Dimension of the invariant subspace

Question

Let $\Gamma \subseteq GL_{n}(\mathbb{C})$ be a finite matrix group. Let this finite matrix group act on $f(x_1,...,x_n) \in \mathbb{C}[x_1,...,x_n]$ like so: $$\Gamma \cdot f(x_1,...,x_n) = f(\Gamma \textbf{x})$$ where $\textbf{x}$ is to be thought of as the column vector of the variables $x_1,...,x_n$.

Define the invariant subspace $\mathbb{C}[x_1,...,x_n]^{\Gamma} = {\{f \in \mathbb{C}[x_1,...,x_n] : A \cdot f = f \hspace{2mm} \forall \hspace{2mm} A \in \Gamma}\}$.

Now, define the Reynold's operator $R_{\Gamma} : \mathbb{C}[x_1,...,x_n] \rightarrow \mathbb{C}[x_1,...,x_n]$ by: $$R_{\Gamma} (f)(\textbf{x}) = \frac{1}{|\Gamma|} \sum_{A \in \Gamma} f(A \textbf{x})$$

Now, the number of linearly independent invariants of $\Gamma$ of degree $1$ is given by $$a_1 = \frac{1}{|\Gamma|} \sum_{A \in \Gamma} trace(A)$$ But I'm not sure why this is so? I know that $R_{\Gamma}$ is projection on to $\mathbb{C}[x_1,...,x_n]$ and $im(R_{\Gamma}) = \mathbb{C}[x_1,...,x_n]^{\Gamma}$, and so this would imply that $trace(R_{\Gamma}) = dim(\mathbb{C}[x_1,...,x_n]^{\Gamma})$, but where do I go from here? What is the trace of this Reynold's Operator?

I'm not even sure if I'm going in the right direction here, because I'm not sure why this $dim(\mathbb{C}[x_1,...,x_n]^{\Gamma})$ would even give the number of linearly independent invariants of $\Gamma$ of degree $1$. Where does the degree $1$ bit come from?

Answer 1

In those cases I need elementary discussions

For any representation of finite group $\rho : G \to GL(V)$ to inversible linear maps of a $\Bbb{C}$-vector space, then $P=\frac{1}{|G|}\sum_{g \in G}\rho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v \in V$ then $Pv \in V^G$ and if $v \in V^G$ then $Pv=v$)

in some basis $B$ you'll have $P = B \pmatrix{I_m & 0 \\ 0 & 0} B^{-1}$ where $m = \dim V^G$ so $trace(P) = trace( \pmatrix{I_m & 0 \\ 0 & 0}) = \dim V^G$.

You need to make clear you are considering $V =\Bbb{C}^n$ and the corresponding $trace$, no polynomial ring.

From there you can construct other representations on $\Bbb{C}[x_1,\ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $\pi(g)(f(x))= f(\rho(g)x)$ is called $\pi = Sym^d\rho$, and what you defined is the natural infinite dimensional rep. $\bigoplus_d Sym^d\rho$ of $G=\Gamma$ on $\Bbb{C}[x_1,\ldots,x_n] = \bigoplus_d \Bbb{C}[x_1,\ldots,x_n]_d$.

Then the point is that $V = V^G \oplus W$ where $W = \ker(P)$ and $W$ is sent to itself by the $A\in \Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,\ldots,y_m,z_1,\ldots,z_{n-m}) = B(x_1,\ldots,x_n)$ : $\Bbb{C}[x_1,\ldots,x_n]= \Bbb{C}[y_1,\ldots,y_m,z_1,\ldots,z_{n-m}]$ and $A.f(y_1,\ldots,y_m,z_1,\ldots,z_{n-m}) = f((y_1,\ldots,y_m,0,\ldots)+BA B^{-1} (0,\ldots,z_1,\ldots,z_{n-m}))$.

If $G$ is a finite group then $\Bbb{C}[x_1,\ldots,x_n]/\Bbb{C}[x_1,\ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/\ker(\rho)$ so $\Bbb{C}[x_1,\ldots,x_n]^G=\Bbb{C}[y_1,\ldots,y_m,f_1,\ldots,f_{n-m}]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $\Bbb{C}[x_1,\ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.