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I am certain that the answer to this question exists somewhere. It might be a classical exercise.

Let $G$ be a finite group. Its table of characters is a square matrix, whose rows are indexed by the conjugacy classes and the columns are indexed by the irreducible characters. It is well defined, up to the order of rows and columns. In particular, its determinant if well-defined up to the sign. Let us define $\Delta$ to be the square of this determinant (this is well-defined). Because the characters form a basis of the space of class functions, we know that $\Delta\ne0$. When $G={\mathbb Z}/n{\mathbb Z}$, $\Delta=n^n$.

Is there a close formula for $\Delta$ for a general group? Is it always an integer?

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    $\begingroup$ If $G={\bf Z}/n{\bf Z}$ then $\Delta$ is the square of a Vandermonde determinant, and can be computed in closed form, but the answer is not $n^2$ but $\pm n^n$, with the sign depending on $n \bmod 4$ if I did this right. $\endgroup$
    – Noam D. Elkies
    Commented Oct 13, 2011 at 20:04
  • $\begingroup$ @Noam. Right. I edit. $\endgroup$
    – Denis Serre
    Commented Oct 13, 2011 at 20:06
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    $\begingroup$ @Denis thanks, but you still need a $\pm$ sign (also for the general formula, since the nice argument you give computes only the absolute value $|\Delta|$, not $\Delta$ itself). $\endgroup$
    – Noam D. Elkies
    Commented Oct 13, 2011 at 20:14
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    $\begingroup$ I think you should post the answer as an answer. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 13, 2011 at 21:53
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    $\begingroup$ One can also ask about determinants of submatrices of the character table, a much more subtle question. See front.math.ucdavis.edu/1110.0818. $\endgroup$
    – Richard Stanley
    Commented Oct 14, 2011 at 0:22

4 Answers 4

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If $A$ is the character table and $A^\ast$ is its conjugate transpose, then the orthogonality relations tell us that $A A^\ast = \text{diag}\{|C_G(g)|\} $, where the enties run over a fixed choice of elements of $G$, one from each conjugacy class. Thus $|\Delta| = \det A A^\ast = \prod |C_G(g)|$ is an integer. On the other hand, $\Delta$ must be rational. This follows from the fact that the action of $\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$ permutes the columns of $A$, hence fixes $\Delta = (\det A)^2$. Thus $\Delta=\pm |\Delta|$ is an integer.

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  • $\begingroup$ I am not convinced with your identity $A^\ast A = \text{diag}\{|C_G(g)|\} $, unless you consider that the columns are indexed by $G$ instead of the set of conjugacy classes; but then $A$ is no longer a square matrix. See my answer. $\endgroup$
    – Denis Serre
    Commented Oct 15, 2011 at 12:11
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    $\begingroup$ Sorry, it should be $AA^\ast$, since your columns are indexed by characters. I'll edit my answer. Anyway, the proof goes as follows. Choose representatives $g_1,\ldots,g_r$ for the conjugacy classes of $G$ and let $\chi_1,\ldots,\chi_r$ be the irreducible characters. Then $A_{ij}=\chi_j(g_i)$ so $A^\ast_{ij}=\overline{\chi_i(g_j)}$ and therefore $(AA^\ast)_{ij}=\sum_k \chi_k(g_i) \overline{\chi_k(g_j)}$. By the orthogonality relations, this sum is equal to $0$ if $i\neq j$ and to $|C_G(g_i)|$ otherwise. $\endgroup$
    – Faisal
    Commented Oct 15, 2011 at 18:39
  • $\begingroup$ I don't get last statement in answer: for any $\sigma\in {\rm Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, $\det(\sigma(A))=\pm \det(A)$; but how this implies integrality of determinant? $\endgroup$
    – Soluble
    Commented Nov 14, 2019 at 12:38
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I found the following answer after posting it: $$\Delta=\epsilon\prod_c\frac{|G|}{|c|},\qquad\epsilon=(-1)^m,$$ where the product is taken over the conjugacy class. And $m$ is the number of pairs of complex conjugate irreducible characters.

Proof. On the one hand, the complex conjugate of the table is itself, up to $m$ transpositions of rows. This is because the conjugate of an IC is an IC. Therefore $$\overline{\det(TC)}=\epsilon\det(TC)$$ ($TC$ stands for ``table of characters''.) Hence $\det(TC)$ is real if $m$ is even, pure imaginary if $m$ is odd. hence $\Delta$ is real and its sign is $\epsilon$.

Now the characters form a unitary basis. Because a unitary matrix has a unit determinant, we may compute $|\Delta|$ by taking any unitary basis. Take $\phi_c(g)$ to be $0$ if $g\not\in c$ and $|G|^{1/2}/|c|^{1/2}$ if $g\in c$. In particular $|\Delta|$ is an integer because $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$

Another Proof: Let $D$ be the diagonal matrix whose diagonal entries are the cardinals of the congugacy classes. We may assume that the first rows of $TC$ are the real characters and the $2m$ last ones are the pairs of complex conjugate characters. Then the $(i,j)$-entry of $M:=(TC)D(TC)^T$ is $|G|\langle\overline{\chi_i},\chi_j\rangle$. From the orthogonality relations, we see that $M={\rm diag}(1,\ldots,1,J,\ldots,J)$ where $$J=\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}.$$ The number of blocks $J$ is precisely $m$. Now take the determinant; we obtain $\Delta\det D=(-1)^m|G|^r$ where $r\times r$ is the size of $TC$. Hence the formula.

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  • $\begingroup$ What is $TC$? What is that curly Z with a line through the middle? $\endgroup$
    – Gerry Myerson
    Commented Oct 14, 2011 at 20:46
  • $\begingroup$ Gerry, I believe $TC$ is the matrix product of the character table $T$ and its conjugate transpose $C$. And $\mathcal{Z}(a)$ is the centralizer of $a$ in $G$. Also, just in case anyone else is wondering (I was), "IC" = "irreducible character". $\endgroup$
    – Faisal
    Commented Oct 14, 2011 at 21:18
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Hasn't this been discussed at length in the recent Annals paper

An Elementary Exposition of Frobenius's Theory of Group-Characters and Group-Determinants Leonard Eugene Dickson (1902)?

The Determinant the OP discusses is NOT the group determinant per se, but comes up shortly after the definition of the group determinant.

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    $\begingroup$ Where are the OP's questions discussed in that paper? $\endgroup$
    – Faisal
    Commented Oct 13, 2011 at 23:12
  • $\begingroup$ It's 1902, pp. 25--49. $\endgroup$
    – KConrad
    Commented Oct 14, 2011 at 13:33
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Corrected version

I believe it is always an integer. We can assume that all our representations are over the algebraic closure $\overline{\mathbb Q}$ of $\mathbb Q$. If $\Gamma$ is the absolute Galois group, then clearly $\Gamma$ acts on the characters of $G$ (if you have a representation, then twist it by the action of $\Gamma$). It thus follows that the determinant squared is fixed by $\Gamma$ (since $\Gamma$ permutes the rows) and so is a rational number. But it is also an algebraic integer so it is an integer.

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    $\begingroup$ According to www.combinatorics.org/Volume_10/PDF/v10i1n3.pdf the answer for the symmetric group is nice. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 13, 2011 at 19:42
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    $\begingroup$ Oops, the link was bad. The answer for $S_n$ is the product of all parts of partitions of $n$. I doubt there is a closed form in general. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 13, 2011 at 19:45
  • $\begingroup$ I made a correction because I forgot the action of $\Gamma$ permutes the rows and so could change the sign of the determinant (e.g. complex conjugation switches two rows). Thus I changed the answer to saying it is an integer. I don't know if it is a square. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 13, 2011 at 19:58
  • $\begingroup$ Actually $\Gamma$ permutes the columns since Denis uses columns as characters. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 13, 2011 at 19:58

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