Let $T$ be an algebraic group. $T$ is called an algebraic torus over $k$, if $T(E)$ is isomorphic to a finite direct product of copies of $G_m(E)$ for some finite finite extension $E$ of $k$, where $G_m$ is the multiplicative group. If $E$ can be $k$, then $T$ is called a split torus over $k$; otherwise, $T$ is called a non split torus over $k$. In this post, I am going to talk something about $SO$ to give examples of non split and split tori.

Definition. Let $k$ be a field, and $V$ be a vector space over $k$. Let $q$ be a quadratic form on $V$. Define $$SO(V,q;k)=\{\gamma \in SL(V) : q(\gamma v)=q(v), \forall v \in V\}.$$

Let $k$ be a field. Let $d \in k\backslash k^2$. Then $E=k(\sqrt{d})$ is a quadratic extension of $k$. If we regard $E$ as a $2$ dimensional vector space over $k$, there is a natural quadratic form q on $E$, given by the field norm. If $e=a+b\sqrt{d} \in E$ for some $a,b \in k$, then $q(e)=a^2-b^2 d$. We denote $E^1$ for the collections of all elements $e \in E$ such that $q(e)=1$, i.e., $E^1 = \{e=a+b\sqrt{d}\in E: a^2-b^2d=1\}.$ If we use $\{1, \sqrt{d}\}$ as a basis for $E$ over $k$, we can represent $q$ by the matrix $M_q=\left(\begin{array}{cc} 1 & \\ & -d \end{array}\right)$ Now, if we tensor the extension tower $k \subset E$ by $E$, we get a vector space $E \oplus E$ over $E$, with a quadratic form $\tilde{q}$ on $E \oplus E$ given also by the matrix $M_q$.

Claim 1. $SO(E,q;k)$ is a non split torus over $k$. More precisely, $SO(E,q;k) \cong E^1$ and $SO(E\oplus E, \tilde{q};E) \cong G_m(E)$.

Proof. Representing $\gamma \in GL(E)$ as a matrix with coefficients in $k$ under the basis $\{1, \sqrt{d}\}$, say, $M_\gamma = \left(\begin{array}{cc} x & y \\ z & w \end{array}\right)$ Then $\gamma \in SO(E,q;k)$ is equivalent to $M_\gamma^T M_q M_\gamma=M_q$, where $M_\gamma^T$ is the transpose of $M_\gamma$. If we write down the relations explicitly, we have a system of equations:

$\left\{$$\begin{split} & x^2-z^2d=1 \\ & xy-zwd=0 \\ & y^2-w^2d=-d \\ & xw-yz = 1 \end{split}\label{eq:system_of_equations}$$\right.$

Now, if we fix $x, y \in k$ with $x^2-z^2d=1$, which is to say $x+z\sqrt{d} \in E^1$. It’s easy to see from $\eqref{eq:system_of_equations}$ that $y$ and $z$ can be solved. Hence, the solutions to $\eqref{eq:system_of_equations}$ is parametrized by $E^1$, which means $SO(E,q;k) \cong E^1$.

If we consider $SO(E \oplus E, \tilde{q}; E)$, since the corresponding matrices to $q$ and $\tilde{q}$ are the same, we will have the same system of equations as $\eqref{eq:system_of_equations}$. However, instead of finding solutions in $k$, we now need to solve $\eqref{eq:system_of_equations}$ in $E$. In $E$, we can factor $x^2-z^2d=1$: $(x+z\sqrt{d})(x-z\sqrt{d})=1.$ Then $x+z\sqrt{d} \in G_m(E)$. If we set $x+z\sqrt{d}=e \in G_m(E)$, then $x-z\sqrt{d}=e^{-1}$. We can then easily solve $\eqref{eq:system_of_equations}$ in terms of $e$. Hence, $SO(E \oplus E, \tilde{q}; E) \cong G_m(E)$. $\square$

The Claim 1 above shows that $SO(E,q;k)$ is a non split torus over $k$. Below, I will construct a quadratic form on $E$ such that $SO(E,q; k)$ is a split torus.

Let $k$ be a field with $\text{char } k \ne 2$. Let $E=k \oplus k$. Let $q$ be the quadratic form on $E$ define by $q(e)=2xy,$ where $x, y \in k$ and $e=(x,y) \in E$.

Claim 2. $SO(E, q;k) \cong G_m(k)$. Hence, $SO(E, q; k)$ is a split torus over $k$.

Proof. Let $e_1=(1,0)$ and $e_2=(0,1)$. Then $\{e_1, e_2\}$ is a basis for $E$ over $k$. Under this basis, the quadratic form $q$ corresponds to the matrix $M_q=\left(\begin{array}{cc} & 1 \\ 1 & \end{array}\right).$

Let $\gamma \in SO(E, q; k)$. Let $M_\gamma = \left(\begin{array}{cc} x & y \\ z & w \end{array}\right)$ be the matrix of $\gamma$ under the basis.

Similar to the procedure in the proof of Claim 1, we can get another system of equations:

$\left\{$$\begin{split} & xz=yw=0 \\ & xw+yz=1 \\ & xw-yz = 1 \end{split}\label{eq:system_of_equations_2}$$\right.$

Solving $\eqref{eq:system_of_equations_2}$, we get $y=z=0$ and $xw=1$. Hence, all the elements in $SO(E, q; k)$ have the form $\left(\begin{array}{cc} x & \\ & x^{-1} \end{array}\right).$ Therefore, $SO(E, q; k) \cong G_m(k)$. $\square$