オイラーの4平方恒等式

実数の場合の証明
\(a_{0},\cdots,a_{3}\)と\(\beta_{0},\cdots,\beta_{3}\)を実数として、\(\alpha,\beta\text{を4元数}\)
\[ \begin{cases} \alpha=a_{\mu}e_{\mu}\\ \beta=b_{\mu}e_{\mu} \end{cases} \] とする。
\begin{align*} \left(a_{0}^{\;2}+a_{1}^{\;2}+a_{2}^{\;2}+a_{3}^{\;2}\right)\left(b_{0}^{\;2}+b_{1}^{\;2}+b_{2}^{\;2}+b_{3}^{\;2}\right) & =\alpha\overline{\alpha}\beta\overline{\beta}\\ & =\alpha\beta\overline{\beta}\overline{\alpha}\\ & =\alpha\beta\overline{\alpha\beta}\\ & =\left|\alpha\beta\right|^{2}\\ & =\left|a_{\mu}e_{\mu}b_{\nu}e_{\nu}\right|^{2}\\ & =\left|a_{0}b_{0}e_{0}e_{0}+a_{0}b_{i}e_{0}e_{i}+a_{i}b_{0}e_{i}e_{0}+a_{i}b_{j}e_{i}e_{j}\right|^{2}\\ & =\left|a_{0}b_{0}+a_{0}b_{i}e_{i}+a_{i}b_{0}e_{i}+a_{i}b_{j}\left(-\delta_{ij}e_{0}+\epsilon_{ijk}e_{k}\right)\right|^{2}\\ & =\left|a_{0}b_{0}-a_{i}b_{j}\delta_{ij}+\left(a_{0}b_{k}+a_{k}b_{0}+a_{i}b_{j}\epsilon_{ijk}\right)e_{k}\right|^{2}\\ & =\left|\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)e_{1}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)e_{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)e_{3}\right|^{2}\\ & =\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)^{2}+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)^{2}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)^{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)^{2} \end{align*}