I ran into the following question; let $x,y$ be two points in $\mathbb{R}^d$. Let $(\psi\_t)\_{t\geq 0}$ be the mapping from $\mathbb{R}^{2d}$ to $\mathbb{R}^{2d}$ defined, for all $t\geq 0$, by
$$
\psi\_t(x,y) = \Big(xe^{-t}+\sqrt{1-e^{-2t}}y, -\sqrt{1-e^{-2t}}x+ye^{-t}\Big).
$$
Re-parameterizing this family of mappings by setting $e^{-t}=\cos\theta$ induces a flow on $\mathbb{R}^{2d}$. I was wondering if this type of re-parameterization is natural and if this group structure is induced in a general context. Indeed, it is linked to the arc-length parameterization but I would like to know if this group structure is specific to the situation or if there is a general theory behind. To add another example, one could have considered the following mapping $\phi\_t$ from $\mathbb{R}^{2d}$ to $\mathbb{R}^{2d}$, for all $t\in \mathbb{R}$,
$$
\phi\_t(x,y) = \Big((1-\frac{t}{\sqrt{2}}) x + \frac{t}{\sqrt{2}} y\;,\; -\frac{t}{\sqrt{2}} x+(1+\frac{t}{\sqrt{2}})y\Big),
$$
which is a flow on $\mathbb{R}^{2d}$ and which is linked to the arc-length parameterization of the curve
$$
\gamma(t)=(e^{-t},(1-e^{-t})),\quad t\geq 0.
$$



Reparameterization and group structure
I am not sure precisely what you want, but I am intrigued that in the two cases that you display you replace an apparently arbitrary parametrisation by one of a very special kind, namely where the $y$-coordinate is a primitive of the $x$ one. (There are many reasons why this is useful but there is no point in my going into details until I know whether this is of interest to you). There is a standard method to do this for “any” curve (scare quotation marks because there can be problems as you hit the $y$-axis). I have worked the details out in the more general case you have posted. The required reparametrisation is explicit only in a very weak sense of the word—it involves taking the inverse of a hypergeometric function.


I will be happy to provide references and further information if this is the direction that you are interested in. If this is the case, I would strongly advise you to take a screen-shot of your question and this answer.


