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changed 1_g to g
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Amritanshu Prasad
Amritanshu Prasad
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The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$ (see details below).

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})1_g, $$$$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$.

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})1_g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$ (see details below).

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

Made correction pointed out by Geoff Robinson.
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Amritanshu Prasad
Amritanshu Prasad
  • 5.7k
  • 1
  • 38
  • 54

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$.

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \frac{d_\pi}{|G|}\chi_\pi, $$$$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})1_g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$.

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \frac{d_\pi}{|G|}\chi_\pi, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$.

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})1_g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

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Amritanshu Prasad
Amritanshu Prasad
  • 5.7k
  • 1
  • 38
  • 54

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$.

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \frac{d_\pi}{|G|}\chi_\pi, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).