# Character of $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$

I've been given the following two questions, and for both I'm really unsure what to do (I'm a beginner with the theory of characters).

Let $G$ be a finite group, $\mathbb{C}G$ its group algebra over $\mathbb{C}$. $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$ forms a $\mathbb{C}G$-module with the action defined, for $a,b \in \mathbb{C}G$, $f \in \text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, by $(af)(b) = f(ba)$.

With this module structure defined on $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, firstly I want to find $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ for all $g \in G$. Secondly, in the case that $G = S_3$, I want to express $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ in terms of simple characters. Could someone please clearly describe how to do this.

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Interesting question, I myself am also currently learning about characters. Don't think I can start a bounty just yet but I would also appreciate it if someone could provide an answer for this. –  HJSprime Oct 17 '12 at 18:59
This is not a clear solution, but something good to keep in mind: Group algebra is selfinjective (in fact symmetric), so $K$-linear dual (here $K$ is an algebraically closed field) of left $KG$-module $KG$ is isomorphic itself as right $KG$-module. Over charactersitic zero, $KG$ is completely reducible (semisimple), hence it is easy to see the characters of the $K$-linear dual is the same as the original one. Over other characteristic, the final result is the same by tweaking the argument using the property that $KG$ is symmetric. –  Aaron Oct 18 '12 at 16:11

The character is just the character $$\rho(g) = \begin{cases} 0 & g\neq1\\ |G| & g=1\end{cases}$$

In other words, the character of the regular representation. Two ways to see this:

1. Pick the basis for $Hom_\mathbb{C}(\mathbb{C}G,\mathbb{C})$ which consists of the functions $f_g$, where for $h\in G$ $$f_g(h) = \begin{cases} 1 & g=h\\ 0 & g\neq h\end{cases}$$ Now to compute the trace $\chi(k)$ with $k\in G$, you need to know when $f_g(hk)=f_g(h)$, and that's easy.
2. $\mathbb{C}G$ has character $\rho$, the regular representation. $\mathbb{C}$ has character $\psi$, the trivial representation. We also have the isomorphism $Hom_\mathbb{C}(\mathbb{C}G,\mathbb{C})\cong \mathbb{C}G^\ast\otimes\mathbb{C}$, with associate character $\bar{\rho}\cdot\psi=\rho$.
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The Group algebra $\mathbb{C}[G]$ can be seen as a Hilbert space of dimension equal to $\# G$. It has a $G$-bimodule structure by right and left translation. Now $\mathbb{C}[G]$ is isomorphic as a $G$-bimodule to the dual $\mathbb{C}[G]$. The only thing to be careful about is that right multiplication gets mapped to left multiplication and vice versa.

Having this in mind, we see that the character to be computed equals $$g \mapsto \sum\limits_{\pi \in \widehat{G}} \dim(\pi) \; tr\; \pi(g) ,$$ where $\widehat{G}$ is the set of iso.-classes of irred. reps of $G$.

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