=x^{2}+y^{2}$ is an open surface; the sphere
$\mathbb{S}^{2}$ is a closed surface.

Note that any proper surface without boundary is either closed or open.

The following claim is a three-dimensional analog of the plane separation
theorem (1F). Despite it might look obvious, its proof is not trivial at all;
a standard proof uses the so-called _Alexander’s duality_ which is a classical
technique in algebraic topology [[, see]]hatcher. We omit its proof since it
would take us far away from the main subject.

7.1. Claim. The complement of any proper topological surface without boundary
(or, equivalently any open or closed topological surface) has exactly two
connected components.

###  E Implicitly defined surfaces

7.2. Proposition. Let $f\colon\mathbb{R}^{3}\to\mathbb{R}$ be a smooth
function. Suppose that $0$ is a regular value of $f$; that is,
$\nabla_{p}f\neq 0$ at any point $p$ such that $f(p)=0$. Then any connected
component $\Sigma$ of the set of solutions of the equation $f(x,y,z)=0$ is a
smooth surface.

Proof. Fix $p\in\Sigma$. Since $\nabla_{p}f\neq 0$ we have

$f_{x}(p)\neq 0,\quad f_{y}(p)\neq 0,\quad\text{or}\quad f_{z}(p)\neq 0.$

We may assume that $f_{z}(p)\neq 0$; otherwise permute the coordinates
$x,y,z$.

The implicit function theorem (0F) implies that a neighborhood of $p$ in
$\Sigma$ is the graph $z=h(x,y)$ of a smooth function $h$ defined on an open
domain in $\mathbb{R}^{2}$. It remains to apply the definition of smooth
surface (Section 7B). ∎

7.3. Exercise. For which constants $\ell$ does the following equation

$\displaystyle x^{2}+y^{2}-z^{2}$ $\displaystyle=\ell$

describe a smooth regular surface.

###  F Local parametrizations

Let $U$ be an open domain in $\mathbb{R}^{2}$ and $s\colon U\to\mathbb{R}^{3}$
be a smooth map. We say that $s$ is regular if its Jacobian has maximal rank;
in this case it means that the vectors $s_{u}$ and $s_{v}$ are linearly
independent at any $(u,v)\in U$; equivalently $s_{u}\times s_{v}\neq 0$, where
$\times$ denotes the vector product.

7.4. Proposition. If $s\colon U\to\mathbb{R}^{3}$ is a smooth regular
embedding of an open connected set $U\subset\mathbb{R}^{2}$, then its image
$\Sigma=s(U)$ is a smooth surface.

Proof. Set

$s(u,v)=(x(u,v),y(u,v),z(u,v)).$

Since $s$ is regular, its Jacobian matrix

$\mathop{\rm Jac}\nolimits s=\begin{pmatrix}x_{u}&x_{v}\\\ y_{u}&y_{v}\\\
z_{u}&z_{v}\end{pmatrix}$

has rank two at any pint $(u,v)\in U$.

Choose a point $p\in\Sigma$; by shifting the $(x,y,z)$ and $(u,v)$ coordinate
systems we may assume that $p$ is the origin and $p=s(0,0)$. Permuting the
coordinates $x,y,z$ if necessary, we may assume that the matrix

$\begin{pmatrix}x_{u}&x_{v}\\\ y_{u}&y_{v}\end{pmatrix},$

is invertible at the origin. Note that this is the Jacobian matrix of the map

$(u,v)\mapsto(x(u,v),y(u,v)).$

The inverse function theorem (0F) implies that there is a smooth regular map
$w\colon(x,y)\mapsto(u,v)$ defined on an open set $W\ni 0$ in the
$(x,y)$-plane such that $w(0,0)=(0,0)$ and $s\circ w(x,y)=(x,y,f(x,y))$ for
some smooth function $f$. That is, the graph $z=f(x,y)$ for $(x,y)\in\nobreak
W$ is a subset in $\Sigma$. By the inverse function theorem this graph is open
in $\Sigma$.

Since $p$ is arbitrary, we get that $\Sigma$ is a surface. ∎

If we have $s$ and $\Sigma$ as in the proposition, then we say that $s$ is a
_smooth parametrization_ of the surface $\Sigma$.

Not all the smooth surfaces can be described by such a parametrization; for
example the sphere $\mathbb{S}^{2}$ cannot. However, any smooth surface
$\Sigma$ admits a local parametrization; that is, any point $p\in\Sigma$
admits an open neighborhood $W\subset\Sigma$ with a smooth regular
parametrization $s$. In this case any point in $W$ can be described by two
parameters, usually denoted by $u$ and $v$, which are called _local
coordinates_ at $p$. The map $s$ is called a _chart_ of $\Sigma$.

If $W$ is a graph $z=h(x,y)$ of a smooth function $h$, then the map

$s\colon(u,v)\mapsto(u,v,h(u,v))$

is a chart. Indeed, $s$ has an inverse $(u,v,h(u,v))\mapsto(u,v)$ which is
continuous; that is, $s$ is an embedding. Further, $s_{u}=(1,0,h_{u})$ and
$s_{v}=(0,1,h_{v})$. Whence the partial derivatives $s_{u}$ and $s_{v}$ are
linearly independent; that is, $s$ is a regular map.

Note that from 7F, we obtain the following corollary.

7.5. Corollary. A connected set $\Sigma\subset\mathbb{R}^{3}$ is a smooth
regular surface if and only if a neighborhood of any point in $\Sigma$ can be
covered by a chart.

7.6. Exercise. Consider the following map

$s(u,v)=(\tfrac{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}u}{1+u^{2}+v^{2}},\tfrac{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}v}{1+u^{2}+v^{2}},\tfrac{2}{1+u^{2}+v^{2}}).$

Show that $s$ is a chart of the unit sphere centered at $(0,0,1)$; describe
the image of $s$.

The map $s$ in the exercise can be visualized using the following map

$(u,v,1)\mapsto(\tfrac{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}u}{1+u^{2}+v^{2}},\tfrac{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}v}{1+u^{2}+v^{2}},\tfrac{2}{1+u^{2}+v^{2}})$

which is called _stereographic projection_ from the plane $z=1$ to the unit
sphere with center at $(0,0,1)$. Note that the point $(u,v,1)$ and its image
lie on the same half-line emerging from the origin.

Let $\gamma(t)=(x(t),y(t))$ be a plane curve. Recall that the _surface of
revolution_ of the curve $\gamma$ around the $x$-axis can be described as the
image of the map

$(t,s)\mapsto(x(t),y(t){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\cos
s,y(t){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin s).$

For fixed $t$ or $s$ the obtained curves are called _meridians_ or _parallels_
of the surface, respectively; note that parallels are formed by circles in the
plane perpendicular to the axis of rotation. The curve $\gamma$ is called the
_generatrix_ of the surface.

7.7. Exercise. Assume $\gamma$ is a closed simple smooth regular plane curve
that does not intersect the $x$-axis. Show that the surface of revolution
around the $x$-axis with generatrix $\gamma$ is a smooth regular surface.

###  G Global parametrizations

A surface can be described by an embedding from a known surface to the space.

For example, consider the ellipsoid

$\Sigma_{a,b,c}=\left\\{\,{(x,y,z)\in\mathbb{R}^{3}}\,:\,{\tfrac{x^{2}}{a^{2}}+\tfrac{y^{2}}{b^{2}}+\tfrac{z^{2}}{c^{2}}=1}\,\right\\}$

for some positive numbers $a$, $b$, and $c$. Note that by 7E, $\Sigma_{a,b,c}$
is a smooth regular surface. Indeed, set
$f(x,y,z)=\tfrac{x^{2}}{a^{2}}+\tfrac{y^{2}}{b^{2}}+\tfrac{z^{2}}{c^{2}}$,
then

$\nabla f(x,y,z)=(\tfrac{2}{a^{2}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x,\tfrac{2}{b^{2}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}y,\tfrac{2}{c^{2}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}z).$

Therefore $\nabla f\neq 0$ if $f=1$; that is, $1$ is a regular value of $f$.

Note that $\Sigma_{a,b,c}$ can be defined as the image of the map
$s\colon\mathbb{S}^{2}\to\mathbb{R}^{3}$, defined as the restriction of the
following map to the unit sphere $\mathbb{S}^{2}$:

$(x,y,z)\mapsto\nobreak(a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x,b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y,c{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}z).$

For a surface $\Sigma$, a map $s\colon\Sigma\to\mathbb{R}^{3}$ is called a
_smooth parametrized surface_ if $s$ is injective and for any chart $f\colon
U\to\Sigma$ the composition $s\circ f$ is smooth and regular; that is, all
partial derivatives $\frac{\partial^{m+n}}{\partial u^{m}\partial
v^{n}}(s\circ f)$ exist and are continuous in the domain of definition and the
two vectors $\frac{\partial}{\partial u}(s\circ f)$ and
$\frac{\partial}{\partial v}(s\circ f)$ are linearly independent.

Note that in this case the image $\Sigma^{*}=s(\Sigma)$ is a smooth surface.
The latter follows since for any chart $f\colon U\to\Sigma$ the composition
$s\circ f\colon U\to\Sigma^{*}$ is a chart of $\Sigma^{*}$. The map $s$ is
called a _diffeomorphism_ from $\Sigma$ to $\Sigma^{*}$; the surfaces $\Sigma$
and $\Sigma^{*}$ are said to be _diffeomorphic_ if there is a diffeomorphism
$s\colon\Sigma\to\Sigma^{*}$.

7.8. Advanced exercise. Show that the surfaces $\Sigma$ and $\Theta$ are
diffeomorphic if

  * (a)

$\Sigma$ and $\Theta$ obtained from the plane by removing $n$ points.

  * (b)

$\Sigma$ and $\Theta$ are open convex subsets of a plane bounded by smooth
curves.

  * (c)

$\Sigma$ and $\Theta$ are open convex subsets of a plane.

  * (d)

$\Sigma$ and $\Theta$ are open star-shaped subsets of a plane.

##  Chapter 8 First order structure

###  A Tangent plane

8.1. Definition. Let $\Sigma$ be a smooth surface. A vector w is a _tangent
vector_ of $\Sigma$ at $p$ if and only if there is a curve $\gamma$ that runs
in $\Sigma$ and has w as a velocity vector at $p$; that is, $p=\gamma(t)$ and
$\text{\sc{w}}=\gamma^{\prime}(t)$ for some $t$.

8.2. Proposition-Definition. Let $\Sigma$ be a smooth surface and
$p\in\Sigma$. Then the set of tangent vectors of $\Sigma$ at $p$ forms a
plane; this plane is called the _tangent plane_ of $\Sigma$ at $p$.

Moreover if $s\colon U\to\Sigma$ is a local chart and $p=s(u_{p},v_{p})$, then
the tangent plane of $\Sigma$ at $p$ is spanned by vectors
$s_{u}(u_{p},v_{p})$ and $s_{v}(u_{p},v_{p})$.

The tangent plane to $\Sigma$ at $p$ is usually denoted by $\mathrm{T}_{p}$ or
$\mathrm{T}_{p}\Sigma$. This plane $\mathrm{T}_{p}$ might be considered as a
linear subspace of $\mathbb{R}^{3}$ or as a parallel plane passing thru $p$;
the latter is sometimes called the _affine tangent plane_. The affine tangent
plane can be interpreted as the best approximation at $p$ of the surface
$\Sigma$ by a plane. More precisely, it has _first order of contact_ with
$\Sigma$ at $p$; that is, $\rho(q)=\nobreak o(|p-q|)$, where $q\in\Sigma$ and
$\rho(q)$ denotes the distance from $q$ to $\mathrm{T}_{p}$.

Proof. Fix a chart $s$ at $p$. Assume $\gamma$ is a smooth curve that starts
at $p$. Without loss of generality, we can assume that $\gamma$ is covered by
the chart; in particular, there are smooth functions $u(t)$ and $v(t)$ such
that

$\gamma(t)=s(u(t),v(t)).$

Applying chain rule, we get

$\gamma^{\prime}=s_{u}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}u^{\prime}+s_{v}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}v^{\prime};$

that is, $\gamma^{\prime}$ is a linear combination of $s_{u}$ and $s_{v}$.

Since the smooth functions $u(t)$ and $v(t)$ can be chosen arbitrarily, any
linear combination of $s_{u}$ and $s_{v}$ is a tangent vector at $p$. ∎

8.3. Exercise. Let $f:\mathbb{R}^{3}\to\mathbb{R}$ be a smooth function with
$0$ as a regular value and $\Sigma$ be a surface described as a connected
component of the set of solutions $f(x,y,z)=0$. Show that the tangent plane
$\mathrm{T}_{p}\Sigma$ is perpendicular to the gradient $\nabla_{p}f$ at any
point $p\in\Sigma$.

8.4. Exercise. Let $\Sigma$ be a smooth surface and $p\in\Sigma$. Choose
$(x,y,z)$-coordinates. Show that a neighborhood of $p$ in $\Sigma$ is a graph
$z=f(x,y)$ of a smooth function $f$ defined on an open subset in the
$(x,y)$-plane if and only if the tangent plane $\mathrm{T}_{p}$ is not
_vertical_ ; that is, if $\mathrm{T}_{p}$ is not perpendicular to the
$(x,y)$-plane.

8.5. Exercise. Show that if a smooth surface $\Sigma$ meets a plane $\Pi$ at a
single point $p$, then $\Pi$ is tangent to $\Sigma$ at $p$.

###  B Directional derivative

In this section we extend the definition of directional derivative to smooth
functions defined on smooth surfaces.

First let us recall the standard definition of directional derivative.

Suppose $f$ is a function defined at a point $p$ in the space, and w a vector.
Consider the function

$h(t)=f(p+t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{w}}).$

Then the directional derivative of $f$ at $p$ along w is defined as

$D_{\text{\sc{w}}}f(p)\mathrel{:=}h^{\prime}(0).$

8.6. Proposition-Definition. Let $\Sigma$ be a smooth regular surface and $f$
a smooth function defined on $\Sigma$. Suppose $\gamma$ is a smooth curve in
$\Sigma$ that starts at $p$ with velocity vector
$\text{\sc{w}}\in\mathrm{T}_{p}$; that is, $\gamma(0)=p$ and
$\gamma^{\prime}(0)=\text{\sc{w}}$. Then the derivative
$(f\circ\gamma)^{\prime}(0)$ depends only on $f$, $p$ and w; it is called the
_directional derivative of $f$ along w at $p$_ and denoted by

$D_{\text{\sc{w}}}f,\quad D_{\text{\sc{w}}}f(p),\quad\text{or}\quad
D_{\text{\sc{w}}}f(p)_{\Sigma}$

— we may omit $p$ and $\Sigma$ if it is clear from the context.

Moreover, if $(u,v)\mapsto s(u,v)$ is a local chart at $p$, then

$D_{\text{\sc{w}}}f=a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}f_{u}+b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}f_{v},$

where $\text{\sc{w}}=a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}s_{u}+b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s_{v}$.

Note that our definition agrees with the standard definition of directional
derivative if $\Sigma$ is a plane. Indeed, in this case
$\gamma(t)=p+\text{\sc{w}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}t$ is a curve
in $\Sigma$ that starts at $p$ with velocity vector w. For a general surface
the point $p+\text{\sc{w}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}t$ might not
lie on the surface; therefore the function $f$ might be undefined at this
point; therefore the standard definition does not work.

Proof. Without loss of generality, we may assume that $\gamma$ is covered by
the chart $s$; if not we can chop $\gamma$. In this case

$\gamma(t)=s(u(t),v(t))$

for some smooth functions $u,v$ defined in a neighborhood of $0$ such that
$u(0)=u_{p}$ and $v(0)=v_{p}$.

Applying the chain rule, we get that

$\displaystyle\gamma^{\prime}(0)$ $\displaystyle=u^{\prime}(0){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s_{u}+v^{\prime}(0){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s_{v}$

at $(u_{p},v_{p})$. Since $\text{\sc{w}}=\gamma^{\prime}(0)$ and the vectors
$s_{u}$, $s_{v}$ are linearly independent, we get that $a=u^{\prime}(0)$ and
$b=v^{\prime}(0)$.

Applying the chain rule again, we get that

$(f\circ\gamma)^{\prime}(0)=a{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}f_{u}+b{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}f_{v}.$ ➊

at $(u_{p},v_{p})$.

Notice that the left hand side in ➊ ‣ 8B does not depend on the choice of the
chart $s$ and the right hand side depends only on $p$, w, $f$, and $s$. It
follows that $(f\circ\gamma)^{\prime}(0)$ depends only on $p$, w and $f$.

The last statement follows from ➊ ‣ 8B. ∎

###  C Tangent vectors as functionals

In this section we introduce a more conceptual way to define tangent vectors.
We will not use this approach in the sequel, but it is better to know about
it.

A tangent vector $\text{\sc{w}}\in\mathrm{T}_{p}$ to a smooth surface $\Sigma$
defines a linear functional111Term _functional_ is used for functions that
take a function as an argument and return a number. $D_{\text{\sc{w}}}$ that
swallows a smooth function $\varphi$ defined in a neighborhood of $p$ in
$\Sigma$ and spits its directional derivative $D_{\text{\sc{w}}}\varphi$. It
is straightforward to check that the functional $D$ obeys the product rule:

$D_{\text{\sc{w}}}(\varphi{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\psi)=(D_{\text{\sc{w}}}\varphi){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\psi(p)+\varphi(p){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(D_{\text{\sc{w}}}\psi).$ ➊

It is not hard to show that the tangent vector w is completely determined by
the functional $D_{\text{\sc{w}}}$. Moreover tangent vectors at $p$ can be
_defined_ as linear functionals on the space of smooth functions that satisfy
the product rule ➊ ‣ 8C.

This definition grabs the key algebraic property of tangent vectors. It might
be a less intuitive way to think about tangent vectors, but it is often
convenient to use in the proofs. For example 8B becomes a tautology.

###  D Differential of map

Any smooth map $s$ from a surface $\Sigma$ to $\mathbb{R}^{3}$ can be
described by its coordinate functions $s(p)=(x(p),y(p),z(p))$. To take a
directional derivative of the map we should take the directional derivative of
each of its coordinate functions.

$D_{\text{\sc{w}}}s\mathrel{:=}(D_{\text{\sc{w}}}x,D_{\text{\sc{w}}}y,D_{\text{\sc{w}}}z).$

Assume $s$ is a smooth map from one smooth surface $\Sigma_{0}$ to another
$\Sigma_{1}$ and $p\in\Sigma_{0}$. Note that
$D_{\text{\sc{w}}}s(p)\in\mathrm{T}_{s(p)}\Sigma_{1}$ for any
$\text{\sc{w}}\in\mathrm{T}_{p}$. Indeed, choose a curve $\gamma_{0}$ in
$\Sigma_{0}$ such that $\gamma_{0}(0)=p$ and
$\gamma_{0}^{\prime}(0)=\text{\sc{w}}$. Observe that
$\gamma_{1}=s\circ\gamma_{0}$ is a smooth curve in $\Sigma_{1}$ and by the
definition of directional derivative, we have
$D_{\text{\sc{w}}}s(p)=\gamma_{1}^{\prime}(0)$. It remains to note that
$\gamma_{1}(0)=\nobreak s(p)$ and therefore its velocity
$\gamma_{1}^{\prime}(0)$ is in $\mathrm{T}_{s(p)}\Sigma_{1}$.

Recall that 8B implies that $d_{p}s\colon\text{\sc{w}}\mapsto
D_{\text{\sc{w}}}s$ defines a linear map
$d_{p}s\colon\mathrm{T}_{p}\Sigma_{0}\to\mathrm{T}_{s(p)}\Sigma_{1}$; that is,

$D_{c{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{w}}}s=c{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}D_{\text{\sc{w}}}s(p)\quad\text{and}\quad
D_{\text{\sc{v}}+\text{\sc{w}}}s=D_{\text{\sc{v}}}s(p)+D_{\text{\sc{w}}}s(p)$

for any $c\in\mathbb{R}$ and $\text{\sc{v}},\text{\sc{w}}\in\mathrm{T}_{p}$.
The map $d_{p}s$ is called the _differential_ of $s$ at $p$.

The differential $d_{p}s$ can be described by a $2{\times}2$-matrix $M$ in
orthonormal bases of $\mathrm{T}_{p}$ and $\mathrm{T}_{s(p)}\Sigma_{1}$. Set
$\mathop{\rm jac}\nolimits_{p}s=|\det M|$; this value does not depend on the
choice of orthonormal bases in $\mathrm{T}_{p}$ and
$\mathrm{T}_{s(p)}\Sigma_{1}$.

Let $s_{1}\colon\Sigma_{1}\to\Sigma_{2}$ be another smooth map between smooth
surfaces $\Sigma_{1}$ and $\Sigma_{2}$. Suppose that
${p_{1}}=s(p)\in\Sigma_{1}$; observe that

$d_{p}(s_{1}\circ s)=d_{p_{1}}s_{1}\circ d_{p}s.$

It follows that

$\mathop{\rm jac}\nolimits_{p}(s_{1}\circ s)=\mathop{\rm
jac}\nolimits_{p_{1}}s_{1}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
jac}\nolimits_{p}s.$ ➊

If $\Sigma_{0}$ is a domain in the $(u,v)$-plane, then the value $\mathop{\rm
jac}\nolimits_{p}s$ can be found using the following formulas

$\displaystyle\mathop{\rm jac}\nolimits s$ $\displaystyle=|s_{v}\times
s_{u}|=$ $\displaystyle=\sqrt{\langle s_{u},s_{u}\rangle{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle s_{v},s_{v}\rangle-\langle
s_{u},s_{v}\rangle^{2}}=$ $\displaystyle=\sqrt{\det[\mathop{\rm
Jac}\nolimits^{\top}s{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
Jac}\nolimits s]}.$

where $\mathop{\rm Jac}\nolimits s$ denotes the Jacobian matrix of $s$; it is
a 2$\times$3 matrix with column vectors $s_{u}$ and $s_{v}$.

The value $\mathop{\rm jac}\nolimits_{p}s$ has the following geometric
meaning: if $P_{0}$ is a region in $\mathrm{T}_{p}$ and
$P_{1}=(d_{p}s)(P_{0})$, then

$\mathop{\rm area}\nolimits P_{1}=\mathop{\rm jac}\nolimits_{p}s{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm area}\nolimits P_{0}.$

This identity will become important in the definition of surface area.

###  E Surface integral and area

Let $\Sigma$ be a smooth surface and $h\colon\Sigma\to\mathbb{R}$ be a smooth
function. Let us define the integral $\iint_{R}h$ of the function $h$ along a
region $R\subset\Sigma$. The definition will be used mostly along surfaces
with boundary, but the definition can be applied to any Borel set
$R\subset\Sigma$.

Recall that $\mathop{\rm jac}\nolimits_{p}s$ is defined in the previous
section. Assume that there is a chart $(u,v)\mapsto s(u,v)$ of $\Sigma$
defined on an open set $U\subset\mathbb{R}^{2}$ such that $R\subset s(U)$. In
this case set

$\iint_{R}h\mathrel{:=}\iint_{s^{-1}(R)}h\circ s(u,v){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm jac}\nolimits_{(u,v)}s{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}du{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}dv.$ ➊

By the substitution rule (0F), the right hand side in ➊ ‣ 8E does not depend
on the choice of $s$. That is, if $s_{1}\colon U_{1}\to\Sigma$ is another
chart such that $s_{1}(U_{1})\supset R$, then

$\iint_{s^{-1}(R)}h\circ s(u,v){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\mathop{\rm jac}\nolimits_{(u,v)}s{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}du{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}dv=\iint_{s_{1}^{-1}(R)}h\circ
s_{1}(u,v){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
jac}\nolimits_{(u,v)}s_{1}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}du{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dv.$

In other words, the defining identity ➊ ‣ 8E makes sense.

A general region $R$ can be subdivided into regions $R_{1},R_{2}\dots$ such
that each $R_{i}$ lies in the image of some chart. After that one could define
the integral along $R$ as the sum

$\iint_{R}h\mathrel{:=}\iint_{R_{1}}h+\iint_{R_{2}}h+\dots$

It is straightforward to check that the value $\iint_{R}h$ does not depend on
the choice of such subdivision.

The area of a region $R$ in a smooth surface $\Sigma$ is defined as the
surface integral

$\mathop{\rm area}\nolimits R=\iint_{R}1.$

The following proposition provides a substitution rule for surface integral.

8.7. Area formula. Suppose $s\colon\Sigma_{0}\to\Sigma_{1}$ is a smooth
parametrization of a smooth surface $\Sigma_{1}$ by a smooth surface
$\Sigma_{0}$. Then for any region $R\subset\Sigma_{0}$ and any smooth function
$f\colon\Sigma_{1}\to\mathbb{R}$ we have

$\iint_{R}(f\circ s){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
jac}\nolimits s=\int_{s(R)}f.$

In particular, if $f\equiv 1$, we have

$\iint_{R}\mathop{\rm jac}\nolimits s=\mathop{\rm area}\nolimits[s(R)].$

Proof. Follows from ➊ ‣ 8D and the definition of surface integral. ∎

Remark. The notion of area of a surface is closely related to the length of a
curve. However, to define length we use a different idea — it was defined as
the least upper bound on the lengths of inscribed polygonal lines. It turns
out that an analogous definition does not work even for very simple surfaces.
The latter is shown by a classical example — the so-called _Schwarz’s boot_.
This example and different approaches to the notion of area are discussed in a
popular article of Vladimir Dubrovsky [26].

###  F Normal vector and orientation

A unit vector that is normal to $\mathrm{T}_{p}$ is usually denoted by
$\nu(p)$; it is uniquely defined up to sign.

A surface $\Sigma$ is called _oriented_ if it is equipped with a unit normal
vector field $\nu$; that is, a continuous map $p\mapsto\nu(p)$ such that
$\nu(p)\perp\mathrm{T}_{p}$ and $|\nu(p)|=1$ for any $p$. The choice of the
field $\nu$ is called the _orientation_ of $\Sigma$. A surface $\Sigma$ is
called _orientable_ if it can be oriented. Note that each orientable surface
admits two orientations: $\nu$ and $-\nu$.

Let $\Sigma$ be a smooth oriented surface with unit normal field $\nu$. The
map $\nu\colon\Sigma\to\mathbb{S}^{2}$ defined by $p\mapsto\nu(p)$ is called
the _spherical map_ or _Gauss map_.

For surfaces, the spherical map plays essentially the same role as the tangent
indicatrix for curves.

The Möbius strip shown on the diagram gives an example of a nonorientable
surface — there is no choice of normal vector field that is continuous along
the middle of the strip (it changes the sign if you try to go around).

Note that each surface is locally orientable. In fact each chart $s(u,v)$
admits an orientation

$\nu=\frac{s_{u}\times s_{v}}{\left|s_{u}\times s_{v}\right|}.$

Indeed, the vectors $s_{u}$ and $s_{v}$ are tangent vectors at $p$; since they
are linearly independent, their vector product does not vanish and it is
perpendicular to the tangent plane. Evidently $(u,v)\mapsto\nu(u,v)$ is a
continuous map. Therefore $\nu$ is a unit normal field.

8.8. Exercise. Let $h:\mathbb{R}^{3}\to\mathbb{R}$ be a smooth function with
$0$ as a regular value and $\Sigma$ a surface described as a connected
component of the set of solutions $h(x,y,z)=0$. Show that $\Sigma$ is
orientable.

Recall that any proper surface without boundary in the Euclidean space divides
it into two connected components (7D). Therefore we can choose the unit normal
field on any smooth proper surface that points into one of the components of
the complement. Therefore we obtain the following observation.

8.9. Observation. Any smooth proper surface in the Euclidean space is
oriented.

In particular it follows that the Möbius strip cannot be extended to a proper
smooth surface without boundary.

###  G Sections

8.10. Advanced exercise. Let $\Pi$ be the $(x,y)$-plane and $A\subset\Pi$ be
any closed subset. Show that there is an open smooth regular surface $\Sigma$
with $\Sigma\cap\Pi=A$.

The exercise above says that plane sections of a smooth regular surface might
look complicated. The following lemma makes it possible to perturb the plane
so that the section becomes nice.

8.11. Lemma. Let $\Sigma$ be a smooth regular surface. Suppose
$f\colon\mathbb{R}^{3}\to\nobreak\mathbb{R}$ is a smooth function. Then for
any constant $r_{0}$ there is an arbitrarily close value $r$ such that each
connected component of the intersection of the level set $L_{r}=f^{-1}\\{r\\}$
with $\Sigma$ is a smooth regular curve.

Proof. The surface $\Sigma$ can be covered by a countable set of charts
$s_{i}\colon U_{i}\to\nobreak\Sigma$. Note that the composition $f\circ s_{i}$
is a smooth function for any $i$. By Sard’s lemma (0F), almost all real
numbers $r$ are regular values for each $f\circ s_{i}$.

Fix such a value $r$ sufficiently close to $r_{0}$ and consider the level set
$L_{r}$ described by the equation $f(x,y,z)=r$. Any point in the intersection
$\Sigma\cap L_{r}$ lies in the image of one of the charts. From above it
admits a neighborhood which is a regular smooth curve; hence the result.∎

8.12. Corollary. Let $\Sigma$ be a smooth surface. Then for any plane $\Pi$
there is a parallel plane $\Pi^{*}$ that lies arbitrary close to $\Pi$ and
such that the intersection $\Sigma\cap\Pi^{*}$ is a union of disjoint smooth
curves.

##  Chapter 9 Curvatures

###  A Tangent-normal coordinates

Fix a point $p$ in a smooth oriented surface $\Sigma$. Consider a coordinate
system $(x,y,z)$ with origin at $p$ such that the $(x,y)$-plane coincides with
$\mathrm{T}_{p}$ and the $z$-axis points in the direction of the normal vector
$\nu(p)$. By 8A, we can present $\Sigma$ locally at $p$ as a graph $z=f(x,y)$
of a smooth function. Note that

$\displaystyle f(0,0)$ $\displaystyle=0,$ $\displaystyle f_{x}(0,0)$
$\displaystyle=0,$ $\displaystyle f_{y}(0,0)$ $\displaystyle=0.$

The first equality holds since $p=(0,0,0)$ lies on the graph and the other two
equalities mean that the tangent plane at $p$ is horizontal.

Set

$\displaystyle\ell$ $\displaystyle=f_{xx}(0,0),$ $\displaystyle m$
$\displaystyle=f_{xy}(0,0)=f_{yx}(0,0),$ $\displaystyle n$
$\displaystyle=f_{yy}(0,0).$

The _Taylor series_ for $f$ at $(0,0)$ up to the second order term can be then
written as

$f(x,y)=\tfrac{1}{2}(\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y^{2})+o(x^{2}+y^{2}).$

Note that values $\ell$, $m$, and $n$ are completely determined by this
equation. The so-called _osculating paraboloid_

$z=\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}(\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}m{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y+n{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}y^{2})$

has _second order of contact_ with $\Sigma$ at $p$.

Note that

$\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y^{2}=\langle M_{p}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}(\begin{smallmatrix}x\\\
y\end{smallmatrix}),(\begin{smallmatrix}x\\\ y\end{smallmatrix})\rangle,$

where $M_{p}$ is the so-called _Hessian matrix_ of $f$ at $(0,0)$,

$M_{p}=\begin{pmatrix}\ell&m\\\ m&n\end{pmatrix}.$ ➊

###  B Principal curvatures

Note that tangent-normal coordinates give an almost canonical coordinate
system in a neighborhood of $p$; it is unique up to a rotation of the
$(x,y)$-plane. Rotating the $(x,y)$-plane results in rewriting the matrix
$M_{p}$ in the new basis.

Since the Hessian matrix $M_{p}$ is symmetric, by the spectral theorem (0D) it
is diagonalizable by orthogonal matrices. That is, by rotating the
$(x,y)$-plane we can assume that $m=0$ in ➊ ‣ 9A; see 0D. In this case

$M_{p}=\begin{pmatrix}k_{1}&0\\\ 0&k_{2}\end{pmatrix},$

the diagonal components $k_{1}$ and $k_{2}$ of $M_{p}$ are called the
_principal curvatures_ of $\Sigma$ at $p$; they might also be denoted as
$k_{1}(p)$ and $k_{2}(p)$, or $k_{1}(p)_{\Sigma}$ and $k_{2}(p)_{\Sigma}$; if
we need to emphasize that we compute them at the point $p$ for the surface
$\Sigma$. We will always assume that $k_{1}\leqslant k_{2}$.

Note that if $x=f(x,y)$ is a local graph representation of $\Sigma$ in these
coordinates, then

$f(x,y)=\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}(k_{1}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+k_{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}y^{2})+o(x^{2}+y^{2}).$

The principal curvatures can be also defined as the eigenvalues of the Hessian
matrix $M_{p}$. The eigendirections of $M_{p}$ are called the _principal
directions_ of $\Sigma$ at $p$. Note that if $k_{1}(p)\neq k_{2}(p)$, then $p$
has exactly two principal directions, which are perpendicular to each other;
if $k_{1}(p)=\nobreak k_{2}(p)$ then all tangent directions at $p$ are
principal.

Note that if we revert the orientation of $\Sigma$, then the principal
curvatures switch their signs and indexes.

A smooth regular curve on a surface $\Sigma$ that always runs in the principal
directions is called a _line of curvature_ of $\Sigma$.

9.1. Exercise. Assume that a smooth surface $\Sigma$ is mirror symmetric with
respect to a plane $\Pi$. Suppose that $\Sigma$ and $\Pi$ intersect along a
smooth regular curve $\gamma$. Show that $\gamma$ is a line of curvature of
$\Sigma$.

###  C More curvatures

Let $p$ be a point on an oriented smooth surface $\Sigma$.

The product

$K(p)=k_{1}(p){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}k_{2}(p)$

is called the _Gauss curvature_ at $p$. We may denote it by $K$, $K(p)$, or
$K(p)_{\Sigma}$ if we need to specify the point $p$ and the surface $\Sigma$.

Since the determinant is equal to the product of the eigenvalues, we get

$K=\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}n-m^{2},$

where $M_{p}=\nobreak(\begin{smallmatrix}\ell&m\\\ m&n\end{smallmatrix})$ is
the Hessian matrix.

9.2. Exercise. Show that any surface with positive Gauss curvature is
orientable.

The sum

$H(p)=k_{1}(p)+k_{2}(p)$

is called the _mean curvature_ 111Some authors define the mean curvature as
$\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}(k_{1}(p)+k_{2}(p))$ — the
mean value of the principal curvatures. It suits the name better, but it is
not as convenient when it comes to calculations. at $p$. We may also denote it
by $H(p)_{\Sigma}$. The mean curvature can be also interpreted as the trace of
the Hessian matrix $M_{p}=\nobreak(\begin{smallmatrix}\ell&m\\\
m&n\end{smallmatrix})$; that is,

$H=\ell+n$

A surface with vanishing mean curvature is called _minimal_.

Note that reverting the orientation of $\Sigma$ does not change the Gauss
curvature, but changes the sign of the mean curvature. In particular, the
Gauss curvature is well defined for a nonoriented surfaces.

###  D Shape operator

In the following definitions we use the notion of directional derivative
defined in 8B.

Let $p$ be a point on a smooth surface $\Sigma$ with orientation defined by
the unit normal field $\nu$. Given $\text{\sc{w}}\in\mathrm{T}_{p}$, its
_shape operator_ is defined by

$\mathop{\rm Shape}\nolimits_{p}\text{\sc{w}}=-D_{\text{\sc{w}}}\nu.$

Equivalently, the shape operator can be defined by

$\mathop{\rm Shape}\nolimits=-d\nu,$ ➊

where $d\nu$ denotes the differential of the spherical map
$\nu\colon\Sigma\to\mathbb{S}^{2}$; that is,
$d_{p}\nu(\text{\sc{v}})=(D_{\text{\sc{v}}}\nu)(p)$.

Recall that $d_{p}\nu$ is a linear map
$\mathrm{T}_{p}\Sigma\to\mathrm{T}_{\nu(p)}\mathbb{S}^{2}$. Note that
$\mathrm{T}_{p}\Sigma$ coincides with $\mathrm{T}_{\nu(p)}\mathbb{S}^{2}$ —
both of them are normal subspaces to $\nu(p)$. Therefore $\mathop{\rm
Shape}\nolimits_{p}$ is indeed a linear operator
$\mathrm{T}_{p}\to\mathrm{T}_{p}$ (the latter also follows from 9D).

For a point $p\in\Sigma$ the shape operator of a tangent vector
$\text{\sc{w}}\in\mathrm{T}_{p}$ will be denoted by $\mathop{\rm
Shape}\nolimits\text{\sc{w}}$ if it is clear from the context which base point
$p$ and which surface we are working with; otherwise we may use the notations

$\mathop{\rm Shape}\nolimits_{p}(\text{\sc{w}})\quad\text{or}\quad\mathop{\rm
Shape}\nolimits_{p}(\text{\sc{w}})_{\Sigma}$

as the situation requires. 222 The following bilinear forms on a tangent plane
$\displaystyle\mathrm{I}(\text{\sc{v}},\text{\sc{w}})$
$\displaystyle=\langle\text{\sc{v}},\text{\sc{w}}\rangle,$
$\displaystyle\mathrm{II}(\text{\sc{v}},\text{\sc{w}})$
$\displaystyle=\langle\mathop{\rm
Shape}\nolimits\text{\sc{v}},\text{\sc{w}}\rangle,$
$\displaystyle\mathrm{III}(\text{\sc{v}},\text{\sc{w}})$
$\displaystyle=\langle\mathop{\rm Shape}\nolimits\text{\sc{v}},\mathop{\rm
Shape}\nolimits\text{\sc{w}}\rangle$ are called the _first, second, and third
fundamental forms_ respectively. These forms were introduced before the shape
operator, but we will not touch them in the sequel.

9.3. Theorem. Suppose that $(u,v)\mapsto s(u,v)$ is a smooth map to a smooth
surface $\Sigma$ with unit normal field $\nu$. Then

$\displaystyle\langle\mathop{\rm Shape}\nolimits(s_{u}),s_{u}\rangle$
$\displaystyle=\langle s_{uu},\nu\rangle,$ $\displaystyle\langle\mathop{\rm
Shape}\nolimits(s_{v}),s_{u}\rangle$ $\displaystyle=\langle
s_{uv},\nu\rangle,$ $\displaystyle\langle\mathop{\rm
Shape}\nolimits(s_{u}),s_{v}\rangle$ $\displaystyle=\langle
s_{uv},\nu\rangle,$ $\displaystyle\langle\mathop{\rm
Shape}\nolimits(s_{v}),s_{v}\rangle$ $\displaystyle=\langle
s_{vv},\nu\rangle,$ $\displaystyle\langle\mathop{\rm
Shape}\nolimits(s_{u}),\nu\rangle$ $\displaystyle=0,$
$\displaystyle\langle\mathop{\rm Shape}\nolimits(s_{v}),\nu\rangle$
$\displaystyle=0$

for any $(u,v)$.

Proof. We will use the shortcut $\nu=\nu(u,v)$ for $\nu(s(u,v))$, so

$\displaystyle\mathop{\rm Shape}\nolimits(s_{u})$
$\displaystyle=-D_{s_{u}}\nu=-\nu_{u},$ $\displaystyle\mathop{\rm
Shape}\nolimits(s_{v})$ $\displaystyle=-D_{s_{v}}\nu=-\nu_{v}.$ ➋

Note that $\nu$ is a unit vector orthogonal to $s_{u}$ and $s_{v}$; therefore

$\displaystyle\langle\nu,s_{u}\rangle$ $\displaystyle\equiv 0,$
$\displaystyle\langle\nu,s_{v}\rangle$ $\displaystyle\equiv 0,$
$\displaystyle\langle\nu,\nu\rangle$ $\displaystyle\equiv 1.$

Taking partial derivatives of these two identities we get

$\displaystyle\langle\nu_{u},s_{u}\rangle+\langle\nu,s_{uu}\rangle$
$\displaystyle=0,$
$\displaystyle\langle\nu_{v},s_{u}\rangle+\langle\nu,s_{uv}\rangle$
$\displaystyle=0,$
$\displaystyle\langle\nu_{u},s_{v}\rangle+\langle\nu,s_{uv}\rangle$
$\displaystyle=0,$
$\displaystyle\langle\nu_{v},s_{v}\rangle+\langle\nu,s_{vv}\rangle$
$\displaystyle=0,$ $\displaystyle 2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\nu_{u},\nu\rangle$ $\displaystyle=0,$ $\displaystyle 2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle\nu_{v},\nu\rangle$ $\displaystyle=0.$

It remains to plug in the expressions from ➋ ‣ 9D. ∎

9.4. Exercise. Show that the shape operator is _self-adjoint_ ; that is,

$\langle\mathop{\rm
Shape}\nolimits\text{\sc{u}},\text{\sc{v}}\rangle=\langle\text{\sc{u}},\mathop{\rm
Shape}\nolimits\text{\sc{v}}\rangle$

for any $\text{\sc{u}},\text{\sc{v}}\in\mathrm{T}_{p}$.

Let us denote by i, j and k the standard basis in the $\mathbb{R}^{3}$. Recall
that the components $\ell$, $m$, and $n$ of the Hessian matrix are defined in
Section 9A.

9.5. Corollary. Let $z=f(x,y)$ be a local representation of a smooth surface
$\Sigma$ in the tangent-normal coordinates at $p$. Suppose that its the
Hessian matrix at $p$ is $(\begin{smallmatrix}\ell&m\\\
m&n\end{smallmatrix})$. Then

$\displaystyle\mathop{\rm Shape}\nolimits\text{\sc{i}}$
$\displaystyle=\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{i}}+m{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{j}},$
$\displaystyle\mathop{\rm Shape}\nolimits\text{\sc{j}}$
$\displaystyle=m{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{i}}+n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{j}};$

that is, the multiplication by the Hessian matrix at $p$ describes its shape
operator.

In particular, the principal curvatures of $\Sigma$ at $p$ are the eigenvalues
of $\mathop{\rm Shape}\nolimits_{p}$ and the principal directions are the
eigendirections of $\mathop{\rm Shape}\nolimits_{p}$.

Since the Hessian matrix is symmetric, the corollary also implies that
$\mathop{\rm Shape}\nolimits$ is self-adjoint which gives another way to solve
9D.

Proof. Note that $s\colon(u,v)\mapsto(u,v,f(u,v))$ is a chart of $\Sigma$ that
covers $p$. Further note that

$\displaystyle s_{u}(0,0)$ $\displaystyle=\text{\sc{i}},$ $\displaystyle
s_{v}(0,0)$ $\displaystyle=\text{\sc{j}},$ $\displaystyle\nu(0,0)$
$\displaystyle=\text{\sc{k}},$ $\displaystyle s_{uu}(0,0)$
$\displaystyle=\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{k}},$
$\displaystyle s_{uv}(0,0)$ $\displaystyle=m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{k}},$ $\displaystyle s_{vv}(0,0)$ $\displaystyle=n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{k}}.$

It remains to apply 9D. ∎

9.6. Corollary. Let $\Sigma$ be a smooth surface with orientation defined by a
unit normal field $\nu$. Suppose the spherical map
$\nu\colon\Sigma\to\mathbb{S}^{2}$ is injective. Then

$\iint_{\Sigma}|K|=\mathop{\rm area}\nolimits[\nu(\Sigma)].$

Proof. Observe that the tangent planes
$\mathrm{T}_{p}\Sigma=\mathrm{T}_{\nu(p)}\mathbb{S}^{2}$ are parallel for any
$p\in\Sigma$. Indeed both of these planes are perpendicular to $\nu(p)$.

Choose an orthonormal basis of $\mathrm{T}_{p}$ consisting of principal
directions, so the shape operator can be expressed by the matrix
$(\begin{smallmatrix}k_{1}&0\\\ 0&k_{2}\end{smallmatrix})$.

Since $\mathop{\rm Shape}\nolimits_{p}=-d_{p}\nu$, 9D implies that

$\mathop{\rm jac}\nolimits_{p}\nu=|\det(\begin{smallmatrix}k_{1}&0\\\
0&k_{2}\end{smallmatrix})|=|K(p)|.$

By the area formula (8E), the statement follows. ∎

9.7. Exercise. Let $\Sigma$ be a smooth surface with orientation defined by a
unit normal field $\nu$. Suppose that $\Sigma$ has unit principal curvatures
at any point.

  * (a)

Show that $\mathop{\rm Shape}\nolimits_{p}(\text{\sc{w}})=\text{\sc{w}}$ for
any $p\in\Sigma$ and $\text{\sc{w}}\in\mathrm{T}_{p}\Sigma$.

  * (b)

Show that $p+\nu(p)$ is constant; that is, the point $c=p+\nu(p)$ does not
depend on $p\in\Sigma$. Conclude that $\Sigma$ is a subset of the unit sphere
centered at $c$.

We define the _angle_ between two oriented surfaces at a point of their
intersection $p$ as the angle between their normal vectors at $p$. The
following exercise is a result by Ferdinand Joachimsthal [44] generalized by
Ossian Bonnet [12].

9.8. Exercise. Assume that two smooth oriented surfaces $\Sigma_{1}$ and
$\Sigma_{2}$ intersect at constant angle along a smooth regular curve
$\gamma$. Show that if $\gamma$ is a curvature line in $\Sigma_{1}$, then it
is also a curvature line in $\Sigma_{2}$.

Conclude that if a smooth surface $\Sigma$ intersects a plane or sphere at
constant angle along a smooth regular curve $\gamma$, then $\gamma$ is a
curvature line of $\Sigma$.

9.9. Exercise. Let $\Sigma$ be a closed smooth surface with orientation
defined by a unit normal field $\nu$. Assume $t$ is sufficiently close to
zero.

  * (a)

Show that the set of points of the form $p+t{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\nu(p)$ with $p\in\Sigma$ forms a smooth surface; denote it by
$\Sigma_{t}$.

  * (b)

Let $a(t)$ be the area of $\Sigma_{t}$. Show that

$a^{\prime}(0)=-\iint_{\Sigma}H,$

where $H$ denotes mean curvature of $\Sigma$.

###  E Curves in a surface

t$\nu$$\mu$

Suppose $\gamma$ is a regular smooth curve in a smooth oriented surface
$\Sigma$. As usual we denote by $\nu$ the unit normal field on $\Sigma$.

Without loss of generality we may assume that $\gamma$ is unit-speed; in this
case $\text{\sc{t}}(s)=\gamma^{\prime}(s)$ is its tangent indicatrix. Let us
use the shortcut notation $\nu(s)=\nobreak\nu(\gamma(s))$. Note that the unit
vectors $\text{\sc{t}}(s)$ and $\nu(s)$ are orthogonal. Тherefore there is a
unique unit vector $\mu(s)$ such that $\text{\sc{t}}(s),\mu(s),\nu(s)$ is an
oriented orthonormal basis; it is called the _Darboux frame_ of $\gamma$ in
$\Sigma$. Since $\mathrm{T}_{\gamma(s)}\perp\nobreak\nu(s)$, the vector
$\mu(s)$ is tangent to $\Sigma$ at $\gamma(s)$. In fact $\mu(s)$ is a
counterclockwise rotation of $\text{\sc{t}}(s)$ by the angle $\tfrac{\pi}{2}$
in the tangent plane $\mathrm{T}_{\gamma(s)}$. This vector can also be defined
as the vector product $\mu(s)=\nu(s)\times\text{\sc{t}}(s)$.

Since $\gamma$ is unit-speed, we have that
$\gamma^{\prime\prime}\perp\gamma^{\prime}$ (see 3A). Therefore the
acceleration of $\gamma$ can be written as a linear combination of $\mu$ and
$\nu$; that is,

$\gamma^{\prime\prime}(s)=k_{g}(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\mu(s)+k_{n}(s){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\nu(s).$

The values $k_{g}(s)$ and $k_{n}(s)$ are called _geodesic_ and _normal
curvatures_ of $\gamma$ at $s$, respectively. Since the frame
$\text{\sc{t}}(s),\mu(s),\nu(s)$ is orthonormal, these curvatures can be also
written as the following scalar products:

$\displaystyle k_{g}(s)$
$\displaystyle=\langle\gamma^{\prime\prime}(s),\mu(s)\rangle=$
$\displaystyle=\langle\text{\sc{t}}^{\prime}(s),\mu(s)\rangle.$ $\displaystyle
k_{n}(s)$ $\displaystyle=\langle\gamma^{\prime\prime}(s),\nu(s)\rangle=$
$\displaystyle=\langle\text{\sc{t}}^{\prime}(s),\nu(s)\rangle.$

Since $0=\langle\text{\sc{t}}(s),\nu(s)\rangle$ we have that

$\displaystyle 0$ $\displaystyle=\langle\text{\sc{t}}(s),\nu(s)\rangle=$
$\displaystyle=\langle\text{\sc{t}}^{\prime}(s),\nu(s)\rangle+\langle\text{\sc{t}}(s),\nu^{\prime}(s)\rangle=$
$\displaystyle=k_{n}(s)+\langle\text{\sc{t}}(s),D_{\text{\sc{t}}(s)}\nu\rangle.$

Applying the definition of shape operator, we get the following:

9.10. Proposition. Assume $\gamma$ is a smooth unit-speed curve in a smooth
surface $\Sigma$. Let $p=\gamma(s_{0})$ and
$\text{\sc{v}}=\gamma^{\prime}(s_{0})$. Then

$k_{n}(s_{0})=\langle\mathop{\rm
Shape}\nolimits_{p}(\text{\sc{v}}),\text{\sc{v}}\rangle,$

where $k_{n}$ denotes the normal curvature of $\gamma$ at $s_{0}$ and
$\mathop{\rm Shape}\nolimits_{p}$ is the shape operator at $p$.

Note that according to the proposition, the normal curvature of a regular
smooth curve in $\Sigma$ is completely determined by the velocity vector v at
the point $p$. By that reason the normal curvature is also denoted by
$k_{\text{\sc{v}}}$; according to the proposition,

$k_{\text{\sc{v}}}=\langle\mathop{\rm
Shape}\nolimits_{p}(\text{\sc{v}}),\text{\sc{v}}\rangle$

for any unit vector v in $\mathrm{T}_{p}$.

Let $p$ be a point on a smooth surface $\Sigma$. Assume we choose tangent-
normal coordinates at $p$ so that the Hessian matrix is diagonalized, then we
have

$M_{p}=\begin{pmatrix}k_{1}(p)&0\\\ 0&k_{2}(p)\end{pmatrix}.$

Consider a vector ${\text{\sc{w}}}=a{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{i}}+b{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{j}}$ in
the $(x,y)$-plane. Then by 9D, we have

$\langle\mathop{\rm
Shape}\nolimits\text{\sc{w}},\text{\sc{w}}\rangle=a^{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}k_{1}(p)+b^{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k_{2}(p).$

If w is unit, then $a^{2}+b^{2}=1$ which implies the following:

9.11. Observation. For any point $p$ on an oriented smooth surface $\Sigma$,
the principal curvatures $k_{1}(p)$ and $k_{2}(p)$ are respectively the
minimum and maximum of the normal curvatures at $p$. Moreover, if $\theta$ is
the angle between a unit vector ${\text{\sc{w}}}\in\mathrm{T}_{p}$ and the
first principal direction at $p$, then

$k_{\text{\sc{w}}}(p)=k_{1}(p){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(\cos\theta)^{2}+k_{2}(p){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(\sin\theta)^{2}.$

The last identity is called _Euler’s formula_.

9.12. Exercise. Let $\Sigma$ be a smooth surface. Show that the sum of the
normal curvatures for any pair of orthogonal directions, at a point
$p\in\Sigma$ equals $H(p)$ — the mean curvature at $p$.

9.13. Meusnier’s theorem. Let $\gamma$ be a regular smooth curve that runs
along a smooth oriented surface $\Sigma$. Let $p=\gamma(t_{0})$,
${\text{\sc{v}}}=\nobreak\gamma^{\prime}(t_{0})$, and
$\alpha=\nobreak\measuredangle(\nu(p),\text{\sc{n}}(t_{0}))$; that is,
$\alpha$ is the angle between the unit normal to $\Sigma$ at $p$ and the unit
normal vector in the Frenet frame of $\gamma$ at $t_{0}$. Then the following
identity holds true

$\kappa(t_{0}){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\cos\alpha=k_{n}(t_{0});$

here $\kappa(t_{0})$ and $k_{n}(t_{0})$ are the curvature and the normal
curvature of $\gamma$ at $t_{0}$, respectively.

Proof. Since $\gamma^{\prime\prime}=\text{\sc{t}}^{\prime}=\kappa{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}$, we get that

$\displaystyle k_{n}(t_{0})$
$\displaystyle=\langle\gamma^{\prime\prime},\nu\rangle=$
$\displaystyle=\kappa(t_{0}){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{n}},\nu\rangle=$ $\displaystyle=\kappa(t_{0}){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos\alpha.$

∎

The theorem above, as well as the statement in the following exercise were
proved by Jean Baptiste Meusnier [59].

9.14. Exercise. Let $\Sigma$ be a smooth surface, $p\in\Sigma$ and
${\text{\sc{v}}}\in\mathrm{T}_{p}\Sigma$ a unit vector. Assume that
$k_{\text{\sc{v}}}(p)\neq 0$; that is, the normal curvature of $\Sigma$ at $p$
in the direction of v does not vanish.

Show that the osculating circles at $p$ of smooth regular curves in $\Sigma$
that run in the direction v sweep out a sphere $S$ with center
$p+\tfrac{1}{k_{\text{\sc{v}}}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\nu$ and
radius $r=\tfrac{1}{|k_{\text{\sc{v}}}|}$.

9.15. Exercise. Let $\gamma(s)=(x(s),y(s))$ be a smooth unit-speed simple
plane curve in the upper half-plane, and $\Sigma$ be the surface of revolution
around the $x$-axis with generatrix $\gamma$.

  * (a)

Show that the parallels and meridians are lines of curvature on $\Sigma$.

  * (b)

Show that

$\frac{|x^{\prime}(s)|}{y(s)}\quad\text{and}\quad\frac{-y^{\prime\prime}(s)}{|x^{\prime}(s)|}$

are the principal curvatures of $\Sigma$ at $(x(s),y(s),0)$ in the direction
of the corresponding parallel and meridian respectively.

  * (c)

Show that $\Sigma$ has Gauss curvature $-1$ at all points if and only if $y$
satisfies the differential equation $y^{\prime\prime}=y$. The case $y=e^{-s}$
is shown; this is the so-called _pseudosphere_.

9.16. Exercise. Show that the _catenoid_ defined implicitly by the equation

$(\operatorname{ch}z)^{2}=x^{2}+y^{2}$

is a minimal surface.

9.17. Exercise. Show that the _helicoid_ defined by the following parametric
equation

$s(u,v)=(u{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin v,u{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos v,v)$

is a minimal surface.

###  F Lagunov’s example

9.18. Exercise. Assume $V$ is a body in $\mathbb{R}^{3}$ bounded by a smooth
surface of revolution with principal curvatures at most 1 in absolute value.
Show that $V$ contains a unit ball.

The following question is a 3-dimensional analog of the moon in a puddle
problem (6E).

9.19. Question. Assume a set $V\subset\mathbb{R}^{3}$ is bounded by a closed
connected surface $\Sigma$ with principal curvatures bounded in absolute value
by 1. Is it true that $V$ contains a ball of radius 1?

According to 9F, the answer is ‘‘yes’’ for surfaces of revolution. Later (see
10B) we will show that the answer is ‘‘yes’’ for convex surfaces. Now we are
going to show by an example that the answer is ‘‘no’’ in the general case;
this example was constructed by Vladimir Lagunov [49].

Construction. Let us start with a body of revolution $V_{1}$ whose cross
section is shown on the diagram below. The boundary curve of the cross section
consists of 6 long vertical line segments included into 3 closed simple smooth
curves. (To make the curves smooth, one has to use cutoffs and mollifiers from
Section 0E.) The boundary of $V_{1}$ has 3 components, each of which is a
smooth sphere.

We assume that the curves have curvature at most 1. Moreover with the
exception of the almost vertical parts, the curve has absolute curvature close
to 1 all the time. The only thick part in $V$ is the place where all three
boundary components come close together; the remaining part of $V$ is assumed
to be very thin. It could be arranged that the radius $r$ of the maximal ball
in $V$ is just a little bit above $r_{2}=\tfrac{2}{\sqrt{3}}-1$. (The small
black disc on the diagram has radius $r_{2}$, assuming that the three big
circles are unit.) In particular, we may assume that $r<\tfrac{1}{6}$.

Exercise 9E gives formulas for the principal curvatures of the boundary of
$V$; which imply that both principal curvatures are at most 1 in absolute
value.

It remains to modify $V_{1}$ to make its boundary connected without increasing
the bounds on its principal curvatures and without allowing larger balls
inside.

Note that each sphere in the boundary contains two flat discs; they come into
pairs closely lying to each other. Let us drill thru two of such pairs and
reconnect the holes by another body of revolution whose axis is shifted but
stays parallel to the axis of $V_{1}$. Denote the obtained body by $V_{2}$;
the cross section of the obtained body is shown on the diagram.

Then repeat the operation for the other two pairs. Denote the obtained body by
$V_{3}$.

Note that the boundary of $V_{3}$ is connected. Assuming that the holes are
large, its boundary can be made so that its principal curvatures are still at
most $1$; the latter can be proved in the same way as for $V_{1}$. ∎

### Remarks

Note that the surface of $V_{3}$ in the Lagunov’s example has genus 2; that
is, it can be parametrized by a sphere with two handles.

Indeed, the boundary of $V_{1}$ consists of three smooth spheres.

When we drill a hole, we make one hole in two spheres and two holes in one
sphere. We reconnect two spheres with a tube and obtain one sphere. By
connecting the two holes of the other sphere with a tube we get a torus; it is
on the right side in the picture of $V_{2}$. That is, the boundary of $V_{2}$
is formed by one sphere and one torus.

To construct $V_{3}$ from $V_{2}$, we make a torus from the remaining sphere
and connect it to the other torus by a tube. This way we get a sphere with two
handles; that is, it has genus 2.

9.20. Exercise. Modify Lagunov’s construction to make the boundary surface a
sphere with 4 handles.

Question 9F can be asked differently: what is the maximal radius $r$ of the
ball that has to be included in any body bounded by a smooth surface with
principal curvatures bounded in absolute value by 1.

One may also consider bodies bounded by more than one smooth surface. In this
case the example of a region between two large concentric spheres with almost
equal radii shows that in the general case there is no bound. Indeed, this
region can be made arbitrarily thin while the curvature of the boundary can be
made arbitrarily close to zero.

Recall that the Lagunov example shows that $r\leqslant r_{2}$, where $r_{2}$
is the radius of the smallest circle tangent to three unit circles that are
tangent to each other, so

$r_{2}=\tfrac{2}{\sqrt{3}}-1<\tfrac{1}{6}.$

The statement in the following exercise is due to Vladimir Lagunov [48]; it
implies that this bound is optimal.

9.21. Advanced exercise. Suppose a connected body $V\subset\mathbb{R}^{3}$ is
bounded by a finite number of closed smooth surfaces with principal curvatures
bounded in absolute value by 1. Assume that $V$ does not contain a ball of
radius $r_{2}$. Show that its boundary has two diffeomorphic connected
components.

Let $r_{3}$ be the radius of the smallest sphere tangent to four unit spheres
that are tangent to each other. Direct calculations show that

$r_{3}=\sqrt{\tfrac{3}{2}}-1>\tfrac{1}{5}.$

In particular $r_{3}>r_{2}$.

The statement in the following exercise is a special case of a theorem by
Vladimir Lagunov and Abram Fet [50, 51].

9.22. Very advanced exercise. Suppose a body $V\subset\mathbb{R}^{3}$ is
bounded by a smooth sphere with principal curvatures bounded in absolute value
by 1. Show that $V$ contains a ball of radius $r_{3}$.

Show that this bound is sharp; that is, show that for every $\varepsilon>0$,
there are examples of $V$ as above not containing a ball of radius
$r_{3}+\varepsilon$.

##  Chapter 10 Supporting surfaces

###  A Definitions

Assume two surfaces $\Sigma_{1}$ and $\Sigma_{2}$ have a common point $p$. If
there is a neighborhood $U$ of $p$ such that $\Sigma_{1}\cap U$ lies on one
side from $\Sigma_{2}$ in $U$, then we say that $\Sigma_{2}$ _locally
supports_ $\Sigma_{1}$ at $p$.

Let us describe $\Sigma_{2}$ locally at $p$ as a graph $z=f_{2}(x,y)$ in
tangent-normal coordinates at $p$. If $\Sigma_{2}$ locally supports
$\Sigma_{1}$ at $p$, then all points of $\Sigma_{1}$ near $p$ lie either above
or below the graph $z=f_{2}(x,y)$.

In both cases the surfaces $\Sigma_{1}$ and $\Sigma_{2}$ have common tangent
plane at $p$, so we can write both as graphs $z=f_{1}(x,y)$ and $z=f_{2}(x,y)$
in the common tangent-normal coordinates at $p$. Note that $\Sigma_{2}$
locally supports $\Sigma_{1}$ at $p$ if and only if

$f_{1}(x,y)\geqslant f_{2}(x,y)\quad\text{or}\quad f_{1}(x,y)\leqslant
f_{2}(x,y)$

for all $(x,y)$ sufficiently close to the origin.

If the surfaces are orientable, we can assume that they are _cooriented_ at
$p$; that is, they have a common unit normal vector at $p$ in the direction of
the $z$-axis. If the normal vectors are opposite, we say that $\Sigma_{1}$ and
$\Sigma_{2}$ are _counteroriented_ at $p$; in this case reverting the
orientation of one of the surfaces makes them cooriented.

If $\Sigma_{2}$ locally supports $\Sigma_{1}$ and they are cooriented at $p$,
then we can say that $\Sigma_{1}$ supports $\Sigma_{2}$ _from inside_ or _from
outside_ , assuming that the normal vector points _inside_ the domain bounded
by the surface $\Sigma_{2}$ in $U$. Using the above notations, $\Sigma_{1}$
locally supports $\Sigma_{2}$ from inside (from outside) if
$f_{1}(x,y)\geqslant f_{2}(x,y)$ (respectively $f_{1}(x,y)\leqslant
f_{2}(x,y)$) for $(x,y)$ in a sufficiently small neighborhood of the origin.

10.1. Proposition. Let $\Sigma_{1}$ and $\Sigma_{2}$ be oriented surfaces.
Assume $\Sigma_{1}$ locally supports $\Sigma_{2}$ from inside at the point $p$
(equivalently $\Sigma_{2}$ locally supports $\Sigma_{1}$ from outside). Then

$k_{1}(p)_{\Sigma_{1}}\geqslant k_{1}(p)_{\Sigma_{2}}\quad\text{and}\quad
k_{2}(p)_{\Sigma_{1}}\geqslant\nobreak k_{2}(p)_{\Sigma_{2}}.$

10.2. Exercise. Give an example of two surfaces $\Sigma_{1}$ and $\Sigma_{2}$
that have a common point $p$ with a common unit normal vector $\nu_{p}$ such
that $k_{1}(p)_{\Sigma_{1}}>\nobreak k_{1}(p)_{\Sigma_{2}}$ and
$k_{2}(p)_{\Sigma_{1}}>\nobreak k_{2}(p)_{\Sigma_{2}}$, but $\Sigma_{1}$ does
not support $\Sigma_{2}$ locally at $p$.

Proof. We can assume that $\Sigma_{1}$ and $\Sigma_{2}$ are graphs
$z=f_{1}(x,y)$ and $z=f_{2}(x,y)$ in common tangent-normal coordinates at $p$,
so we have $f_{1}\geqslant f_{2}$.

Fix a unit vector
${\text{\sc{w}}}\in\mathrm{T}_{p}\Sigma_{1}=\mathrm{T}_{p}\Sigma_{2}$.
Consider the plane $\Pi$ passing thru $p$ and spanned by the normal vector
$\nu_{p}$ and w. Let $\gamma_{1}$ and $\gamma_{2}$ be the curves of
intersection of $\Sigma_{1}$ and $\Sigma_{2}$ with $\Pi$.

Let us orient $\Pi$ so that the common normal vector $\nu_{p}$ for both
surfaces at $p$ points to the left from w. Further, let us parametrize both
curves so that they are running in the direction of w at $p$ and are therefore
cooriented. Note that in this case the curve $\gamma_{1}$ supports the curve
$\gamma_{2}$ from the right.

By 6C, we have the following inequality for the normal curvatures of
$\Sigma_{1}$ and $\Sigma_{2}$ at $p$ in the direction of w:

$k_{\text{\sc{w}}}(p)_{\Sigma_{1}}\geqslant
k_{\text{\sc{w}}}(p)_{\Sigma_{2}}.$ ➊

According to 9E,

$k_{1}(p)_{\Sigma_{i}}=\min\left\\{\,{k_{\text{\sc{w}}}(p)_{\Sigma_{i}}}\,:\,{{\text{\sc{w}}}\in\mathrm{T}_{p},|{\text{\sc{w}}}|=1}\,\right\\}$

for $i=1,2$. Choose w so that
$k_{1}(p)_{\Sigma_{1}}=k_{\text{\sc{w}}}(p)_{\Sigma_{1}}$. Then by ➊ ‣ 10A, we
have that

$\displaystyle k_{1}(p)_{\Sigma_{1}}$
$\displaystyle=k_{\text{\sc{w}}}(p)_{\Sigma_{1}}\geqslant$
$\displaystyle\geqslant k_{\text{\sc{w}}}(p)_{\Sigma_{2}}\geqslant$
$\displaystyle\geqslant\min_{\text{\sc{v}}}\left\\{\,{k_{\text{\sc{v}}}(p)_{\Sigma_{2}}}\,\right\\}=$
$\displaystyle=k_{1}(p)_{\Sigma_{2}};$

here we assume that $\text{\sc{v}}\in\mathrm{T}_{p}$ and $|\text{\sc{v}}|=1$.
That is, $k_{1}(p)_{\Sigma_{1}}\geqslant k_{1}(p)_{\Sigma_{2}}$.

Similarly, by 9E, we have that

$k_{2}(p)_{\Sigma_{i}}=\max_{\text{\sc{w}}}\left\\{\,{k_{\text{\sc{w}}}(p)_{\Sigma_{i}}}\,\right\\}.$

Let us fix w so that
$k_{2}(p)_{\Sigma_{2}}=k_{\text{\sc{w}}}(p)_{\Sigma_{2}}$. Then

$\displaystyle k_{2}(p)_{\Sigma_{2}}$
$\displaystyle=k_{\text{\sc{w}}}(p)_{\Sigma_{2}}\leqslant$
$\displaystyle\leqslant k_{\text{\sc{w}}}(p)_{\Sigma_{1}}\leqslant$
$\displaystyle\leqslant\max_{\text{\sc{v}}}\left\\{\,{k_{\text{\sc{v}}}(p)_{\Sigma_{1}}}\,\right\\}=$
$\displaystyle=k_{2}(p)_{\Sigma_{1}};$

that is, $k_{2}(p)_{\Sigma_{1}}\geqslant k_{2}(p)_{\Sigma_{2}}$. ∎

10.3. Corollary. Let $\Sigma_{1}$ and $\Sigma_{2}$ be oriented surfaces.
Assume $\Sigma_{1}$ locally supports $\Sigma_{2}$ from inside at the point
$p$. Then

  * (a)

$H(p)_{\Sigma_{1}}\geqslant H(p)_{\Sigma_{2}}$;

  * (b)

If $k_{1}(p)_{\Sigma_{2}}\geqslant 0$, then $K(p)_{\Sigma_{1}}\geqslant
K(p)_{\Sigma_{2}}$.

Proof; 10A The statement follows from 10A and the definition of mean curvature

$H(p)_{\Sigma_{i}}=k_{1}(p)_{\Sigma_{i}}+k_{2}(p)_{\Sigma_{i}}.$

10A. Since $k_{2}(p)_{\Sigma_{i}}\geqslant k_{1}(p)_{\Sigma_{i}}$ and
$k_{1}(p)_{\Sigma_{2}}\geqslant 0$, we get that all the principal curvatures
$k_{1}(p)_{\Sigma_{1}}$, $k_{1}(p)_{\Sigma_{2}}$, $k_{2}(p)_{\Sigma_{1}}$, and
$k_{2}(p)_{\Sigma_{2}}$ are nonnegative. By 10A, it implies that

$\displaystyle K(p)_{\Sigma_{1}}$ $\displaystyle=k_{1}(p)_{\Sigma_{1}}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}k_{2}(p)_{\Sigma_{1}}\geqslant$
$\displaystyle\geqslant k_{1}(p)_{\Sigma_{2}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k_{2}(p)_{\Sigma_{2}}=$ $\displaystyle=K(p)_{\Sigma_{2}}.$

∎

10.4. Exercise. Show that any closed surface in a unit ball has a point with
Gauss curvature at least 1.

10.5. Exercise. Show that any closed surface that lies at distance at most 1
from a straight line has a point with Gauss curvature at least 1.

###  B Convex surfaces

A proper surface without boundary that bounds a convex region is called
_convex_.

10.6. Exercise. Show that the Gauss curvature of any convex smooth surface is
nonnegative at each point.

10.7. Exercise. Assume $R$ is a convex body in $\mathbb{R}^{3}$ bounded by a
surface with principal curvatures at most 1. Show that $R$ contains a unit
ball.

Recall that a region $R$ in the Euclidean space is called _strictly convex_ if
for any two points $x,y\in R$, any point $z$ between $x$ and $y$ lies in the
interior of $R$.

Clearly any open convex set is strictly convex; the cube (as well as any
convex polyhedron) gives an example of a non-strictly convex set. Recall that
a closed convex region is strictly convex if and only if its boundary does not
contain a line segment.

10.8. Lemma. Let $z=f(x,y)$ be the local description of a smooth surface
$\Sigma$ in tangent-normal coordinates at some point $p\in\Sigma$. Assume both
principal curvatures of $\Sigma$ are positive at $p$. Then the function $f$ is
strictly convex in a neighborhood of the origin and has a local minimum at the
origin.

In particular the tangent plane $\mathrm{T}_{p}$ locally supports $\Sigma$
from outside at $p$.

Proof. Since both principal curvatures are positive, by 9D, we have

$D^{2}_{\text{\sc{w}}}f(0,0)=\langle\mathop{\rm
Shape}\nolimits_{p}({\text{\sc{w}}}),{\text{\sc{w}}}\rangle\geqslant
k_{1}(p)>0$

for any unit tangent vector ${\text{\sc{w}}}\in\mathrm{T}_{p}\Sigma$ (which is
the $(x,y)$-plane).

Since the set of unit vectors is compact, we have that

$D^{2}_{\text{\sc{w}}}f(0,0)>\varepsilon$

for some fixed $\varepsilon>0$ and any unit tangent vector
${\text{\sc{w}}}\in\mathrm{T}_{p}\Sigma$.

By continuity of the function $(x,y,{\text{\sc{w}}})\mapsto
D^{2}_{\text{\sc{w}}}f(x,y)$, we have that $D^{2}_{\text{\sc{w}}}f(x,y)>0$ if
$\text{\sc{w}}\neq 0$ and $(x,y)$ lies in a sufficiently small neighborhood of
the origin. This property implies that $f$ is a strictly convex function in a
neighborhood of the origin in the $(x,y)$-plane (see Section 0E).

Finally since $\nabla f(0,0)=0$ and $f$ is strictly convex in a neighborhood
of the origin it has a strict local minimum at the origin. ∎

10.9. Exercise. Let $\Sigma$ be a smooth surface (without boundary) with
positive Gauss curvature. Show that any connected component of intersection of
$\Sigma$ with a plane $\Pi$ is either a single point or a smooth regular plane
curve whose signed curvature has constant sign.

The following theorem gives a global description of surfaces with positive
Gauss curvature.

10.10. Theorem. Suppose $\Sigma$ is a proper smooth surface with positive
Gauss curvature. Then $\Sigma$ bounds a strictly convex region.

In fact the statement holds for surfaces with possible self-intersections; it
was stated and proved by James Stoker [73] who attributed it to Jacques
Hadamard, who proved a closely relevant statement in [38, item 23].

Note that in the proof we have to use that the surface is a connected set;
otherwise a pair of disjoint spheres which bound two disjoint balls would give
a counterexample.

Proof. Since the Gauss curvature is positive, we can choose unit normal field
$\nu$ on $\Sigma$ so that the principal curvatures are positive at any point.
Denote by $R$ the region bounded by $\Sigma$ that lies on the side of $\nu$;
that is, $\nu$ points inside of $R$ at any point of $\Sigma$. (The region $R$
exists by 7D.)

Let us show that $R$ is _locally strictly convex_ ; that is, for any point
$p\in R$, the intersection of $R$ with a small ball centered at $p$ is
strictly convex.

Indeed, suppose that $z=f(x,y)$ is a local description of $\Sigma$ in the
tangent-normal coordinates at $p$. By 10B, $f$ is strictly convex in a
neighborhood of the origin. In particular the intersection of a small ball
centered at $p$ with the epigraph $z\geqslant f(x,y)$ is strictly convex.

Since $\Sigma$ is connected, so is $R$; moreover any two points in the
interior of $R$ can be connected by a polygonal line in the interior of $R$.

Assume the interior of $R$ is not convex; that is, there are points $x,y\in R$
and a point $z$ between $x$ and $y$ that does not lie in the interior of $R$.
Consider a polygonal line $\beta$ from $x$ to $y$ in the interior of $R$. Let
$y_{0}$ be the first point on $\beta$ such that the chord $[x,y_{0}]$ touches
$\Sigma$ at some point, say $z_{0}$.

Since $R$ is locally strictly convex, $R\cap\nobreak B(z_{0},\varepsilon)$ is
strictly convex for all sufficiently small $\varepsilon>0$. On the other hand
$z_{0}$ lies between two points in the intersection $[x,y_{0}]\cap
B(z_{0},\varepsilon)$. Since $[x,y_{0}]\subset R$, we arrived to a
contradiction.

Therefore the interior of $R$ is a convex set. Note that the region $R$ is the
closure of its interior, therefore $R$ is convex as well.

Since $R$ is locally strictly convex, its boundary $\Sigma$ contains no line
segments. Therefore $R$ is strictly convex. ∎

Remark. We proved a more general statement. Namely, _any closed connected
locally convex region in the Euclidean space is convex_.

10.11. Exercise. Assume that a closed convex surface $\Sigma$ surrounds a unit
circle. Show that there is a point $p\in\Sigma$ with $K(p)\leqslant 1$.

10.12. Exercise. Let $\Sigma$ be a closed convex smooth surface of diameter at
least $\pi$; that is, there is a pair of points $p,q\in\Sigma$ such that
$|p-q|\geqslant\pi$. Show that $\Sigma$ has a point with Gauss curvature at
most 1.

10.13. Theorem. Suppose $\Sigma$ is a closed smooth convex surface. Then it is
a smooth sphere; that is, $\Sigma$ admits a smooth regular parametrization by
$\mathbb{S}^{2}$.

The following exercise will guide you thru the proof of the theorem.

10.14. Exercise. Assume a convex compact region $R$ contains the origin in its
interior and is bounded by a smooth surface $\Sigma$.

  * (a)

Show that any half-line that starts at the origin intersects $\Sigma$ at a
single point; that is, there is a positive function
$\rho\colon\mathbb{S}^{2}\to\nobreak\mathbb{R}$ such that $\Sigma$ consists of
the points $q=\nobreak\rho(\xi){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\xi$ for
$\xi\in\mathbb{S}^{2}$.

  * (b)

Show that $\rho\colon\mathbb{S}^{2}\to\mathbb{R}$ is a smooth function.

  * (c)

Conclude that $\xi\mapsto\rho(\xi){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\xi$
is a smooth regular parametrization $\mathbb{S}^{2}\to\nobreak\Sigma$.

10.15. Theorem. Suppose $\Sigma$ is an open smooth strictly convex surface.
Then there is a coordinate system in which $\Sigma$ is the graph $z=\nobreak
f(x,y)$ of a convex function $f$ defined on a convex open region $\Omega$ of
the $(x,y)$-plane.

Moreover, $f(x_{n},y_{n})\to\infty$ if $(x_{n},y_{n})\in\Omega$ and
$(x_{n},y_{n})\to(x_{\infty},y_{\infty})\in\nobreak\partial\Omega$.

10.16. Exercise. Assume a strictly convex closed noncompact region $R$
contains the origin in its interior and is bounded by a smooth surface
$\Sigma$.

  * (a)

Show that $R$ contains a half-line $\ell$.

  * (b)

Show that any line $m$ parallel to $\ell$ intersects $\Sigma$ at most at one
point.

  * (c)

Consider an $(x,y,z)$-coordinate system such that the $z$-axis points in the
direction of $\ell$. Show that the projection of $\Sigma$ to the $(x,y)$ plane
is an open convex set; denote it by $\Omega$.

  * (d)

Conclude that $\Sigma$ is a graph $z=f(x,y)$ of a convex function $f$ defined
on $\Omega$. (It proves the main statement in 10B.)

  * (e)

Prove the last statement in 10B.

10.17. Exercise. Show that any open surface $\Sigma$ with positive Gauss
curvature is a topological plane; that is, there is an embedding
$\mathbb{R}^{2}\to\mathbb{R}^{3}$ with image $\Sigma$.

Try to show that $\Sigma$ is a smooth plane; that is, the embedding $f$ can be
made smooth and regular.

10.18. Exercise. Show that any open smooth surface $\Sigma$ with positive
Gauss curvature lies inside an infinite circular cone. In other words, there
is an $(x,y,z)$-coordinate system in which $\Sigma$ lies in the region
$z\geqslant\nobreak m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\sqrt{x^{2}+y^{2}}$ for some $m>0$.

10.19. Exercise. Assume $\Sigma$ is a smooth convex surface with positive
Gauss curvature. Show that

  * (a)

If $\Sigma$ is closed, then the spherical map
$\nu\colon\Sigma\to\mathbb{S}^{2}$ is a bijection. Conclude that

$\iint_{\Sigma}K=4{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

  * (b)

If $\Sigma$ is open, then the spherical map $\nu\colon\Sigma\to\mathbb{S}^{2}$
maps $\Sigma$ bijectively into a subset of a hemisphere. Conclude that

$\iint_{\Sigma}K\leqslant 2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

###  C Saddle surfaces

A surface is called _saddle_ if its Gauss curvature at each point is
nonpositive; in other words the principal curvatures at each point have
opposite signs or at least one of them is zero.

If the Gauss curvature is negative at each point, then the surface is said to
be _strictly saddle_ ; equivalently, this means that the principal curvatures
have opposite signs at each point. Note that in this case the tangent plane
cannot support the surface even locally — moving along the surface in the
principal directions at a given point, one goes above and below the tangent
plane at this point.

10.20. Exercise. Let $f\colon\mathbb{R}\to\mathbb{R}$ be a smooth positive
function. Show that the surface of revolution of the graph $y=\nobreak f(x)$
around the $x$-axis is saddle if and only if $f$ is convex; that is, if
$f^{\prime\prime}(x)\geqslant 0$ for any $x$.

A surface $\Sigma$ is called _ruled_ if for every point $p\in\Sigma$ there is
a line segment $\ell_{p}\subset\Sigma$ passing thru $p$ that is infinite or
its endpoint(s) lie on the boundary line of $\Sigma$.

10.21. Exercise. Show that any ruled surface $\Sigma$ is saddle.

10.22. Exercise. Let $\Sigma$ be an open strictly saddle surface and
$f\colon\mathbb{R}^{3}\to\nobreak\mathbb{R}$ be a smooth convex function. Show
that the restriction of $f$ to $\Sigma$ does not have a point of strict local
maximum.

A tangent direction on a smooth surface with vanishing normal curvature is
called _asymptotic_. A smooth regular curve that always runs in an asymptotic
direction is called an _asymptotic line_.

Recall that a set $R$ in the plane is called _star-shaped_ if there is a point
$p\in R$ such that for any $x\in R$ the line segment $[p,x]$ belongs to $R$.

The statement in the following exercise is due to Dmitri Panov [66].

10.23. Advanced exercise. Let $\gamma$ be a closed smooth asymptotic line in
such graph $z=\nobreak f(x,y)$ of a smooth function $f$. Assume that the graph
is strictly saddle in a neighborhood of $\gamma$. Show that the region in the
$(x,y)$-plane bounded by the projection of $\gamma$ to such plane cannot be
star-shaped.

10.24. Advanced exercise. Let $\Sigma$ be a smooth surface and $p\in\Sigma$.
Assume $K(p)<0$. Show that there is a neighborhood $\Omega$ of $p$ in $\Sigma$
such that the intersection of $\Omega$ with the tangent plane $\mathrm{T}_{p}$
is a union of two smooth curves _intersecting transversally_ at $p$.

###  D Hats

Note that a closed surface cannot be saddle. Indeed, let $\Sigma$ be a closed
surface and consider a smallest sphere that contains $\Sigma$ in its interior;
it supports $\Sigma$ at some point $p$ and at this point the principal
curvatures must have the same sign. The following more general statement is
proved using the same idea.

10.25. Lemma. Assume $\Sigma$ is a compact saddle surface and its boundary
line lies in a closed convex region $R$. Then the entire surface $\Sigma$ lies
in $R$.

Remark. Note that in the case when $\Sigma$ is strictly saddle, the follows
from 10C.

Proof. Arguing by contradiction, assume there is a point $p\in\Sigma$ that
does not lie in $R$. Let $\Pi$ be a plane that separates $p$ from $R$; it
exists by 0C. Denote by $\Sigma^{\prime}$ the part of $\Sigma$ that lies with
$p$ on the same side of $\Pi$.

Since $\Sigma$ is compact, it is surrounded by a sphere; let $\sigma$ be the
circle of intersection of this sphere and $\Pi$. Consider the smallest
spherical dome $\Sigma_{0}$ with boundary $\sigma$ that surrounds
$\Sigma^{\prime}$.

Note that $\Sigma_{0}$ supports $\Sigma$ at some point $q$. Without loss of
generality we may assume that $\Sigma_{0}$ and $\Sigma$ are cooriented at $q$
and $\Sigma_{0}$ has positive principal curvatures. In this case $\Sigma_{0}$
supports $\Delta$ from outside. By 10A, we have
$K(q)_{\Sigma}\geqslant\nobreak K(q)_{\Sigma_{0}}>0$, a contradiction. ∎

10.26. Exercise. Let $\Delta$ be a compact smooth regular saddle surface with
boundary and $p\in\Delta$. Assume the boundary line of $\Delta$ lies in the
unit sphere centered at $p$. Show that if $\Delta$ is a disc, then
$\mathop{\rm length}\nolimits(\partial\Delta)\geqslant 2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi$.

Remark. Show that the statement does not hold without assuming that $\Delta$
is a disc.

If $\Delta$ is as in the exercise, then in fact $\mathop{\rm
area}\nolimits\Delta\geqslant\pi$. The proof of this statement can be obtained
by applying the so-called _coarea fromula_ together with the inequality in the
exercise.

10.27. Exercise. Show that an open saddle surface cannot lie inside of an
infinite circular cone.

A disc $\Delta$ in a surface $\Sigma$ is called a _hat_ of $\Sigma$ if its
boundary line $\partial\Delta$ lies in a plane $\Pi$ and
$\Delta\backslash\partial\Delta$ lies on one side of $\Pi$.

10.28. Proposition. A smooth surface $\Sigma$ is saddle if and only if it has
no hats.

Note that a saddle surface can contain a closed plane curve. For example the
hyperboloid $x^{2}+y^{2}-z^{2}=1$ contains the unit circle in the
$(x,y)$-plane. However, according to the proposition (as well as the lemma), a
plane curve cannot bound a disc (as well any compact set) in a saddle surface.

Proof. Since a plane is convex, the ‘‘only if’’ part follows from 10D; it
remains to prove the ‘‘if’’ part.

Assume $\Sigma$ is not saddle; that is, it has a point $p$ with strictly
positive Gauss curvature; or equivalently, the principal curvatures $k_{1}(p)$
and $k_{2}(p)$ have the same sign.

Let $z=f(x,y)$ be a graph representation of $\Sigma$ in tangent-normal
coordinates at $p$. Consider the set $F_{\varepsilon}$ in the $(x,y)$-plane
defined by the inequality $f(x,y)\leqslant\varepsilon$. By 10B, $f$ is convex
in a small neighborhood of $(0,0)$. Therefore $F_{\varepsilon}$ is convex, for
sufficiently small $\varepsilon>0$. In particular, $F_{\varepsilon}$ is a
topological disc.

Note that $(x,y)\mapsto(x,y,f(x,y))$ is a homeomorphism from $F_{\varepsilon}$
to

$\Delta_{\varepsilon}=\left\\{\,{(x,y,f(x,y))\in\mathbb{R}^{3}}\,:\,{f(x,y)\leqslant\varepsilon}\,\right\\};$

so $\Delta_{\varepsilon}$ is a topological disc for any sufficiently small
$\varepsilon>0$. Note that the boundary line of $\Delta_{\varepsilon}$ lies on
the plane $z=\varepsilon$ and the entire disc lies below it; that is,
$\Delta_{\varepsilon}$ is a hat of $\Sigma$. ∎

The following exercise shows that $\Delta_{\varepsilon}$ is in fact a smooth
disc. This can be used to prove a slightly stronger version of 10D; namely in
the definition of hats one can assume that the disc is smooth.

10.29. Exercise. Let $f\colon\mathbb{R}^{2}\to\mathbb{R}$ be a smooth strictly
convex function with minimum at the origin. Show that the set
$F_{\varepsilon}$ in the graph $z=f(x,y)$ defined by the inequality
$f(x,y)\leqslant\varepsilon$ is a smooth disc for any $\varepsilon>0$; that
is, there is a diffeomorphism $F_{\varepsilon}$ to the unit disc
$\Delta=\nobreak\left\\{\,{(x,y)\in\mathbb{R}^{2}}\,:\,{x^{2}+y^{2}\leqslant
1}\,\right\\}$.

10.30. Exercise. Let $L\colon\mathbb{R}^{3}\to\mathbb{R}^{3}$ be an affine
transformation; that is, $L$ is an invertible map
$\mathbb{R}^{3}\to\mathbb{R}^{3}$ that sends any plane to a plane. Show that
for any saddle surface $\Sigma$ the image $L(\Sigma)$ is also a saddle
surface.

###  E Saddle graphs

The following theorem was proved by Sergei Bernstein [9].

10.31. Theorem. Let $f\colon\mathbb{R}^{2}\to\mathbb{R}$ be a smooth function.
Assume its graph $z=f(x,y)$ is a strictly saddle surface in $\mathbb{R}^{3}$.
Then $f$ is not bounded; that is, there is no constant $C$ such that
$|f(x,y)|\leqslant C$ for any $(x,y)\in\mathbb{R}^{2}$.

The theorem states that a saddle graph cannot lie between two parallel
horizontal planes; applying 10D we get that saddle graphs cannot lie between
two parallel planes, not necessarily horizontal. The following exercise shows
that the theorem does not hold for saddle surfaces which are not graphs.

10.32. Exercise. Construct an open strictly saddle surface that lies between
two parallel planes.

Since $\exp(x-y^{2})>0$, the following exercise shows that there are strictly
saddle graphs with functions bounded on one side; that is, both (upper and
lower) bounds are needed in the proof of Bernshtein’s theorem.

10.33. Exercise. Show that the graph $z=\exp(x-y^{2})$ is strictly saddle.

The following exercise gives a condition that guarantees that a saddle surface
is a graph; it can be used in combination with Bernshtein’s theorem.

10.34. Advanced exercise. Let $\Sigma$ be an open smooth strictly saddle disc
in $\mathbb{R}^{3}$. Assume there is a compact subset $K\subset\Sigma$ such
that the complement $\Sigma\backslash K$ is the graph $z=f(x,y)$ of a smooth
function defined in an open domain of the $(x,y)$-plane. Show that the surface
$\Sigma$ is a graph.

Note that according to 10D, there are no proper saddle surfaces inside a
parallelepiped with boundary line on one of its facets. The following lemma
gives an analogous statement for a parallelepiped with an infinite side.

10.35. Lemma. There is no proper strictly saddle smooth surface that has its
boundary line in a plane $\Pi$ and lies at a bounded distance from a line
contained in $\Pi$.

Proof. Note that by 10D, the statement can be reformulated in the following
way: _There is no proper strictly saddle smooth surface with boundary line in
the $(x,y)$-plane and contained in a region of the form:_

$R=\left\\{\,{(x,y,z)\in\mathbb{R}^{3}}\,:\,{0\leqslant z\leqslant
r,0\leqslant y\leqslant r}\,\right\\}.$

Let us prove this statement.

Assume the contrary, let $\Sigma$ be such a surface. Consider the projection
$\hat{\Sigma}$ of $\Sigma$ to the $(x,z)$-plane. It lies in the upper half-
plane and below the line $z=r$.

Consider the open upper half-plane
$H=\left\\{\,{(x,z)\in\mathbb{R}^{2}}\,:\,{z>0}\,\right\\}$. Let $\Theta$ be
the connected component of the complement $H\backslash\hat{\Sigma}$ that
contains all the points above the line $z=r$.

Note that $\Theta$ is convex. If not, then there is a line segment $[p,q]$ for
some $p,q\in\Theta$ that cuts from $\hat{\Sigma}$ a compact piece.

Consider the plane $\Pi$ thru $[p,q]$ that is perpendicular to the
$(x,z)$-plane. Note that $\Pi$ cuts from $\Sigma$ a compact region $\Delta$.
By a general position argument (see 8G) we can assume that $\Delta$ is a
compact surface with boundary line in $\Pi$ and the remaining part of $\Delta$
lies on one side from $\Pi$. Since the plane $\Pi$ is convex, this statement
contradicts 10D.

Summarizing, $\Theta$ is an open convex set of $H$ that contains all points
above $z=r$. By convexity, together with any point $w$, the set $\Theta$
contains all points on the half-lines that start as $w$ and _point up_ ; that
is, in directions with positive $z$-coordinate. In other words, with any point
$w$, the set $\Theta$ contains all points with larger $z$-coordinates.

Since $\Theta$ is open it can be described by an inequality $z>r_{0}$. It
follows that the plane $z=r_{0}$ supports $\Sigma$ at some point (in fact at
many points). By 10A, the latter is impossible — a contradiction. ∎

Proof of 10E. Denote by $\Sigma$ the graph $z=f(x,y)$. Assume the contrary;
that is, $\Sigma$ lies between two planes $z=\pm C$.

Note that the function $f$ cannot be constant. It follows that the tangent
plane $\mathrm{T}_{p}$ at some point $p\in\Sigma$ is not horizontal.

Denote by $\Sigma^{+}$ the part of $\Sigma$ that lies above $\mathrm{T}_{p}$.
Note that it has at least two connected components which are approaching $p$
from both sides in the principal direction with positive principal curvature.
Indeed, if there was a curve that runs in $\Sigma^{+}$ and approaches $p$ from
both sides, then it would cut a disc from $\Sigma$ with boundary line above
$\mathrm{T}_{p}$ and some points below it; the latter would contradict 10D.

The surface $\Sigma$ seeing from above.

Summarizing, $\Sigma^{+}$ has at least two connected components, denote them
by $\Sigma^{+}_{0}$ and $\Sigma^{+}_{1}$. Let $z=h(x,y)=a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}x+b{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+c$
be the equation of $\mathrm{T}_{p}$. Note that $\Sigma^{+}$ contains all
points in the region

$R_{-}=\left\\{\,{(x,y,f(x,y))\in\Sigma}\,:\,{h(x,y)<-C}\,\right\\}$

which is a connected set and no points in

$R_{+}=\left\\{\,{(x,y,f(x,y))\in\Sigma}\,:\,{h(x,y)>C}\,\right\\}$

Whence one of the connected components, say $\Sigma^{+}_{0}$, lies in

$R_{0}=\left\\{\,{(x,y,f(x,y))\in\Sigma}\,:\,{|h(x,y)|\leqslant
C}\,\right\\}.$

This set lies on a bounded distance from the line of intersection of
$\mathrm{T}_{p}$ with the $(x,y)$-plane.

Let us move $\mathrm{T}_{p}$ slightly upward and cut from $\Sigma^{+}_{0}$ the
piece above the obtained plane, say $\bar{\Sigma}^{+}_{0}$. By the general
position argument (8G), we can assume that $\bar{\Sigma}^{+}_{0}$ is a surface
with smooth boundary line; by construction the boundary line lies in the
plane. Note that the obtained surface $\bar{\Sigma}^{+}_{0}$ still lies on a
bounded distance to a line. The latter is impossible by 10E. ∎

### Remarks

Note that Bernstein’s theorem and the lemma in its proof do not hold for
nonstrictly saddle surfaces; counterexamples can be found among infinite
cylinders over smooth regular curves. In fact it can be shown that these are
the only counterexamples; a proof is based on the same idea, but it is much
more technical.

By 10D, saddle surfaces can be defined as smooth surfaces without hats. This
definition can be used for arbitrary surfaces, not necessarily smooth. Some
results, for example Bernshtein’s characterization of saddle graphs can be
extended to generalized saddle surfaces, but this class of surfaces is far
from being understood; see [5, Chapter 4] and the references therein.

##  Chapter 11 Shortest paths

###  A Intrinsic geometry

We begin to study the _intrinsic geometry_ of surfaces. A property is called
_intrinsic_ if it can be checked by measuring things inside the surface, for
example lengths of curves or angles between the curves that lie in the
surface. Otherwise, if a definition of a property essentially uses the ambient
space, then it is called _extrinsic_.

For instance, the mean curvature as well as the Gauss curvature are defined
via principal curvatures, which are extrinsic. Later (15H) it will be shown
that _remarkably_ the Gauss curvature is actually intrinsic — it can be
calculated via measurements inside the surface. The mean curvature is not
intrinsic, for example, the intrinsic geometry of the $(x,y)$-plane is not
distinguishable from the intrinsic geometry of the graph
$z=\nobreak(x+y)^{2}$. However, while the mean curvature of the former
vanishes at all points, the mean curvature of the latter does not vanish, say
at $p=(0,0,1)$.

The following exercise should help you get in the right mood; it might look
like a tedious problem in calculus, but actually it is an easy problem in
geometry. We learned this problem from Joel Fine, who attributed it so
Frederic Bourgeois [30].

11.1. Exercise. А cowboy stands at the bottom of a frictionless ice-mountain
formed by a cone with a circular base with angle of inclination $\theta$. He
wants to climb the mountain; he throws up his lasso which slips neatly over
the tip of the cone, pulls it tight and starts to climb.

What is the critical angle $\theta$ at which the cowboy can no longer climb
the ice-mountain?

###  B Definition

Let $p$ and $q$ be two points on a surface $\Sigma$. Recall that
$|{p}-\nobreak{q}|_{\Sigma}$ denotes the induced length distance from $p$ to
$q$; that is, the greatest lower bound of the lengths of paths in $\Sigma$
from $p$ to $q$.

Note that if $\Sigma$ is smooth, then any two points in $\Sigma$ can be joined
by a piecewise smooth path. Since any such path is rectifiable, the value
$|{p}-\nobreak{q}|_{\Sigma}$ is finite for any pair $p,q\in\Sigma$.

A path $\gamma$ from $p$ to $q$ in $\Sigma$ that minimizes the length is
called a _shortest path_ from $p$ to $q$.

The image of a shortest path between $p$ and $q$ in $\Sigma$ is usually
denoted by $[p,q]$ or by $[p,q]_{\Sigma}$. In general there might be no
shortest path between two given points on a surface and there might be many of
them; this is shown in the following two examples.

Usually, if we write $[p,q]_{\Sigma}$, we are assuming that such a shortest
path exists and we have made a choice of one of them.

Nonuniqueness. There are plenty of shortest paths between the poles of a round
sphere — each meridian is a shortest path.

Nonexistence. Let $\Sigma$ be the $(x,y)$-plane with the origin removed.
Consider two points $p=(1,0,0)$ and $q=(-1,0,0)$ in $\Sigma$.

11.2. Claim. There is no shortest path from $p$ to $q$ in $\Sigma$.

Proof. Note that $|{p}-\nobreak{q}|_{\Sigma}=2$. Indeed, given
$\varepsilon>\nobreak 0$, consider the point
$s_{\varepsilon}=(0,\varepsilon,0)$. Observe that the polygonal path
$ps_{\varepsilon}q$ lies in $\Sigma$ and its length $2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt{1+\varepsilon^{2}}$ approaches $2$ as
$\varepsilon\to 0$. It follows that $|{p}-\nobreak{q}|_{\Sigma}\leqslant 2$.
Since
$|{p}-\nobreak{q}|_{\Sigma}\geqslant\nobreak|{p}-\nobreak{q}|_{\mathbb{R}^{3}}=2$,
we get $|{p}-\nobreak{q}|_{\Sigma}=2$.

It follows that a shortest path from $p$ to $q$, if it exists, must have
length 2. By the triangle inequality any curve of length 2 from $p$ to $q$
must run along the line segment $[p,q]$; in particular it must pass thru the
origin. Since the origin does not lie in $\Sigma$, there is no shortest from
$p$ to $q$ in $\Sigma$ ∎

11.3. Proposition. Any two points in a proper smooth surface can be joined by
a shortest path.

Proof. Fix a proper smooth surface $\Sigma$ and two points $p$ and $q$. Set
$\ell=|{p}-\nobreak{q}|_{\Sigma}$.

By the definition of induced length-metric (Section 2G), there is a sequence
of paths $\gamma_{n}$ from $p$ to $q$ in $\Sigma$ such that

$\mathop{\rm length}\nolimits\gamma_{n}\to\ell\quad\text{as}\quad n\to\infty.$

Without loss of generality, we may assume that $\mathop{\rm
length}\nolimits\gamma_{n}<\ell+1$ for any $n$ and each $\gamma_{n}$ is
parametrized proportional to its arc-length. In particular each path
$\gamma_{n}\colon[0,1]\to\Sigma$ is $(\ell+1)$-Lipschitz; that is,

$|\gamma(t_{0})-\gamma(t_{1})|\leqslant(\ell+1){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}|t_{0}-t_{1}|$

for any $t_{0},t_{1}\in[0,1]$.

Note that the image of $\gamma_{n}$ lies in the closed ball
$\bar{B}[p,\ell+1]$ for any $n$. It follows that the coordinate functions of
$\gamma_{n}$ are uniformly equicontinuous and uniformly bounded. By the
Arzelá–Ascoli theorem (0E) there is a convergent subsequence of $\gamma_{n}$
and its limit, say $\gamma_{\infty}\colon[0,1]\to\mathbb{R}^{3}$, is
continuous; that is, $\gamma_{\infty}$ is a path. Evidently $\gamma_{\infty}$
runs from $p$ to $q$; in particular

$\mathop{\rm length}\nolimits\gamma_{\infty}\geqslant\ell.$

Since $\Sigma$ is a closed set, $\gamma_{\infty}$ lies in $\Sigma$. Finally,
since length is semicontinuous (2F), we get that

$\mathop{\rm length}\nolimits\gamma_{\infty}\leqslant\ell.$

Therefore $\mathop{\rm length}\nolimits\gamma_{\infty}=\ell$ or, equivalently,
$\gamma_{\infty}$ is a shortest path from $p$ to $q$. ∎

###  C Closest point projection

11.4. Lemma. Let $R$ be a closed convex set in $\mathbb{R}^{3}$. Then for
every point $p\in\mathbb{R}^{3}$ there is a unique point $\bar{p}\in R$ that
minimizes the distance to $R$; that is, $|p-\bar{p}|\leqslant|p-x|$ for any
point $x\in R$.

Moreover the map $p\mapsto\bar{p}$ is short; that is,

$|p-q|\geqslant|\bar{p}-\bar{q}|$ ➊

for any pair of points $p,q\in\mathbb{R}^{3}$.

The map $p\mapsto\bar{p}$ is called the _closest point projection_ ; it maps
the Euclidean space to $R$. Note that if $p\in R$, then $\bar{p}=p$.

Proof. Fix a point $p$ and set

$\ell=\inf\left\\{\,{|p-x|}\,:\,{x\in R}\,\right\\}.$

Choose a sequence $x_{n}\in R$ such that $|p-x_{n}|\to\ell$ as $n\to\infty$.

Without loss of generality, we can assume that all the points $x_{n}$ lie in a
ball of radius $\ell+1$ centered at $p$. Therefore we can pass to a _partial
limit_ $\bar{p}$ of $x_{n}$; that is, $\bar{p}$ is a limit of a subsequence of
$x_{n}$. Since $R$ is closed, $\bar{p}\in R$. By construction

$\displaystyle|p-\bar{p}|$ $\displaystyle=\lim_{n\to\infty}|p-x_{n}|=\ell.$

Hence the existence follows.

Assume there are two distinct points $\bar{p},\bar{p}^{\prime}\in R$ that
minimize the distance to $p$. Since $R$ is convex, their midpoint
$m=\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(\bar{p}+\bar{p}^{\prime})$ lies in $R$. Note that
$|p-\bar{p}|=\nobreak|p-\bar{p}^{\prime}|=\ell$; that is, the triangle
$[p\bar{p}\bar{p}^{\prime}]$ is isosceles and therefore the triangle
$[p\bar{p}m]$ is right with the right angle at $m$. Since a leg of a right
triangle is shorter than its hypotenuse, we have $|p-m|<\ell$, a
contradiction.

It remains to prove ➊ ‣ 11C. We can assume that $\bar{p}\neq\bar{q}$,
otherwise there is nothing to prove.

Note that if $\measuredangle[\bar{p}\,{}^{p}_{\bar{q}}]<\tfrac{\pi}{2}$, then
$|{p}-\nobreak{x}|<\nobreak|{p}-\nobreak{\bar{p}}|$ for some point
$x\in[\bar{p},\bar{q}]$. Since $[\bar{p},\bar{q}]\subset K$, the latter is
impossible.

Therefore $p=\bar{p}$ or
$\measuredangle[\bar{p}\,{}^{p}_{\bar{q}}]\geqslant\tfrac{\pi}{2}$. In both
cases the orthogonal projection of $p$ to the line $\bar{p}\bar{q}$ lies
behind $\bar{p}$, or coincides with $\bar{p}$. The same way we show that the
orthogonal projection of $q$ to the line $\bar{p}\bar{q}$ lies behind
$\bar{q}$, or coincides with $\bar{q}$. It implies that the orthogonal
projection of the line segment $[p,q]$ to the line $\bar{p}\bar{q}$ contains
the line segment $[\bar{p},\bar{q}]$. In particular

$|p-q|\geqslant|\bar{p}-\bar{q}|.$

∎

11.5. Corollary. Assume a surface $\Sigma$ bounds a closed convex region $R$
and $p,q\in\Sigma$. Denote by $W$ the outer closed region of $\Sigma$; in
other words $W$ is the union of $\Sigma$ and the complement of $R$. Then

$\mathop{\rm length}\nolimits\gamma\geqslant|{p}-\nobreak{q}|_{\Sigma}$

for any path $\gamma$ in $W$ from $p$ to $q$. Moreover if $\gamma$ does not
lie in $\Sigma$, then the inequality is strict.

Proof. The first part of the corollary follows from the lemma and the
definition of length. Indeed, consider the closest point projection
$\bar{\gamma}$ of $\gamma$. Note that $\bar{\gamma}$ lies in $\Sigma$ and
connects $p$ to $q$ therefore

$\mathop{\rm length}\nolimits\bar{\gamma}\geqslant|{p}-\nobreak{q}|_{\Sigma}.$

To prove the first statement, it remains to show that

$\mathop{\rm length}\nolimits\gamma\geqslant\mathop{\rm
length}\nolimits\bar{\gamma}.$ ➋

Consider a polygonal line $\bar{p}_{0}\dots\bar{p}_{n}$ inscribed in
$\bar{\gamma}$. Let $p_{0}\dots p_{n}$ be the corresponding polygonal line
inscribed in $\gamma$; that is $p_{i}=\gamma(t_{i})$ if
$\bar{p}_{i}=\bar{\gamma}(t_{i})$. By 11C
$|p_{i}-p_{i-1}|\geqslant\nobreak|\bar{p}_{i}-\bar{p}_{i-1}|$ for any $i$.
Therefore

$\mathop{\rm length}\nolimits p_{0}\dots p_{n}\geqslant\mathop{\rm
length}\nolimits\bar{p}_{0}\dots\bar{p}_{n}.$

Taking the least upper bound of each side of the inequality for all inscribed
polygonal lines $p_{0}\dots p_{n}$ in $\gamma$, we get ➋ ‣ 11C.

It remains to prove the second statement. Suppose there is a point
$w=\nobreak\gamma(t_{1})\notin\Sigma$; note that $w\notin R$. By the
separation lemma (0C) there is a plane $\Pi$ that cuts $w$ from $\Sigma$. The
curve $\gamma$ must intersect $\Pi$ at two points: one point before $t_{1}$
and one after. Let $a=\gamma(t_{0})$ and $b=\gamma(t_{2})$ be these points.
Note that the arc of $\gamma$ from $a$ to $b$ is strictly longer that $|a-b|$;