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Suppose, $C$ is a linear manifold (i.e., manifold which is closed under addition and multiplication) and $\Gamma$ is a Lie group. Can we say in general $C/\Gamma$ is also a linear manifold?

Can we the same about the following case: Let $C$ is space of all parametrized curves $f:[0,1]\rightarrow \mathbb{R}^2$ and $\Gamma$ is the collection of all possible parametrization $\gamma:[0,1]\rightarrow [0,1]$.

In this case is $C/\Gamma$ is a linear manifold?

Janak
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1 Answers1

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In the set-theoretic case, we can't: let $V$ denote a non-trivial real vectorspace. Then $\mathrm{Aut}(V)$ acts on $V$. But $V / \mathrm{Aut}(V)$ consists of two points (namely $\{\{0\},V \setminus \{0\}\}$). So it can't be made into a real vectorspace.

I see no reason the smooth manifold case should be any different. Ideas, anyone?

goblin GONE
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