In my recent number theory seminar on “Hilbert’s 90 and generalizations” (notes here), Professor Goins asked the following interesting question.

Let ${K}$ be a field and ${d\in K^*}$. Define ${T_d}$ to be the torus

$\displaystyle \left\{ \begin{pmatrix} x & dy \\ y & x \end{pmatrix} : x,y \in L, x^2-dy^2=1\right\}.$

What values of ${d}$ give ${K}$-isomorphic tori?

(The question was perhaps motivated by the observation that over the reals, the sign of ${d}$ determines completely whether ${T_d}$ would be split (i.e., isomorphic to ${\mathbb R^*}$) or anisotropic (i.e., isomorphic to ${S^1}$).

Here are two ways of looking at the answer.

• For ${d,e \in K^*}$, we determine when two matrices ${\displaystyle\begin{pmatrix} x & dy \\ y & x \end{pmatrix}}$ and ${\displaystyle\begin{pmatrix} u & ev \\ v & u \end{pmatrix}}$ are conjugate in ${\text{SL}_2(K)}$. Solving the system

$\displaystyle \begin{pmatrix} x & dy \\ y & x \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} . \begin{pmatrix} u & ev \\ v & u \end{pmatrix} . \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

gives ${\displaystyle \frac{e}{d} = \left(\frac{b}{c}\right)^2, de = \left(\frac{d}{a}\right)^2}$.

Thus ${e \in d.(K^*)^2}$ i.e., the ${T_d}$‘s are classified by ${\displaystyle \frac{K^*}{(K^*)^2}}$. (For ${K=\mathbb R}$, this is isomorphic to ${\{\pm 1\}}$ so the sign of ${d}$ determines ${T_d}$ upto conjugation.) By Kummer theory, ${\displaystyle \frac{K^*}{(K^*)^2} \cong H^1(\text{Gal}(\overline K / K), \mu_2)}$, where ${\mu_2 = \{\pm 1\}}$ are the second roots of unity. Thus there is a correspondence between isomorphism classes of tori ${T_d \; (d \in K^*)}$ and quadratic extensions of ${K}$.

• Another way to look at the same thing is as follows. Fix ${d \in K^*}$. Let ${L}$ be an extension of ${K}$ wherein ${T_d}$ splits. Now ${T_d(L)}$ is a split torus of rank 1. For an algebraic group ${G}$ over an algebraically closed field, we have the exact sequence

$\displaystyle 1 \rightarrow \text{Inn}(G) \rightarrow \text{Aut}(G) \rightarrow \text{Aut}(\Psi_0(G)) \rightarrow 1,$

where ${\Psi_0(G)}$ is the based root datum ${(X,\Delta,X\;\check{}, \Delta\,\check{})}$ associated to ${G}$. (Here, ${X = X^*(G)}$ and ${\Delta}$ is the set of simple roots of ${X}$ corresponding to a choice of a Borel subgroup of ${G}$.) For details, see Corollary 2.14 of Springer’s paper “Reductive Groups” in Corvallis.

In our case, ${G = T_d}$ so ${\Psi_0(G) = ( \mathbb Z, \emptyset, \mathbb Z \;\check{}, \emptyset)}$.

$\displaystyle \text{Aut}(\Psi_0(G)) \cong \text{Aut}(\mathbb Z) \cong \{ \pm 1\}.$

Now ${L/K}$ forms of ${T_d}$ are in bijective correspondence with

$\displaystyle H^1(\text{Gal}(L/K), \text{Aut}(\Psi_0(G))) = H^1(\text{Gal}(L/K), \{\pm 1\}) \cong \text{Hom}_{\mathbb Z}(\text{Gal}(L/K), \{\pm 1\});$

the last isomorphism because the Galois group acts trivially on the split torus ${T_d(L)}$. ${\blacksquare}$