Realizability of irreducible representations of dihedral groups

Question

Consider a dihedral group of degree n and order 2n. Its two-dimensional irreducible representations can be realized over the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$, with the usual action by rotations and reflections. Also, any splitting field of characteristic zero for this group must contain $\mathbb{Q}(\cos(2\pi/n))$, because this is the field generated by the characters of the irreducible representations. (Here, splitting field for a finite group means a field over which all the irreducible representations are realized).

When n is a multiple of 4, these two fields are the same; otherwise, they are not. My question: when n is not a multiple of 4, what conditions would ensure that the smaller subfield $\mathbb{Q}(\cos(2\pi/n))$ is a splitting field for the dihedral group? I think we can restrict attention to n odd.

For instance, when $n = 3$, the dihedral group of degree 3, order 6, has $\cos(2\pi/3) = -1/2$, $\sin(2\pi/3) = \sqrt{3}/2$. So, any splitting field must contain $\mathbb{Q}(1/2) = \mathbb{Q}$. Also, $\mathbb{Q}(1/2,\sqrt{3}/2) = \mathbb{Q}(\sqrt{3})$ is a splitting field since it contains the usual representation given by rotations and reflections.

However, $\mathbb{Q}$ is also a splitting field. To see this, we think of the group as the symmetric group of degree three and take the standard representation. So in this case, we see that $\mathbb{Q}(\cos(2\pi/n))$ works as splitting field even though it is smaller than the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$.

NOTE: It is not true in general that if the characters all take values in a field, the representation can be realized over that field. The standard counterexample is the quaternion group, whose characters all take rational values but whose irreducible representations are realized only when we go to $\mathbb{Q}(i)$. However, some weaker variant of the result may be true for the groups that we are interested in here, namely, the dihedral groups.

Answer 1

The name of the concept you are looking for is the Schur index. The Schur index is $1$ iff the representation can be realized over the field of values. The Schur index divides the degree of the character.

In your case, the the Schur index is either $1$ or $2$. You can use a variety of tests to eliminate $2$, but for instance:

Fein, Burton; Yamada, Toshihiko, The Schur index and the order and exponent of a finite group, J. Algebra 28, 496-498 (1974). ZBL0243.20008.

shows that if the Schur index was $2$, then $4$ divides the exponent of $G$.

In other words, all of your representations are realizable over the field of values.

Isaacs's Character Theory of Finite Groups has most of this in it, and I found the rest of what I needed in Berkovich's Character Theory collections. Let me know if you want more specific textbook references.

Edit: I went ahead and looked up the Isaacs pages, and looks like textbook is enough here: Lemma 10.8 on page 165 handles induced irreducible characters from complemented subgroups, and shows that the Schur index divides the order of the original character. Taking the subgroup to be the rotation subgroup and the original character to be faithful (or whichever one you need for your particular irreducible when $n$ isn't prime), you get that the Schur index divides $1$. The basics of the Schur index are collected in Corollary 10.2 on page 161.

At any rate, Schur indices are nice to know about, and if Isaacs's book doesn't have what you want, then Berkovich (or Huppert) has just a silly number of results helping to calculate it.

Edit: Explicit matrices can be found too. If $n=4k+2$ is not divisible $4$, and $G$ is a dihedral group of order $n$ with presentation $\langle a,b \mid aa=b^n=1, ba=ab^{n-1} \rangle$, then one can use companion polynomials to give an explicit representation (basically creating an induced representation from a complemented subgroup). Send $a$ to $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, also known as multiplication by $x$. Send $b$ to $\begin{pmatrix}0 & -1\\1 & \zeta + \frac{1}{\zeta}\end{pmatrix}$, also known as the companion matrix to the minimum polynomial of $\zeta$ over the field $\mathbb{Q}(\zeta+\frac{1}{\zeta})$, where $\zeta$ is a primitive $(2k+1)$st root of unity.

Compare this to the more direct choice of $a = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ and $b = \begin{pmatrix} \zeta & 0\\0 & \frac{1}{\zeta}\end{pmatrix}$. If you conjugate this by $\begin{pmatrix}1 & \zeta \\\zeta & 1\end{pmatrix}$ then you get my suggested choice of a representation.

In general, finding pretty, (nearly-)integral representations over a minimal splitting field is hard (and there may not be a unique minimal splitting field), but in some cases you can do it nicely.

Let me know if you continue to find this stuff interesting. I could ramble on quite a bit longer, but I think MO prefers focused answers.

Answer 2

I have already voted up Jack Schmidt's answer, and I promise that he is far more expert in this area than I.

However, I spent a while searching the internet for an answer. Since I knew the keyword -- Schur index -- I thought it would be easy. To my dismay, I couldn't find anywhere with a discussion of what the Schur indices of the irreducible complex representations of the simplest and most familiar families of finite groups should be!

Finally I found the following, which I think is an alternate proof, but someone else will have to confirm it:

By Proposition 7.1.1 of Benson's Polynomial Invariants of Finite Groups, any complex representation which is generated by pseudoreflections can be realized over the field generated by its character values.

[N.B.: If $V$ is a finite dimensional vector space over any field $F$, then a pseudoreflection is a linear automorphism $T$ of $V$ of finite order whose fixed point set is a hyperplane in $V$.]

I believe it is true that the two-dimensional complex irreducible representations of $D_n$ can all seen to be generated by pseudoreflections. Right?

Answer 3

Since, as Pete Clark pointed out, it is one way to answer the original question, it may be worth explaining with Schur index theory, rather than invariant theory, why a finite irreducible complex linear group $G$ in dimension $n$ which contains even a single pseudoreflection is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,F)$, where $F$ is the field generated by the traces of the elements of $G$. Let $\chi$ be the character of the given representation of $G$. Let $x$ be a (genuine, ie non-scalar) pseudo-reflection of $G$. Then ${\rm Res}^{G}_{\langle x \rangle}(\chi) = (n-1)1 + \lambda$ for a non-trivial character $\lambda$ of $\langle x \rangle$.

Recall that the Schur index $m_{F}(\chi)$ is the smallest positive multiplicity with which $\chi$ occurs as a constituent of a character afforded by a representation of $G$ over $F$. A key property of $m_{F}(\chi)$, from the general theory, is that $m_{F}(\chi)$ even divides the multiplicity with which $\chi$ occurs in the character afforded by a representation of $G$ realised over $F$.

Now certainly the regular representation of $G$ can be realized over $F$, and $\chi$ occurs with multiplicity $n = \chi(1)$ in the regular character. Hence $m_{F}(\chi)$ divides $n$. But by Frobenius reciprocity, $\chi$ occurs with multiplicity $n-1$ in the permutation representation of $G$ induced from the trivial representation of $\langle x \rangle$, and this representation is certainly also realised over the field $F$. Hence $m_{F}(\chi)$ also divides $n-1$, so we conclude that $m_{F}(\chi) = 1$, that is, the given representation may be realised over $F$.