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:

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

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:

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

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, \(SO(E, q; k) \cong G_m(k)\). \(\square\)