# Dimension of the invariant subspace

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?

• The left action of a $\gamma$ on $f$ is $\gamma \cdot f(\textbf{x} )= f(\gamma^{-1} \textbf{x})$. Otherwise it is a right-action.
– user598294
Commented Mar 29, 2019 at 23:12
• If $R_\Gamma$ is a projection, then its restriction $r_\Gamma : \mathbb{C}_{\leqslant 1}[\textbf{x}] \to \mathbb{C}_{\leqslant 1}[\textbf{x}]$ (the polynomials of degree $\leqslant 1$) is the projection on $\mathbb{C}_{\leqslant 1}[\textbf{x}]^\Gamma$ and $a_1=\dim(\mathbb{C}_{\leqslant 1}[\textbf{x}]^\Gamma)=rk(r_\Gamma)=trace(r_\Gamma)$. To calcultate $trace(r_\Gamma)$, use the basis $(x_1,\cdots,x_n)$ of $\mathbb{C}_{\leqslant 1}[\textbf{x}]$. It should lead you to the result
– user598294
Commented Mar 29, 2019 at 23:12
• How do you mean use the basis? Commented Mar 29, 2019 at 23:49
• Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M \cdot f(x_1,...,x_n) = f(M \textbf{x})$? Commented Mar 29, 2019 at 23:57
• @theman Try computing $MN\cdot f$ and $M\cdot (N\cdot f)$ with that definition. You'll notice that $(MN\cdot f)x=f(MNx)$, whereas $(M\cdot (N\cdot f))x=(N\cdot f)(Mx)=f(NMx)$. These are not equal.
– jgon
Commented Mar 30, 2019 at 0:04

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.

• So, where I've written $trace(R_{\Gamma}) = dim(\mathbb{C}[x_1,...,x_n]^G)$, does this not maky any sense? Commented Mar 30, 2019 at 10:42
• No, except if you meant the Krull dimension of the ring $\mathbb{C}[x_1,...,x_n]^G$ (the size of its transcendental basis over $\Bbb{C}$) in which case yes, which is what I did with $\mathbb{C}[x_1,...,x_n]^G = \Bbb{C}[y_1,\ldots,y_m]$ Commented Mar 30, 2019 at 12:47
• Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_{\mathbb{C}}(\mathbb{C}[x_1,...,x_n]_{1}^{\Gamma})$ right? Commented Mar 30, 2019 at 13:00
• @theman $\mathbb{C}[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $\mathbb{C}[x_1,...,x_n]^G$ always contains more than $\Bbb{C}[y_1,\ldots,y_m]$, $\mathbb{C}[x_1,...,x_n]/\mathbb{C}[x_1,...,x_n]^G$ is a Galois extension of degree $|G/\ker(\rho)|$ thus the transcendental degree of $\mathbb{C}[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,\ldots,f_{|G/\ker(\rho)|}$ then $\mathbb{C}[x_1,...,x_n]=\sum_{j=1}^{|G/\ker(\rho)|} \mathbb{C}[x_1,...,x_n]^G f_j(x)$) Commented Mar 30, 2019 at 13:08
• @ Oh yes, I see. Thank you very much! Commented Mar 30, 2019 at 13:10