1
$\begingroup$

Let $\mathcal{C}$ be the space of all parametric curves $x:[0,1]\rightarrow \mathbb{R}^2$. Let the set of all re-parameterizations of curves is $\Gamma = \{\gamma : [0, 1] \rightarrow [0, 1]| \gamma (0) = 0, \gamma (1) = 1, \gamma$ is a diffeomorphism$\}$ and reparametrization of $x$ with $\gamma$: $x*\gamma=x(\gamma).\sqrt(\gamma')$. Here the group action $\gamma$ is isometric.

We define the equivalent class $[x]=\{x*\gamma: \gamma\in \Gamma\}$ and space of all class $\mathcal{S}=\{[x]:x\in \mathcal{C}\}$. So $\mathcal{S}$ is a quotient space of $\mathcal{C}$.

Let us define a proper in $\mathcal{S}$ as $d([x_1],[x_2])=\inf_{\gamma_1,\gamma_2\in \Gamma}||x_1*\gamma_1-x_2*\gamma_2||$.

Problem: Whether addition of two class in $\mathcal{S}$ is well defined under the above metric!!

My attempt: To show this I need to first show that, $[x_1]+[x_2]=[x_1+x_2]$ and the above operation does not depends on representative members of the class. Let $x_1,x'_1\in [x_1]$ and $x_2,x'_2\in [x_2]$, then \begin{eqnarray} &&d([x_1+x_2],[x'_1+x'_2])\nonumber\\ &=&\inf_{\gamma_1,\gamma_2\in \Gamma}||(x_1+x_2)*\gamma_1-(x'_1+x'_2)*\gamma_2||\nonumber\\ &\leq& \inf_{\gamma_1,\gamma_2\in \Gamma}\{||x_1*\gamma_1-x'_1*\gamma_2||+||x'_2*\gamma_2-x_2*\gamma_1||\}\nonumber\\ &=&\inf_{\gamma_1,\gamma_2\in \Gamma}\{||x'_1-x_1*\gamma_2*\gamma_1^{-1}||+||x'_2-x_2*\gamma_1*\gamma_2^{-1}||\}\;\mbox{using isometries of $\Gamma$}\nonumber \\ &=&\inf_{\alpha,\delta\in \Gamma}\{||x_1-x'_1*\alpha||+||x'_2-x_2*\delta||\}\;\;\mbox{where}\;\alpha=\gamma_2*\gamma_1^{-1},\;\delta= \gamma_1*\gamma_2^{-1}\nonumber\\ &=&\inf_{\alpha\in \Gamma}||x_1-x_1*\alpha||+\inf_{\delta\in \Gamma}||x'_2-x_2*\delta||,\;\mbox{using isometries of the group action $\Gamma$}\nonumber\\ &=&0 \end{eqnarray} The last equality hold because of the fact that $x_1,x'_1 \in [x_1]$ and $x_2,x'_2 \in [x_2]$.

So addition is well defined in $\mathcal{S}$.

Please feel free to comment on the proof I have given. Or perhaps there are another easy way to prove it. Thanks.

$\endgroup$
  • $\begingroup$ Not sure about the equality in the second last line. Both the quantity involve same $\gamma$'s, one is inverse of other. Can the group action permits me to split the infimum in that line. $\endgroup$ – Janak Apr 13 '16 at 5:10

Your Answer

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

Browse other questions tagged or ask your own question.