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I would like to understand the definition of "$K$-conjugacy classes" I found in an article by G. Pazderski

  • Pazderski, Gerhard. "On the number of irreducible representations of a finite group." Arch. Math. (Basel) 44 (1985), no. 2, 119–125. MR780258 DOI:10.1007/BF01194075

What is the diference of "conjugacy classes" and "$K$-conjugacy classes"?

Especially I am interested in the case where $G$ is the dihedral group of oder $2p$, $p \neq 2$ a prime and $K=\mathbb{Z} / q \mathbb{Z}$, $q$ a prime.

I know just the usual cunjugacy classes of the dihedral group: Let $x$ be the rotation and $y$ the reflection. Then the conjugacy classes are $\{1\}, \{y,xy,\cdots, x^{p-1}y \}$ and $(p-1)/2$ conjugacy classes with two elements (all rotations). What are the $\mathbb{Z} / q \mathbb{Z}$-conjugacy classes?

Thank you for help

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1 Answer 1

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The definition of $K$-conjugacy is given on the first page of the paper. If $K$ is not algebraically closed, it might be missing some roots of unity. For instance, if $q=13$ and $p=5$, then the elements of $G$ have eigenvalues that are $5$th roots of unity, but the field $\mathbb{Z}/13\mathbb{Z}$ has no $5$th roots of unity. If we try to adjoin a 5th root of unity $\zeta_5$ to $\mathbb{Z}/13\mathbb{Z}$, then we end up adjoin all of its Frobenius conjugates too: $\zeta_5^{13}, \zeta_5^{13^2}, \zeta_5^{13^3}, \zeta_5^{13^4} = \zeta_5$. This last equality follows since $13^4 \equiv 1 \mod 5$.

Since we can't tell the eigenvalue $\zeta_5$ from $\zeta_5^{13}$ using the field $\mathbb{Z}/13\mathbb{Z}$, we also cannot tell the difference between $g$ and $g^{13}$ using $\mathbb{Z}/13\mathbb{Z}$-representations. In particular, if $g$ is a rotation of order 5, then it is $K$-conjugate to each of $g^{13}, g^{13^2}, g^{13^3}, g^{13^4}=g$.

You can imagine the field automorphisms of $K$ act not only on character values, but also on group elements.

Of course, this only works on $K$-regular classes, so don't take $q \in \{2,p\}$ or don't count the reflections (when $q=2$) or the rotations (when $q=p$) at all.

As suggested in the Math Review, see Curtis–Reiner's Representation Theory (not Methods) page 306, §42.8.

Realizing a dihedral group as a matrix group

The standard realization of a dihedral group as a matrix group views it as a group of symmetries of a convex polygon centered at the origin.

$$\left\langle \begin{bmatrix} -1 & . \\ . & 1 \end{bmatrix}, \begin{bmatrix} \cos(2\pi/p) & -\sin(2\pi/p) \\ \sin(2\pi/p) & \cos(2\pi/p) \end{bmatrix}\right\rangle$$

The matrix corresponding to a rotation has eigenvalues $\zeta_p$ and $\bar \zeta_p = 1/\zeta_p$. The trace of the matrix is just $\zeta_p + \zeta_p^{-1} = 2\cos(2\pi/p)$. It turns out to be easier to change fields if we diagonalize the rotation matrix over $\mathbb{C}$ first.

$$\left\langle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} \zeta_p & 0 \\ 0 & \bar \zeta_p \end{bmatrix} \right\rangle$$

Notice the rotation matrix still has trace $2\cos(2\pi/2) = \zeta_p + \zeta_p^{-1} \in \mathbb{R}$. This is another example of $K$-conjugacy, where a real matrix cannot have just one complex eigenvalue. The complex eigenvalue come in pairs, related by the Galois group. Notice also that on roots of unity, the action of the Galois groups is $z \mapsto z^{-1}$. In other words, here $q=-1$. The trace of the reflection is always going to be 0, which also is a real number.

Before trying to write our group as matrices over $\mathbb{Z}/q\mathbb{Z}$, let's just try to get our matrices to have trace inside $\mathbb{Z}/q\mathbb{Z}$. Inside the algebraic closure of $\mathbb{Z}/q\mathbb{Z}$ there is a $\zeta_p$, a primitive $p$th root of unity (assuming $p\neq q$). Any polynomial in $\zeta_p$ is also in the algebraic closure. When is one of those polynomials in $\mathbb{Z}/q\mathbb{Z}$ itself? Exactly when raising it to the $q$th power does not change it. In particular, when we include $\zeta_p$, we have to include $\zeta_p^{q}, \zeta_p^{q^2}, \ldots$.

If we do this for $p=5$, $q=13$, we get

$$\left\langle \left[\begin{array}{rr|rr} . & . & 1 & . \\ . & . & . & 1 \\ \hline 1 & . & . & . \\ . & 1 & . & . \end{array}\right], \left[\begin{smallmatrix} \zeta_p & . & . & . \\ . & \zeta_p^q & . & . \\ . & . & \zeta_p^{q^2} & . \\ . & . & . & \zeta_p^{q^3} \end{smallmatrix}\right] \right\rangle$$

Notice how conjugating by the reflection swaps the top 2 $\zeta_p^{q^i}$ with the bottom two. Since this is the same as applying $x \mapsto x^{q^2}$ it sure is nice that $-1 \equiv q^2 \mod p$. Hence we still get a dihedral group over an algebraic closure of $K$, and all of the traces of the matrices lie in $K$ itself. Of course we need to write this over $K$ itself, but replacing the rotation matrix with the companion matrix of its minimal polynomial and the reflection with the permutation matrix that swaps the right basis vectors, we get:

$$\left\langle \begin{bmatrix} . & . & . & 1 \\ . & . & 1 & . \\ . & 1 & . & . \\ 1 & . & . & . \end{bmatrix}, \begin{bmatrix} . & 1 & . & . \\ . & . & 1 & . \\ . & . & . & 1 \\ -1 & -1 & -1 & -1 \end{bmatrix} \right\rangle$$

We've used up all the $\zeta_p^i$ there are, and so this is the only faithful $\mathbb{Z}/q\mathbb{Z}$ representation.

Things work almost identically for any $q$ such that $q$ has multiplicative order $p-1$ mod $p$. The left matrix is always the "reversal" permutation, and the right matrix is always the "1s on the super, -1s on the bottom row" except of course we get $(p-1) \times (p-1)$ matrices in order make sure $\zeta_p$ being used implies $\zeta_p^q$ is used. Since the order of $q$ mod $p$ is $p-1$, that uses up all of the $\zeta_p$.

If $q$ has smaller order mod $p$, say $(p-1)/k$, then you only need to include the $\zeta_p^{q^i}$, and that doesn't use up all of the $\zeta_p^i$. You'll get $k$ bunches of $(p-1)/k$ roots of unity. Each of the $k$ bunches produces its own irreducible $K$-representation. I believe the formulas for the matrix entries over $K$ are a little more complicated, but the "diagonal" version over the algebraic closure (with trace still in $K$) is unchanged and very easy.

If you only care about traces, then more or less everything I said holds and is very practical for every finite group of order coprime to $q$. When the order is not coprime to $q$, you need to worry about $K$-regular classes.

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Ok, then this way would't help to get the irreducible representations of G over K. Here mathoverflow.net/questions/102518/… someone mentioned the Mackey's theorem. But I'm sure it works just for algebraically closed fields. Do you know any other ways? –  guest Jul 25 '12 at 19:10
    
These K-conjugates make it easier to find the representations in some sense. In fact, my description of the reason K-conjugates matter is basically what lets you write down the representations. You don't need any theorems to find these really, though it helps to know the count so you know when you are done. I can write down the matrices if you want, but they barely change from $\mathbb{C}$. –  Jack Schmidt Jul 25 '12 at 19:21
    
I would be very grateful if you could give me this irred. representations (and a hint how to get them). I cannot find a method which works. –  guest Jul 25 '12 at 19:31
    
I added the faithful irreducible reps. The non-faithful ones don't change: trivial and sign. –  Jack Schmidt Jul 25 '12 at 20:09
    
I should also mention: if your fields are not finite and not algebraically closed, then you need a little more than just the trace. What I said still basically works, but sometimes you need to take even more of the $\zeta_p$ than it took to get the trace right. The number of bunches of $\zeta_p$ that get squished is called the "Schur index". Over a finite field, the Schur index (of all groups) is just 1: every bunch gets left alone. –  Jack Schmidt Jul 25 '12 at 20:21

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