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I have a question about some contradictory instructions provided in group theory. This website says that the character table for the group $C_3$ is as shown below.

Character Table for C_3

where the cyclic group is $A, A^2, A^3=E$, and $\Gamma^{(i)}$ are the irreducible representations of the group. To my understanding, the rows and columns in a character table must be orthogonal, which works for all rows and columns except when multiplying $\Gamma^{(2)}$ by $\Gamma^{(3)}$, where we get $1+1+1=3\neq0$.

The orthogonality of rows just means that multiplying entries in a row together and summing them up gives zero. So for $\Gamma^{(1)}$ and $\Gamma^{(2)}$, the orthogonality is that $(1)(1)+(1)\text{e}^{2\pi i/3}+(1)\text{e}^{-2\pi i/3}=0$. The same applies to columns.

We also know that the characters are just the traces of the irreducible representations. The dimensionalities of the irreducible representations are all 1, so the characters must just be group elements of the irreducible representations. Correct me if I am wrong on that point.

I've seen several character tables where the orthogonality does no seem to hold and its very confusing because that is how you find a character table, by finding characters which make it orthogonal.

Edit: I was thrown off because on this website the character table at the bottom is not orthogonal and it has all real characters, so I was unsure if I was doing it correctly.

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    $\begingroup$ Orthogonality is with respect to the Hermitian inner product, not the symmetric one. $\endgroup$ – Lord Shark the Unknown May 29 at 21:17
  • $\begingroup$ This is a very important question. Some people get confused with the orthogonality relations, so when I teach it, I usually show several formulations. I also emphasize strongly that the inner product is in $\mathbb{C}^n$, that is, $xy= \sum x_i\overline{y_i}$. $\endgroup$ – A. Pongrácz May 29 at 21:21
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We define the inner product of characters of a finite group $G$ as follows: $$\left<\chi_1,\chi_2\right>=\frac1{|G|}\sum_{g\in G}\chi_1(g)\overline{\chi_2(g)}.$$ Actually that's the same as $$\left<\chi_1,\chi_2\right>=\frac1{|G|}\sum_{g\in G}\chi_1(g)\chi_2(g^{-1}).$$ Then if $\chi_1$ and $\chi_2$ are irreducible characters, $\left<\chi_1,\chi_2\right>=1$ or $0$ according to whether $\chi_1=\chi_2$ or not.

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  • $\begingroup$ That's it! I completely forgot about complex conjugating, this has been a mess for me all day. $\endgroup$ – Kraig May 29 at 21:37

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