I have some trouble understanding the decomposition of representations into irreducible ones.
For example, take $G = S_3$, the symmetric group.
Then $G$ has three irreducible representations, namely, the trivial one, the sign one, and the two-dimensional represention given by the planar symmetries of a triangle.
Let $V$ be the $\mathbb C G$-module given by the two-dimensional irreducible representation, $T$ be the trivial module, and $S$ be the module given by the sign representation.
It is easy to see from the characters that $V\otimes V \cong V\oplus T\oplus S$ as $\mathbb C G$-modules. However, what explicitly is this $G$-invariant isomorphism?
Best Answer
Say that we have irreducible $\mathbb{C}G$-modules $V_i$ and an isomorphism $W \cong \oplus n_i V_i$. Consider $E_i = Hom_{\mathbb{C}G}(V_i,W)$ and let $W_i$ be the space generated in $W$ by the $f(v)$ for $f$ in $E_i$ and $v$ in $V_i$. Remember that $W_i$ is called the $V_i$-isotypic component of $W$.
What you have is a true decomposition $W = \oplus W_i$ and the isomorphism $W \cong \oplus n_i V_i$ is obtained by giving isomorphisms $W_i \cong n_i V_i$. There is no natural choice for this isomorphism, even when $n_i = 1$. In order to construct one, take any $w_1$ in $W_i$. Its orbit generates a copy $W_{i,1}$ of $V_i$ in $W_i$. Then take another $w_2$ not in $W_{i,1}$ and continue. In this way, you obtain a preferred isomorphism $W_i \cong n_iV_i$.
The real difficulty is to understand $W_i$ in $W$; for instance in your case to understand which line in $V\otimes V$ is the $S$-isotypic component. For this, there is a projection formula: denote by $\chi_i$ the character of $V_i$ and by $\rho(t)$ the endomorphism of $W$ given by the action of $t \in G$. Then, the projection of $W$ on the $W_i$ parallel to the other isotypic components is given by:
$$ p_i = \frac{n_i}{g}\sum_{t \in G} \chi_i(t)^\ast \rho_t,$$ with $n_i$ the degree of $V_i$ and $g$ the cardinal of $G$. And of course $W_i = p_i(W)$.