Let $\mathcal{C}$ be the space of all parametric curves $x:[0,1]\rightarrow \mathbb{R}^2$. Also let $\mathcal{C}$ is a linear manifold in the sense that $x_1,x_2\in \mathcal{C}$ implies that $cx_1+dx_2\in \mathcal{C}$ for $c,d\in \mathcal{\mathbb{R}}$.
Now define the equivalent class $[x]=\{x*\gamma: \gamma\in \Gamma\}$ and 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$\}$, where $\gamma:[0,1]\rightarrow [0,1]$, is re-parametrization function and $\Gamma$ is the space of all re-parametrization function.
Now $\mathcal{S}=\{[x]:x\in \mathcal{C}\}$ is a quotient space of $\mathcal{C}$ and we define distance measure $d([x_1],[x_2])=\inf_{\gamma_1,\gamma_2\in \Gamma}||x_1*\gamma_1-x_2*\gamma_2||$ in $\mathcal{S}$ and the group action $\gamma$ is isometric.
Problem: Whether $\mathcal{S}$ is a linear space or not !!
My attempt: To show this is a linear space I need to first show that, $[x_1]+[x_2]=[x_1+x_2]$. To show this I need to show 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 $\mathcal{S}$ is closed under addition and similarly it can be shown that this is closed under scalar multiplication.
Any helpful comments are welcome. Thanks.