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Let $\chi$ be an irreducible (complex) character of a finite group, $G$. The Schur index $m_{K}(\chi)$ of $\chi$ over the field $K$ is the smallest positive integer $m$ such that $m\chi$ is afforded by a representation over the field $K(\chi)$. The most interesting case is $K=\mathbb{Q}$. Given the character table, or only the particular character one is interested in, one can usually derive bounds for $m(\chi)=m_{\mathbb{Q}}(\chi)$. For example, $m(\chi)$ divides $\chi(1)$ and $n[\chi^n,1_G]$ for all $n\in \mathbb{N}$ (Fein), and the Benard-Schacher Theorem tells us that $\mathbb{Q}(\chi)$ contains a primitive $m(\chi)$-th root of unity.
On the other hand, the example of the quaternion group $Q_8$ and the dihedral group $D_8$ shows that two groups might have identical character tables, but corresponding characters with different Schur indices. I am curious wether there are examples that are even worse than this.

Notation: To state this more precisely, I'll make the following assumptions: We are given two finite groups $G$ and $H$, such that there is a bijection $\tau\colon {\rm Cl}(G) \to {\rm Cl}(H)$ from the classes of $G$ to the classes of $H$, and such that $\psi \mapsto \psi \circ \tau$ is a bijection ${\rm Irr}(H)\to {\rm Irr}(G)$. Now:

Is there an example with $m(\chi)/m(\chi\circ\tau)\notin \{1,2,1/2\}$ for some $\chi\in {\rm Irr}(H)$?

Is there an example with $G$ of odd order and $m(\chi) / m(\chi\circ\tau)\neq 1$ for some $\chi \in {\rm Irr}(H)$?

Now let us assume that we know the power maps of the character table. These are the maps $\pi_n^G\colon {\rm Cl}(G)\to {\rm Cl}(G)$ induced by $g\mapsto g^n$. (These maps are stored in the tables of the character table library of GAP.) Given these maps, one can compute $[\chi_C, 1_C]$ for cyclic subgroups $C\leq G$, for example. Also we can compute the Frobenius-Schur Indicator and thus the Schur index over $\mathbb{R}$.
Now assume that $\tau\circ \pi_n^G = \pi_n^H\circ \tau$ in the above situation (then $(G,H)$ is called a Brauer pair).

Is there a Brauer pair $(G,H)$ such that $m(\chi)/m(\chi\circ\tau)\neq 1$ for some $\chi\in {\rm Irr}(H)$?

I would appreciate any examples or (pointers to) results that show the impossibility of such examples.

Thanks

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    $\begingroup$ These are serious questions but technically difficult. On the optimistic side, extensive study of Schur indices for finite simple groups over several decades has established mainly very small indices in spite of the great complexity of the character theory (especially as developed by Lusztig rom the Deligne-Lusztig foundations). But for arbitrary groups I have no idea what to expect, since the Schur index tends to go beyond the bounds of usual character information. $\endgroup$
    – Jim Humphreys
    Nov 22, 2010 at 23:09
  • $\begingroup$ The following paper of Walter Feit may be helpful: Schur indices of characters of groups related to finite sporadic simple groups. Israel J. Math. 93 (1996), 229–251. Another person who has worked extensively on such questions, and pointed out some necessary invariants needed to determine Schur indices is Alex Turull. $\endgroup$
    – Geoff Robinson
    Apr 17, 2011 at 19:55
  • $\begingroup$ Nice question ! When you write "ψ↦ψ∘τ is a bijection Irr(H)→Irr(G)" what does it mean ? Is ψ element of Irr(H) ? How does you combine it with tau ( ψ∘τ ) ? $\endgroup$
    – Alexander Chervov
    Sep 8, 2012 at 17:23
  • $\begingroup$ @AlexanderChervov, I think that Ladisch means to identify irreducible representations with their characters, viewed as functions on the set of conjugacy classes, and so to transport a representation from $H$ to $G$ via a bijection between their conjugacy classes. For a 'bare' such bijection, not guaranteed to preserve any interesting information (such as the size of the class), I'm not sure why the result should be another irreducible representation. $\endgroup$
    – LSpice
    Oct 13, 2015 at 20:10
  • $\begingroup$ @AlexanderChervov: What L. Spice said. $\psi$ is in $\operatorname{Irr}(H)$, that is, an irreducible character, and I view $\psi$ as a function on the conjugacy classes. Thus $\psi\circ \tau$ is a function on the conjugacy classes of $G$. In general, this will not be a character, of course, but here the assumption is just that it induces a bijection between $\operatorname{Irr}(H)$ and $\operatorname{Irr}(G)$. This formalizes the idea that the character tables of $G$ and $H$ are "the same" and that the bijection $\tau$ says how these character tables are "the same". $\endgroup$
    – Frieder Ladisch
    Oct 16, 2015 at 11:40

1 Answer 1

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The following is a theorem of K. Kronstein:

Theorem: for $k$ a number field or a nonarchimedean completion of a number field, if it is possible to detect the Schur index $m_k$ of all finite groups from their character table and power maps, then $m_k(\chi) \leq 2$ for all characters $\chi$ of finite groups.

In particular, for a finite group $G$, the map $\tau: G \times G \to G \times G$ defined by $\tau(x,y) = (x,y^{-1})$ induces a bijection on conjugacy classes preserving the power maps $x \mapsto x^n$ and a bijection on characters, but does not necessarily preserve Schur indices larger than 2. Thus, the map $\tau$ provides a positive answer to the first and third questions. It also provides a positive answer to the second question once one produces an example of a group of odd order with a character with Schur index greater than 2.

I provide Kronstein's proof here:

Suppose $\chi$ is a character of $G$, $k$ is a number field or nonarchimedean completion of a number field, and the Schur index $m = m_k(\chi)$ is at least $3$. Let $K = k(\chi)$, let $V$ be an irreducible $KG$-module affording the character $m\chi$, and let $D = End_{KG}(V)$. It is a division algebra with order $m$ in the Brauer group of $K$.

Consider the characters $\chi \boxtimes \chi$ and $\chi \boxtimes \chi^\vee$ on $G \times G$. Then $$End_{K(G\times G)}(V\boxtimes V) = D \otimes_K D.$$ Since $m>2$, $D \otimes_K D$ is not split, so $\chi\boxtimes \chi$ has Schur index greater than 1. However, $$End_{K(G \times G)}(V \boxtimes V^\vee) = D \otimes_K D^{op},$$ which splits over $K$, and thus $\chi \boxtimes \chi^\vee$ has Schur index 1.

Reference:

Karl Kronstein. Character tables and the Schur index. In Representation Theory of Finite Groups and Related Topics, volume 21 of Proceedings of Symposia in Pure Mathematics, pages 97–98. American Mathematical Soc., 1971

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    $\begingroup$ Thank you very much. (I think you meant "positive answer"?) There is a group $U$ of order $63$ with a character with Schur index $3$, and thus we can take $G= U\times U = H$ (a group of order $63^2$) as an example for all three questions. It feels a bit like cheating since we have $G=H$ here, but we can also fabricate non-isomorphic Brauer pairs by forming direct products $G\times P_1$ and $G\times P_2$, where $(P_1,P_2)$ is a Brauer pair. $\endgroup$
    – Frieder Ladisch
    Dec 8, 2022 at 18:19

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