We know that for a representation $V$ of a Lie algebra or a quantum group, we can define character of $V$ as $ch(V)=\sum_{\mu} dim(V_{\mu})e^{\mu}$, where $V_{\mu}$ is the weight space of $V$ with weight $\mu$. I didn't find the corresponding definition for representations of finite dimensional algebras (for example, associative algebras). Is there some kind of character theory for representations of finite dimensional algebras? Thank you very much.


In ordinary group representation theory, a character is a homomorphism $\chi:G\to \mathbb{C}^\times$. A character associated to a representation $\rho: G\to GL(V)$ is $\chi_\rho: g\mapsto trace(\rho(g))$. Character of Lie algebra arising from differentiating a Lie group is the same thing in the sense that character of Lie algebra representation is character of corresponding Lie group representation.

However, things are complicated once we switch the field from $\mathbb{C}$ to something else (in particular, over field of positive characteristic). For example, the modular character and the Brauer character are two different definitions for characters of the group algebra $kG$ where char$(k)>0$. When we say 'things are complicated', it means many useful properties of ordinary characters are lost when we just generalise the concept to extending to the other fields (that's why we have Brauer characters in place of modular characters in positive characteristic for group algebras). A 'good' character theory need some sort of nice behaviour of the corresponding algebra. e.g. divison in to conjugacy classes of a group and semisimplicity of group algebras over $\mathbb{C}$, and the notion of weight spaces for Lie algebras

Also since the 20's, after the work of, say Dade and Green, one sees module theorectic approach to representation is more convenient than character theorectic one. That's the reason why you don't see character being mentioned for general associative algebras. On the other hand, a representation for algebra $A$ is $\rho:A\to End(V)$, so in theory you can also define characters of algebras.

Nevertheless, character theory is not completely forgotten, one can still do character theory (although not as well-behaved as group/lie algebra representation one) for some special algebras to get some insight. The example I know of is the Brauer algebra, which is built by 'stacking' layers of symmectric group algebras, and exhibit a nice basis which uses large part of symmectric group elements, hence one can use characters to see some elementary properties, but probably not very much can be said in general.


Sure. If $A$ is a finite-dimensional algebra over a field $k$ and $V$ a finite-dimensional $A$-module (so we have a morphism $\rho : A \to \text{End}_k(V)$), we can define a character $\chi_V(a) = \text{tr}(\rho(a)) : A \to k$ just as in the case of finite groups (where $A = k[G]$). Characters are additive in short exact sequences, so if $\chi_V$ has a composition series consisting of $n_i$ copies of the simple $A$-module $V_i$, then $$\chi_V = \sum n_i \chi_{V_i}.$$

Moreover, if $k$ has characteristic zero, the characters corresponding to distinct simple modules are linearly independent. (I don't know a short way to prove this although I am sure there is one; it follows from Artin-Wedderburn applied to $A/J(A)$.) So the character of a module tells you precisely what the simple modules in its composition series are.

Note that any character $\chi_V$ above is among other things a linear functional $A/[A, A] \to k$ (where $[A, A]$ is the subspace, not the ideal, generated by commutators), so $\dim A/[A, A]$ is an upper bound on the number of simple modules. I don't know if this upper bound is attained in general.

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    $\begingroup$ Simple proof of linear independence, if I'm not missing anything: Let $S_i$ be a simple module; let $\mathcal{m}_i$ be the annihilator of $S_i$. Then $\mathcal{m}_i$ is a maximal $2$-sided ideal. (Because, if $N \supsetneq \mathcal{m}_i$, then $N S_i$ is a submodule.) So $\mathcal{m}_1 + \mathcal{m}_2 = A$. Let $f_i \in \mathcal{m}_i$ such that $1=f_1+f_2$. Let $\chi_i$ be the character of $S_i$. Then $\chi_i(f_j) = \delta_{ij} \dim_{k} S_i$. $\endgroup$ – David E Speyer Apr 11 '12 at 16:24

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