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It can be shown, via the construction of the representations of the symmetric group, that every representation of $S_n$ is equivalent to a representation with values in $\mathbb{Q}.$ Presumably, this is a fairly rare phenomenon: it clearly doesn't hold for cyclic groups ($\mathbb{Z}/p\mathbb{Z}$ has one-dimensional representations given by the $p$th roots of unity, hence $p-1$ of its representations lie outside of $\mathbb{Q}$).

Moreover, there is a formula which constructs the rational characters of a group (due to Artin: see Curtis and Reiner section 15), but it doesn't seem to give an answer to the following question:

Are there other "classes" of groups such that every irreducible representation is realizable over $\mathbb{Q}$?

Take "classes" to mean whatever you think appropriate (so long as it doesn't mean the collection of all groups with only rational irreps).

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    $\begingroup$ At the minimum each element must be conjugate to its inverse. $\endgroup$ Apr 25, 2012 at 2:27
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    $\begingroup$ More than that: If $g$ has order $n$, and $GCD(m,n)=1$, then $g$ must be conjugate to $g^m$. But neither of these are sufficient conditions. See also groupprops.subwiki.org/wiki/Rational-representation_group . $\endgroup$ Apr 25, 2012 at 2:42
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    $\begingroup$ To subsume the list in David Speyer's comment: any direct product of symmetric groups and copies of $D_4$ (also called $D_8$) $\endgroup$
    – Will Sawin
    Apr 25, 2012 at 3:53
  • $\begingroup$ $S_6(2)$, perhaps? It's index 2 in $W(E_7)$, and its characters are rational. $\endgroup$
    – S. Carnahan
    Apr 25, 2012 at 10:00
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    $\begingroup$ @Will: In fact every Weyl-Group and therefore every product of Weyl groups has this property. The Dieder group of order 8 is just the Weyl group $W(B_2)$. $\endgroup$ Apr 25, 2012 at 14:01

2 Answers 2

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Finite groups all of whose ordinary complex representations have rational-valued characters are sometimes called Q-groups (and sometimes called rational groups). There is a monograph by Denis Kletzing (Structure and Representations of Q-Groups, Springer Lecture Notes in Mathematics 1084, 1984) which might be of interest. Symmetric groups are Q-groups, as you mention, as are Weyl groups. I can't remember if Kletzing constructs any other infinite families although I do remember that $D_n$ is a Q-group only when $n=1,2,3,4$ or $6$. It's also worth mentioning that homomorphic images and direct products of Q-groups are also Q-groups.

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    $\begingroup$ In fact, all irreps of Weyl groups can be realized over $\mathbb{Q}$, which is slightly stronger than having rational-valued characters. For the exceptional Weyl groups, this result is due to M. Benard (On the Schur indices of characters of exceptionel Weyl groups, Ann. Math 94) $\endgroup$ Apr 25, 2012 at 12:55
  • $\begingroup$ Artin's Theorem actually provides an explicit formula for the values of rational characters of a general group (see: Curtis and Reiner sect. 15) but I'm more interested in groups of the type F. Ladish mentions, i.e., those which are realizable over $\mathbb{Q}$ $\endgroup$ Apr 25, 2012 at 14:20
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This is closely related to two previous questions.

See the answer by M.Z. to Proving interesting theorems about S_n using its character table.. This gives a necessary and sufficient condition for the entries of the character table to be integers. This is David Speyer's remark above.

The following question then discusses the Schur index What can be said about Schur indices, given only the character table?. However the conclusion is that this is difficult to determine.

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