If $\operatorname{tr}(g)=\operatorname{tr}(h)\neq\pm 2$ in $\operatorname{SL}_2(q)$, then is $g$ conjugate to $h$ for all $q$?

Question

This is somewhat of a revival of this question.

The Question:

If $\operatorname{tr}(g)=\operatorname{tr}(h)\neq\pm 2$ in $\operatorname{SL}_2(q)$, then is $g$ conjugate to $h$ in $\operatorname{SL}_2(q)$ for all $q$?

Write $s\sim_G t$ for "$s$ is conjugate to $t$ in $G$".

Almost A Proof?

Let $q$ be odd. The characteristic polynomial for $g$ is

$$\begin{align} \chi_g(x)&=x^2-\operatorname{tr}(g)x+\det(g)\\ &=x^2-(\lambda +\lambda^{-1})x+1 \end{align}$$

for not necessarily distinct eigenvalues $\lambda,\lambda^{-1}$ of $g$.

Since $\operatorname{tr}(g)=\operatorname{tr}(h)\neq\pm 2$, we have that the rational canonical form of $g$ and $h$ in $\color{red}{\operatorname{GL}_2(q)}$ is

$$\operatorname{RCF}_{\color{red}{\operatorname{GL}_2(q)}}(g)=\begin{pmatrix} 0 & -1 \\ 1 & \operatorname{tr}(g) \end{pmatrix}.$$

So $$g\sim_{\color{red}{\operatorname{GL}_2(q)}}\operatorname{RCF}_{\color{red}{\operatorname{GL}_2(q)}}(g)=\operatorname{RCF}_{\color{red}{\operatorname{GL}_2(q)}}(h)\sim_{\color{red}{\operatorname{GL}_2(q)}} h.$$ So there is some $P\in\color{red}{\operatorname{GL}_2(q)}$ with $g=PhP^{-1}$.

I don't know whether there is such a $P$ in $\operatorname{SL}_2(q)$.

I don't know what to do when $q$ is even. (Did I need to assume $q$ is odd?)

---

No obvious counterexamples come to mind.

Possible counterexamples might be found when $\operatorname{tr}(g)= 0$.

---

Idea:

We could let $P\mapsto \mu P$ for $\mu^2=\det(P)\in \Bbb F_q^*$. Then

$$\begin{align} \det((\mu P)h(\mu P)^{-1})&=\det(\mu P)\det(h)\det(\mu^{-1}P^{-1})\\ &=\mu^2\mu^{-2}\det(h)\\ &=\det(h)\\ &=1 \end{align}$$

and $\det(\mu P)=1$.

Does this work? What if the $\det(P)$ is not a square?

Answer 1

$\def\Conj{\text{Conj}}\def\Tr{\text{Tr}}\def\SL{\text{SL}}\def\GL{\text{GL}}\def\FF{\mathbb{F}}$The answer is "yes". For a group $G$ and an element $h \in G$, we'll write $\Conj_G(h)$ for the conjugacy class $\{ ghg^{-1} : g \in G\}$. Basically, your question is, for $g$ with $\SL_2$ with $\Tr(g) =\Tr(h)\neq \pm 2$, is $\Conj_{\GL_2}(g)$ the same as $\Conj_{\SL_2}(h)$? The answer, as we will show, is yes.

Proposition Let $G$ be a group, let $N$ be a normal subgroup and let $n \in N$. Let $C_G(n)$ be the centralizer $\{ g \in G : gn=ng \}$. Then the $G$-conjugacy class of $n$ is the same as the $N$-conjugacy class of $n$ if and only if the composite $C_G(n) \subseteq G \to G/N$ is surjective.

Proof: First, suppose that $\Conj_G(n) = \Conj_N(n)$. Let $\overline{g}$ be a class in $G/N$ and lift $\overline{g}$ to $g \in G$. Then there is some $m \in N$ with $mnm^{-1} = gng^{-1}$, so $m^{-1} g$ centralizes $n$. This shows that $m$ represents the class $\overline{g}$ in $G/N$.

Now, suppose that $C_G(n) \subseteq G \to G/N$ is surjective. Let $g \in G$, we must find an $m \in N$ such that $mnm^{-1} = gng^{-1}$. By hypothesis, we can find $c \in C_G(n)$ with $cN = gN$. Then $(gc^{-1}) n (gc^{-1})^{-1} = gng^{-1}$ and $gc^{-1} \in N$, so $g c^{-1}$ is our $m$. $\square$

We now apply this to $\SL_2$ inside $\GL_2$. For $g \in \SL_2$, we must compute $C_{\GL_2}(g)$, and then determine whether or not $\det: C_{\GL_2}(g) \to \mathbb{F}_q^{\ast}$ is surjective. Note that, when working with $C_{\GL_2}(g)$, we may conjugate $g$ by $\GL_2$ without changing the truth of our criterion.

We use the hypothesis that $\Tr(g) \neq \pm 2$, so the characteristic polynomial of $g$ is separable.

Case 1: The characteristic polynomial of $g$ splits over $\FF_q$. Then we can diagonalize $g$ as $\begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{bmatrix}$ over $\mathbb{F}_q$. Then $g$ is centralized by $\begin{bmatrix} u & 0 \\ 0 & v \end{bmatrix}$, which has determinant $uv$, so $\det: C_{\GL_2}(g) \to \mathbb{F}_q^{\ast}$ is surjective.

Case 2: The characteristic polynomial of $g$ is irreducible over $\FF_q$. In that case, the algebra $\FF_q[g]$ inside the $2 \times 2$ matrices is isomorphic to $\FF_{q^2}$, so the centralizer of $g$ includes (and, in fact, is equal to) $\FF_{q^2}^{\ast}$. The determinant map $\FF_{q^2}^{\ast} \to \mathbb{F}_q^{\ast}$ is the norm map, which is surjective, see, for example, here.

This conclusion is special to finite fields. For example, over the real numbers $\begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$ is not $\SL_2(\mathbb{R})$ conjugate to $\begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix}$. That's because the centralizer of $\begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$ is the set of matrices of the form $\begin{bmatrix} u&v \\ -v&u \end{bmatrix}$, with determinant $u^2+v^2$, and negative reals are not of the form $u^2+v^2$.

Answer 2

Let me try and give a slightly more 'high-brow' explanation of David Speyer's beautiful answer, particularly the last comment about the difference between $\mathbb{F}_q$ and $\mathbb{R}$. Considering you tagged this with algebraic groups, I assume this is not totally out of your field of view.

Let $G$ be a reductive group over a (perfect) field $k$. Then, one can consider the conjugation action of $G$ on $G$ and take the GIT quotient $G\mathbin{/\mkern-6mu/}G\simeq \mathrm{Spec}(\mathcal{O}(G)^G)$. I can't really discuss GIT quotients in full generality here, but maybe see this answer of mine for a basic idea. In particular, one would 'like' to think that $G\mathbin{/\mkern-6mu/}G$ parameterizes conjugacy classes in $G$ but this is not quite true--more on this later.

In any case, there is the following famous theorem of Chevalley.

Theorem (Chevalley--Steinberg restriction theorem): Let $T\subseteq G$ be a maximal torus and let $W=N_G(T)/T$ its Weyl group (scheme). Then, the natural inclusion $T\to G$ induces an isomorphism $$T\mathbin{/\mkern-6mu/}W\xrightarrow{\sim}G\mathbin{/\mkern-6mu/}G.$$

Remark: I say Weyl group scheme here as $W$ is a finite etale group scheme which need not be constant in general, but is when $G$ is a split reductive group and $T$ is a maximal split torus, which is the situation we are in. So you should feel free to assume that. $\blacksquare$

Example ($G=\mathrm{SL}_2)$: taking $G=\mathrm{SL}_2$ and $T=\left\{\begin{pmatrix}a & 0\\ 0 & a^{-1}\end{pmatrix}\right\}$ we see that $W=\mathbb{Z}/2$ which acts on $W$ by permuting the diagonal entries. In particular, it's easy to see $\mathcal{O}(T)^W$ consists precisely of polynomials in the function

$$\begin{pmatrix}a & 0\\ 0 & a^{-1}\end{pmatrix}\mapsto a+a^{-1},$$

i.e., $\mathcal{O}(T)^W=k[\mathrm{tr}]$. So $T\mathbin{/\mkern-6mu/} W\simeq \mathbb{A}^1_k$. Together with Chevalley's theorem we deduce that the map

$$\mathrm{tr}\colon G\mathbin{/\mkern-6mu/} G\to \mathbb{A}^1_k$$

is actually an isomorphism. $\blacksquare$

But, the piper is due his payment. By passing to $k$-points in the above example we deduce a bijection

$$\mathrm{tr}\colon (G\mathbin{/\mkern-6mu/} G)(k)\xrightarrow{\sim}k.$$

But, what exactly is $(G\mathbin{/\mkern-6mu/}G)(k)$? As mentioned before it is tempting to think that $(G\mathbin{/\mkern-6mu/} G)(k)$ is just the set of $G(k)$-conjugacy classes in $G(k)$. This is not correct.

In fact, this is not correct on two counts, one of which is easier to understand:

• (more difficult to intuit) Because $G\mathbin{/\mkern-6mu/} G$ is a 'coarse quotient' involving distinction on functions, it can actually only parameterize Zariski closed orbits/conjugacy classes. But, it turns out that this closedness is equivalent to semisimple, so this means that $(G\mathbin{/\mkern-6mu/}G)(\overline{k})$ is parameterizing $G(\overline{k})$-conjugacy classes of semi-simple elements of $G(\overline{k})$,
• (simpler to intuit) Because $G\mathbin{/\mkern-6mu/}G$ is meant to be algebraic we should have, in particular, that the map $(G\mathbin{/\mkern-6mu/}G)(k)\to (G\mathbin{/\mkern-6mu/}G)(\overline{k})$ is injective and identifies with the Galois-fixed points of $(G\mathbin{/\mkern-6mu/}G)(\overline{k})$. In particular, $(G\mathbin{/\mkern-6mu/}G)(k)$ can't be studying $G(k)$-conjugacy (i.e., rational conjugacy) in $G(k)$ but actually $G(\overline{k})$-conjguacy (i.e., geometric conjugacy) in $G(k)$.

Putting these two observations together, we see that we have

$$\begin{aligned}\left\{\begin{matrix}\text{Semi-simple elements of }G(k)\\\text{up to }G(\overline{k})\text{-conjugacy}\end{matrix}\right\} &\hookrightarrow \left\{\begin{matrix}\text{Semi-simple conjugacy classes in }G(\overline{k})\\ \text{fixed by }\mathrm{Gal}(\overline{k}/k)\end{matrix}\right\}\\ &\simeq (G\mathbin{/\mkern-6mu/}G)(k).\end{aligned}$$

Note that, in full generality, this first inclusion is actually not surjective (not every Galois-fixed conjugacy class contains a Galois-fixed element!), but it is true (by a deep result of Kottwitz) if $G$ is quasi-split and $G^\mathrm{der}$ is simply connected (e.g., $G=\mathrm{SL}_2$).

That said, there is a set which exactly measures how many rational conjugacy classes are contained in a given geometric conjugacy class. Namely, let's fix $g$ in $G(k)$ (back in the general setting again). Then, if $h$ is another element of $G(k)$ and $\gamma\in G(\overline{k})$ is such that $g=\gamma h\gamma^{-1}$, then one can check that the association $$\mathrm{Gal}(\overline{k}/k)\to Z_G(g)(\overline{k}),\quad \sigma\mapsto \gamma\sigma(\gamma)^{-1},$$ actually defines a $1$-cocycle in $Z^1(\mathrm{Gal}(\overline{k}/k),Z_G(g)(\overline{k}))$ (which I shorten to $Z^1(k,Z_G(g))$) which is clearly a $1$-coboundary for $G(\overline{k})$ but not for $Z_G(\overline{k})$, i.e., it lies in $$\mathrm{ker}(H^1(k,Z_G(g))\to H^1(k,G)).$$ One can soup this up to in fact show that the set of rational conjugacy classes of $G(k)$ living in the same geometric conjugacy class as $g\in G(k)$ is parameterized by $$\mathrm{ker}(H^1(k,Z_G(g))\to H^1(k,G)).$$

Example cont. ($G=\mathrm{SL}_2)$: Let's finally return to your actual question. As observed by David Speyer, the key thing is that $\mathrm{tr}(g)=\mathrm{tr}(h)\ne \pm 2$ implies, first and foremost, that $g$ and $h$ have to be semi-simple. Thus, by the above discussion $g$ and $h$ define the same element of $(\mathrm{SL}_2\mathbin{/\mkern-6mu/}\mathrm{SL}_2)(k)$, and thus define the same geometric conjugacy class.

To show they are actually rationally conjugate we must show that $h$ defines the trivial class in

$$\mathrm{ker}\left(H^1(k,Z_{\mathrm{SL}_2}(g))\to H^1(k,\mathrm{SL}_2)\right).$$

But, because $\mathrm{SL}_2$ is simply connected (or just checking by hand over $\overline{k}$) one can actually check that $Z_{\mathrm{SL}_2}(g)$ is a connected reductive group.

So, if we assume that $k=\mathbb{F}_q$ we deduce from Lang's theorem tells us then that $$H^1(\mathbb{F}_q,Z_{\mathrm{SL}_2}(g))=H^1(\mathbb{F}_q,\mathrm{SL}_2)=\{\ast\}.$$

So, in fact, the geometric conjugacy class of $g$ coincides with the rational conjugacy class for $g$ and so, in fact, $g$ and $h$ are rationally conjugate as desired. $\blacksquare$

Non-example ($G=\mathrm{SL}_2$, $k=\mathbb{R}$): Let us now use this reasoning to also explain David Speyer's counterexample over $\mathbb{R}$.

Consider the elements

$$g=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix},\qquad h=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}$$

of $\mathrm{SL}_2(\mathbb{R})$. As these have the same trace, we see by our GIT considerations that they are geometrically conjugate (i.e., conjugate in $\mathrm{SL}_2(\mathbb{C})$. That said, they are not rationally conjugate. Indeed, if $\begin{pmatrix}a & b\\ c & d\end{pmatrix}\in\mathrm{SL}_2(\mathbb{R})$ then

$$\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}^{-1}=\begin{pmatrix} a c + b d & -a^2 - b^2\\ c^2 + d^2 & -a c - b d\end{pmatrix}$$

So, for this to equal $\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}$ one can check one would need $a=-d$ and $b=c$ which would mean that the determinant of $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $-(a^2+b^2)$ which is not possible.

The culprit we see is that $Z_{\mathrm{SL}_2}(g)$ is isomorphic to the unit circle torus $T$ (i.e., the unique non-split rank $1$ torus over $\mathbb{R}$), and so

$$\begin{aligned}\ker\left(H^1(\mathbb{R},Z_{\mathrm{SL}_2}(g))\to H^1(\mathbb{R},\mathrm{SL}_2)\right) &= \ker\left(H^1(\mathbb{R},T)\to \{\ast\}\right)\\ &= H^1(\mathbb{R},T)\\ &\simeq \mathbb{Z}/2,\end{aligned}$$

and so we get geometric conjugacy classes that decompose into two distinct rational conjugacy classes. $\blacksquare$

Remark: Let me finish by more clearly enunciating where the finite field case was saved where the real case was not, and where this showed up in David Speyer's answer. Namely, for both $\mathbb{F}_q$ and $\mathbb{R}$ there is a unique non-split rank $1$ torus, let me write it $T_{\mathbb{F}_q}$ and $T_{\mathbb{R}}$, respectively. Both of these tori can occur as the centralizer of a torus in $\mathrm{SL}_2$ (exercise!), and in fact the same is true (correctly interpreted) for any field.

But, as $H^1(k,\mathrm{SL}_2)$ is trivial for any field, we see that the only hope for your question to hold is that $H^1(k,T_k)$ is trivial. For $\mathbb{R}$ this is false, and for $\mathbb{F}_q$ this is true, thus the situation stands as we saw. But, we claimed that $H^1(k,T_k)$ is trivial by Lang's theorem above when $k=\mathbb{F}_q$. While true, in this case, there is a more concrete way to calculate $H^1(k,T_k)$. Namely, if $k'$ is the unique degree $2$ extension of $k$, then $H^1(k,T_k)=k^\times/\mathrm{Nm}_{k'/k}((k')^\times)$. This is $\mathbb{Z}/2$ when $k=\mathbb{R}$ and trivial when $k=\mathbb{F}_q$. Thus, this type of reasoning also secretly appeared in David Speyer's answer in the consideration of Case 2. $\blacksquare$