# Characters of (distinct) irreducible finite-dimensional representations of $A$

I need help to understand the proof of this theorem. The theorem can be found in the book Introduction to representation theory by Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina. Page 27

Characters of (distinct) irreducible finite-dimensional representations of $A$ are linearly independent.

Proof: If $V_1,...,V_r$ are nonisomorphic irreducible finite-dimensional representations of $A$, then $\rho_{V_1}\oplus...\oplus\rho_{V_r}:A\rightarrow \text{End}V_1\oplus...\oplus\text{End}V_r$ is surjective by the density theorem, so $\chi_{V_1},...,\chi_{V_r}$ are linearly independent.

I can't understand this part that suggest for a more detailed proof:

(Indeed, if $\Sigma_{i=1}^{r} \lambda_i\chi_{V_i}(a)=0$ for all $a \in A$, then $\Sigma_{i=1}^{r}\lambda_i\text{Tr}(M_i)=0$ for all $M_i \in \text{End}_k V_i$. But each $Tr(M_i)$ can range independently over $k$, so it must be that $\lambda_1,...,\lambda_r=0$)

How I can see this part? But each $Tr(M_i)$ can range independently over $k$, so it must be that $\lambda_1,...,\lambda_r=0$)

Any help would be greatly appreciated.

Thanks.

• I don't know the context, but if it's really true that $\mathrm{Tr}(M_i)$ ranges over all of $k$, choose $M$'s such that $\mathrm{Tr}(M_1) = 1$ and the others are zero. – rogerl Mar 18 '15 at 0:23
• Ok, but I can't see how this works. Did you mean that I have to choose $M's=M'_i$? – JimmyJP Mar 18 '15 at 0:44
• Is $A$ a group algebra over an algebraically closed field, or some more general algebra? You haven't told us. – Geoff Robinson Mar 18 '15 at 10:59
• We consider this result over a algebraically closed field in the context of the theorem. – JimmyJP Mar 18 '15 at 12:52

Each $\mathrm{Tr}(M_i)$ can range independently over $k$ because $k$ is algebraically closed so if $\dim V_i = n$ then for any $a \in k$ we may take $M_i = \sqrt[n]{a}I$. Then $\mathrm{Tr}(M_i) = a$.
I am not sure what $$A$$ is, but if you are over charactersitic $$0$$ field $$k$$, you do not need to use the fact that the ground field is algebraically closed. Let $$V_{1}, \dots, V_{l}$$ be distinct irreducible $$G$$-modules. We use the notation $$\rho_{i} : G \rightarrow \text{GL}(V_{i})$$ for defining group maps. Write $$\rho = \rho_{1} \oplus \cdots \oplus \rho_{l}$$. Consider the following map of (non-commutative) rings $$k[\rho(g): g \in G] \rightarrow k[\rho_{1}(g_{1}): g_{1} \in G] \times \cdots \times k[\rho_{l}(g_{l}): g_{l} \in G]$$ given by $$\rho(g) \mapsto (\rho_{1}(g), \dots, \rho_{l}(g))$$, which is well-defined by definition of $$\rho$$. An application of Chinese Remainder Theorem shows that this map is surjective; here the fact that $$V_{1}, \dots, V_{l}$$ are distinct is used.
Now, let $$\chi_{1}, \dots, \chi_{l}$$ be characters of $$V_{1}, \dots, V_{l}$$. To show that they are linearly independent, let $$a_{1}\chi_{1} + \cdots + a_{l}\chi_{l} = 0$$ for $$a_{i} \in k$$. It is enough to find $$y = c_{1}g_{1} + \cdots + c_{r}g_{r} \in kG$$ such that $$\rho_{j}(y) = 0$$ for all $$j \neq i$$ and $$\rho_{i}(y) = \text{id}_{V_{i}}$$, because
$$0 = \sum_{j=1}^{r}c_{j}(a_{1}\chi_{1}(g_{j}) + \cdots + a_{l}\chi_{l}(g_{j})) = a_{i}\dim(V_{i}),$$
so the fact that $$\text{char}(k) = 0$$ would imply that $$a_{i} = 0$$.
Finding such $$y$$ would have been done if $$k[\rho(g) : g \in G]$$ is $$k$$-linearly generated by $$\rho(g)$$'s. But in fact it is! This is because their monomial looks like $$\rho(x_{1}) \cdots \rho(x_{r}) = \rho(x_{1} \cdots x_{r})$$. This should be enough.