# Linearisation of a group

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 $f_i=0.$

-
Being classical or semisimple has nothing to do with the first statement; it suffices that $G$ be connected. – Qiaochu Yuan Nov 3 '13 at 8:19
@Qiaochu Yuan. Corrected. Thank you. – Leox Nov 3 '13 at 8:49
What are the quantifiers? Do you allow $L$ to depend only on $G$ or on $G$ and $V$? On a related note, could you elaborate on "it is easy to see that it is impossible for the symmetric group"? Finally, the case of the symmetric group suggests that maybe a sensible question is whether in addition to all that there is a functor $\mathcal{F}$ of $V$ for which $\mathbb{C}[V]^G\simeq\mathbb{C}[\mathcal{F}(V)]^L$? – Vladimir Dotsenko Nov 3 '13 at 14:38
@Vladimir Dotsenko. 1. $L$ may depend on $G$ and $V.$ 2. I have added the proof for the symmetric group. – Leox Nov 3 '13 at 15:32
Ah I see, you mean the symmetric group in standard representation. I was wondering if you meant that or the symmetric group and (more or less) arbitrary representation. – Vladimir Dotsenko Nov 3 '13 at 15:46

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.

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 $v$.

-
Great! Thank you. – Leox Nov 3 '13 at 21:26
Where did you use that $G$ is a finite group? – Leox Nov 3 '13 at 21:30
@Leox: the first line, $\mathbb{C}(V)$ being a Galois extension of $\mathbb{C}(V)^G$ (hence they have the same transcendence degree, hence $\mathbb{C}[V]^G$ has the same Krull dimension as $\mathbb{C}[V]$, etc.). – Qiaochu Yuan Nov 3 '13 at 22:04
This is very nice, Qiaochu. – Mariano Suárez-Alvarez Nov 3 '13 at 23:25
There should be a reasonably elementary if possibly tedious argument involving Hilbert series (there is a nice lower bound on the terms of the Hilbert series of $\mathbb{C}[V]^G$ and so it suffices to find a suitable upper bound on the terms of the Hilbert series of $\mathbb{C}[V]^D$ for $D$ some derivation) but I'm not seeing it at the moment. Anyone want to take a shot? – Qiaochu Yuan Nov 4 '13 at 4:57

You asked for it! Here's the Hilbert series argument. I need to assume that the derivation $D$ is homogeneous, or equivalently comes from extending a linear map $V \to V$. This covers the case of a Lie algebra action on $V$.

First, recall that $\dim S^n(V^{\ast}) = {\dim V + n - 1 \choose \dim V - 1}$ has asymptotic growth rate $\Theta(n^{\dim V - 1})$ for $\dim V$ fixed and $n \to \infty$.

Let $g_n = \dim S^n(V^{\ast})^G$ denote the dimension of the $n^{th}$ graded component of $\mathbb{C}[V]^G$. An explicit expression for the Hilbert series $\sum g_n t^n$ can be obtained from Molien's theorem which shows in particular that the leading term in the asymptotic expansion of $g_n$ is at least $\frac{1}{|G|} n^{\dim V - 1}$ (hence, as above, that $\mathbb{C}[V]^G$ has Krull dimension $\dim V$). There should be a more elementary way to obtain this estimate.

On the other hand, let $D$ be a derivation induced from a linear map $V \to V$. Let $e_1, ..., e_n$ be a Jordan basis of $D$ with generalized eigenvalues $\lambda_1, ..., \lambda_n$. Then the monomial $e_1^{k_1} ... e_n^{k_n}$ in $\mathbb{C}[V]$ is a generalized eigenvector of $D$ with generalized eigenvalue $\sum k_i \lambda_i$. In particular, the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has dimension at most the number of monomials of degree $n$ of generalized eigenvalue $0$. The exponents of these monomials are determined by the kernel of the map

$$\mathbb{Z}^n \ni (k_1, ..., k_n) \mapsto \sum_{i=1}^n k_i \lambda_i \in \mathbb{C}.$$

If at least one of the $\lambda_i$ is nonzero, the kernel of the above map is a submodule of rank at most $n-1$, from which it follows that the dimension $d_n$ of the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has asymptotic growth rate at most as fast as $O(n^{\dim V - 2})$.

If all of the $\lambda_i$ are equal to zero, then $D$ is nilpotent. In this case we need to inspect the action of $D$ on monomials more carefully. Permute the Jordan basis so that $D(e_i)$ is either $e_{i-1}$ or $0$ and so that $D(e_2) = e_1$, and consider the lexicographic order on $\mathbb{C}[V]$ with respect to the $e_i$. Then $D(e_1^{m_1} ... e_k^{m_k})$ has leading term $m_2 e_1^{m_1+1} e_2^{m_2-1} ... e_k^{m_k}$, from which it follows that the image of $D$ acting on $S^n(V^{\ast})$ has dimension at least the number of monomials with $m_1 \ge 1$, or the dimension of $S^{n-1}(V^{\ast})$. The two dimensions are polynomials in $n$ with the same leading term and $d_n$ is their difference, so we again find that the asymptotic growth rate of $d_n$ is at most $O(n^{\dim V - 2})$.

-
Thank you. If I understood correctly you estimate an asymptotic growth rate for the leading term of the two Hilbert series. The first one for the graded algebra $C[V]^G$ and the second one is for $C[V]^D.$ The rates are different, therefore $C[V]^G \neq C[V]^D.$ Please explane again where did you use that $G$ is a finite group? Why these reasoning does't pass for $G=SL_2(C).$? – Leox Nov 4 '13 at 6:49
@Leox: when I appeal to Molien's theorem. – Qiaochu Yuan Nov 4 '13 at 6:52
Oh Yes, the Krull dimension of $C[V]^{SL_2}$ equals $\dim V-1$ and then the asymptotic growth rate for its Hilbert series leading term is almost $O(n^{\dim V-2}),$ like as for one derivation. In other words, a finite group has not enough elements (hence $C[V]^G$ is too big) and it is main reason why $C[V]^G \neq C[V]^D$. Thank you for the clarification. – Leox Nov 4 '13 at 7:42