Definition:Quaternion/Mistake

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

• 1974: Robert Gilmore: Lie Groups, Lie Algebras and Some of their Applications ... (previous) ... (next): Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $3$. FIELD