# Character of representation of an associative algebra

Let $A$ be an associative algebra over a field $k$ , and $V$ is a finite representation of $A$.

Character $\chi_V(a) = tr|_V\rho(a)$ where $a\in A$

There exists a filtration $0= V_0\subset V_1\subset \cdots \subset V_n =V$ such that $V_i/V_{i-1}$ is irreducible for all nonegative integer $i\leq n$

If there exists another filtration $0= V'_0\subset V'_1\subset \cdots \subset V'_n =V$

Let $W_i = V_i/V_{i-1} , W'_i = V'_i/V'_{i-1}$ , then there exists a permutation $\sigma$ such that $W_i = W'_{\sigma (i)}$ for all nonegative integer $i\leq n$

Here is a proof from the lecture notes introduction to representation theory by Etingof

(Suppose $Char(k)=0$) It's obvious that $\sum_{i=1}^n \chi_{W_i}=\sum_{i=1}^n \chi_{W'_i}$ . And we know that the characters of irreducible representations are linear independent , so the multiplicity of every irreducible representation $W$ of $A$ among $W_i$ and among $W'_i$ are the same. This implies the theorem.

What does the multiplicity of every irreducible representation $W$ of $A$ means?

The number of $W_i$ (or $W_i'$) that are isomorphic to $W$.
Since $\sum_{i=1}^n \chi_{W_i}=\sum_{i=1}^n \chi_{W'_i}$ (because both are equal to the character of $V$), and so, grouping together the terms involving isomorphic $W_i$ (or $W_i'$), $$\sum_Wm_W\chi_W=\sum_{i=1}^n \chi_{W_i}=\sum_{i=1}^n \chi_{W'_i}=\sum_Wm'_W\chi_W,$$ where the first and last sums are over the isomorphism classes of irrreducible representations $W$ and where $m_W$ is the multiplicity of $W$ among the $W_i$ (and $m'_W$ the multiplicity among the $W'_i$), the linear idependence of the $\chi_W$ implies that $m_W=m'_W$ for every $W$.
• @JeremyRickard Why are all $\chi_W$ linearly independent? I know that it is true for finite group when the characteristic of $F$ does not divide the carnality of the group. However it seems that the proof for group case is not applicable to associative algebra case. I shall be grateful if you give any hint on this point. Thank you! – Hebe Dec 17 '15 at 14:44