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Let $\gamma=\partial K_1(0,0)$ be the circle with radius $r=1$ and origin $(0,0)$ in $\mathbb R^2$. Then for any $t_0$ we have $\gamma'(t_0)\neq \begin{pmatrix} 0 \\ 0\end{pmatrix}$. Let $v=\begin{pmatrix} \gamma_1'(t_0) \\ \gamma_2'(t_0) \end{pmatrix}$ be the tangent vector to the curve at $\gamma(t_0)$. Choose $$\gamma:[0,2\pi]\rightarrow \mathbb R^2,~t\mapsto \begin{pmatrix} \cos(t) \\ \sin(t) \end{pmatrix}$$ as a parametric equation of the circle. This yields $$v=\begin{pmatrix}-\sin(t_0) \\ \cos(t_0)\end{pmatrix}$$ for the tangent vector. Thus $$n_1=\begin{pmatrix} \cos(t_0) \\ \sin(t_0) \end{pmatrix},~n_2=\begin{pmatrix} -\cos(t_0) \\ -\sin(t_0) \end{pmatrix}$$ are the two "obvious" choices for vectors which are perpendicular to $v$. For $t_0=0$ we would get $$\gamma(0)=\begin{pmatrix} 1 \\ 0\end{pmatrix}~v=\begin{pmatrix} 0 \\ 1 \end{pmatrix},~n_1=\begin{pmatrix} 1 \\ 0 \end{pmatrix},~n_2=\begin{pmatrix} -1 \\ 0 \end{pmatrix}.$$

Now: if we attach $n_1$ and $n_2$ to $\gamma(0)$, only one vector will point into the circle (in this case $n_2$) while the other vector will point out of the circle. For a circle with positiv orientation it is easy to get the one vector pointing inside, as the curvature doesn't change. Same goes for a circle with negative orientation. But is it possible to choose this vector for an arbitrary jordan curve? In other words:

Let $\gamma:[0,1]\rightarrow\mathbb R^2$ be a smooth jordan curve, i.e. $\gamma$ is a differentiable simple closed curve with $\gamma'(t)\neq (0,0)$. Let $$v=\begin{pmatrix} \gamma_1'(t_0) \\ \gamma_2'(t_0)\end{pmatrix}$$ be the tangent vector to the curve at $\gamma_(t_0)$ and $$n_1=\begin{pmatrix} \gamma_2'(t_0) \\ -\gamma_1'(t_0)\end{pmatrix},~n_2=\begin{pmatrix} -\gamma_2'(t_0) \\ \gamma_1'(t_0)\end{pmatrix}$$ be vectors perpendicular to $v$. Is it possible to determine which vector $\varepsilon \cdot n_1$ or $\varepsilon \cdot n_2$ will lie completely in the interior of $\gamma$ if we attach it to $\gamma(t_0)$?

($\varepsilon>0$ is used to shrink the vector if necessary so it doesn't come out "on the other side")

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  • $\begingroup$ Basically you need a tool to test whether a point is 'inside' or 'outside'. Winding number is one of these tools. en.wikipedia.org/wiki/Winding_number $\endgroup$ – Xipan Xiao Jul 3 '15 at 15:30
  • $\begingroup$ @XipanXiao, I guess one could take both endpoints of $n_1$ and $n_2$ and calculate the windung of $\gamma$ around these two points. As $\gamma$ is a jordan curve this will work...but not quite what I was thinking of. ;) Thanks for your input, it solves the question of "if", any other suggestions are still welcome. :) $\endgroup$ – Hirshy Jul 3 '15 at 15:41
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You cannot decide this locally. A Jordan curve in Euclidean two space is an oriented $1$-submanifold, and chosing the normal amounts to the choice of an orientation. This, on the other hand, means, that if you know an interior normal at one point and have a global regular parametrization, you can figure out at that single point whether it is orientation pre- or reversing, and then you know it for the whole curve.

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  • $\begingroup$ Thanks for your input! I might have to think of some other way for my proof, then. $\endgroup$ – Hirshy Jul 3 '15 at 16:18

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