0
$\begingroup$

I have just began studying the lecture notes introduction to representation theory by Etingof for several days.

I learn that every finite representation $V$ of an associative algebra $A$ admits 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 positive integer $i\leq n$

From that , I encounter a confusion.

Let $W = V_1\oplus V_2/V_1\oplus \cdots \oplus V/V_{n-1}$ , then $W$ has character as $\chi_W = \chi_V+\chi_{V_2/V_1} + \cdots +\chi_{V/V_{n-1}}$ which is the same as $\chi_V$. Hence $V \cong W$

However , $V = V_1\oplus V_2/V_1\oplus \cdots \oplus V/V_{n-1}$ shows $V$ is completely reducible. In other words , every finite representation is completely reducible.

I have already known there exists finite representation which is not completely reducible , but I don't know where I make mistakes.

Please help me , thank you.

$\endgroup$
1
$\begingroup$

For representations of arbitrary associative algebras it is not true that $\chi_V=\chi_W\Rightarrow V\cong W$.

$\endgroup$
  • $\begingroup$ Oh, $V \cong W \Rightarrow \chi_V = \chi_W$ is true for associative algebra , but the inverse holds only for finite group , right? But that means two nonisomorphical finite representations can have the same character. It sounds odd. $\endgroup$ – Syuizen Nov 1 '15 at 14:53
  • $\begingroup$ @Syuizen That's right. The fact that it holds for representations of finite groups (over $\mathbb{C}$ at least) is connected with the fact that in that case all representations are completely reducible. $\endgroup$ – Jeremy Rickard Nov 1 '15 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.