Let $G$ be a finite group acting on itself by the action $g \ast x := xg^{-1}$. Then this corresponds to an representation $\rho : G \to GL(L^2(G))$, where $L^2(G)$ denotes the space of function on $G$ (this is called the right regular representation). This is given for $f \in L^2(G)$ by $$ (\rho(g)f)x := f(xg) $$ i.e. the arguments of $f$ are "translated" by $g$. That this is a representation could be found here. Now we have $L^2(G) \cong F_G$, where $F_G$ denotes the free vector space on $G$, where the element $$ \sum_{x \in G} a_x g \in F_G $$ corresponds to the function $f(x) = a_x$. With this $$ g \ast \left( \sum_{x \in G} a_x g \right) = \sum_{x\in G} a_x \cdot xg^{-1} = \sum_{x \in G} a_{xg} \cdot x $$ which corresponds to the function $f(xg)$, hence $g \ast f = \rho(g)f$.

Now I have a question on the character of the right regular representation, we have $$ \chi_{reg} = \sum_{i}^d d_i \chi_i $$ where $d_i = \dim V_i = \chi_i(1)$ and $\chi_i$ are the characters for the irreducible representations $V_i$ of $G$. This is stated in these lecture notes, and in the proof it is just mentioned that this follows from $$ L^2(G) \cong \bigoplus_i R_i, \quad R_i \cong d_i \cdot V_i \quad \mbox{(i.e. $V_i\oplus\cdots\oplus V_i$ with $d_i = \dim V_i$ copies)} $$ I do not see the connection between the direct sum of $L^2(G)$ and the linear combination of $\chi_{reg}$? Can someone please explain?

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    $\begingroup$ I am completely unclear what it is you do not understand. The theorem about the decomposition of $L^2(G)$ is a moderately deep result, but if you believe that, then the result about $\chi_{\rm reg}$ follows just by taking the character. $\endgroup$ – Derek Holt Oct 12 '15 at 14:12
  • $\begingroup$ I do not see it, why does $L^2(G) \cong \oplus_i R_i$ implies that the character $\chi_{reg}(g) = \chi_{\rho}(g) = \mbox{tr} \;\rho(g)$ is $\chi_{\rho} = \sum_i d_i \chi_i$. We have $\rho(g) \in GL(L^2(G)) \cong GL(\bigoplus_i R_i) = \bigoplus_i GL(R_i) = \bigoplus_i \bigoplus_j^{d_i} GL(V_i)$. That means that $\rho(g)$ is a unique sum $\sum_i^d d_i \rho_i$ where $\rho_i$ is a presententation $G \to V_i$, but does this implies $\chi_i = \mbox{tr} \; \rho_i$ for the irreducible representations, or could it be any other character as well? $\endgroup$ – StefanH Oct 12 '15 at 16:33
  • $\begingroup$ I think you are just getting confused by the notation. The $V_i$ are not representations, they are representatives of the isomorphism classes of simple ${\mathbb C}G$-modules. Each $V_i$ affords a representation $\rho_i:G \to {\rm GL}(V_i)$, and $\chi_i = {\rm tr} \rho_i$ by definition. In your last comment, you wrote "presentation $G \to V_i$", which is meaningless. You mean "representation $\rho_i:G \to {\rm GL}(V_i)$". $\endgroup$ – Derek Holt Oct 12 '15 at 17:15
  • $\begingroup$ But could we not have different representations for one $V_i$, i.e. $\rho_i G \to GL(V_i)$ and $\rho_i' : G \to GL(V_i)$ with $\rho_i \ne \rho_i'$, so the sum should have different summands, but it has $\chi_i$ for each factor $V_i$... $\endgroup$ – StefanH Oct 12 '15 at 17:18
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    $\begingroup$ No, $V_i$ is not just a vector space, it is a (simple) ${\mathbb C}G$-module. It is a basic result from Representation Theory that a ${\mathbb C}G$-module $V_i$ determines the associated representation $\rho_i$ up to equivalence of representations, and equivalent representations have the same character. $\endgroup$ – Derek Holt Oct 12 '15 at 19:08

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