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Algebra-II: Companion notes

Abhay Soman

Section 1.8 Quaternion algebra

In this section, we assume that \(F\) is a field with the property that \(2\neq 0\text{.}\) We define a Quaternion algebra over \(F\text{.}\)

Definition 1.8.1. (Quaternion).

Let \(a,b\in F\) be nonzero elements. Consider a \(4\)-dimensional vector space \(Q\) over \(F\) with basis \(\{1,i,j,k\}\text{.}\) We write \(\alpha\cdot1=\alpha\) for any \(\alpha\in F\text{.}\) Thus \(Q=\{x_0+x_1i+x_2j+x_3k:x_i\in F\}\text{.}\) We make \(Q\) a ring by defining
\begin{equation} i^2=a,\;j^2=b,\;\text{and}\; ij=-ji=k\tag{1.8.1} \end{equation}
and
\begin{equation} \alpha i=i\alpha,\;\alpha j=j\alpha,\;\text{and}\;\alpha k=k\alpha\quad\text{for all}\;\alpha\in F.\tag{1.8.2} \end{equation}
Hence
\begin{equation*} ik=i(ij)=(i^2)j=aj\quad\text{and}\quad jk=j(ij)=j(-ji)=-(j^2)i=-bi \end{equation*}
and
\begin{equation*} ki=(ij)i=-(ji)i=-aj\quad\text{and}\quad kj=(ij)j=i(j^2)=bi. \end{equation*}
Given \(x_0+x_1i+x_2j+x_3k\in Q\) and \(y_0+y_1i+y_2j+y_3k\in Q\) their multiplication is defined as follows.
\begin{align*} (x_0+x_1i+x_2j+x_3k)(y_0+y_1i+y_2j+y_3k)\amp=(x_0y_0)+(x_0y_1)i+(x_0y_2)j+(x_0y_3)k\\ \amp\;\quad+ (x_1y_0)i+(x_1y_1) a+(x_1y_2)k+(x_1y_3)a j\\ \amp\;\quad\quad +(x_2y_0)j-(x_2y_1)k+(x_2y_2)b-(x_2y_3)bi\\ \amp\;\quad\quad\quad +(x_3y_0)k-(x_3y_1)aj+(x_3y_2)bi-(x_3y_3)ab.\\ \amp=(x_0y_0+x_1y_1a+x_2y_2b-x_3y_3ab)\\ \amp\;\quad+(x_0y_1+x_1y_0-x_2y_3b+x_3y_2b)i\\ \amp\;\quad\quad+(x_0y_2+x_1y_3a+x_2y_0-x_3y_1a)j\\ \amp\;\quad\quad\quad+(x_0y_3+x_1y_2-x_2y_1+x_3y_0)k \end{align*}
This also shows that \(1\) is the unity of \(Q\text{,}\) and \(Q\) is an \(F\)-algebra.
The algebra \(Q\) is called a quaternion algebra over \(F\) and it is denoted by \((a,b)_F\text{.}\)

Convention 1.8.2.

We say that a quaternion algebra \(Q=(a,b)_F\) is generated by \(i\) and \(j\) over \(F\) with relations.
\begin{equation} i^2=a,\;j^2=b,\;\text{and}\; ij=-ji=k\tag{1.8.3} \end{equation}
The Hamiltonian quaternion algebra is the quaternion algebra over \(\R\) generated by \(i\) and \(j\) such that
\begin{equation} i^2=-1,\;j^2=-1,\;\text{and}\;ij=-ji=k.\tag{1.8.4} \end{equation}
The Hamiltonian quaternion is denoted by \(\mathbb{H}=(-1,-1)_{\R}\text{.}\)
Let \(F\) be a field and \(b\in F\) a nonzero element. Consider the following matrices.
\begin{equation*} \mathcal{I}=\begin{pmatrix}1\amp 0\\0\amp -1\end{pmatrix}\quad\text{and}\quad\mathcal{J}=\begin{pmatrix}0\amp b\\1\amp 0\end{pmatrix} \end{equation*}
The quaternion algebra generated by \(\mathcal{I}\) and \(\mathcal{J}\) is all \(2\times 2\) matrices over \(F\text{,}\) \(M_2(F)\text{.}\) This follows because
\begin{equation*} I_2=\begin{pmatrix}1\amp 0\\0\amp 1\end{pmatrix},\quad \mathcal{I}=\begin{pmatrix}1\amp 0\\0\amp -1\end{pmatrix},\quad\mathcal{J}=\begin{pmatrix}0\amp b\\1\amp 0\end{pmatrix}\quad\text{and}\quad\mathcal{IJ}=\begin{pmatrix}0\amp b\\-1\amp 0\end{pmatrix} \end{equation*}
is an \(F\)-vector space basis of \(M_2(F)\text{.}\) Furthermore,
\begin{equation*} \mathcal{I}^2=I_2,\quad\mathcal{J}^2=b I_2,\quad\text{and}\quad\mathcal{IJ}=-\mathcal{JI}. \end{equation*}
Therefore, \(M_2(F)\) is a quaternion algebra.
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