Let $\gamma:[0;1]\to\mathbb{R}^2$ be regular Jordan

Question

Let $\gamma:[0;1]\to\mathbb{R}^2$ be regular Jordan

I don't understand what is regular Jordan?? Maybe $\forall t\in[0;1],\gamma^{'}(t)\neq 0$.

And why $\mathrm{Im}\gamma$ is regular curve in the plane $\mathbb{R}^2$?

Answer 1

Yes, regular Jordan should be precisely that. More concretely, if your mapping $\gamma \colon [0,1] \rightarrow \mathbb{R}^2$ is said to be a Jordan curve then it means $\gamma$ is continuous, $\gamma(0) = \gamma(1)$ and $\gamma(x) = \gamma(y)$ cannot happen for $x\neq 0$ and $y\neq 1$, i.e. the mapping is injective everywhere on $(0,1)$ (see the definition in this wiki article). A curve is said to be regular if it is differentiable and the first derivative does not vanish (so wee need to assume that $\gamma$ is not only continuous but also differentiable, see this wiki article where the definition can be found in the section called Differential geometry).

What this geometrically means is that the image $\operatorname{Im}(\gamma) $ will be a (continuous) loop which does not intersect itself. If we want to interprete the regularity condition then, e.g. if $\gamma$ is understood as a parametrization of a point particle moving along the curve $\operatorname{Im}(\gamma) $ then its velocity is always non-zero.

Example of a regular Jordan mapping would be a classical parametrization of a unit circle $$ \gamma(t) = (\cos(2\pi t), \sin(2\pi t) ) , t \in [0,1] $$ which is continuous, differentiable and the first derivative $$\gamma^{\prime} (t) = (-2\pi \sin(2\pi t), 2\pi \cos(2\pi t) ) $$ never vanishes.