Characters determine Representations over characteristic 0 field k

Projection Operator

Let $G$ be a finite group with representation $\rho :G\to G{L}_n(k)=Aut(k^n)$ and $\rho :G\to G{L}_m(k)=Aut(k^m)$ where $k^n$ and $k^m$ are $kG$-modules.

If $tr(\rho (g))=tr({\rho }^{\prime }(g))$ for all $g$, then $E{\cong }_{R-mod}F$ as $kG$-modules.

What does it mean “determine”?

Let $k$ be a characteristic 0 field and $G$ be a finite group. Then by the Maschke’s Theorem and Artin-Weddenburn Theorem,

$$kG{\cong }_{Ring}{M}_{n_1}(D_1)\times \cdots M_{n_s}(D_s)\cong R_1\times \cdots \times R_s.$$

We know at least one component $k$ and let denote as $R_1=M_{n_1}(D_1)$. So that $n_1=1$.

If we can determine $s$ and $n_i$‘s, we can say we can determine the representation of $G$ over $k$.

How?

We have the decomposition of the unity $1_{kG}$ using Different Point of View, i.e. we have simple rings $(R_i,e_i)$ for $i=1,\dots ,s$.

$$1_{kG}=e_1+\cdots +e_s.$$

Let $L_i=D_i^{n_i}$ be a $kG$-mod and ${\chi }_i$ be the simple character of the representation on $L_i$. By Module,

$$E{\cong }_{kG-mod}{\oplus }_{i=1}^s{n}_i{L}_i.$$

Since ${\chi }_i$‘s are simple characters, and $e_i$‘s are unit of $R_i$‘s. So, ${\chi }_i(e_i)=di{m}_k{L}_i=n_{i}di{m}_k{D}_i$, and $\chi (e_j)=0$.

Therefore we can determine $\chi$ defined on $G$ as $\chi ={\sum }_{i=1}^n{n}_i{\chi }_i$ where $a_i\in {\mathbb{Z}}_{\ge 0}$.

We call $n_i$ as multiplicity of ${\chi }_i$ in $\chi$.

Wait! Isn’t Character defined on $G$?

Yes, it is defined on $G$, but we are using $kG$ to determine. We can eaily extend as $\chi :kG\to k$ such as

$$\sum a_{g}g\mapsto \sum a_g\chi (g).$$

Determine Again

Obviously, if we have two isomorphic $kG$-modules $V$ and $V^{\prime }$, then corresponding characters are same.

On the other hand, we know that Corollary, so if we have two same characters, two $R$-modules are isomorphic.