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I seem to be a bit confused about this. Let $G$ be a Lie group and let us define the left $G$ action on a manifold $M$: $$\triangleright:G\times M\to M$$

Let $M=V$, where $V$ is the representation space (a vector space), so that for $g\in G$ and $v\in V$: $$g\triangleright v:=R(g)(v)$$ where $R(g)\in GL(V)$ is an automorphism. $R(g)$ maps the zero vector to the zero vector, i.e this action can never be free, because the stabilizer of the zero vector is the whole $G$ and not the identity.

Suppose now we want to get rid of the zero vector, so we choose for example $M=\mathbb R^d\backslash\{(0,\cdots,0)\}$. This is no longer a vector space, because it does not contain the zero vector, nor is it a vector subspace for the same reason. The above defined action could now be a free action and here is where I lost track. I have seen this in a series of lectures I am watching and I cannot understand it.

Obviously, $R(g)$ would definetely be non-linear, but by definition: $$R:G\to GL(V)$$ that is $R(g)$ is an invertible map from $V$ to $V$. How can I initially define this representation if V is no longer a vector space? Obviously I'm lacking some other definitions or something.

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You can define the action of $G$ on $V-\{0\}$ because for every $v\in V, v\neq 0$, and $g\in G, g(v)\neq 0$, so $G$ preserves $V-\{0\}$ and acts on it by diffeomorphisms.

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    ok, but the problem is that $G$ is acting on V in terms of its group representation $R(G)$, so that the action is $R(g)(v)$. Could you also explain the "acts on it by diffeomorphisms" part. – 2017-02-08
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    because it is the restriction of a linear map on an open subset. – 2017-02-08
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    I see, but a restriction does intuitively mean that I don't care how this $R(g) $ behaves for all $v$ that are not in the restriction subspace. But here our chosen subspace is structurally different, it is not a vector space, so how can $R(g)$ act on its elements (which are no longer vectors), as $R(g)$ is defined to act on vectors. Is it valid to just say "ok we take this part of the space, its elements are still vectors and we examine the action of $R(g)$ on them"? Sorry, I'm new to this. – 2017-02-08