# Twisted forms of $\mathrm{SL}(2,q)$

$$\DeclareMathOperator\SL{SL}$$Let $$q = p^r$$ be a prime power. Let $$H$$ denote the subgroup of $$\SL(2,\overline{\mathbb{F}}_q)$$ consisting of matrices of the form $$\begin{pmatrix}a & b\\ b^q & a^q\end{pmatrix}$$, where $$a,b\in\mathbb{F}_{q^2}$$. Some notes I'm reading claim that $$H$$ is conjugate to $$\SL(2,\mathbb{F}_q)$$.

This is a very naive question — How does one classify the conjugates of $$\SL(2,\mathbb{F}_q)$$ inside $$\SL(2,\overline{\mathbb{F}}_q)$$? How do you generally classify the conjugates of $$G(\mathbb{F}_q)$$ inside $$G(\overline{\mathbb{F}}_q)$$ when $$G$$ is say a split reductive algebraic group over $$\mathbb{F}_p$$?

Also, since I expect this is easy to answer for the experts here, what is the explicit matrix that conjugates $$H$$ to $$\SL(2,\mathbb{F}_q)$$?

References would be appreciated!

• What do you mean by "classify"? A somewhat unhelpful answer is that they are classified by elements of $G( \overline{\mathbb F}_q)$ modulo the centralizer of $G(\mathbb F_q)$ inside $G( \overline{\mathbb F}_q)$, and this centralizer is almost always the center of $G(\overline{\mathbb F}_q)$. The "classification" is the bijection that sends an element $g$ to $g G(\mathbb F_q) g^{-1}$. Oct 9 at 23:43
• The main question might be why this given subgroup can be conjugate in such a way. The answer might follow from a general principle, but most likely not from a classification of conjugates in the suggested way.
– YCor
Oct 10 at 6:51
• But I don't see why the set of such matrices (assuming they have determinant 1) should form a subgroup. Actually, it seems it is not: if $a\in\mathbf{F}_4-\mathbf{F}_2$, then $\begin{pmatrix}1 & a\\1&a+1\end{pmatrix}\begin{pmatrix}1 & a+1\\1&a\end{pmatrix}=\begin{pmatrix}1+a&0\\a&a\end{pmatrix}$.
– YCor
Oct 10 at 7:01
• @WillSawin You're right. This is a badly phrased question. I will likely delete this question in a few minutes and perhaps ask a better question later. Oct 10 at 19:37
• @YCor I actually made a typo. The bottom row should be $b^q,a^q$. Anyway, I've realized this is a bad question. I will delete this in a few minutes and perhaps ask a better question later. Oct 10 at 19:37

If $$\theta$$ lies in $$\mathbb F_{q^2} \setminus \mathbb F_q$$, then $$\operatorname{Int}\begin{pmatrix} 1 & 1 \\ \theta & \overline\theta \end{pmatrix}\begin{pmatrix} a & b \\ \overline b & \overline a \end{pmatrix}$$, where $$\overline\cdot$$ is the non-trivial Galois conjugation, lies in $$\operatorname{SL}_2(\mathbb F_q)$$. We can easily bring $$\begin{pmatrix} 1 & 1 \\ \theta & \overline\theta \end{pmatrix}$$ into $$\operatorname{SL}(2, \overline{\mathbb F_q})$$ by multiplying it by $$\delta^{-1}$$, where $$\delta^2 = \overline\theta - \theta$$.
$$\DeclareMathOperator\tr{tr}$$If $$q$$ is an odd prime power, then we may choose $$\theta$$ so that $$\theta^2$$ lies in $$\mathbb F_q$$. Then $$\operatorname{Int}\begin{pmatrix} 1 & 1 \\ \theta & -\theta \end{pmatrix}\begin{pmatrix} a & b \\ \overline b & \overline a \end{pmatrix}$$ equals $$\dfrac1 2\begin{pmatrix} \tr(a + b) & \tr(\theta^{-1}(a - b)) \\ \tr(\theta(a + b)) & \tr(a - b) \end{pmatrix}$$ for all $$a, b \in \mathbb F_{q^2}$$.