This question concerns character tables. They seem to have so high number of algebraic properties that it seems to be impossible to construct a believable one that is not in fact a character table of any particular group. Question arises:

Are there any character-table-like matricies that are not actual character tables?

More precisely:

Let extended character table mean $n \times n$ matrix $M$ with associated $\mathbb{Z}^n$ vector $v$ (corresponding to class sizes) with following properties:

Let $S$ denote sum of entries of $v$.

  • First entry of $v$ is equal to $1$.
  • All entries of $v$ are positive
  • $v_i$ divides $S$.
  • All entries in the matrix $M$ are algebraic integers
  • Rows of the matrix obey orthogonality relation as in usual character table (that is $M_{ij}v_j \overline{M_{kj}} = \delta_{ik}$)
  • Columns of the matrix obey orthogonality relation as in usual character table (that is $M_{ij}M_{ik} = (v_j)^{-1} S \delta_{jk}$)
  • Rows of the matrix are bounded by the first entry (that is $\forall_j |M_{ij}| \leq |M_{i1}|$)
  • First column consists of positive integers (edit suggested by @Joppy)


Are there any extended character tables that do not arise from a group?

  • 1
    $\begingroup$ @Thomas: needn’t all group character tables have entries in the subring generated by the roots of unity? That’s much smaller than the ring of algebraic integers. So I’d expect many character tables to not come from groups. $\endgroup$
    – Aphelli
    Nov 2, 2020 at 9:16
  • $\begingroup$ @Mindlack there are a couple of orthogonality relations we need to satisfy -- it seems that entries can't be too large, but maybe You are on the right track. $\endgroup$
    – Radost
    Nov 2, 2020 at 10:41
  • $\begingroup$ Should you add the condition that the entries in the first column are positive integers? Otherwise something like $M = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}$ works I think $\endgroup$
    – Joppy
    Nov 3, 2020 at 3:50
  • $\begingroup$ @Joppy Yeah, I guess your example works but that's not very satisfying. I'll add the additional criterion. $\endgroup$
    – Radost
    Nov 3, 2020 at 17:40
  • 1
    $\begingroup$ Also the algebraic integers are cyclotomic integers specifically. $\endgroup$
    – anon
    Nov 3, 2020 at 17:58

1 Answer 1


What about this one: $$\begin{pmatrix} 1&1&1&1&1&1\\ 1&1&1&1&-1&-1\\ 1&1&1&-1&1&-1\\ 1&1&1&-1&-1&1\\ 2&2&-2&0&0&0\\ 4&-2&0&0&0&0 \end{pmatrix}?$$ Taken from the paper "Character table sudokus" https://link.springer.com/article/10.1007/s00013-023-01859-w See also the reference therein.


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