Definition:Quaternion/Mistake

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Source Work

1974: Robert Gilmore: Lie Groups, Lie Algebras and Some of their Applications:

Chapter $1$: Introductory Concepts
$1$. Basic Building Blocks


Mistake

Every quaternion can be represented in the form
$q = q_0 1 + q_1 \lambda_1 + q_2 \lambda_2 + q_3 \lambda_3$
where the $q_i \, \paren {i = 0, 1, 2, 3}$ are real numbers and the $\lambda_1$ have multiplicative properties defined by
\(\ds \lambda_0 \lambda_i\) \(=\) \(\ds \lambda_i \lambda_0 = \lambda_i\) \(\ds i = 0, 1, 2, 3\)
\(\ds \lambda_i \lambda_i\) \(=\) \(\ds -\lambda_0\)
\(\ds \lambda_1 \lambda_2\) \(=\) \(\ds -\lambda_2 \lambda_1 = \lambda_3\)
\(\ds \lambda_2 \lambda_3\) \(=\) \(\ds -\lambda_3 \lambda_2 = \lambda_1\)
\(\ds \lambda_3 \lambda_1\) \(=\) \(\ds -\lambda_1 \lambda_3 = \lambda_2\)


Correction

The representation of $q$ should read:

$q = q_0 \lambda_0 + q_1 \lambda_1 + q_2 \lambda_2 + q_3 \lambda_3$

and the line:

$\lambda_i \lambda_i = -\lambda_0$

holds only for $i = 1, 2, 3$ because:

$\lambda_0 \lambda_0 = \lambda_0$


One suspects that the author conflated this presentation with one where $\lambda_0$ has been identified with the number $1$, and only partially has this been translated into its current form.


Sources