Hopf turning-tangent theorem with ordinary corners

Assume first that $\gamma$ has no corners, so $T$ is continuous and $T(0)=T(1)$. Let $K$ be the convex hull of the compact curve ${\gamma }^{\ast }$, let $p_0$ be an extreme point of $K$, and note $p_0=\gamma (t_0)$ for some $t_0$; after translating $p_0$ to the origin and rotating, ${\gamma }^{\ast }\subseteq \{y\ge 0\}$ while $D\subseteq K\subseteq \{y\ge 0\}$ because $D$ is a compact connected region whose interior is the connected bounded component of ${\mathbb{R}}^2\setminus {\gamma }^{\ast }$ and ${\mathbb{R}}^2\setminus K$ is connected and unbounded.

Now let $\gamma$ have its finitely many ordinary corners at parameters $t_1<\cdots <t_m$, none at the identified endpoint. At corner $p_j$, let $T_j^{-},T_j^{+}$ be the incoming and outgoing unit tangents. Their signed principal turn ${\alpha }_j$ lies in $(-\pi ,\pi )$, so they are not antipodal. Use the unit vector $(T_j^{-}+T_j^{+})\mathrm{/}\mathrm{\mid }{T}_j^{-}+T_j^{+}\mathrm{\mid }$ as a positive $x$-axis. Both one-sided tangents have positive $x$-component; shrinking about $p_j$ makes both incident regular $C^2$ arcs a single positively traversed continuous piecewise $C^2$ graph $y=f_j(x)$, with $p_j=(0,0)$ and $x$ strictly increasing through $0$. Choose pairwise disjoint closed graph disks $B_j$ that meet ${\gamma }^{\ast }$ only in these two arcs; around a smaller compact graph subarc containing $p_j$, choose a fixed thin vertical tube with closure inside $B_j$. For sufficiently small $\delta >0$, the points $P_j^{\pm }=(\pm \delta ,f_j(\pm \delta ))$ lie in that tube.

Reparametrizing cyclically so that $t_0=0$ gives a cornerless closed curve $\gamma :[0,1]\to {\mathbb{R}}^2$ with $\gamma (0)=\gamma (1)=0$, $\gamma (t)\mathrm{\ne }0$ for $0<t<1$ and $T(0)=T(1)=(1,0)$: indeed the tangent line at $p_0$ is a supporting line of $K$, since $⟨\gamma (t)-p_0,\nu ⟩\ge 0$ with equality at $t_0$ forces the derivative at $t_0$ to be orthogonal to the inward normal $\nu =(0,1)$, and the local positive-orientation condition of [F5] rules out the direction $(-1,0)$.

Replace $f_j$ on $[-\delta ,\delta ]$ by the cubic Hermite polynomial $H_{j,\delta }$ matching both $f_j(\pm \delta )$ and $f_j^{\mathrm{\prime }}(\pm \delta )$, and leave $f_j$ unchanged outside that interval. In the rescaled variable $u=(x+\delta )\mathrm{/}(2\delta )$, the four fixed Hermite basis polynomials have bounded absolute values; since $f_j(\pm \delta )=O(\delta )$ and $f_j^{\mathrm{\prime }}(\pm \delta )=O(1)$, both $H_{j,\delta }$ and $H_{j,\delta }-f_j$ are uniformly $O(\delta )$ on the replacement interval. The resulting arc is a regular embedded $C^1$ graph, piecewise $C^2$, with no tangent jump at its two joining points. Its positive unit tangent angle lies in the coordinate interval $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$ and its signed curvature integral is exactly ${\alpha }_j(\delta )=\arctan f_j^{\mathrm{\prime }}(\delta )-\arctan f_j^{\mathrm{\prime }}(-\delta )$ by the tangent-angle derivative formula on the polynomial piece. Continuity of the one-sided tangent germs gives ${\alpha }_j(\delta )\to {\alpha }_j$.

Define the chord path $w:[0,1]\to S^1$ by $w(t)=\gamma (t)\mathrm{/}\mathrm{\mid }\gamma (t)\mathrm{\mid }$ for $0<t<1$ and $w(0)=T(0)$, $w(1)=-T(0)$; it is continuous because the two limits are the tangent directions at $0$ and $1$, and it takes values in the closed upper semicircle. The map $\theta \mapsto (\cos \theta ,\sin \theta )$ is a homeomorphism of $[0,\pi ]$ onto that semicircle, so $w$ has a continuous lift with values in $[0,\pi ]$ starting at $0$ and ending at $\pi$; hence $w$ has total angle change $\pi$.

On the triangle $\mathrm{\Delta }=\{(s,t):0\le s\le t\le 1\}$ define $H(s,t)=(\gamma (t)-\gamma (s))\mathrm{/}\mathrm{\mid }\gamma (t)-\gamma (s)\mathrm{\mid }$ for $s<t$, $H(t,t)=T(t)$, and $H(0,1)=-T(0)$. Then $H$ is continuous: away from the diagonal it is a quotient of continuous nonzero vectors; at a diagonal point $(t,t)$ with $0<t<1$ the secants tend to $T(t)$; at $(0,0)$ and $(1,1)$ they tend to $T(0)$; and at $(0,1)$ one has $\gamma (t)-\gamma (s)=-((1-t)+s){\gamma }^{\mathrm{\prime }}(0)+o((1-t)+s)$ with $(1-t)+s>0$, so the direction tends to $-T(0)$.

Replace each graph subarc by step 2.2's Hermite graph. Because its $x$-coordinate increases strictly, each new arc is embedded; its $O(\delta )$ displacement keeps it in its disjoint graph disk and away from the rest of the closed curve. More explicitly, the graph interpolation $f_{j,s}=(1-s)f_j+s{H}_{j,\delta }$ on $[-\delta ,\delta ]$, equal to $f_j$ elsewhere, consists of embedded arcs with the same endpoints. Choose a vertical cutoff $\chi$ equal to $1$ near the original graph and $0$ outside its fixed thin tube. The map $(x,y)\mapsto (x,y+s\chi (y-f_j(x))(H_{j,\delta }(x)-f_j(x)))$, extended by the identity outside $B_j$, carries the old arc to the interpolated arc. Its derivative in $y$ is $1+O(\delta )>0$ uniformly, so each map is an orientation-preserving homeomorphism fixed near $\mathrm{\partial }{B}_j$. Combining the disjoint supports gives an ambient isotopy and a simple closed piecewise $C^2$ curve ${\gamma }_{\delta }$ that is $C^1$ at the joining points, bounds a positively oriented disk region, and equals $\gamma$ outside the $B_j$.

The reversed chord path $v(s)=-w(s)$ takes values in the closed lower semicircle and has a continuous lift with values in $[\pi ,2\pi ]$ running from $\pi$ to $2\pi$; hence its total angle change is also $\pi$, and the reversed paths $s\mapsto w(1-s)$ and $s\mapsto v(1-s)$ have total angle changes $-\pi$.

The restriction of $H$ to the boundary of the closed disk $\mathrm{\Delta }$ is a closed loop in $S^1$ whose degree, defined by the endpoint difference of a lift, equals $0$. Since $\mathrm{\Delta }$ is compact and $H$ is continuous, there is a finite triangulation of $\mathrm{\Delta }$ so fine that the image of every triangle lies in an open semicircle of $S^1$. A loop lying in an open semicircle has degree zero, because the single-valued angle coordinate of that semicircle is a continuous lift and returns to its starting value at the endpoint; degree is additive under concatenation of loops. Traversing the triangle boundaries presents $\mathrm{\partial }\mathrm{\Delta }$ with each interior edge occurring twice with opposite orientations, and the two corresponding degree contributions cancel, so the degree of $H{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Delta }}$ is the sum of the degrees of the triangle boundary loops, each of which is zero.

Reading the three boundary sides of step 3.2 in the positive orientation of $\mathrm{\Delta }$ gives total angle changes $2\pi \mathrm{rot}(\gamma )$ along the diagonal, $-\pi$ along the side $t=1$, and $-\pi$ along the side $s=0$, by steps 3.1, 4.1 and [F1]; combined with step 4.2 this gives $0=2\pi \mathrm{rot}(\gamma )-2\pi$, that is $\mathrm{rot}(\gamma )=1$.

For the cornerless case the tangent-angle lift of [F1] satisfies ${\theta }^{\mathrm{\prime }}=⟨{\gamma }^{\mathrm{\prime }\mathrm{\prime }},J{\gamma }^{\mathrm{\prime }}⟩=k_{\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}$ along each unit-speed piece by [F4], so the fundamental theorem of calculus gives ${\int }_{\gamma }{k}_{\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}ds=2\pi \mathrm{rot}(\gamma )=2\pi$.

Applying steps 1.1–6.1 to the cornerless closed curve ${\gamma }_{\delta }$ gives ${\int }_{{\gamma }_{\delta }}{k}_{\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}ds=2\pi$. The integral splits into the integrals over the retained subarcs of $\gamma$ and the Hermite-graph contributions, each equal to ${\alpha }_j(\delta )$ by step 2.2.

Letting $\delta \to 0$, the removed original graph intervals have total arclength $O(m\delta )$ because their derivatives are bounded on the finitely many regular $C^2$ germs. Their curvature integrals therefore tend to zero, so the integrals over the retained subarcs tend to ${\int }_{\gamma }{k}_{\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}ds$. Also ${\alpha }_j(\delta )\to {\alpha }_j$ by step 2.2. Taking the limit in step 7.1 gives ${\int }_{\gamma }{k}_{\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}ds+{\sum }_j{\alpha }_j=2\pi$.