Elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$

Question

For some research work, I need to know the classification of elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$, up to conjugation.

Since I essentially need $n\le 4$, I think that I can show it by hand, using cyclotomic extensions and Galois theory, but is there some work in the literature on this?

EDIT: Looking at the possible orders is essentially trivial in $\mathrm{GL}(n,\mathbb{Q})$, by just looking at the cyclotomic polynomials. The conjugacy classes require a little more work but are easy exercises, at least in low dimension. For $\mathrm{PGL}(n,\mathbb{Q})$, the case of orders prime to $n$ follows essentially from the case of $\mathrm{GL}(n,\mathbb{Q})$, the orders are more interesting.

Answer 1

I figured I'd write up the elementary observations here, since no one else has: If $g^k = \mathrm{Id}$ in $PGL_n$, then $g^k = a \mathrm{Id}$ in $GL_n$ for some nonzero $a$. So the minimal polynomial of $g$ divides $x^k-a$. Let $p_1 p_2 \cdots p_r$ be the factorization of $x^k-a$ over $\mathbb{Q}$. We can use rational canonical form to write down a $\deg p_i \times \deg p_i$ matrix $g_i$ (over $\mathbb{Q}$) with characteristic polynomial $p_i$. Then, for any nonnnegative integers $a_i$ such that $\sum a_i \deg p_i = n$, we can take the block diagonal matrix whose entries are $a_i$ copies of $g_i$. Conversely, if $g^k = a$, we can break $g$ into blocks according to the irreducible factors of $x^k-a$.

So, what remains is to analyze the degrees of the $p_i$. Let $K$ be the splitting field of $x^k-a$ over $\mathbb{Q}$ and let $G$ be the Galois group. Write $\zeta$ for a primitive $k$-th root of $1$ and $\alpha$ for a chosen $k$-th root of $a$ inside $K$. Then every element of $G$ is of the form $\zeta^i \alpha \mapsto \zeta^{ui+v}$ for $u \in (\mathbb{Z}/k)^{\times}$ and $v \in \mathbb{Z}/k$. So $G$ is a subgroup of $(\mathbb{Z}/k)^{\times} \ltimes (\mathbb{Z}/k)$. Also, $G$ surjects onto $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$, so, for every $u \in \mathbb{Z}/k^{\times}$, the group $G$ contains an element of the form $i \mapsto ui+v$. The problem is to describe the orbits of such a group on $\mathbb{Z}/k$.

For example, when $k=4$, then $G$ is a subgroup of $\{ \pm 1 \} \ltimes (\mathbb{Z}/4)$. If it isn't the whole group (in which case $x^4-a$ is irreducible), then it is either the group generated by $i \mapsto -i$ (in which case $x^4-a$ factors as $(\mbox{linear}) (\mbox{linear})(\mbox{quadratic})$, like $x^4-1$), or the group generated by $i \mapsto -i+1$ (in which case $x^4-a$ factors as $(\mbox{quadratic}) (\mbox{quadratic})$, like $x^4+4=(x^2+2x+2)(x^2-2x+2)$), or else $\{ \pm 1 \} \times 2 \mathbb{Z}/4$, in which case (in which case $x^4-a$ factors as $(\mbox{quadratic}) (\mbox{quadratic})$, like $x^4-4$).

At this point, it isn't clear what to do next, and the question is also a bit unfocused. Here are some (in my opinion) natural questions:

• Is is true that $x^k-a$ always has a factor of degree $\geq \phi(k)$? UPDATE: No. $x^8-16 = (x^2-2)(x^2+2)(x^2-2x+2)(x^2+2x+2)$, and $\phi(8) = 4$.

• For fixed $k$, what is the smallest $n$ for which $PGL_n(\mathbb{Q})$ has an element of order $k$? It is NOT $\phi(k)$: We can build elements of order $15$ in $GL_6$ as the direct sum of elements of orders $3$ and $5$ in $GL_2$ and $GL_4$, even though $\phi(15) = 8$.

• I don't have an example of a case where $PGL_n(\mathbb{Q})$ has an element of order $k$ but $GL_n(\mathbb{Q})$ doesn't. I imagine such a thing exists, but it would be good to have an example.

Answer 2

Let $\lambda^k=a$. Consider how the Galois group of the splitting field of the minimal polynomial of $\lambda$ acts on the roots. As David Speyer points out, this is a subgroup of $(\mathbb Z/k)^\times \ltimes (\mathbb Z/k)$. If this subgroup has a nontrivial intersection with $ (\mathbb Z/k)$, then there is a $l$ dividing $k$ sucht hat for any root $\lambda$ of the minimal polynomial, and $l$th root of unity $\mu$, $\mu \lambda$ is also a root.

So the degree of the minimal polynomial of $\lambda$ is $l$ times the degree of the minimal polynomial of $\lambda^l$. Now we may reduce to the case where the Galois group has trivial intersection with $\mathbb Z/k$.

Observation 1: In this case the Galois group is abelian, so the size of the orbit is the size of the Galois group, so the size of the orbit is at least $\phi$ of the order of the element. So no new orders of elements of $PGL_n(\mathbb Q)$ appear that aren't orders of elements of $GL_n(\mathbb Q)$.

Observation 2: If $k$ is odd, then there is some element of the form $x \to 2x+b$ in the Galois group. This has the unique fixed point $x=-b$. Because the Galois group is abelian, every other element fixes this point. So the $k$th roots of $a$ are some rational number times the $k$th roots of unity.

Full classification: Let $m \in (\mathbb Z/k)^\times$ be an element such that $m-1 \in 2 (\mathbb Z/k)^\times$. Then there is an element $x \to mx+b$ in the Galois group. If $b$ is even, then this element has two fixed points, the solutions to $(m-1)x+b=0$, so every other element either fixes or reverses those points. What this means is that there is a quadratic extension of $\mathbb Q$ fixed by the index $2$ subgroup of the Galois group that fixes those two points, and that those two points are of the form $q \sqrt{D}$ where $q$ is rational and $\sqrt{D}$ generates that quadratic extension. One can easily classify all sets of roots of this type.

I see my argument in the case $b$ odd doesn't actually work so I don't have a full classification.

Answer 3

It seems to me that the question is equivalent to classifying up to conjugacy in ${\rm GL}(n,\mathbb{Q}),$ those matrices $M$ which satisfy $M^{d} = \frac{u}{v}I,$ where $d$ is a positive integer, $u$ is a positive integer, $v$ is anon-zero integer, ${\rm gcd}(u,v) = 1$ and neither $u$ nor $v$ is divisible by the $d$-th power of any prime, and furthermore, no lower power of $M$ is scalar. Then it becomes a question of determining what the possible rational canonical forms for $M$ can be. Maschke's theorem still goes through in this situation, suitably interpreted, and the minimum polynomial of $M$ is multiplicity free. The possible factors for the characteristic polynomial of such an $M$ are easy to determine.

Answer 4

See https://math.temple.edu/~lorenz/papers/sizes/sizes.pdf

Look for The Minkowski sequence