that $p$ is not a limit point of any sequence $q_{n}\in Q$. That is, any limit
of a sequence in $Q$ lies in $Q$; that is, by the definition, $Q$ is closed.

Now suppose $\Omega=\mathcal{X}\backslash Q$ is not open. Show that there is a
point $p\in\Omega$ and a sequence $q_{n}\in Q$ such that such that
$|{p}-\nobreak{q_{n}}|_{\mathcal{X}}<\nobreak\tfrac{1}{n}$ for any $n$.
Conclude that $q_{n}\to p$ and $n\to\infty$; therefore $Q$ is not closed.

1A; 1A. Use that a continuous injection defined on a compact domain is an
embedding (0H).

parit1A. The image of $\gamma$ might have the shape of the digit $8$ or $9$.

1B. Let $\alpha$ be a path, connecting $p$ to $q$.

Passing to an arc of $\alpha$ if necessary, we can assume that $\alpha(t)\neq
p,q$ for $t\neq 0,1$.

An open set $\Omega$ in $(0,1)$ will be called _suitable_ if for any connected
component $(a,b)$ of $\Omega$ we have $\alpha(a)=\alpha(b)$. Show that the
union of nested suitable sets is suitable. Therefore we can find a maximal
suitable set $\hat{\Omega}$.

Define $\beta(t)=\alpha(a)$ for any $t$ in a connected component
$(a,b)\subset\hat{\Omega}$, and $\beta(t)=\alpha(t)$ for $t$ not in
$\hat{\Omega}$. Note that for any $x\in[0,1]$ the set
$\beta^{-1}\\{\beta(x)\\}$ is connected.

It remains to reparametrize $\beta$ to make it injective. In order to do that,
we need to construct a non-decreasing surjective function
$\tau\colon[0,1]\to\nobreak[0,1]$ such that $\tau(t_{1})=\tau(t_{2})$ if and
only if there is a connected component $(a,b)\subset\hat{\Omega}$ such that
$t_{1},t_{2}\in\nobreak[a,b]$.

The required function $\tau$ can be constructed in a similar way to the so-
called _devil’s staircase_ — learn this construction and modify it.

The simple arc we are looking for is $\beta\circ\sigma$, where
$\sigma\colon[0,1]\to[0,1]$ is any right inverse of $\tau$.

1C. Denote the union of two half-axis by $L$.

Recall that the function $f$ is smooth; see Section 0E. Therefore
$t\mapsto\alpha(t)=(f(t),f(-t))$ is smooth as well.

Observe that $f(t)\to\infty$ as $t\to\infty$. Since $f(0)=0$, the intermediate
value theorem implies that $f(t)$ takes all nonnegative values for $t\geqslant
0$. Use it to show that $L$ is the range of $\alpha$.

Further, show that the function $f$ is strictly increasing for $t>0$. Use this
to show that the map $t\mapsto\alpha(t)$ is injective.

Summarizing, we get that $\alpha$ is a smooth parametrization of $L$.

Now suppose $\beta\colon t\mapsto(x(t),y(t))$ is a smooth parametrization of
$L$. Without loss of generality we may assume that $x(0)=y(0)=0$. Note that
$x(t)\geqslant 0$ for any $t$ therefore $x^{\prime}(0)=0$. The same way we get
that $y^{\prime}(0)=0$. That is, $\beta^{\prime}(0)=0$; so $L$ does not admit
a smooth regular parametrization.

1C. Apply the definitions. For 1C you need to check that
$\gamma^{\prime}_{\ell}\neq 0$. For 1C you need to check that
$\gamma_{\ell}(t_{0})=\nobreak\gamma(t_{1})$ only if $t_{0}=t_{1}$.

1E. This is the so-called _semicubical parabola_ ; it is shown on the diagram.
Try to argue similarly to 1C.

1E. For $\ell=0$ the system describes a pair of points $(0,0,\pm 1)$, so we
can assume that $\ell\neq 0$. Note that the first equation describes the unit
sphere centered at the origin and the second equation describes a cylinder
over the circle in the $(x,y)$-plane with center $(-\tfrac{\ell}{2},0)$ and
radius $|\tfrac{\ell}{2}|$.

$x$$y$$z$

For $\ell\neq 0$, find the gradients $\nabla f$ and $\nabla h$ for the
functions

$\displaystyle f(x,y,z)$ $\displaystyle=x^{2}+y^{2}+z^{2}-1$ $\displaystyle
h(x,y,z)$ $\displaystyle=x^{2}+\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x+y^{2}$

and show that they are linearly dependent only on the $x$-axis. Conclude that
for $\ell\neq\pm 1$ each connected component of the set of solutions is a
smooth regular curve.

Show that

  * $\diamond$

if $|\ell|<1$, then the set has two connected components with $z>0$ and $z<0$.

  * $\diamond$

if $|\ell|\geqslant 1$, then the set is connected.

Note that the linear independence of the gradients provides only a sufficient
condition. Therefore the case $\ell=\pm 1$ has to be checked by hand. In this
case a neighborhood of $(\pm 1,0,0)$ does not admit a smooth regular
parametrization — try to prove it. The case $\ell=1$ is shown on the diagram.

Remark. The case $\ell=\pm 1$ is called _Viviani’s curve_. It admits the
following smooth regular parametrization with a self-intersection:

$t\mapsto(\pm(\cos t)^{2},\cos t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin
t,\sin t).$

1F. Assume contrary, then there is a sequence of real numbers
$t_{n}\to\pm\infty$ such that $\gamma(t_{n})$ converges; denote its limit by
$p$. Let $K$ be a closed ball centered at $p$. Observe that $\gamma^{-1}(K)$
is not compact. Conclude that $\gamma$ is not proper.

1F. Show and use that a set $C\subset\mathbb{R}^{3}$ is closed if and only if
the intersection $K\cap C$ is compact for any compact
$K\subset\mathbb{R}^{3}$.

1F. Without loss of generality we may assume that the origin does not lie on
the curve.

Show that inversion of the plane
$(x,y)\mapsto(\tfrac{x}{x^{2}+y^{2}},\tfrac{y}{x^{2}+y^{2}})$ maps our curve
to a closed curve with the origin removed. Apply Jordan’s theorem for the
obtained curve and use the inversion again.

2A. Show that the least upper bound in 2A can be taken for all sequences
$a=t_{0}\leqslant t_{1}\leqslant\dots\nobreak\leqslant t_{k}=b$.

Suppose that $\gamma_{2}$ is reparametrization of $\gamma_{1}$ by
$\tau\colon[a_{1},b_{1}]\to[a_{2},b_{2}]$; without loss of generality we may
assume that $\tau$ is nondecreasing. Set $\theta_{i}=\tau(t_{i})$. Observe
that
$a_{2}=\theta_{0}\leqslant\nobreak\theta_{1}\leqslant\dots\nobreak\leqslant\theta_{k}=b_{2}$
if and only if $a_{1}=t_{0}\leqslant t_{1}\leqslant\dots\nobreak\leqslant
t_{k}=b_{1}$. Make a conclusion.

2A. Choose a partition $0=t_{0}<\dots<t_{n}=1$ of $[0,1]$. Set $\tau_{0}=0$
and

$\tau_{i}=\max\left\\{\,{\tau\in[0,1]}\,:\,{\beta(\tau_{i})=\alpha(t_{i})}\,\right\\}$

for $i>0$. Show that $(\tau_{i})$ is a partition of $[0,1]$; that is,
$0=\tau_{0}<\tau_{1}<\dots<\tau_{n}=1$.

By construction

$\displaystyle|\alpha(t_{0})$
$\displaystyle-\alpha(t_{1})|+|\alpha(t_{1})-\alpha(t_{2})|+\dots+|\alpha(t_{n-1})-\alpha(t_{n})|=$
$\displaystyle=|\beta(\tau_{0})-\beta(\tau_{1})|+|\beta(\tau_{1})-\beta(\tau_{2})|+\dots+|\beta(\tau_{n-1})-\beta(\tau_{n})|.$

Since the partition $(t_{i})$ is arbitrary, we get

$\mathop{\rm length}\nolimits\beta\geqslant\mathop{\rm
length}\nolimits\alpha.$

Remark. Note that the partition $(\tau_{i})$ is not arbitrary, therefore the
inequality might be strict; it might happen if $\beta$ runs back and forth
along $\alpha$.

2A. For 2A, apply the fundamental theorem of calculus for each segment in a
given partition. For 2A consider a partition such that the velocity vector
$\alpha^{\prime}(t)$ is nearly constant on each of its segments.

2A. Use the theorems of Rademacher and Lusin (0E and 0E).

2B; 2B.

Look at the diagram and guess the parametrization of an arc of the snow flake
by $[0,1]$. Extend it to the whole snow flake. Show that it indeed describes
an embedding of the circle in the plane.

2B. Suppose that $\gamma\colon[0,1]\to\mathbb{R}^{2}$ is a rectifiable curve
and let $\gamma_{k}$ be a scaled copy of $\gamma$ with factor $k>0$; that is,
$\gamma_{k}(t)=k{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\gamma(t)$ for any $t$.
Show that

$\mathop{\rm length}\nolimits\gamma_{k}=k{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\mathop{\rm length}\nolimits\gamma.$

Now suppose the arc $\gamma$ of the Koch snowflake shown on the diagram is
rectifiable; denote its length by $\ell$. Observe that $\gamma$ can be divided
into 4 arcs such that each one is a scaled copy of $\gamma$ with factor
$\tfrac{1}{3}$. It follows that $\ell=\tfrac{4}{3}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\ell$. Evidently $\ell>0$ — a contradiction.

2C. We have to assume that $a\neq 0$ or $b\neq 0$; otherwise we get a constant
curve.

Show that the curve has constant velocity
$|\gamma^{\prime}(t)|\equiv\sqrt{a^{2}+b^{2}}$. Therefore
$s=t/\sqrt{a^{2}+b^{2}}$ is an arc-length parameter. It remains to substitute
$s{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt{a^{2}+b^{2}}$ for $t$.

2D. Choose a closed polygonal line $p_{1}\dots p_{n}$ inscribed in $\beta$. By
2D, we can assume that its length is arbitrary close to the length of $\beta$;
that is, given $\varepsilon>0$

$\mathop{\rm length}\nolimits(p_{1}\dots p_{n})>\mathop{\rm
length}\nolimits\beta-\varepsilon.$

Show that we may assume in addition that each point $p_{i}$ lies on $\alpha$.

Observe that since $\alpha$ is simple, the points $p_{1},\dots,p_{n}$ appear
on $\alpha$ in the same cyclic order; that is, the polygonal line $p_{1}\dots
p_{n}$ is also inscribed in $\alpha$. In particular

$\mathop{\rm length}\nolimits\alpha\geqslant\mathop{\rm
length}\nolimits(p_{1}\dots p_{n}).$

It follows that

$\mathop{\rm length}\nolimits\alpha>\mathop{\rm
length}\nolimits\beta-\varepsilon.$

for any $\varepsilon>0$. Whence

$\mathop{\rm length}\nolimits\alpha\geqslant\mathop{\rm
length}\nolimits\beta.$

If $\alpha$ has self-intersections, then the points $p_{1},\dots,p_{n}$ might
appear on $\alpha$ in a different order, say $p_{i_{1}},\dots,p_{i_{n}}$.
Apply the triangle inequality to show that

$\mathop{\rm length}\nolimits(p_{i_{1}}\dots p_{i_{n}})\geqslant\mathop{\rm
length}\nolimits(p_{1}\dots p_{n})$

and use it to modify the proof above.

2E. Denote by $\ell_{\text{\sc{u}}}$ the line segment obtained by orthogonal
projection of $\gamma$ to the line in the direction u. Since
$\gamma_{\text{\sc{u}}}$ runs along $\ell_{\text{\sc{u}}}$ back and forth, we
get

$\mathop{\rm length}\nolimits\gamma_{\text{\sc{u}}}\geqslant 2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
length}\nolimits\ell_{\text{\sc{u}}}.$

Applying the Crofton formula, we get that

$\mathop{\rm length}\nolimits\gamma\geqslant\pi{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\overline{\mathop{\rm
length}\nolimits\ell_{\text{\sc{u}}}}.$

In the case of equality, the curve $\gamma_{\text{\sc{u}}}$ runs exactly back
and forth along $\ell_{\text{\sc{u}}}$ without additional zigzags for almost
all (and therefore for all) u.

Let $K$ be a closed set bounded by $\gamma$. Observe that the last statement
implies that every line may intersect $K$ only along a closed segment or a
single point. It follows that $K$ is convex.

2E. The proof is identical to the proof of the standard Crofton formula. To
find the coefficient it is sufficient to check it on a unit interval. The
latter can be done by integration:

$\displaystyle\frac{1}{k_{a}}$ $\displaystyle=\frac{1}{\mathop{\rm
area}\nolimits\mathbb{S}^{2}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\iint_{\mathbb{S}^{2}}|x|;$ $\displaystyle\frac{1}{k_{b}}$
$\displaystyle=\frac{1}{\mathop{\rm area}\nolimits\mathbb{S}^{2}}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\iint_{\mathbb{S}^{2}}\sqrt{1-x^{2}}.$

The answers are $k_{a}=2$ and $k_{b}=\tfrac{4}{\pi}$.

2G. The ‘‘only-if’’ part is trivial. To show the ‘‘if’’ part, assume $A$ is
not convex; that is, there are points $x,y\in A$ and a point $z\notin A$ that
lies between $x$ and $y$.

Since $A$ is closed, its complement is open. That is, the ball
$B(z,\varepsilon)$ does not intersect $A$ for some $\varepsilon>0$.

Show that there is $\delta>0$ such that any path of length at most
$|{x}-\nobreak{y}|_{\mathbb{R}^{3}}+\delta$ passes thru $B(z,\varepsilon)$. It
follows that
$|{x}-\nobreak{y}|_{A}\geqslant|{x}-\nobreak{y}|_{\mathbb{R}^{3}}+\delta$, in
particular $|{x}-\nobreak{y}|_{A}\neq|{x}-\nobreak{y}|_{\mathbb{R}^{3}}$.

2H. The spherical curve shown on the diagram does not have antipodal pairs of
points. However it has three points $x,y,z$ on one of its sides and their
antipodal points $-x,-y,-z$ on the other. Show that this property is
sufficient to conclude that the curve does not lie in any hemisphere.

2H. Assume contrary, then by the hemisphere lemma (2H) $\gamma$ lies in an
open hemisphere. In particular it cannot divide $\mathbb{S}^{2}$ into two
regions of equal area — a contradiction.

2H. The very first sentence is wrong — it is _not_ sufficient to show that
diameter is at most 2. For example an equilateral triangle with circumradius
slightly above $1$ may have diameter (which is defined as the maximal distance
between its points) slightly bigger than $\sqrt{3}$, so it can be made smaller
that $2$.

On the other hand, it is easy to modify the proof of the hemisphere lemma (2H)
to get a correct solution. That is, (1) choose two points $p$ and $q$ on
$\gamma$ that divide it into two arcs of the same length; (2) set $z$ to be a
midpoint of $p$ and $q$, and (3) show that $\gamma$ lies in the unit disc
centered at $z$.

2H. For 2H, modify the proof of the original Crofton formula (Section 2E).

2H. Assume $\mathop{\rm length}\nolimits\gamma<2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. By 2H,

$\overline{\mathop{\rm length}\nolimits\gamma_{\text{\sc{u}}}}<2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

Therefore we can choose u so that

$\mathop{\rm length}\nolimits\gamma_{\text{\sc{u}}}<2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi.$

Observe that $\gamma_{\text{\sc{u}}}$ runs in a semicircle $h$ and therefore
$\gamma$ lies in a hemisphere with $h$ as a diameter.

3B. For a unit-speed parametrization $\gamma$, we have
$\gamma^{\prime\prime}\equiv 0$. Therefore the velocity
$\text{\sc{v}}=\gamma^{\prime}$ is constant. Hence
$\gamma(t)=p+(t-t_{0}){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{v}}$,
where $p=\gamma(t_{0})$.

3B. Observe that $\alpha(t)\mathrel{:=}\gamma_{\lambda}(t/\lambda)$ is a unit-
speed parametrization of the curve $\gamma_{\lambda}$. Apply the chain rule
twice.

3B. Differentiate the identity $\langle\gamma(s),\gamma(s)\rangle=1$ a couple
of times.

3C. Set $\text{\sc{t}}=\tfrac{\gamma^{\prime}}{|\gamma|}$. Prove and use the
following identities:

$\displaystyle\gamma^{\prime\prime}-(\gamma^{\prime\prime})^{\perp}$
$\displaystyle=\text{\sc{t}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\gamma^{\prime\prime},\text{\sc{t}}\rangle,$
$\displaystyle|\gamma^{\prime}|$
$\displaystyle=\sqrt{\langle\gamma^{\prime},\gamma^{\prime}\rangle}.$

3C. Apply 3C for the parametrization $t\mapsto(t,f(t))$.

3C. Without loss of generality we may assume that $\gamma$ has a unit-speed
parametrization.

Consider the tangent indicatrix $\text{\sc{t}}(s)=\gamma^{\prime}(s)$. Note
that t is a spherical curve and $|\text{\sc{t}}^{\prime}|\leqslant 1$. Use it
to construct a sequence of unit-speed spherical curves
$\text{\sc{t}}_{n}\colon\mathbb{I}\to\mathbb{S}^{2}$ such that
$\text{\sc{t}}_{n}(s)\to\text{\sc{t}}(s)$ as $n\to\infty$ for any $s$.

It remains to show that the following sequence of curves solves the problem:

$\gamma_{n}(s)=\gamma(a)+\int_{a}^{s}\text{\sc{t}}_{n}(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dt.$

3E. Show that $\gamma_{a,b}^{\prime\prime}\perp\gamma^{\prime}_{a,b}$ and
apply 3C.

3E. Apply Fenchel’s theorem.

3E. We can assume that $\gamma$ is a unit-speed curve. Set
$\theta(s)=\measuredangle(\gamma(s),\gamma^{\prime}(s))$. Since
$\langle\text{\sc{t}},\text{\sc{t}}\rangle=1$, we have
$\text{\sc{t}}^{\prime}\perp\text{\sc{t}}$. Observe that
$\langle\text{\sc{t}},\sigma\rangle=\cos\theta$; therefore

$\displaystyle\kappa{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin\theta$
$\displaystyle=|\text{\sc{t}}^{\prime}|{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\sin\theta\geqslant-\langle\text{\sc{t}}^{\prime},\sigma\rangle=\langle\text{\sc{t}},\sigma^{\prime}\rangle-\langle\text{\sc{t}},\sigma\rangle^{\prime}=(|\sigma^{\prime}|+\theta^{\prime}){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\sin\theta.$

Whence $\kappa\geqslant|\sigma^{\prime}|+\theta^{\prime}$ if $\theta\neq
0,\pi$. It remains to integrate this inequality and show that the set with
$\theta\neq 0$ or $\pi$ does not create a problem.

An alternative solution can be built on 3G.

3E; 3E. Observe that

$\langle\gamma^{\prime}(s),\gamma^{\prime}(s)\rangle=1,\quad\text{and}\quad\langle\gamma(s),\gamma(s)\rangle\leqslant
1.$

Therefore

$\displaystyle\langle\gamma^{\prime\prime}(s),\gamma(s)\rangle$
$\displaystyle\geqslant-|\gamma^{\prime\prime}(s)|{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}|\gamma(s)|\geqslant-\kappa(s)$ ➊

for all $s$.

3E. Since $\gamma$ is unit-speed, we have $\ell=\mathop{\rm
length}\nolimits\gamma$. Therefore

$\displaystyle\int_{0}^{\ell}\langle\gamma(s),\gamma^{\prime}(s)\rangle^{\prime}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}ds$
$\displaystyle=\int_{0}^{\ell}\langle\gamma^{\prime}(s),\gamma^{\prime}(s)\rangle{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}ds+\int_{0}^{\ell}\langle\gamma(s),\gamma^{\prime\prime}(s)\rangle{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}ds\geqslant\mathop{\rm
length}\nolimits\gamma-\Phi({\gamma}).$

3E. By the fundamental theorem of calculus, we have

$\displaystyle\int_{0}^{\ell}\langle\gamma(s),\gamma^{\prime}(s)\rangle^{\prime}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}ds$
$\displaystyle=\langle\gamma(\ell),\gamma^{\prime}(\ell)\rangle-\langle\gamma(0),\gamma^{\prime}(0)\rangle.$

Since $\gamma(0)=\gamma(\ell)$ and $\gamma^{\prime}(0)=\gamma^{\prime}(\ell)$,
the right hand side vanishes.

Note that without loss of generality we can assume the curve in 3E is
described by a loop $\gamma\colon[0,\ell]\to\mathbb{R}^{3}$ parametrized by
arc-length. We can also assume that the origin is the center of the ball; that
is $|\gamma|\leqslant 1$. Since $\gamma$ is a smooth closed curve, we have
$\gamma^{\prime}(0)=\gamma^{\prime}(\ell)$ and $\gamma(0)=\gamma(\ell)$.
Therefore 3E and 3E imply 3E.

3F. Set $\alpha=\measuredangle(\text{\sc{w}},\text{\sc{u}})$ and
$\beta=\measuredangle(\text{\sc{v}},\text{\sc{u}})$. Try to guess the example
from the diagram.

The shown curve is divided into three arcs: I, II, and III. Arc I turns from v
to u; it has total curvature $\alpha$. Similarly, the arc III turns from u to
w and has total curvature $\beta$. Arc II goes very close and almost parallel
to the chord $pq$ and its total curvature can be made arbitrarily small.

3G. Use the fact that an exterior angle of a triangle equals the sum of the
two remote interior angles; for the second part apply induction on the number
of vertices.

3G. By 3G, $\Phi({\gamma})\geqslant\Phi({\beta})$; it remains to show that
$\Phi({\gamma})\leqslant\sup\\{\Phi({\beta})\\}$. In other words, given
$\varepsilon>0$ and any polygonal line
$\sigma=\text{\sc{u}}_{0}\dots\text{\sc{u}}_{k}$ inscribed in the tangent
indicatrix t of $\gamma$, we need to construct a polygonal line $\beta$
inscribed in $\gamma$ such that

$\mathop{\rm length}\nolimits\sigma<\Phi({\beta})+\varepsilon.$ ➋

Suppose $\text{\sc{u}}_{i}=\text{\sc{t}}(s_{i})$. Choose an inscribed
polygonal line $\beta=p_{0}\dots p_{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k+1}$ such that $p_{2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}i}$ and
$p_{2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}i+1}$ lie sufficiency close to
$\gamma(s_{i})$; so we can assume that the direction of $p_{2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}i+1}-p_{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}i}$ is sufficiency close to $\text{\sc{u}}_{i}$ for each $i$. Conclude
that ➋ ‣ Semisolutions holds true for the constructed polygonal line $\beta$.

3G. An example for 3G is shown on the diagram.

3G. Assume $x$ is a point of self-intersection. Show that we may choose two
points $y$ and $z$ on $\gamma$ so that the triangle $xyz$ is nondegenerate. In
particular, $\measuredangle[y\,{}^{x}_{z}]+\measuredangle[z\,{}^{y}_{x}]<\pi$,
or, equivalently, for a nonclosed polygonal line $xyzx$ we have
$\Phi({xyzx})>\pi$. It remains to apply 3G.

3G. Consider the closed polygonal line $acbd$. Observe that
$\Phi({acbd})=4{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. It remains to
apply 3G.

3H. Start with the curve $\gamma_{1}$ shown on the diagram. To obtain
$\gamma_{2}$, slightly unbend (that is, decrease the curvature of) the dashed
arc of $\gamma_{1}$.

3H. Choose a value $s_{0}\in[a,b]$ that splits the total curvature of $\gamma$
into two equal parts. Observe that
$\measuredangle(\gamma^{\prime}(s_{0}),\gamma^{\prime}(s))\leqslant\theta$ for
any $s$. Use this inequality in the same way as in the proof of the bow lemma.

3H. Let $\ell=\mathop{\rm length}\nolimits\gamma$. Suppose
$\ell_{1}<\ell<\ell_{2}$. Let $\gamma_{1}$ be an arc of unit circle with
length $\ell$.

Show that the distance between the ends of $\gamma_{1}$ is smaller than
$|p-q|$ and apply the bow lemma (3H).

3H. If $\mathop{\rm length}\nolimits\gamma<2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi$, apply the bow lemma (3H) to $\gamma$ and an arc of the unit circle
of the same length.

4B. The arc-length parameter $s$ is already found in 2C. It remains to find
the Frenet frame and calculate curvature and torsion. The latter can be done
by straightforward calculations; the answers are

$\displaystyle\text{\sc{t}}(t)$
$\displaystyle=\tfrac{1}{\sqrt{a^{2}+b^{2}}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(-a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin t,a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos t,b),$ $\displaystyle\text{\sc{n}}(t)$
$\displaystyle=(-\cos t,-\sin t,0),$ $\displaystyle\text{\sc{b}}(t)$
$\displaystyle=\tfrac{1}{\sqrt{a^{2}+b^{2}}}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(b{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin t,-b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos t,a),$ $\displaystyle\kappa$
$\displaystyle\equiv\tfrac{a}{a^{2}+b^{2}},\qquad\tau\equiv\tfrac{b}{a^{2}+b^{2}}.$

nbt

It remains to show that the map
$(a,b)\mapsto(\frac{a}{a^{2}+b^{2}},\frac{b}{a^{2}+b^{2}})$ sends bijectively
the half plane $a>0$ onto itself.

4C. By the product rule, we get

$\displaystyle\text{\sc{b}}^{\prime}$
$\displaystyle=(\text{\sc{t}}\times\text{\sc{n}})^{\prime}=\text{\sc{t}}^{\prime}\times\text{\sc{n}}+\text{\sc{t}}\times\text{\sc{n}}^{\prime}.$

It remains to substitute the values from ➊ ‣ 4C and ➋ ‣ 4C and simplify.

4C. This is a consequence of the equation $\text{\sc{b}}^{\prime}=-\tau{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}$.

4C. Observe that
$\tfrac{\gamma^{\prime}\times\gamma^{\prime\prime}}{|\gamma^{\prime}\times\gamma^{\prime\prime}|}$
is a unit vector perpendicular to the plane spanned by $\gamma^{\prime}$ and
$\gamma^{\prime\prime}$, so, up to sign, it has to be equal to b. It remains
to check that the sign is right.

4D; 4D. Observe that $\langle\text{\sc{w}},\text{\sc{t}}\rangle^{\prime}=0$.
Show that it implies that $\langle\text{\sc{w}},\text{\sc{n}}\rangle=0$.

By Frenet formulas, $\langle\text{\sc{w}},\text{\sc{n}}\rangle^{\prime}=0$
implies that $\langle\text{\sc{w}},-\kappa{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{t}}+\tau{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{b}}\rangle=0$.

4D. Show that $\text{\sc{w}}^{\prime}=0$; it implies that
$\langle\text{\sc{w}},\text{\sc{t}}\rangle=\tfrac{\tau}{\kappa}$. In
particular, the velocity vector of $\gamma$ makes a constant angle with w;
that is, $\gamma$ has constant slope.

4D. Suppose $\alpha$ is an evolvent of $\gamma$ and w is a fixed vector. Show
that $\langle\text{\sc{w}},\alpha\rangle$ is constant if $\gamma$ makes
constant angle with w.

4E. Part 4E follows from the fact that
$(\text{\sc{t}},\text{\sc{n}},\text{\sc{b}})$ is an orthonormal basis. For 4E
take the first and second derivatives of the identity
$\langle\gamma,\gamma\rangle=1$ and simplify using the Frenet formulas. Part
4E follows from 4E and the Frenet formulas. By 4E,
$\int\tfrac{\tau}{\kappa}=0$, hence 4E follows. Part 4E is proved by algebraic
manipulations.

4E. Use the Frenet formulas to show that $(\gamma+\tfrac{1}{\kappa}{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{n}}+\tfrac{\kappa^{\prime}}{\kappa^{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\tau}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\text{\sc{b}})^{\prime}=0$.

4F. Use the second statement in 4B.

4F. Note that the function

$\rho(\ell)=|\gamma(t+\ell)-\gamma(t)|^{2}=\langle\gamma(t+\ell)-\gamma(t),\gamma(t+\ell)-\gamma(t)\rangle$

is smooth and does not depend on $t$. Express speed, curvature and torsion of
$\gamma$ in terms of the derivatives $\rho^{(n)}(0)$. Be patient, you will
need two derivatives for the speed, four for the curvature and six for the
torsion. Once it is done, apply 4F.

5A. Without loss of generality, we may assume that $\gamma_{0}$ is
parametrized by its arc-length. Then

$\displaystyle|\gamma_{1}^{\prime}|$
$\displaystyle=|\gamma_{0}^{\prime}+\text{\sc{t}}^{\prime}|=|\text{\sc{t}}+\kappa{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{n}}|=\sqrt{1+\kappa^{2}}\geqslant
1=|\gamma_{0}^{\prime}|;$

that is, $|\gamma_{1}^{\prime}(t)|\geqslant|\gamma_{0}^{\prime}(t)|$ for any
$t\in[a,b]$. It remains to integrate this inequality and apply 2A.

5C. Observe that $\gamma^{\prime}_{a}(t)=(1+a{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\cos t,-a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin t)$; that is,
$\gamma^{\prime}_{a}$ runs clockwise along a circle with center at $(1,0)$ and
radius $|a|$.

Case $|a|>1$, Note that
$\text{\sc{t}}_{a}(t)=\gamma^{\prime}_{a}/|\gamma^{\prime}_{a}|$ runs
clockwise and makes a full turn in time $2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi$. Therefore $\Psi({\gamma_{a}})=-2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi$ and $\Phi({\gamma})_{a}=|\Psi({\gamma_{a}})|=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi$.

Case $|a|<1$. Set $\theta_{a}=\arcsin a$. Show that
$\text{\sc{t}}_{a}(t)=\gamma^{\prime}_{a}/|\gamma^{\prime}_{a}|$ starts with
the horizontal direction $\text{\sc{t}}_{a}(0)=(1,0)$, turns monotonically to
angle $\theta_{a}$, then monotonically to $-\theta_{a}$ and then monotonically
to back to $\text{\sc{t}}_{a}(2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi)=(1,0)$. It follows that if $|a|>1$, then $\Psi({\gamma_{a}})=0$ and
$\Phi({\gamma_{a}})=4{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\theta_{a}$.

Case $a=-1$. The velocity $\gamma^{\prime}_{-1}(t)$ vanishes at $t=0$ and
$2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. Nevertheless, the curve admits
a smooth regular parametrization — find it. In this case
$\Psi({\gamma_{-1}})=-\pi$ and $\Phi({\gamma_{-1}})=\pi$.

Case $a=1$. The velocity $\gamma^{\prime}_{1}(t)$ vanishes at $t=\pi$. At
$t=\pi$ the curve has a cusp; that is, $\gamma_{1}$ turns exactly back at the
time $\pi$. So $\gamma_{1}(t)$ has undefined total signed curvature. The curve
is a joint of two smooth arcs with external angle $\pi$, and the total
curvature of each arc is $\tfrac{\pi}{2}$, so
$\Phi({\gamma_{1}})=\tfrac{\pi}{2}+\pi+\tfrac{\pi}{2}=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi$.

5C. Look at the drawings; we assume that the two marked points in the last
example have parallel tangent lines.

5C; 5C. Show that

$\gamma_{\ell}^{\prime}(t)=(1-\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k(t)){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\gamma^{\prime}(t).$

By regularity of $\gamma$, $\gamma^{\prime}\neq 0$. So if
$\gamma_{\ell}^{\prime}(t)=0$, then $\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k(t)=1$.

5C. Observe that we can assume that $\gamma$ is parametrized by its arc-
length, so $\gamma^{\prime}(t)=\text{\sc{t}}(t)$. Suppose
$|\ell|<\frac{1}{\kappa(t)}$ for any $t$. Then

$|\gamma_{\ell}^{\prime}(t)|=(1-\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k(t)).$

Therefore

$\displaystyle L(\ell)$ $\displaystyle=\int_{a}^{b}(1-\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}k(t)){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}dt=\int_{a}^{b}1{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}dt-\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\int_{a}^{b}k(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dt=L(0)+\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\Psi({\gamma}).$

5C. Consider the unit circle $\gamma(t)=(\cos t,\sin t)$ for $t\in[0,2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\pi]$ and $\gamma_{\ell}$ for $\ell=2$.

5D. Use the definition of osculating circle via order of contact and use the
fact that inversions send circles to circlines.

5E. Find $\text{\sc{t}}(t)$ and $\text{\sc{n}}(t)$. Use the formula in 3C to
calculate curvature $\kappa(t)$. Apply the formula given right before the
exercise.

5E. Start with a plane spiral curve as shown on the diagram. Increase the
torsion of the dashed arc without changing the curvature until a self-
intersection appears.

5E. Observe that if a line or circle is tangent to $\gamma$, then it is
tangent to the osculating circle at the same point. Then apply the spiral
lemma (5E).

6B. Apply the spiral lemma (5E).

Direct solution. We will assume that the curvature does not vanish at $p$, the
remaining case is simpler. We may assume that $\gamma$ is a unit-speed curve
and $p=\gamma(0)$. Further, we may assume that be the center of curvature of
$\gamma$ at $p$ is the origin; in other words, $\kappa(0){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\gamma(0)+\text{\sc{n}}(0)=0$.

Since the osculating circle supports $\gamma$ at $p$, we get that the function
$f\colon t\mapsto\langle\gamma(t),\gamma(t)\rangle$ has a local minimum or
maximum at $0$.

Direct calculations show that

$\displaystyle f^{\prime}$
$\displaystyle=\langle\gamma,\gamma\rangle^{\prime}=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle\text{\sc{t}},\gamma\rangle,$
$\displaystyle f^{\prime\prime}$ $\displaystyle=2{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{t}},\gamma\rangle^{\prime}=2{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{t}}^{\prime},\gamma\rangle+2{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{t}},\text{\sc{t}}\rangle=2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\kappa{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{n}},\gamma\rangle+2,$ $\displaystyle
f^{\prime\prime\prime}$ $\displaystyle=2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\kappa^{\prime}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{n}},\gamma\rangle+2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\kappa{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{n}}^{\prime},\gamma\rangle+\cancel{2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\kappa{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle\text{\sc{n}},\text{\sc{t}}\rangle}.$

Observe that $\text{\sc{n}}^{\prime}(0)\perp\gamma(0)$. Therefore
$f^{\prime}(0)=0$, $f^{\prime\prime}(0)=0$, and
$f^{\prime\prime\prime}(0)=\nobreak-2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\kappa^{\prime}/\kappa$. Since $f$ is a maximum or minimum at $0$, we
get that $f^{\prime\prime\prime}(0)=0$ and therefore $\kappa^{\prime}=0$.

6C. Consider the coordinate system with $p$ as the origin and $x$-axis as the
common tangent line to $\gamma_{1}$ and $\gamma_{2}$. We may assume that
$\gamma_{i}$ are defined in $(-\varepsilon,\varepsilon)$ for some small
$\varepsilon>0$, so that they run almost horizontally to the right.

Given $t\in[0,1]$ consider the curve $\gamma_{t}$ that is tangent and
cooriented to the $x$ axis at $\gamma_{t}(0)$ and has signed curvature defined
by $k_{t}(s)=(1-\nobreak t){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}k_{0}(s)+t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}k_{1}(s)$. It exists by
5B.

Choose $s\approx 0$. Consider the curve $\alpha_{s}\colon
t\mapsto\gamma_{t}(s)$. Show that $\alpha_{s}$ moves almost vertically up. Use
the fact that $\gamma_{t}$ moves almost horizontally to the right to show that
in a small neighborhood of $p$, the curve $\gamma_{1}$ lies above
$\gamma_{0}$, whence the statement follows.

6C. Reduce the radius of the circle until it touches $\gamma$. Observe that
the circle supports $\gamma$ and apply 6C.

6C \+ 6C \+ 6C. Observe that one of the arcs of curvature 1 in the families
shown on the diagram supports $\gamma$ and apply 6C.

To do the second part in 6C, use the shown family and another family of arcs
curved in the opposite direction. (Compare to 6E.)

6D. Note that we can assume that $\gamma$ bounds a convex figure $F$,
otherwise by 6D its curvature changes sign and therefore it has zero curvature
at some point.

Choose two points $x$ and $y$ surrounded by $\gamma$ such that $|x-\nobreak
y|>\nobreak 2$. Look at the maximal lens bounded by two arcs with a common
chord $xy$ that lies in $F$. Apply the supporting test (6C).

6D; 6D. Apply the lens lemma to show that if $p_{2}$ lies between $p_{1}$ and
$p_{3}$, then the curvature of $\gamma$ switches its sign.

6D. Show that $p_{4}$ lies between $p_{2}$ and $p_{3}$; further $p_{5}$ lies
between $p_{3}$ and $p_{4}$; and so on.

6D. According to 6D, the point $p_{3}$ might lie between $p_{1}$ and $p_{2}$
and then further order is determined uniquely or $p_{1}$ lies between $p_{2}$
and $p_{3}$. In the latter case we have two choices, either $p_{4}$ lies
between $p_{2}$ and $p_{3}$ and then further order is determined uniquely or
$p_{2}$ lies between $p_{3}$ and $p_{4}$. In the latter case we get a choice
again.

Assume we make the first choice on the step number $k$. Without loss of
generality we may assume that $p_{k}$ lies to the right from $p_{k-2}$. Then
we have the following order:

$p_{k-2},p_{p-4},\dots,p_{p-5},p_{k-3},p_{k},p_{k+2},\dots,p_{k+1},p_{k-1}.$

The case $k=7$ is shown on the diagram.

6E. Note that $\gamma$ contains a simple loop; apply to it 6E.

6F. Repeat the proof of the theorem for each cyclic concatenation of an arc of
$\gamma$ from $p_{i}$ to $p_{i+1}$ with a large arc of the circle.

An example for the second part can be guessed from the diagram.

7E. Denote the set of solutions by $\Sigma_{\ell}$.

Show that $\nabla_{p}f=0$ if and only if $p=(0,0,0)$. Use 7E to conclude that
if $\ell\neq 0$, then $\Sigma_{\ell}$ is a union of disjoint smooth regular
surfaces.

Show that $\Sigma_{\ell}$ is connected if and only if $\ell\leqslant 0$. It
follows that if $\ell<0$, then $\Sigma_{\ell}$ is a smooth regular surface,
and if $\ell>0$ — it is not.

The case $\ell=0$ has to be done by hand — it does not satisfy the sufficient
condition in 7E, but that does not solely imply that $\Sigma_{0}$ is not a
smooth surface.

Show that any neighborhood of origin in $\Sigma_{0}$ cannot be described by a
graph in any coordinate system; so by the definition (Section 7B) $\Sigma_{0}$
is not a smooth surface.

7F. First check that the image of $s$ lies in the unit sphere centered at
$(0,0,1)$; that is, show that

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

for any $u$ and $v$.

Furthermore, show that the map $(x,y,z)\mapsto(\tfrac{2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}x}{x^{2}+y^{2}+z^{2}},\tfrac{2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y}{x^{2}+y^{2}+z^{2}})$ describes the inverse
of $s$, which is continuous away from the origin. In particular, $s$ is an
embedding that covers the entire sphere except for the origin.

It remains to show that $s$ is regular; that is, $s_{u}$ and $s_{v}$ are
linearly independent at all points of the $(u,v)$-plane.

7F. Set $s\colon(t,\theta)\mapsto(x(t),y(t){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\cos\theta,y(t){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin\theta)$. Show
that $s$ is regular; that is, $s_{t}$ and $s_{\theta}$ are linearly
independent. (It might help to observe that $s_{t}\perp s_{\theta}$).

Show that $s$ is a embedding; that is, any $(t_{0},\theta_{0})$ admits a
neighborhood $U$ in the $(t,\theta)$-plane such that the restriction $s|_{U}$
has a continuous inverse. It remains to apply 7F.

7G. The solutions of these exercises are built on the following general
construction known as the _Moser trick_.

Suppose that $\text{\sc{u}}_{t}$ is a smooth time-dependent vector field on a
plane. Consider the ordinary differential equation
$x^{\prime}(t)=\text{\sc{u}}_{t}(x(t))$. Consider the map $\iota\colon
x(0)\mapsto x(1)$ where $t\mapsto x(t)$ is a solution of the equation. The map
$\iota$ is called the _flow_ of the vector field $\text{\sc{u}}_{t}$ for the
time interval $[0,1]$. Observe that according to 0G the map $\iota$ is smooth
in its domain of definition. Moreover the same holds for its inverse
$\iota^{-1}$; indeed, $\iota^{-1}$ is the flow of the vector field
$-\text{\sc{u}}_{1-t}$. That is, $\iota$ is a diffeomorphism from its domain
of definition to its image.

Therefore, in order to construct a diffeomorphism from one open subset of the
plane to another it is sufficient to construct a smooth vector field such that
its flow maps one set to the other; such a map is automatically a
diffeomorphism.

7G. Suppose $\Sigma=\mathbb{R}^{2}\backslash{p_{1},\dots,p_{n}}$ and
$\Theta=\mathbb{R}^{2}\backslash{q_{1},\dots,q_{n}}$. Choose smooth paths
$\gamma_{i}\colon[0,1]\to\nobreak\mathbb{R}^{2}$ such that
$\gamma_{i}(0)=p_{i}$, $\gamma_{i}(1)=q_{i}$, and
$\gamma_{i}(t)\neq\gamma_{j}(t)$ if $i\neq j$.

Choose a smooth vector field $\text{\sc{v}}_{t}$ such that
$\text{\sc{v}}_{t}(\gamma_{i}(t))=\gamma^{\prime}_{i}(t)$ for any $i$ and $t$.
We can assume in addition that $\text{\sc{v}}_{t}$ vanishes outside of a
sufficiently large disc; this can be arranged by multiplying the vector field
by a function $\sigma_{1}(R-|x|)$; see page 0E.

It remains to apply the Moser trick to the constructed vector field.

7G–7G. Without loss of generality we can assume that the origin belongs to
both $\Sigma$ and $\Theta$.

In each case show that there is a vector field v defined on
$\mathbb{R}^{2}\backslash\\{0\\}$ that flows $\Sigma$ to $\Theta$. In fact one
can choose radial fields of that type, but be careful with the cases 7G and 7G
— they are not as easy as one might think.

8A. Let $\gamma$ be a smooth curve in $\Sigma$. Observe that
$f\circ\gamma(t)\equiv 0$. Differentiate this identity and apply the
definition of tangent vector (8A).

8A. Assume a neighborhood of $p$ in $\Sigma$ is a graph $z=f(x,y)$. In this
case $s\colon(u,v)\mapsto\nobreak(u,v,f(u,v))$ is a smooth chart at $p$. Show
that the plane spanned by $s_{u}$ and $s_{v}$ is not vertical; together with
8A, this proves the if part.

For the only-if part, fix a chart $s\colon(u,v)\mapsto(x(u,v),y(u,v),z(u,v))$,
and apply the inverse function theorem to the map
$(u,v)\mapsto\nobreak(x(u,v),y(u,v))$.

8A. Choose $(x,y,z)$-coordinates so that $\Pi$ is the $(x,y)$-plane and $p$ is
the origin. Let $(u,v)\mapsto s(u,v)$ be a small chart of $\Sigma$ such that
$p=s(0,0)$. Denote by k the unit vector in the direction of the $z$ axis.

Show that we can assume that $\langle s(u,v),\text{\sc{k}}\rangle$ has
constant sign in a punctured neighborhood of $0$. Conclude that
$s_{u}\perp\text{\sc{k}}$ and $s_{v}\perp\text{\sc{k}}$ at $0$, hence the
result.

8F. By 8A, $\nu=\tfrac{\nabla h}{|\nabla h|}$ defines a unit normal field on
$\Sigma$.

8G. Use cutoffs and mollifiers from the analysis section (0E) to construct a
smooth nonnegative function $f$ on the $(x,y)$-plane such that $f(x,y)=0$ if
and only if $(x,y)\in A$. Observe that the graph $z=f(x,y)$ describes the
required surface.

9B. Fix a point $p$ on $\gamma$. Since $\Sigma$ is mirror symmetric with
respect to $\Pi$, so is the tangent plane $\mathrm{T}_{p}$.

Choose $(x,y)$-coordinates on $\mathrm{T}_{p}$ so that the $x$-axis is the
intersection $\Pi\cap\mathrm{T}_{p}$. Suppose that the osculating paraboloid
is described by the graph $z=\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y^{2})$. Since $\Sigma$ is mirror symmetric, so
is the paraboloid; that is, changing $y$ to $(-y)$ does not change the value
$\ell{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x^{2}+2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}m{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}x{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}y+n{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}y^{2}$. In other words $m=0$, or equivalently,
the $x$-axis points in a direction of curvature.

9C. Note that the principal curvatures have the same sign at each point.
Therefore we can choose a unit normal $\nu$ at each point so that both
principal curvatures are positive. Show that it defines a global field on the
surface.

9D. Apply 9D to a map $s$ such that $s_{u}(0,0)=\text{\sc{u}}$ and
$s_{v}(0,0)=\text{\sc{v}}$.

9D; 9D. Observe that $\Sigma$ has unit Hessian matrix at each point and apply
the definition of shape operator.

9D. Choose a chart $s$ in $\Sigma$. Show that $\tfrac{\partial}{\partial
u}(s+\nu)=\tfrac{\partial}{\partial v}(s+\nu)=0$. Make a conclusion.

9D. We can assume that $\gamma$ is parametrized by arc-length. Denote by
$\nu_{1}(s)$ and $\nu_{2}(s)$ the unit normal vectors to $\Sigma_{1}$ and
$\Sigma_{2}$ at $\gamma(s)$. Since $\gamma$ is a curvature line in
$\Sigma_{1}$, $\nu_{1}^{\prime}$ is proportional to $\gamma^{\prime}$; in
particular $\langle\nu_{1}^{\prime},\nu_{2}\rangle=0$.

By the assumption, $\langle\nu_{1}(t),\nu_{2}(t)\rangle$ is constant. By
taking its derivative and applying the above identity, show that
$\langle\nu_{1},\nu_{2}^{\prime}\rangle=0$. Conclude that $\nu_{2}^{\prime}$
is proportional to $\gamma^{\prime}$ and therefore $\gamma$ is a curvature
line in $\Sigma_{2}$.

9D; 9D. Fix $t$. Set $f_{t}\colon p\mapsto p+t{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\nu(p)$; it maps $\Sigma$ to $\Sigma_{t}$.

Apply the definition of shape operator to show that
$d_{p}f_{t}(\text{\sc{v}})=\text{\sc{v}}-t{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\mathop{\rm Shape}\nolimits_{p}(\text{\sc{v}})$. Since $\Sigma$ is
closed, the norm of $\mathop{\rm Shape}\nolimits$ is bounded. Whence $f_{t}$
is regular for $t$ sufficiently close to $0$.

9D. By the area formula (8E), we have

$a(t)=\int_{p\in\Sigma}\mathop{\rm jac}\nolimits_{p}f_{t}.$

Show and use that for fixed $p$, the function $t\mapsto\mathop{\rm
jac}\nolimits_{p}f_{t}$ has derivative $-H(p)$ at zero.

9E. Apply 9E and the definition of mean curvature.

9E. Use Meusnier’s theorem (9E), to find the center and radius of curvature of
$\gamma$ in terms of its normal curvature at $p$; make a conclusion.

9E. Use 9B and Meusnier’s theorem (9E).

9E. Use 9E.

9E. Apply Meusnier’s theorem (9E) to show that the coordinate curves
$\alpha_{v}\colon u\mapsto s(u,v)$ and $\beta_{u}\colon v\mapsto s(u,v)$ are
asymptotic; that is, they have vanishing normal curvature.

Observe that these two families are orthogonal to each other. Therefore the
Hessian matrix in the frame $s_{u}/|s_{u}|$ and $s_{v}/|s_{v}|$ will have
zeros on the diagonal. Use the fact that the mean curvature is the trace of
the Hessian matrix.

9F. Use 9B and 6E.

9. Drill an extra hole or combine two examples together.

9. Denote by $L$ the _cut locus_ of $V$; that is, $L$ is a closure of the set of points $x\in V$ such that there are at least two points on $\partial V$ that minimize the distance to $x$.

Choose a connected component $\Sigma$ of the boundary $\partial V$. Show that
$L$ is a smooth surface and the closest point projection $L\to\Sigma$ is a
smooth regular parametrization of $\Sigma$. In particular there is a unique
point on $\Sigma$ that minimizes the distance to a given point $x\in L$. It
follows that there is another component $\Sigma^{\prime}$ with the same
property.

Finally show that $\partial V$ cannot be more than two components.

9. Read about Bing’s two-room house. Try to thicken it to construct the needed example.

Assume $V$ does not contain a ball of radius $r_{3}$. Show that its cut locus
$L$ is formed by a few smooth surfaces that meet by three along their boundary
curves. Use this description to show that $L$ is not _simply-connected_ ; that
is, there is a loop in $L$ that cannot be deformed continuously to a trivial
loop. Conclude that $V$ is not simply-connected.

Finally, arrive at a contradiction by showing that if $V$ is bounded by a
smooth sphere, then $V$ is simply-connected.

10A. Choose curvatures such that $k_{2}(p)_{\Sigma_{1}}>\nobreak
k_{2}(p)_{\Sigma_{2}}>k_{1}(p)_{\Sigma_{1}}>k_{1}(p)_{\Sigma_{2}}$ and suppose
that the first principal direction of $\Sigma_{1}$ coincides with the second
principal direction of $\Sigma_{2}$ and vice-versa.

10A \+ 10A. Apply the same reasoning as in the problems 6C–6C, but use
families of spheres instead.

10B. Show and use that any tangent plane $\mathrm{T}_{p}$ supports $\Sigma$ at
$p$.

10B. Assume a maximal ball in $V$ has radius $r$ and touches the boundary of
$R$ at the points $p$ and $q$. Consider the projection $R$ of $V$ to a plane
thru $p$, $q$ and the center of the ball. Show that $R$ is bounded by a smooth
closed convex curve with curvature at most $1$. Argue as in 6C to show that
$r\geqslant 1$.

10B. Suppose that a point $p$ lies in the intersection $\Pi\cap\Sigma$.

Show that if the tangent plane $\mathrm{T}_{p}\Sigma$ is parallel to $\Pi$,
then $p$ is an isolated point of the intersection $\Pi\cap\Sigma$.

It follows that if $\gamma$ is a connected component of the intersection
$\Pi\cap\Sigma$ that is not an isolated point, then $\Pi$ _intersects $\Sigma$
transversally_ along $\gamma$; that is, at each point $p\in\gamma$ the tangent
plane $\mathrm{T}_{p}\Sigma$ is not parallel to $\Pi$. Apply the implicit
function theorem to show that $\gamma$ is a smooth regular curve.

Finally, observe that the curvature of $\gamma$ cannot be smaller than the
normal curvature of $\Sigma$ in the same direction. Whence $\gamma$ has no
points with vanishing curvature, and therefore its signed curvature has
constant sign.

10B. Look for a supporting spherical dome with the unit circle as the
boundary.

10B. Note that we can assume that the surface has positive Gauss curvature,
otherwise the statement is evident. By 10B, the surface bounds a convex region
that contains a line segment of length $\pi$.

Use 3C and 9E to show that the Gauss curvature of the surface of revolution of
the graph $y=a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin x$ for $x\in(0,\pi)$
cannot exceed $1$. Try to support the surface $\Sigma$ from inside by a
surface of revolution of the described type. (Compare to 6D.)

Remark. The exercise can be reduced to the following deeper result: _if the
Gauss curvature of $\Sigma$ is at least $1$, then the intrinsic diameter of
$\Sigma$ cannot exceed $\pi$_. The latter means that any two points in
$\Sigma$ can be connected by a path that lies in $\Sigma$ and has length at
most $\pi$. This theorem was proved by Heinz Hopf and Willi Rinow [40] and
named after Sumner Myers who generalized it [62].

10B. To prove 10B use the convexity of $R$. To prove 10B observe that the map
$\Sigma\to\mathbb{S}^{2}$ is smooth and regular, then apply the inverse
function theorem to show that its inverse is smooth as well.

10B; 10B We can assume that the origin lies on $\Sigma$. Consider a sequence
of points $x_{n}\in\Sigma$ such that $|x_{n}|\to\nobreak\infty$ as
$n\to\infty$. Denote by $\text{\sc{u}}_{n}$ the unit vector in the direction
of $x_{n}$; that is $\text{\sc{u}}_{n}=\tfrac{x_{n}}{|x_{n}|}$.

Since the unit sphere is compact, we can pass to a subsequence of $(x_{n})$
such that $\text{\sc{u}}_{n}$ converges to a unit vector u. Show that the
half-line from the origin in the direction of u can be taken as $\ell$.

10B \+ 10B \+ 10B. Since $R$ is convex, so is its projection $\Omega$.

Note and use that for any $q\in\Sigma$, the directions
$\text{\sc{v}}_{n}=\tfrac{x_{n}-q}{|x_{n}-q|}$ converge to u as well.

Show that $\Sigma$ has no vertical tangent planes. Conclude that the
projection map from $\Sigma$ to the $(x,y)$-plane is regular. Use the inverse
function theorem to show that $\Omega$ is open.

10B. Arguing by contradiction, suppose that for some sequence
$(x_{n},y_{n})\to\nobreak(x_{\infty},y_{\infty})\in\partial\Omega$ the
sequence $f(x_{n},y_{n})$ stays bounded above. We can pass to a subsequence
such that either $f(x_{n},y_{n})$ converges to some finite value $z_{\infty}$
or it diverges to $-\infty$.

In the first case, show that the point $(x_{\infty},y_{\infty},z_{\infty})$
does not lie on $\Sigma$, but it has arbitrarily close points on $\Sigma$.
That is $\Sigma$ is not proper — a contradiction.

If $f(x_{n},y_{n})\to-\infty$, use the convexity of $f$ to show that
$f(\tfrac{x_{n}}{2},\tfrac{y_{n}}{2})\to\nobreak-\infty$. Note that origin
belongs to $\Omega$; use it to show that
$(\tfrac{x_{\infty}}{2},\tfrac{y_{\infty}}{2})\in\Omega$. Arrive to a
contradiction.

10B. By 10B, $\Sigma$ is parametrized by an open convex plane domain $\Omega$.
It remains to show that $\Omega$ can parametrize the whole plane.

We may assume that the origin of the plane lies in $\Omega$. Show that in this
case the boundary of $\Omega$ can be written in polar coordinates as
$(\theta,f(\theta))$ where $f\colon\mathbb{S}^{1}\to\mathbb{R}$ is a positive
continuous function. Then homeomorphism to the plane can be described in polar
coordinates by changing only the radial coordinate; for example as
$(\theta,r)\mapsto\nobreak(\theta,\tfrac{r}{1-r/f(\theta)})$.

To do the second part, one may apply 7G.

10B. Choose a coordinate system so that the $(x,y)$-plane supports $\Sigma$ at
the origin, so $\Sigma$ lies in the upper half-space.

Show that there is $\varepsilon>0$ such that any line starting from the origin
with slope at most $\varepsilon$ may intersect $\Sigma$ only in the unit ball
centered at the origin; we may assume that $\varepsilon$ is small, say
$\varepsilon<1$. Consider the cone formed by the half-lines from the origin
with slope $\varepsilon$ shifted down by $1$ and observe that the entire
surface lies inside this cone.

10B. Choose distinct points $p,q\in\Sigma$. Apply 10B to show that the angle
$\measuredangle(\nu(p),p-q)$ is acute and $\measuredangle(\nu(q),p-q)$ is
obtuse. Conclude that $\nu(p)\neq\nu(q)$; that is,
$\nu\colon\Sigma\to\mathbb{S}^{2}$ is injective.

10B. Given a unit vector u, consider a point $p\in\Sigma$ that maximizes the
scalar product $\langle p,\text{\sc{u}}\rangle$. Show that
$\nu(p)=\text{\sc{u}}$. Conclude that the spherical map
$\nu\colon\Sigma\to\mathbb{S}^{2}$ is onto, and therefore it is a bijection.

Applying 9D, we get that the integral is $4{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi=\mathop{\rm area}\nolimits\mathbb{S}^{2}$.

10B. Choose an $(x,y,z)$-coordinate system provided by 10B. Observe that for
any $p$ the normal vector $\nu(p)$ forms an obtuse angle with the $z$-axis. It
follows that the image $\nu(\Sigma)$ lies in the south hemisphere.

Applying 9D, we get that the integral is $2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\pi=\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathop{\rm
area}\nolimits\mathbb{S}^{2}$.

10C. Apply 9E.

10C. Prove and use that each point $p\in\Sigma$ has a direction with vanishing
normal curvature.

10C. Suppose $p\in\Sigma$ is a point of strict local maximum of $f$. Arrive to
a contradiction by showing that $\Sigma$ is supported by its tangent plane at
$p$.

10C. Denote by $\Pi_{t}$ the tangent plane to $\Sigma$ at $\gamma(t)$ and by
$\ell_{t}$ the tangent line to $\gamma$ at time $t$.

Note that $\Pi_{t}$ is the graph of a linear function, say $h_{t}$, defined on
the $(x,y)$-plane. Denote by $\bar{\ell}_{t}$ the projection of $\ell_{t}$ to
the $(x,y)$-plane. Show that the derivative $\tfrac{d}{dt}h_{t}(w)$ vanishes
at the point $w$ if and only if $w\in\bar{\ell}_{t}$ and the derivative
changes sign if $w$ goes from one side of $\bar{\ell}_{t}$ to the other.

Denote by $\bar{\gamma}$ the projection of $\gamma$ to the $(x,y)$-plane. If
$\bar{\gamma}$ is star shaped with respect to a point $w$, then $w$ cannot
cross $\bar{\ell}_{t}$. Therefore the function $t\mapsto h_{t}(w)$ is
monotonic on $\mathbb{S}^{1}$. Observe that this function cannot be constant,
and arrive at a contradiction.

10C. This problem follows easily from the so-called _Morse lemma_. The
following sketch is a slightly stripped version of it. A more conceptual proof
can be built on the Moser’s trick [65].

Choose tangent-normal coordinates at $p$ so that the coordinate axes point in
the principal directions of $\Sigma$ at $p$; let $z=f(x,y)$ be the local graph
representation of $\Sigma$. We need to show that the solution of the equation
$f(x,y)=0$ is a union of two smooth curves that intersect transversely at $p$.

Prove the following claim:

  * $\diamond$

Suppose that $x\mapsto h(x)$ is a smooth function defined in an open interval
$\mathbb{I}\ni 0$ such that $h(0)=h^{\prime}(0)=0$ and
$h^{\prime\prime}(0)>0$. Then, for a smaller interval $\mathbb{J}\ni 0$ there
is a unique smooth function $a\colon\mathbb{J}\to\mathbb{R}$ such that
$h=a^{2}$, $a(0)=0$ and $a^{\prime}(0)>0$.

Note that by passing to a small rectangular domain $|x|,|y|<\varepsilon$, we
can assume that $f_{xx}>\varepsilon$ and $f_{yy}<-\varepsilon$. Show that if
$\varepsilon$ is small, then for every $x$ there is unique $y(x)$ such that
$f_{x}(x,y(x))=0$; moreover the function $x\mapsto y(x)$ is smooth.

Set $h(x)=f(x,y(x))$. Note that $h(0)=h^{\prime}(0)=0$ and
$h^{\prime\prime}>0$. Applying the claim, we get a function $a$ such that
$h=a^{2}$, where $a(0)=0$ and $a^{\prime}(0)>0$.

Observe that $g(x,y)=h(x)-f(x,y)\geqslant 0$ and $g_{y}(x,y(x))=\nobreak
g(x,y(x))=\nobreak 0$ and $g_{yy}>0$. Applying the claim to each function
$y\mapsto g(x,y)$ with fixed $x$, we get that $g(x,y)=b(x,y)^{2}$ for a smooth
function $b$ such that $b(x,y(x))=0$ and $b_{y}(x,y(x))>0$.

It follows that

$\displaystyle f(x,y)$ $\displaystyle=a(x)^{2}-b(x,y)^{2}=(a(x)-b(x,y)){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}(a(x)+b(x,y)).$

That is $f(x,y)=0$ if $a(x)\pm b(x,y)=0$.

It remains to observe that the two functions $g_{\pm}(x,y)=a(x)\pm b(x,y)$
have distinct non-zero gradients at $0$. Therefore each equation $a(x)\pm
b(x,y)=0$ defines a smooth regular curve in a neighborhood of $p$; see Section
1E.

10D. Use 10D and the hemisphere lemma (2H).

10D. Assume $\Sigma$ is an open saddle surface that lies in a cone $K$. Show
that there is a plane $\Pi$ that cuts $\Sigma$ and cuts from $K$ a compact
region. Conclude that $\Pi$ cuts from $\Sigma$ a compact region as well.

By 8G one can move the plane $\Pi$ slightly so that it cuts from $\Sigma$ a
compact surface with boundary. Apply 10D.

10D. Consider the radial projection of $F_{\varepsilon}$ to the sphere
$\Sigma$ with center at $p=(0,0,\varepsilon)$; that is, a point $q\in
F_{\varepsilon}$ is mapped to a point $s(q)$ on the sphere that lies on the
ray $pq$.

Show that $s$ is a diffeomorphism from $F_{\varepsilon}$ to the south
hemisphere of $\Sigma$. It remains to observe that the unit disc is
diffeomorphic to the hemisphere.

10D. Apply 10D.

10E. Find an example among the surfaces of revolution. Use 9E in the proof.

10E. Look at the sections of the graph by planes parallel to the $(x,y)$-plane
and to the $(x,z)$-plane, then apply Meusnier’s theorem (9E).

10E. Suppose that orthogonal projection of $\Sigma$ to the $(x,y)$-plane is
not injective. Show that there is a point $p\in\Sigma$ with a vertical tangent
plane; that is, $\mathrm{T}_{p}\Sigma$ is perpendicular to the $(x,y)$-plane.

Let $\Gamma_{p}$ be the connected component of $p$ in the intersection of
$\Sigma$ and the affine tangent plane $\mathrm{T}_{p}\Sigma$. Use 10C to show
that $\Gamma_{p}$ is a union of smooth regular curves that can cross each
other transversely. Moreover two of these curves pass thru $p$ and
$\Gamma_{p}$ does not bound a compact region on $\Sigma$.

Show that $\Gamma_{p}$ must have at least 4 ways to escape to infinity. On the
other hand, since $\Sigma$ is a graph outside of compact set $K$, we have that
$\Gamma_{p}\backslash K$ is a graph of a real-to-real function that has only
two ways to escape to infinity — a contradiction.

11A. Cut the lateral surface of the ice-mountain by a line from the cowboy to
the tip; unfold it on the plane (see the picture) and try to figure out what
is the image of the strained lasso.

11C. Note that by 10B, $\Sigma$ bounds a strictly convex region. Therefore we
can assume that $\nu(p)\neq\nu(q)$, otherwise $p=q$ and the inequality is
evident.

Further, we can assume that $\nu(p)+\nu(q)\neq 0$, otherwise the right hand
side is undefined.

In the remaining case the tangent planes $\mathrm{T}_{p}$ and $\mathrm{T}_{q}$
intersect along a line, say $\ell$. Set $\alpha=\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\measuredangle(\nu(p),\nu(q))$; observe that
$2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\cos\alpha=|\nu(p)+\nu(q)|$. Let
$x\in\ell$ be the point that minimizes the sum $|p-x|+\nobreak|x-q|$. Show
that $\measuredangle[x\,{}^{p}_{q}]\geqslant\pi-2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\alpha$. Conclude that

$|p-x|+|x-q|\leqslant\frac{|p-q|}{\cos\alpha}.$

Finally, apply 11C to show that

$|{p}-\nobreak{q}|_{\Sigma}\leqslant|p-x|+|x-q|.$

11C. Suppose there is a minimizing curve $\gamma\not\subset\Delta$ with
endpoints $p$ and $q$ in $\Delta$.

Without loss of generality, we may assume that only one arc of $\gamma$ lies
outside of $\Delta$. Reflection of this arc with respect to $\Pi$ together
with the remaining part of $\gamma$ forms another curve $\hat{\gamma}$ from
$p$ to $q$; it runs partly along $\Sigma$ and partly outside $\Sigma$, but
does not get inside $\Sigma$. Note that

$\mathop{\rm length}\nolimits\hat{\gamma}=\mathop{\rm length}\nolimits\gamma.$

Denote by $\bar{\gamma}$ the closest point projection of $\hat{\gamma}$ on
$\Sigma$. Note that the curve $\bar{\gamma}$ lies in $\Sigma$, it has the same
ends as $\gamma$, and by 11C

$\mathop{\rm length}\nolimits\bar{\gamma}<\mathop{\rm length}\nolimits\gamma.$

This means that $\gamma$ is not length minimizing, a contradiction.

11C. Show that the closest point projection $\partial B\to\Sigma$ is
surjective and apply 11C.

12A. We can assume that the origin lies on the axis of revolution and i points
in the direction of such axis. Use 12A to show that it is sufficient to prove
that $\langle\gamma^{\prime}\times\gamma,\text{\sc{i}}\rangle$ is constant.

Since $\gamma^{\prime\prime}(t)\perp\mathrm{T}_{\gamma(t)}$, the three vectors
i, $\gamma$, and $\gamma^{\prime\prime}$ lie in the same plane. In particular
$\langle\gamma^{\prime\prime}\times\gamma,\text{\sc{i}}\rangle=0$. Therefore
$\langle\gamma^{\prime}\times\gamma,\text{\sc{i}}\rangle^{\prime}=\langle\gamma^{\prime}\times\gamma^{\prime},\text{\sc{i}}\rangle+\langle\gamma^{\prime\prime}\times\gamma,\text{\sc{i}}\rangle=0$.

12A. By Lemma 12A, we can assume that $\gamma$ is parametrized by arc-length.
By the definition of geodesic, we have that
$\gamma^{\prime\prime}(s)\perp\mathrm{T}_{\gamma(s)}$. Therefore

$\gamma^{\prime\prime}(s)=k_{n}(s){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\nu(\gamma(s)),$

where $k_{n}(s)$ is the normal curvature of $\gamma$ at time $s$. Since
$\gamma$ is asymptotic, $k_{n}(s)\equiv 0$; that is,
$\gamma^{\prime\prime}(s)\equiv 0$, therefore $\gamma^{\prime}$ is constant
and $\gamma$ runs along a line segment; see 3B.

12A. Denote by $\mu$ a unit vector perpendicular to $\Pi$. Since $\gamma$ lies
in $\Pi$, we have that $\gamma^{\prime\prime}$ is parallel to $\Pi$, or
equivalently $\gamma^{\prime\prime}\perp\mu$. Since $\gamma$ is unit speed, 3A
implies that $\gamma^{\prime\prime}\perp\gamma^{\prime}$.

Since $\Sigma$ is mirror symmetric with respect to the plane $\Pi$, the
tangent plane $\mathrm{T}_{\gamma(t)}\Sigma$ is also mirror symmetric with
respect to $\Pi$ as well. It follows that $\mathrm{T}_{\gamma(t)}\Sigma$ is
spanned by $\mu$ and $\gamma^{\prime}(t)$. Hence
$\gamma^{\prime\prime}\perp\mu$ and
$\gamma^{\prime\prime}\perp\gamma^{\prime}$ imply
$\gamma^{\prime\prime}\perp\mathrm{T}_{\gamma(t)}\Sigma$; that is, $\gamma$ is
a geodesic.

12B. By 12A, any meridian of $\Sigma$ is a closed geodesic. Consider an
arbitrary closed geodesic $\gamma$.

In the case when $\gamma$ is tangent to a meridian at some point, then by the
uniqueness part of Proposition 12B, $\gamma$ runs along that meridian; in
particular it is non-contractible.

In the remaining case, $\gamma$ can intersect meridians only transversely.
Therefore the longitude of $\gamma$ is monotonic. Whence again $\gamma$ is
non-contractible.

12D. Show that $\nu(\gamma(t))=(\cos t,\sin t,0)$. Calculate
$\gamma^{\prime\prime}(t)$ and show that it is proportional to
$\nu(\gamma(t))$. The latter is equivalent to
$\gamma^{\prime\prime}(t)\perp\mathrm{T}_{\gamma(t)}$. Note that the line
segment from $\gamma(0)$ to $\gamma(2\pi)$ is contained in $\Sigma$.

12D. Assume that two shortest paths $\alpha$ and $\beta$ have two common
points $p$ and $q$. Denote by $\alpha_{1}$ and $\beta_{1}$ the arcs of
$\alpha$ and $\beta$ from $p$ to $q$. Suppose that $\alpha_{1}$ is distinct
from $\beta_{1}$.

Note that $\alpha_{1}$ and $\beta_{1}$ are shortest paths with the same
endpoints; in particular they have the same length. Exchanging $\alpha_{1}$ in
$\alpha$ to $\beta_{1}$ produces a shortest path, say $\hat{\alpha}$, that is
distinct from $\alpha$. By 12D, $\hat{\alpha}$ is a geodesic.

Suppose $\alpha_{1}$ is a proper subarc of $\alpha$; that is,
$\alpha_{1}\neq\alpha$, or equivalently, either $p$ or $q$ is not an endpoint
of $\alpha$. Then $\alpha$ and $\hat{\alpha}$ share one point and velocity
vector at this point. By 12B $\alpha$ coincides with $\hat{\alpha}$ — a
contradiction.

It follows that $p$ and $q$ are the endpoints of $\alpha$. Analogously, $p$
and $q$ are the endpoints of $\beta$.

For the second part, one could consider two distinct curves of the form

$\gamma_{b}(t)=(\cos t,\sin t,bt),t\in\mathbb{R}$

in the cylinder $x^{2}+y^{2}=1$.

12D. Assume a shortest path $\alpha$ crosses $\Pi$ at least twice. In this
case there is an arc $\alpha_{1}$ of $\alpha$ that lies entirely on one side
on $\Pi$, has ends on $\Pi$, and these ends are distinct from the ends of
$\alpha$.

Let us remove the arc $\alpha_{1}$ from $\alpha$ and replace it with the
reflection of $\alpha_{1}$ across $\Pi$. Note that the obtained curve, say
$\beta$, lies in the surface; it has the same length as $\alpha$, and it
connects the same pair of points, say $p$ and $q$. Therefore $\beta$ is
another shortest path from $p$ to $q$ in $\Sigma$ that is distinct from
$\alpha$.

By 12D, $\alpha$ and $\beta$ are geodesics. Since $\alpha$ and $\beta$ have a
common subarc, they share one point and velocity vector at this point; by 12B
$\alpha$ coincides with $\beta$ — a contradiction.

12D. Let $W$ be the closed region outside of $\Sigma$. Show that the distance
$|{p^{t}}-\nobreak{q}|_{W}$ is constant with respect to $t$. This implies that
the concatenation of the line segment $[p^{t},\gamma(t)]$ and the arc
$\gamma|_{[t,\ell]}$ is a shortest path from $p^{t}$ to $q$ in $W$. Since
$\Sigma$ is strictly convex,

$|{\gamma(t)}-\nobreak{q}|_{W}>|{\gamma(t)}-\nobreak{q}|_{\mathbb{R}^{3}}$

for all $t<\ell$. Hence

$\displaystyle|{p^{t}}-\nobreak{q}|_{W}$
$\displaystyle=|{p^{t}}-\nobreak{\gamma(t)}|_{W}+|{\gamma(t)}-\nobreak{q}|_{W}$
$\displaystyle>|{p^{t}}-\nobreak{\gamma(t)}|_{\mathbb{R}^{3}}+|{\gamma(t)}-\nobreak{q}|_{\mathbb{R}^{3}}$
$\displaystyle\geqslant|{p^{t}}-\nobreak{q}|_{\mathbb{R}^{3}}.$

If the segment $[p^{t},q]$ was entirely contained in $W$ for some $t<\ell$,
then $|{p^{t}}-\nobreak{q}|_{W}=|{p^{t}}-\nobreak{q}|_{\mathbb{R}^{3}}$, which
would be a contradiction.

12E. Equip $\Sigma$ with unit normal field $\nu$ that points inside. Denote by
$k(t)$ the normal curvature of $\Sigma$ at $\gamma(t)$ in the direction of
$\gamma^{\prime}(t)$. Since $\Sigma$ is convex, $k(t)\geqslant 0$ for any $t$.

Since $\gamma$ is a geodesic, we have $\gamma^{\prime\prime}(t)=k(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\nu(\gamma(t))$.

Since $\gamma$ has unit speed,
$\langle\gamma^{\prime}(t),\gamma^{\prime}(t)\rangle=1$ for any $t$.

Without loss of generality, we can assume that $p$ is the origin of
$\mathbb{R}^{3}$. Since $p$ is inside $\Sigma$, we have that
$\langle\gamma(t),\nu(\gamma(t))\rangle\leqslant 0$ for any $t$. It follows
that

$\langle\gamma^{\prime\prime}(t),\gamma(t)\rangle=k(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle\gamma(t),\nu(\gamma(t))\rangle\leqslant
0$

for any $t$.

Applying the above estimates, we get

$\displaystyle\rho^{\prime\prime}(t)$
$\displaystyle=\langle\gamma(t),\gamma(t)\rangle^{\prime\prime}=2{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\langle\gamma^{\prime\prime}(t),\gamma(t)\rangle+2{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\langle\gamma^{\prime}(t),\gamma^{\prime}(t)\rangle\leqslant 2.$

12F. We can assume $\gamma$ is parametrized by arc-length. If we write
$\nu(t)=\nu(\gamma(t))$, then $\langle\gamma^{\prime}(t),\nu(t)\rangle\equiv
0$. Also, since $\gamma$ is a geodesic, we have that
$\gamma^{\prime\prime}(t)\parallel\nu(t)$. Then

$\displaystyle|\gamma^{\prime\prime}(t)|$
$\displaystyle=|\langle\gamma^{\prime\prime}(t),\nu(t)\rangle|$
$\displaystyle=|\langle\gamma^{\prime}(t),\nu(t)\rangle^{\prime}-\langle\gamma^{\prime}(t),\nu^{\prime}(t)\rangle|$
$\displaystyle=|\langle\gamma^{\prime}(t),\nu^{\prime}(t)\rangle|$
$\displaystyle\leqslant|\gamma^{\prime}(t)||\nu^{\prime}(t)|$
$\displaystyle=|\nu^{\prime}(t)|.$

Integrating both sides with respect to $t$, we get

$\mathop{\rm length}\nolimits(\nu\circ\gamma)\geqslant\Phi({\gamma}).$

12F. Suppose $\gamma(t)=(x(t),y(t),z(t))$. Show that

$|\gamma^{\prime\prime}(t)|=z^{\prime\prime}(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt{1+\ell^{2}}$ ➊

for any $t$.

Observe that $z^{\prime}(t)\to\pm\tfrac{\ell}{\sqrt{1+\ell^{2}}}$ as
$t\to\pm\infty$. Conclude that

$\int_{-\infty}^{+\infty}z^{\prime\prime}(t){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}dt=\frac{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\ell}{\sqrt{1+\ell^{2}}}.$ ➋

By ➊ ‣ Semisolutions and ➋ ‣ Semisolutions, we have

$\displaystyle\Phi({\gamma})$
$\displaystyle=\int_{-\infty}^{+\infty}|\gamma^{\prime\prime}(t)|{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dt=\sqrt{1+\ell^{2}}{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}\int_{-\infty}^{+\infty}z^{\prime\prime}(t){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}dt=2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\ell.$

12F. Use 12F and 3G. For the second part, consider a geodesic on a cone with
slope $2$ and mollify its tip.

Remark. The statement still holds for $\sqrt{3}$-Lipschitz functions, and
$\sqrt{3}=\operatorname{tg}\tfrac{\pi}{3}$ is the optimal bound; see 14A. It
is the same slope as in the problem about the cowboy and the lasso (11A).

13A; 13A. Show and use that
$\langle\text{\sc{v}}(t),\text{\sc{v}}^{\prime}(t)\rangle=0$.

13A Show that $|\text{\sc{v}}(t)|$, $|\text{\sc{w}}(t)|$, and
$\langle\text{\sc{v}}(t),\text{\sc{w}}(t)\rangle=0$, are constants; it can be
done the same way as 13A. Then use the formula
$\langle\text{\sc{v}}(t),\text{\sc{w}}(t)\rangle=|\text{\sc{v}}(t)|{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}|\text{\sc{w}}(t)|{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos\theta$.

13B. Observe that $\Sigma_{1}$ supports $\Sigma_{2}$ at any point of $\gamma$.
Conclude that $\gamma$ has identical spherical images as a curve in
$\Sigma_{1}$ and in $\Sigma_{2}$. Apply 13B.

13C. Consider the right angled spherical triangle that an octant of
$\mathbb{R}^{3}$ cuts form the sphere. Argue that parallel transport around it
rotates the tangent plane by the angle $\tfrac{\pi}{2}$.

14A. By 10B, $\Sigma$ is a smooth sphere. By Jordan theorem (0H) the curve
$\gamma$ divides $\Sigma$ in two discs. Let us denote by $\Delta$ the disc
that lies on the left from $\gamma$.

Observe that $\Psi({\gamma})=\mathop{\rm length}\nolimits\gamma$ and apply the
Gauss–Bonnet formula (14A) for $\Delta$.

14A. Apply 9D, 10B, and the Gauss–Bonnet formula (14A).

For the last part, apply 2H.

14A. For the first part apply the Gauss–Bonnet formula (14A).

For the second part, arguing by contradiction, assume that two closed
geodesics $\gamma_{1}$ and $\gamma_{2}$ do not intersect. This would imply
that $\gamma_{2}$ lies in one of the regions, say $R_{1}$, that $\gamma_{1}$
cuts from $\Sigma$. Similarly $\gamma_{1}$ lies in one of the regions, say
$R_{2}$, that $\gamma_{2}$ cuts from $\Sigma$.

Observe that $R_{1}$ and $R_{2}$ cover $\Sigma$ with overlap. Therefore the
first part implies that the integral of the Gauss curvature over $\Sigma$ is
less than $4{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. The latter
contradicts 10B.

14A; (easy). Consider the 4 regions bounded by loops. Apply Gauss–Bonnet
formula (14A) to show that the integral of the Gauss curvature on each of
these regions exceeds $\pi$. The latter contradicts 10B.

(tricky). Denote by $\alpha$, $\beta$, and $\gamma$ the angles of the
triangle. Apply the Gauss–Bonnet formula (14A) to show that the loops surround
regions over which the integral of the Gauss curvature is $\pi+\alpha$,
$\pi+\beta$, and $\pi+\gamma$ respectively.

Apply the Gauss–Bonnet formula to the triangular region to show that
$\alpha+\beta+\gamma>\pi$. Use 10B to get at a contradiction.

14A. Note that it is sufficient to show that the surface has no geodesic
loops. Estimate the integral of the Gauss curvature over the entire surface
and over a disc in it surrounded by a geodesic loop.

14A; 14A. By 11B and 12D, any two points in $\Sigma$ can be connected by a
geodesic. Suppose that points $p$ and $q$ can be connected by two distinct
geodesics $\gamma_{1}$ and $\gamma_{2}$. By 12D, $\gamma_{1}$ and $\gamma_{2}$
share only their endpoints. Since the surface is simply-connected,
$\gamma_{1}$ and $\gamma_{2}$ together bound a disc, say
$\Delta\subset\Sigma$. It remains to apply Gauss–Bonnet formula to $\Delta$
and make a conclusion.

14A. Use part 14A and 12C.

14C. Apply 13D and 14C.

14E. Repeat the end of the proof of 14E for a one-parameter family of
geodesics $\gamma_{\tau}$ defined by $\gamma_{\tau}(0)=\alpha(\tau)$ and
$\gamma^{\prime}(0)=\alpha^{\prime}(\tau)$.

15C. By the Gauss lemma (15A), polar coordinates with respect to $q$ produce a
semigeodesic chart at any nearby point. Therefore, it is sufficient to find a
point $q\neq p$ such that polar coordinates on $\Sigma$ with respect to $q$
cover $p$. By 12C, any $q$ sufficiently close to $p$ does the job.

15C; 15C. Show that we can choose an orientation on $\Sigma$ so that
$b_{r}(0,\theta)=\nobreak 1$ for any $\theta$. Conclude that we can assume
that $b(r,\theta)>0$ for small positive values $r$.

Suppose $b(r_{1},\theta_{1})<0$ at some pair $(r_{1},\theta_{1})$ with
$0<r_{1}<r_{0}$. Observe that if $\theta_{2}$ is sufficiency close to
$\theta_{1}$ then the radial curves $r\mapsto b(r,\theta_{1})$ and $r\mapsto
b(r,\theta_{2})$ defined on the interval $(0,r_{1})$ intersect. Therefore
$\exp_{p}$ is not injective in $B$ — a contradiction.

15C. Suppose that $b(r_{1},\theta_{1})=0$. Apply 15C to show that
$b_{r}(r_{1},\theta_{1})=0$.

Apply 15C to conclude $b(r,\theta_{1})=0$ for any $r$. The latter contradicts
that $b_{r}(0,\theta_{1})=1$.

15C. We need to show that $\exp_{p}$ is regular in $B$. Suppose that vector
$\text{\sc{v}}\in B$ has polar coordinates $(r,\theta)$ for some $r>0$. Show
that $\exp_{p}$ is regular at v if $b(r,\theta)\neq 0$. Conclude that
$\exp_{p}$ is regular in $B\backslash\\{0\\}$.

By 12C, $\exp_{p}$ is regular at $0$. Whence $\exp_{p}|_{B}$ is a regular
injective smooth map. Use the inverse function theorem (0F) to show that the
restriction $\exp_{p}|_{B}$ is a diffeomorphism to its image.

15C; 15C. Since the frame $\text{\sc{u}},\text{\sc{v}},\nu$ is orthonormal,
the first two vector identities are equivalent to the following six real
identities:

$\displaystyle\langle\text{\sc{u}}_{u},\text{\sc{u}}\rangle$
$\displaystyle=0,$
$\displaystyle\langle\text{\sc{u}}_{u},\text{\sc{v}}\rangle$
$\displaystyle=-\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}a_{v},$
$\displaystyle\langle\text{\sc{u}}_{u},\nu\rangle$ $\displaystyle=a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\ell,$ ➊
$\displaystyle\langle\text{\sc{v}}_{u},\text{\sc{v}}\rangle$
$\displaystyle=0,$
$\displaystyle\langle\text{\sc{v}}_{u},\text{\sc{u}}\rangle$
$\displaystyle=\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}a_{v},$
$\displaystyle\langle\text{\sc{v}}_{u},\nu\rangle$ $\displaystyle=a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}m.$

Taking the partial derivatives of the identities
$\langle\text{\sc{u}},\text{\sc{u}}\rangle=1$ and
$\langle\text{\sc{v}},\text{\sc{v}}\rangle=1$ with respect to $u$ we get the
first two identities in ➊ ‣ Semisolutions.

Further, observe that

$\displaystyle\text{\sc{v}}_{u}$ $\displaystyle=\tfrac{\partial}{\partial
v}(\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}s_{v})=\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}s_{uv}+\tfrac{\partial}{\partial u}(\tfrac{1}{b}){\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}s_{v}.$ ➋

Since $s_{u}\perp s_{v}$, it follows that

$\displaystyle\langle\text{\sc{v}}_{u},\text{\sc{u}}\rangle$
$\displaystyle=\tfrac{1}{a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}b}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle s_{vu},s_{u}\rangle=\tfrac{1}{2{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}a{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}b}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tfrac{\partial}{\partial
v}\langle s_{u},s_{u}\rangle=\tfrac{1}{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}b}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\tfrac{\partial a^{2}}{\partial
v}=\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}a_{v}.$

Taking the partial derivative of $\langle\text{\sc{u}},\text{\sc{v}}\rangle=0$
with respect to $u$ we get

$\displaystyle\langle\text{\sc{v}}_{u},\text{\sc{u}}\rangle+\langle\text{\sc{v}},\text{\sc{u}}_{u}\rangle$
$\displaystyle=0.$

Hence we get two more identities in ➊ ‣ Semisolutions.

Since $\text{\sc{u}},\text{\sc{v}}$ is an orthonormal frame, by 9D we have

$\displaystyle\tfrac{1}{a^{2}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\langle
s_{uu},\nu\rangle$ $\displaystyle=\ell,$ $\displaystyle\tfrac{1}{a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}b}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\langle s_{uv},\nu\rangle$ $\displaystyle=m,$ ➌
$\displaystyle\tfrac{1}{a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}b}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\langle s_{vu},\nu\rangle$ $\displaystyle=m,$
$\displaystyle\tfrac{1}{b^{2}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\langle
s_{vv},\nu\rangle$ $\displaystyle=n.$

Applying ➋ ‣ Semisolutions, ➌ ‣ Semisolutions, and $s_{v}\perp\nu$ we get

$\displaystyle\langle\text{\sc{u}}_{u},\nu\rangle$
$\displaystyle=\tfrac{1}{a}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\langle
s_{uu},\nu\rangle=a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\ell,$
$\displaystyle\langle\text{\sc{v}}_{u},\nu\rangle$
$\displaystyle=\tfrac{1}{a}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\langle
s_{uv},\nu\rangle=a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}m,$

that imply the last two equalities in ➊ ‣ Semisolutions.

Therefore the first two identities in 15C are proved; the remaining two
identities can be proved along the same lines.

15C. Recall that the Gauss curvature equals the determinant of the matrix
$(\begin{smallmatrix}\ell&m\\\ m&n\end{smallmatrix})$; that is, $K=\ell{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}n-m^{2}$. Therefore

$\displaystyle a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}b{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}K$ $\displaystyle=a{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}b{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(\ell{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}n-m^{2})=\langle\text{\sc{u}}_{u},\text{\sc{v}}_{v}\rangle-\langle\text{\sc{u}}_{v},\text{\sc{v}}_{u}\rangle$
$\displaystyle=\left(\tfrac{\partial}{\partial
v}\langle\text{\sc{u}}_{u},\text{\sc{v}}\rangle-\cancel{\langle\text{\sc{u}}_{uv},\text{\sc{v}}\rangle}\right)-\left(\tfrac{\partial}{\partial
u}\langle\text{\sc{u}}_{v},\text{\sc{v}}\rangle-\cancel{\langle\text{\sc{u}}_{uv},\text{\sc{v}}\rangle}\right)=\tfrac{\partial}{\partial
v}(-\tfrac{1}{b}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}a_{v})-\tfrac{\partial}{\partial u}(\tfrac{1}{a}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}b_{u}).$

15C. Apply 15C assuming that $a=b$ and simplify.

15G. Note and use that by 15F, $\exp_{p}$ is length-preserving.

15G. Modify the proof of 15F to show that $K\equiv 1$ implies that
$b(\theta,r)=\nobreak\sin r$ for all small $r\geqslant 0$.

15G. To find $x(t)$, one may solve the differential equation
$x^{\prime}(t)=\sqrt{1-|y^{\prime}(t)|^{2}}$. Its solution is defined on any
interval where $|y^{\prime}(t)|=|a{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sin
t|<1$ holds.

Apply 9E to calculate Gauss curvature of $\Sigma_{a}$.

By 15G, small discs $\Delta_{a}$ in $\Sigma_{a}$ admit intrinsic isometry to a
disc in a unit sphere. To show that $\Delta_{a}$ is not congruent to a
sphericsl disc $\Delta_{1}$ for $a\neq 1$, show that its principle curvature
is not 1 at some point.

16A. Apply the triangle inequality $c_{n}\leqslant a_{n}+b_{n}$ and the bound
$a_{n},b_{n}>\varepsilon$ to show that the sequence

$\frac{a_{n}+b_{n}+c_{n}}{2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}a_{n}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}b_{n}}$

is bounded. Conclude that

$\frac{(a_{n}+b_{n})^{2}-c_{n}^{2}}{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}a_{n}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}b_{n}}=\frac{(a_{n}+b_{n}+c_{n}){\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}(a_{n}+b_{n}-c_{n})}{2{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}a_{n}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}b_{n}}\to 0$

as $n\to\infty$.

Use the last statement together with the cosine rule

$a_{n}^{2}+b_{n}^{2}-2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}a_{n}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}b_{n}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\cos\tilde{\theta}_{n}-c_{n}^{2}=0$

to show that $\cos\tilde{\theta}_{n}\to-1$ as $n\to\infty$; conclude that
$\theta_{n}\to\pi$.

16B. Observe that
$|{p}-\nobreak{x}|_{\Sigma}\leqslant|{p}-\nobreak{q}|_{\Sigma}$ and
$|{q}-\nobreak{x}|_{\Sigma}\leqslant|{p}-\nobreak{q}|_{\Sigma}$. Conclude that
$\tilde{\measuredangle}(x\,{}^{p}_{q})\geqslant\tfrac{\pi}{3}$. Apply 16B.

16B. Observe that
$\measuredangle[p\,{}^{x}_{y}]+\measuredangle[p\,{}^{y}_{z}]+\measuredangle[p\,{}^{z}_{x}]\leqslant
2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\pi$. Apply 16B.

16E. Choose vectors u, and w such that $|\text{\sc{w}}|=|{p}-\nobreak{x}|$,
$|\text{\sc{u}}|=1$, and
$\measuredangle(\text{\sc{u}},\text{\sc{w}})=\measuredangle[x\,{}^{p}_{q}]$.
Consider the function

$f\colon t\mapsto t+|\text{\sc{w}}|-|\text{\sc{w}}-t{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\text{\sc{u}}|$

is concave.

Set $t_{0}=|{x}-\nobreak{y}|$ and $t_{1}=|{x}-\nobreak{q}|$ Observe that

$\displaystyle f(t_{0})$
$\displaystyle=|{x}-\nobreak{p}|+|{x}-\nobreak{y}|-\tilde{\curlyvee}[x\,{}^{p}_{y}]$
$\displaystyle f(t_{1})$
$\displaystyle=|{x}-\nobreak{p}|+|{x}-\nobreak{q}|-\tilde{\curlyvee}[x\,{}^{p}_{q}]$

Show and use that $f$ is concave and $f(0)=0$.

16E; 16E. Apply Rauch comparison 15F and the property of exponential map in
12C.

16E. Argue by contradiction; if the statement does not hold then for any
$p\in\Sigma$ there is a point $q=q(p)\in\Sigma$ such that
$|{q}-\nobreak{p}|_{\Sigma}<100{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}R_{p}$
and $R_{q}<(1-\tfrac{1}{100}){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}R_{p}$.

Start with any point $p_{0}$ and consider a sequence $p_{n}$ defined by
$p_{n+1}=\nobreak q(p_{n})$. Show that $p_{n}$ converges and $R_{p_{n}}\to 0$
as $n\to\infty$. Arrive to a contradiction using 16E.

16E. Repeat the proof assuming that $p$ is provided by 16E.

16E; 16E. Apply the Gauss–Bonnet formula for each region that $\gamma$ cuts
from $\Sigma$ and simplify the obtained inequalities.

16E. Consider the pentagon $\Delta$ with induced length-metric. Note that all
its angles cannot exceed $\pi$. Repeat the proof of 16B to show that the
comparison holds in $\Delta$.

Choose a vertex $v$ of $\Delta$, subdivide its sides so that each arc of the
subdivision is a minimizing geodesic, so the boundary of $\Delta$ is a broken
geodesic. Divide $\Delta$ into triangles by joining $v$ to other vertices of
the broken geodesic. Take a model triangle for each so that they shared sides
as in $\Delta$. Use the comparison to show that these plane triangles form a
required polygon.

16E. It remains to show that the plane pentagon provided by 16E cannot exist.
Imagine that there is air pressure in the pentagon, it should not move it; so
the total force should vanish. Use part 16E to show that this is not the case.
(Equivalently, show that the sum of its side vectors oriented counterclocwise
does not vanish.)

16G. Apply 16G and 16G.

16G. Use 16G or 16G twice: first for the triangle $[pxy]$ and
$\bar{x}\in[p,x]$ and second for the triangle $[p\bar{x}y]$ and
$\bar{y}\in[p,y]$. Then apply the angle monotonicity (0B).

16G. It is sufficient to prove the Jensen inequality; that is,

$|{\gamma_{1}(\tfrac{1}{2})}-\nobreak{\gamma_{2}(\tfrac{1}{2})}|\leqslant\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}|{\gamma_{1}(0)}-\nobreak{\gamma_{2}(0)}|+\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}|{\gamma_{1}(1)}-\nobreak{\gamma_{2}(1)}|.$

Let $\delta$ be the geodesic path from $\gamma_{2}(0)$ to $\gamma_{1}(1)$.
From 16G, we have

$\displaystyle|{\gamma_{1}(\tfrac{1}{2})}-\nobreak{\delta(\tfrac{1}{2})}|$
$\displaystyle\leqslant\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}|{\gamma_{1}(0)}-\nobreak{\delta(0)}|,$
$\displaystyle|{\delta(\tfrac{1}{2})}-\nobreak{\gamma_{2}(\tfrac{1}{2})}|$
$\displaystyle\leqslant\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}|{\delta(1)}-\nobreak{\gamma_{2}(1)}|,$

Sum it up and apply the triangle inequality.

Remark. Note that modulo the comparison theorem, the case of the Euclidean
plane is just as hard.

16G; 16G. Choose two points $x,z\in\Sigma$; let $y$ be the midpoint of
$[x,z]$. Denote by $\bar{x}$ and $\bar{z}$ the points on $\delta$ that
minimize the distances to $x$ and $z$ respectively; so we have
$f(x)=|{x}-\nobreak{\bar{x}}|_{\Sigma}$, and
$f(z)=|{z}-\nobreak{\bar{z}}|_{\Sigma}$.

Since $\Delta$ is convex, the midpoint of $[\bar{x},\bar{z}]$, say $\bar{y}$,
lies in $\Delta$. By 16G, we have

$\displaystyle f(y)$
$\displaystyle\leqslant|{y}-\nobreak{\bar{y}}|_{\Sigma}\leqslant\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip
0.5pt}(|{x}-\nobreak{\bar{x}}|_{\Sigma}+|{z}-\nobreak{\bar{z}}|_{\Sigma})=\tfrac{1}{2}{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}(f(x)+f(z)).$

Observe that this inequality implies convexity of $f$.

16G. Suppose $\gamma$ is a unit-speed geodesic in $\Delta$ and $\gamma(0)=p$.
Let $q$ be a point on $\delta$ that minimize the distance to $p$. Denote by
$\varphi$ the angle between $\gamma$ and $[p,q]$ at $p$. Note that it is
sufficient to show that

$f\circ\gamma(t)\leqslant|{p}-\nobreak{q}|_{\Sigma}+t{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos\varphi$ ➊

for $t$ sufficiently close to $0$.

Chose small $t>0$ and set $x=\nobreak\gamma(t)$. Consider a model triangle
$[\tilde{p}\tilde{q}\tilde{x}]=\nobreak\tilde{\triangle}(pqx)$. Let
$\tilde{\delta}$ be the line thru $\tilde{q}$ perpendicular to
$[\tilde{p},\tilde{q}]$. Denote by $\tilde{y}$ the orthogonal projection of
$\tilde{x}$ to $\tilde{\delta}$. By comparison we have that

$\displaystyle\varphi=\measuredangle[q\,{}^{p}_{x}]$
$\displaystyle\geqslant\tilde{\measuredangle}(q\,{}^{p}_{x})\mathrel{=:}\tilde{\varphi},$
$\displaystyle\psi\mathrel{:=}\measuredangle[p\,{}^{q}_{x}]$
$\displaystyle\geqslant\tilde{\measuredangle}(p\,{}^{q}_{x})\mathrel{=:}\tilde{\psi}$

Since $t=|{\tilde{p}}-\nobreak{\tilde{x}}|_{\mathbb{R}^{2}}$ is small, so is
the distances $|{\tilde{q}}-\nobreak{\tilde{y}}|_{\mathbb{R}^{2}}$. Therefore
we can choose a point $y$ on $\delta$ such that
$|{q}-\nobreak{y}|_{\Sigma}=|{\tilde{q}}-\nobreak{\tilde{y}}|_{\mathbb{R}^{2}}$.
Moreover, if we choose in on the _right side_ , then
$\tfrac{\pi}{2}-\psi\geqslant\measuredangle[q\,{}^{x}_{y}]$. By construction
and comparison, we have

$\displaystyle\measuredangle[\tilde{q}\,{}^{\tilde{x}}_{\tilde{y}}]$
$\displaystyle=\tfrac{\pi}{2}-\tilde{\psi}\geqslant\tfrac{\pi}{2}-\psi\geqslant\measuredangle[q\,{}^{x}_{y}]\geqslant\tilde{\measuredangle}(q\,{}^{x}_{y}).$

Therefore
$|{x}-\nobreak{y}|_{\Sigma}\leqslant|{\tilde{x}}-\nobreak{\tilde{y}}|_{\mathbb{R}}$.
Since $\tilde{\varphi}\leqslant\varphi$, we get that

$\displaystyle f\circ\gamma(t)$
$\displaystyle\leqslant|{x}-\nobreak{y}|_{\Sigma}\leqslant|{\tilde{x}}-\nobreak{\tilde{y}}|_{\mathbb{R}}=$
$\displaystyle=|{p}-\nobreak{q}|_{\Sigma}+t{\hskip 0.5pt\cdot\nobreak\hskip
0.5pt}\cos\tilde{\varphi}\leqslant|{p}-\nobreak{q}|_{\Sigma}+t{\hskip
0.5pt\cdot\nobreak\hskip 0.5pt}\cos\varphi.$

That is, we proved ➊ ‣ Semisolutions for small $t>0$. The case $t<0$ can be
done along the same lines.

16H; 16H. Choose a sequence of points $x_{n}\in K$ that escapes to infinity;
that is, such that $|{p}-\nobreak{x_{n}}|_{\Sigma}\to\infty$ as $n\to\infty$.
Denote by $\text{\sc{u}}_{n}\in\mathrm{T}_{p}$ the unit vector in the
direction of $[p,x_{n}]_{\Sigma}$. Let $\text{\sc{u}}_{\infty}$ be a partial
limit of $\text{\sc{u}}_{n}$.

Show that the geodesic that starts from $p$ in the direction of
$\text{\sc{u}}_{\infty}$ is a half-line that runs in $K$.

16H. The concavity of $f$ follows from 16H. Whence the convexity of $S_{c}$
follows.

Suppose that $S_{c}$ is not compact. Note that we can assume that $p\in
S_{c}$. Then by 16H there is a half-line $\lambda$ that starts at $p$ and runs
in $S_{c}$ for some $c$. It remains to observe that $f\circ\lambda(t)+t=0$ and
use it to arrive at a contradiction.

16H. Show that $f(x)-c=\mathop{\rm dist}\nolimits_{\partial S_{c}}x$ for any
$x\in S_{c}$. Conclude that $S_{s}$ has no interior points. Since $S_{s}$ is
convex, the first part follows.

In the Euclidean plane $S_{s}$ is a single point. In the cylinder, $S_{s}$ is
a circle. The surface of $r$-neighborhood of an infinite rectangle provides an
example with a line segment as $S_{s}$; this example is not smooth, but it can
be smoothed keeping this property. One can use the cutoffs and mollifiers to
do this (Section 0E).

16I. Show that a cylinder does not contain a line and half-line that meet at
exactly one point. It remains to apply the line splitting theorem (16I).

## Afterword

For further study, you need to read a text on tensor calculus; the book of
Richard Bishop and Samuel Goldberg [10] is one of our favorites. Once it is
done, you are ready to do Riemannian geometry.

You may go with ‘‘Comparison geometry’’ [20] — the good old classics from Jeff
Cheeger and David Ebin. In this case you might be surprised to see that you
already know half of material in the book; altho it is only written using a
different language. For example proofs of the comparisons theorem require only
cosmetic modifications.

A safer option would be another classical book ‘‘Riemannsche Geometrie im
Großen’’ [35] by Detlef Gromoll, Wilhelm Klingenberg, and Werner Meyer (it is
available in German and Russian).

Mikhael Gromov’s ‘‘Sign and geometric meaning of curvature’’ [37] is more
challenging, but worth trying.

Good luck.

Anton Petrunin and

Sergio Zamora Barrera.

## Index

  * $|{x}-\nobreak{y}|$, $|{x}-\nobreak{y}|_{\mathcal{X}}$, 8
  * $\lvert x-y\rvert_{A}$, 35
  * $\measuredangle[y\,{}^{x}_{z}]$, 11
  * $\tilde{\measuredangle}(x\,{}^{y}_{z})$, 155
  * $[p,q]$, $[p,q]_{\Sigma}$, 116
  * $[y\,{}^{x}_{z}]$, 155
  * $[xyz]$, $[xyz]_{\Sigma}$, 155
  * $\tilde{\triangle}$, 155
  * $B(p,R)_{\mathcal{X}}$, $\bar{B}[p,R]_{\mathcal{X}}$, 9
  * $D_{\text{\sc{w}}}f$, 83
  * $d_{p}s$, 85
  * $f_{x}$, 25
  * $\mathop{\rm GB}\nolimits$, 137
  * $H$, 92
  * i, j, k, 94
  * $\mathop{\rm Jac}\nolimits$, 16
  * $\mathop{\rm jac}\nolimits$, 17
  * $K$, 92
  * $\kappa$, 39
  * $k$, 57
  * $k_{1}$, $k_{2}$, 91
  * $k_{g}$, $k_{n}$, 96
  * $k_{\text{\sc{v}}}$, 96
  * $\mathop{\rm length}\nolimits\gamma$, 27
  * $\ell$, $m$, $n$, 90
  * $M_{p}$, 91
  * $\nu$, 87
  * $\Phi({\gamma})$, 44
  * $\Psi({\gamma})$, 60
  * $\sigma_{s}$, 64
  * $\tau$, 52
  * t, $\mu$, $\nu$, 95
  * t, n, b, 51
  * absolute curvature, 57
  * Alexandrov’s lemma, 164
  * almost all, 16
  * alternating quadrisecants, 48
  * arc, 23
  * arc-length parametrization, 29
  * asymptotic direction and line, 109
  * base of loop, 23
  * bell function, 15
  * Bernstein theorem, 111
  * binormal, 51
  * Borel subsets, 17
  * boundary line, 77
  * bow lemma, 48
  * broken geodesic, 133
  * Busemann function, 168
  * catenoid, 98
  * center of curvature, 64
  * chart, 80
  * circline, 63
  * Clairaut’s relation, 120
  * closed
    * ball, 8
    * curve, 23
    * set, 10
    * surface, 78
  * closest point projection, 117
  * coarea fromula, 110
  * compact, 20
  * comparison theorem, 156
  * conformal chart, 147
  * conformal factor, 147
  * connected component, 21
  * connected set, 21
  * constant slope, 53
  * continuous, 9
  * controriented
    * curves, 67
    * surfaces, 102
  * convergence of points, 9
  * convex
    * curve, 30
    * function, 18
    * hull, 12
    * set, 12, 119
    * surface, 104
  * cooriented
    * curves, 67
    * surfaces, 102
  * Crofton’s formula, 32
  * cross product, 40
  * curvature, 38, 57, 60, 92, 96
  * curve, 23
  * cusp, 43
  * cut locus, 186
  * Darboux frame, 95
  * diffeomorphism, 81
  * differential, 85
  * directional derivative, 18, 83
  * embedding, 20
  * endpoint, 23
  * Euler characteristic, 141
  * Euler’s formula, 97
  * evolute, 65
  * evolvent, 54
  * excess of triangle, 11
  * explicitly defined curve, 25
  * exponential map, 123
  * external angle, 43
  * extrinsic property, 115
  * Fáry–Milnor theorem, 48
  * fat triangle, 165
  * Fenchel’s theorem, 41
  * flow, 184
  * Four-vertex theorem, 74
  * Frenet formulas, 53
  * Frenet frame, 51, 57
  * fundamental form, 93
  * Gauss curvature, 92
  * Gauss map, 88
  * Gauss–Bonnet formula, 135, 141
  * generatrix, 80
  * geodesic, 120
    * curvature, 96
    * triangle, 155
  * geodesically complete, 121
  * global support, 68
  * Green formula, 149
  * half-line, 167
  * hat, 110
  * helicoid, 98
  * Hessian matrix, 91
  * hinge, 155
  * homeomorphism, 10, 20
  * Hopf–Rinow theorem, 121
  * hyperplane separation theorem, 12
  * implicit function theorem, 16
  * implicitly defined curve, 25
  * induced length-metric, 35
  * injectivity radius, 124
  * inscribed polygonal line, 28
  * intrinsic diameter, 119
  * intrinsic distance, 155
  * intrinsic geometry, 115
  * intrinsic isometry, 153
  * intrinsic property, 115
  * inverse function theorem, 16
  * inversion, 64
  * Ionin–Pestov theorem, 72
  * Jacobi field, 145
  * Jensen’s inequality, 18
  * Jordan’s theorem, 21
  * Koch snowflake, 28
  * Lagunov’s example, 98
  * laplacian, 148
  * length of curve, 27
  * length-metric, 35
  * length-preserving, 153
  * Liberman’s lemma, 127
  * line, 169
  * line of curvature, 91
  * Lipschitz function, 13
  * local
    * coordinates, 80
    * support, 68, 102
    * to global theorem, 72
  * locally flat surface, 153
  * logarithmic map, 126
  * loop, 23
  * mean curvature, 92
  * meridian, 80
  * metric space, 8
  * Meusnier’s theorem, 97
  * minimal surface, 92
  * minimizing geodesic, 125
  * model
    * side, 160
    * triangle, 155
  * model angle, 155
  * monotone map, 23
  * Morse lemma, 188
  * Moser trick, 184
  * natural parametrization, 29
  * neighborhood, 10
  * normal, 51
    * curvature, 96
    * vector, 51
  * open
    * ball, 8
    * curve, 26
    * set, 10
    * surface, 78
  * order of contact, 40, 52, 55, 64, 82, 90
  * oriented surface, 87
  * orthogonal chart, 147
  * osculating
    * circline, 64
    * paraboloid, 90
    * plane, 52
    * sphere, 55
  * parallel, 80
    * field, 130
    * transport, 131
  * parametrization, 22, 79, 81
  * parametrized curve, 23
  * partial limit, 117
  * partition, 27
  * path, 21, 23
  * path-connected set, 21
  * piecewise smooth regular curve, 43
  * plane curve, 24
  * point, 8
    * of tangency, 67
  * polar coordinates, 142
  * principal curvatures and directions, 91
  * proper
    * curve, 26
    * surface, 77
  * pseudosphere, 98
  * radius of curvature, 64
  * Rauch comparison theorem, 151
  * real interval, 22
  * rectifiable curve, 27
  * regular
    * curve, 24
    * point, 16
    * value, 16, 25
  * Remarkable theorem, 154
  * reparametrization, 23
  * rotation, 148
  * ruled surface, 108
  * saddle surface, 108
  * self-adjoint operator, 93
  * self-intersections, 23
  * semicubical parabola, 174
  * semigeodesic, 144, 147
  * shape operator, 92
  * shortest path, 116
  * signed curvature, 57
  * simple curve, 22
  * simply-connected, 136, 186
  * smooth
    * curve, 24
    * function, 24, 76
    * map, 16
    * parametrization, 79
    * surface, 76
  * space curve, 24
  * spherical
    * curve, 35
    * map, 88
  * Spiral lemma, 64
  * star-shaped, 109, 124
  * stereographic projection, 80
  * support from inside, 68, 102
  * support from outside, 68, 102
  * surface, 76
    * of revolution, 80
    * with boundary, 77
  * tangent, 51
    * curves, 40
    * field, 130
    * indicatrix, 39
    * line, 40
    * plane, 82
    * vector, 82
  * tangent-normal coordinates, 90
  * Theorema Egregium, 154
  * thin triangle, 165
  * topological, 10
    * surface, 76
  * torsion, 52
  * total
    * curvature, 41
    * signed curvature, 60
  * totally convex/geodesic, 172
  * transversality, 109, 186
  * trivial curve, 136
  * trochoid, 60
  * uniform convergence, 14
  * uniformly continuous, 14
  * unit-speed curve, 29
  * vector product, 40
  * velocity vector, 24
  * vertex of curve, 68
  * Viviani’s curve, 174

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