This is an exercise from the book "Groups and Representations" by Alperin & Bell. This quantity is well defined upto a sign. By column orthogonality relations, its squared norm is $\displaystyle\prod_{\substack{g}} |C_G(g)|$, where g runs over the representatives of the conjugacy classes. If the group is cyclic, the determinant is just a Vandermonde determinant.

I wonder if there is a nice explanation for an arbitrary group.

  • $\begingroup$ Are you asking for a proof or for a reinterpretation such as the Vandermonde in the cyclic case? The former should come out a study of $A^HA$ and column orthogonality, but it sounds like you know that. $\endgroup$ – Jyrki Lahtonen Jun 9 '12 at 16:57
  • $\begingroup$ A nice explanation of what? $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 17:01
  • $\begingroup$ I am just looking for a solution of the exercise. $\endgroup$ – user33321 Jun 9 '12 at 17:23
  • $\begingroup$ @JanDvorak Ok, sorry I didn't know that. $\endgroup$ – user33321 Nov 5 '13 at 14:40

Let $A$ be the character table as a square matrix. In other words $A_{ij}=\chi_i(g_j)$, where $\chi_i$ the distinct irreducible characters, and $g_j$ are representatives of conjugacy classes. Let $A^H$ be the conjugate transpose. Let us compute the matrix product $B:=A^HA$. At position $(i,j)$ we get $$ B_{ij}=\sum_k A^H_{ik}A_{kj}=\sum_k \overline{A_{ki}}A_{kj}=\sum_k\overline{\chi_k(g_i)}\chi_k(g_j)=|C_G(g_j)|\delta_{ij} $$ by the second orthogonality relation. Therefore $B$ is diagonal, and $$ \det B=\prod_j|C_G(g_j)|. $$ So by the multiplicativity of determinant we also have $$ \det(A^H)\det(A)=\det B=\prod_j|C_G(g_j)|. $$ Let us study the relation between $A^H$ and the transpose $A^T$. Let us define a permutation $s$ of the conjugacy classes by the mapping $[g]\mapsto [g^{-1}]$. It is obviously a product of disjoint 2-cycles, and its fixed points are exactly the conjugacy classes stable under taking the inverse element. Let us denote by $\ell$ the number of orbits of size two. If we denote by $\tilde{A}$ the matrix that we get from $A$ by permuting the columns according to $s$, then the general fact $\chi(g^{-1})=\overline{\chi(g)}$ allows us to identify $A^H$ as $\tilde{A}^T$. Clearly $$ \det\tilde{A}=(-1)^{\ell}\det A,$$ so we get the equation $$ (-1)^{\ell}(\det A)^2=\det B=\prod_j|C_G(g_j)|. $$ The sign of $\det A$ will always remain ambiguous, because we have no natural ordering neither for the conjugacy classes nor for the characters, so all we can say is that $$ \det A=\pm i^{\ell}\sqrt{\prod_j|C_G(g_j)|}. $$

  • $\begingroup$ Thank you for the answer but the exercise asks $det(A)$ as I wrote in the title of the question. In the text I summarized what I obtained so far. I am sorry if what I wrote misled you. $\endgroup$ – user33321 Jun 9 '12 at 18:41
  • $\begingroup$ @Serkan, ok. I would have been surprised, if you had not figured out this much:-). Thinking... $\endgroup$ – Jyrki Lahtonen Jun 9 '12 at 18:43
  • $\begingroup$ @Serkan, I think I solved it. $\endgroup$ – Jyrki Lahtonen Jun 9 '12 at 19:07
  • $\begingroup$ I think this is a really nice solution, thank you :) $\endgroup$ – user33321 Jun 9 '12 at 21:36

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