If $G$ is a connected Lie group acting on a vector $\mathbb{C}$-space $V$ then it is well known that the algebra of invariants $\mathbb{C}[V]^G$ coincides with the algebra of invariants $\mathbb{C}[V]^L$ of corresponding Lie algebra $L.$


**Question.** Let now $G$ be a finite group, $V$ be its representation. Is there exists a Lie algebra $L$ and its representation on $V$ such that $\mathbb{C}[V]^G=\mathbb{C}[V]^L$?


It is easy to see that it impossible for symmetric group $S\_n$. But maybe there are classes of finite groups for which can be found the positive answer? 


**Edit.** Let $V=< v\_1,v\_2,\ldots,v\_n >$ be standard representation of the symmetric group $S\_n.$ Suppose that there exist a derivation $D=f\_1 \frac{\partial}{\partial x\_1}+\cdots+ f\_n \frac{\partial}{\partial x\_n}, f\_i \in S(V)$ of the symmetric algebra $S(V)$ such that $D(I)=0$ for all $I \in \mathbb{C}[V]^{S\_n}$.
Let $I\_1,I\_2,\ldots, I\_n \in S(V)$ be a minimal generating set for $\mathbb{C}[V]^{S\_n}.$ Since $D(I\_k)=0, \forall k,$ we get the system of polynomial equations on $f\_1,f\_2,\ldots,f\_n:$
$$
f\_1 \frac{\partial I\_1}{\partial x\_1}+\cdots+ f\_n \frac{\partial I\_1}{\partial x\_n}=0,\\
f\_1 \frac{\partial I\_2}{\partial x\_1}+\cdots+ f\_n \frac{\partial I\_2}{\partial x\_n}=0,\\
\ldots \\
f\_1 \frac{\partial I\_n}{\partial x\_1}+\cdots+ f\_n \frac{\partial I\_n}{\partial x\_n}=0.
$$
The Jacobian of system of invariants $I\_1,I\_2,\ldots, I\_n$ is not zero. It follows the system has only trivial solution $f\_i=0.$



Linearisation of a group
Geometric considerations suggest the answer is always no: we expect $\text{Spec } \mathbb{C}[V]^G \cong V/G$ to have dimension $\dim V$, but we expect $\text{Spec } \mathbb{C}[V]^L \cong V/L$ to have dimension at most $\dim V - \dim L$, at least if the action is sufficiently nontrivial.


One proof is as follows. First recall that $\mathbb{C}(V)$ is a Galois extension of $\mathbb{C}(V)^G$, which is in turn the fraction field of $\mathbb{C}[V]^G$. 


**Lemma:** Let $L/K$ be a separable extension and let $D : L \to L$ be a derivation. Then $D$ is uniquely determined by its restriction to $K$. In particular, if $D |\_K = 0$, then $D = 0$.


*Proof.* Let $x \in L$ and let $p(x) = \sum p\_i x^i$ be its minimal polynomial over $K$. Then


$$D(\sum p\_i x^i) = \sum \left( D(p\_i) x^i + i p\_i x^{i-1} D(x) \right) = 0$$


hence $D(x)$ is uniquely determined by the $D(p\_i)$ as long as $\sum i p\_i x^{i-1} \neq 0$. But this is guaranteed by separability. $\Box$


Now, any derivation on $\mathbb{C}[V]$ extends uniquely to a derivation on $\mathbb{C}(V)$, and by the lemma if such a derivation vanishes on $\mathbb{C}(V)^G$ (in particular, if it vanishes on $\mathbb{C}[V]^G$) then it vanishes identically. 




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**Edit:** Here is a reinterpretation of the above argument which is closer to the geometric intuition. The above lemma can be adapted to show that the inclusion $\mathbb{C}[V]^G \to \mathbb{C}[V]$ induces an isomorphism on Zariski tangent spaces at any point $v \in V$, where by the Zariski tangent space at a point $p$ of a variety $X$ I mean the space of linear functions $d : \mathcal{O}\_{X, p} \to \mathbb{C}$ such that $d(ab) = d(a) b(p) + a(p) d(b)$. On the other hand, if $D : \mathbb{C}[V] \to \mathbb{C}[V]$ is a nonzero derivation, then there exists a point $v \in V$ such that composing $D$ with the evaluation map at $v$ gives a nonzero Zariski tangent vector at $v$, from which it follows that the inclusion $\mathbb{C}[V]^D \to \mathbb{C}[V]$ cannot induce an isomorphism on Zariski tangent spaces at $v$. 


