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Let $\chi_{R}$ denote the character of the right regular representation $R$. Compute the inner product $(\chi_{R},\chi_{R})$ directly and also by decomposing $R$ into a sum of irreducible representations.

But I have the following difficulties?

1- what is the character of the right regular representation? I searched my book (Ernest Vinberg "Linear representations of groups ") but I did not find the definition.

2-How can I decompose $R$ into a sum of irreducible representations? and I think the value of the inner product should be the same .... correct?

EDIT :

I got the answer for my first question, it is 0 if $g \neq e$ and its n (order of the group) if $g = e$.

Still I am working on the second question. but I appreciate any help on it.

EDIT 2: I have known that I will use that $\chi_{R} = \sum_{i = 1}^{q} d_{\pi_{i}} \chi_{\pi_{i}} $ but still I do not know how to reach the results I obtained in the first part, could anyone help me please?

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    $\begingroup$ My guess is that the goal of the exercise is for you to prove the formula $$n=\sum_{i=1}^qd_{\pi_i}^2$$ by computing this inner product in two different ways and equating the results. $\endgroup$ – Jyrki Lahtonen Mar 15 at 5:25
  • $\begingroup$ You mean that the two inner products will not give the same result?@JyrkiLahtonen $\endgroup$ – Intuition Mar 15 at 5:33
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    $\begingroup$ No, they definitely give the same result. But they give the result in terms of different quantities we associate with the group. So we have derived a non-obvious conclusion. Like, you are not expected to get $n$ as the answer from the second calculation. You are expected to get an equation that the numbers $d_{\pi_i}$ must satisfy. $\endgroup$ – Jyrki Lahtonen Mar 15 at 5:35
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    $\begingroup$ Do check that the formula $n=\sum_i d_i^2$ holds for any group where you have a complete list of the irreducible characters. $\endgroup$ – Jyrki Lahtonen Mar 15 at 5:36
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    $\begingroup$ So you should calculate $$\langle\sum_i d_{\pi_i}\chi_{\pi_i},\sum_jd_{\pi_j}\chi_{\pi_j}\rangle=\sum_i\sum_j d_{\pi_i}d_{\pi_j}\langle \chi_{\pi_i},\chi_{\pi_j}\rangle.$$ $\endgroup$ – Jyrki Lahtonen Mar 15 at 6:00

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