# What can be said about Schur indices, given only the character table?

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

• 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. – Jim Humphreys Nov 22 '10 at 23:09
• 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. – Geoff Robinson Apr 17 '11 at 19:55
• 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 ) ? – Alexander Chervov Sep 8 '12 at 17:23
• @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. – LSpice Oct 13 '15 at 20:10
• @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". – Frieder Ladisch Oct 16 '15 at 11:40