Decompose a 1D polynomial into irreducible representations I am taking an introductory course for group- and representation theory and struggle with this problem:
Consider the integral:
$\int_{-1}^1 P_2(x)dx$ where $P_2(x)=\alpha + \beta x + \gamma x^2$ 
Identify the symmetry group of the integration domain and decompose the
polynomial $P_2(x)$ into irreducible representations of this group.

I guess the symmetry group has to be $\mathbb{Z}_{2}$, right? I have absolutely no idea how this relates to the polynomial though.
It might be very obvious, but a simple explanation would be appreciated.  
 A: Let restrict ourselves to the linear maps that preserve $[-1,1]$. As you know already we have then a $\mathbb{Z}/2\mathbb{Z}$ generated by $\phi(x) = -x$. 
Let $\phi$ act on the space $V$ of polynomials (of degree at most two) by $\phi \cdot f = f\circ \varphi$. This gives us a (linear) representation $\rho\colon\mathbb{Z}/2\mathbb{Z}\to {\rm GL}(V)$. Fix the basis $\{1, x, x^2\}$. Since $\phi\cdot 1 = 1$, $\phi\cdot x = -x$, $\phi\cdot x^2 = x^2$ we decompose $V= V_0\oplus V_1$ where $V_0= \left<1,x^2\right>$ is the "even part" and $V_1= \left<x\right>$ is the "odd part". Let $R_0$ and $R_1$ be the respective Reynolds maps given by 
$$
R_0(v) = \frac{v+\phi\cdot v}{2}, \, R_1(v) = \frac{v-\phi\cdot v}{2}, 
$$
for any $v\in V$. It follows that $V_0$ and $V_1$ are their respective images, $R_0+R_1$ is the identity and $R_i^2 =R_i$ for $i=0,1$. Therefore,
$$
P_2(x) = (R_0+R_1)(P_2(x)) = \frac{P_2(x)+P_2(-x)}{2} + \frac{P_2(x)-P_2(-x)}{2} 
$$
is the required decomposition.
For the integral, we have that 
$$
\int\limits_{-1}^1 \frac{P_2(x)-P_2(-x)}{2}dx  = - \int\limits_{1}^{-1} \frac{P_2(x)-P_2(-x)}{2}dx 
$$
Hence it must be zero, which gives us 
$$
\int\limits_{-1}^1 P_2(x) dx  =  \int\limits_{1}^{-1} \frac{P_2(x)+P_2(-x)}{2}dx = \int\limits_{1}^{-1}(\alpha +\gamma x^2) dx
$$
This is really simple but I hope it clarifies you how the general case works. In the two dimensional question you'll do the same with the group $S_3$ of symmetries of the triangle.
