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 $$ \gamma(t)=(e^{-t},(1-e^{-t})),\quad t\geq 0. $$

  • $\begingroup$ @YCor: why the -1 ? Is it completely trivial ? I am not a mathematician but an engineer and I don't know much about differential structure and group.. $\endgroup$ – b57 Mar 2 at 20:39
  • $\begingroup$ Your post currently has 19 views, this doesn't tell you who did the downvote. $\endgroup$ – YCor Mar 2 at 20:42
  • 1
    $\begingroup$ You twice refer to "this group structure". What is the group? $\endgroup$ – LSpice Mar 2 at 20:50
  • 1
    $\begingroup$ I don't understand the question. The initial mapping you write down appears unmotivated, then you ask if this associated flow is natural. How did you come up with the original mapping? If the original mapping came up in some context maybe there is an answer to your question, but without the context it's difficult to say anything. $\endgroup$ – Ryan Budney Mar 2 at 21:31
  • 2
    $\begingroup$ Ok. Let $p\in [1,2]$. I start from the following mapping on $\mathbb{R}^d$ defined by $xe^{-t}+(1-e^{-pt})^{\frac{1}{p}}y$. Then, the question is: how to lift it to $\mathbb{R}^{2d}$ and re-parameterized it in order to obtain a one parameter group? I know how to do it for $p=2$ and for $p=1$. But not in between. This seems linked to the arc-length parametrization of the corresponding curve $\endgroup$ – b57 Mar 2 at 21:36

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.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your mysterious answer! I have worked out the details as well in the general case and arrived more or less at the same analytical conclusion. Since I am not a professional mathematician, I am not sure to understand your warning at the end of the answer. $\endgroup$ – b57 Mar 3 at 15:32
  • $\begingroup$ The warning hints at the fact that your question might be in danger of being deleted! $\endgroup$ – user131781 Mar 3 at 15:37
  • $\begingroup$ Is that so ? Why ? $\endgroup$ – b57 Mar 3 at 19:53
  • $\begingroup$ Questions which get hit by an instant downvote are in a very parlous state. Firstly they are then less likely to be read and so less likely get upvotes by later readers who might appreciate them and/or provide useful answers (which then might get upvotes). Questions with no upvotes and no answers are susceptible to deletion. Of course, I am not hinting that this is a deliberate tactic of those who slap on instant downvotes—honi soit qui mal y pense. $\endgroup$ – user131781 Mar 5 at 8:05
  • $\begingroup$ Ok, thank you for your answer and for this precision. Being new on this site affects the way one perceives instant downvotes. Best $\endgroup$ – b57 Mar 5 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.