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Given an algebra $A$ with finite dimensional representation $V$ with action $\rho$, I want to prove the following statement:

  • If $W\subset V$ are finite dimensional representations of A, then $\chi_V=\chi_W+\chi_{W/V}$

I think you have to use the dimension theorem of quotient spaces, namely: $\dim(V/W)=\dim(V)-\dim(W)$, but then what? Can someone help me? Thank you

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I still think the proof should be elementary and use the observation on the trace... but I'm too green at representation theory :) I had been thinking about semisimple algebras the whole time. –  rschwieb Jan 15 '13 at 15:57
    
@rschwieb: in my opinion your answer was perfectly fine. But since you've deleted it, I've posted one myself in the same spirit. I hope you don't mind. –  Marek Jan 15 '13 at 17:00
    
@Marek I think that I was stuck thinking in semsimple algebras, and I agree with Tobias that one cannot guarantee an invariant complement. –  rschwieb Jan 15 '13 at 18:09

1 Answer 1

Pick a basis for $V$ such that first $k = \dim (W)$ vectors form a basis for $W$. In this basis, every element $\rho(g)$ is represented by the matrix $$\pmatrix{A & B \cr 0 & C}$$ where $A$ and $C$ are square matrices which represent $\rho$ restricted to $W$ and $V/W$ respectively and $B$ is a projector to $W$. Indeed, we have $$\pmatrix{A & B \cr 0 & C} \pmatrix{w \cr 0} = \pmatrix {Aw \cr 0}$$ for every $w \in W$. Similarly, for $u \in V / W$, which we can represent as $v + w$ with unique $v \in V$ and arbitrary $w \in W$, we have $$\pmatrix{A & B \cr 0 & C} \pmatrix{w \cr v} = \pmatrix {Aw + Bv \cr Cv}$$ and because $Aw + Bv \in W$ this is independent of the choice of $w$. Now, just take trace of this matrix and you're done (because trace does not depend on the choise of basis).

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Ah, of course, you don't need a direct sum decomposition: you just need upper triangular matrices :) –  rschwieb Jan 15 '13 at 18:10

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