Quaternions Defined by Ordered Pairs

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Theorem

Consider the quaternions $\Bbb H$ as numbers in the form:

$a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$

where:

$a, b, c, d$ are real numbers;
$\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are entities related to each other in the following way:
\(\ds \mathbf i \mathbf j = - \mathbf j \mathbf i\) \(=\) \(\ds \mathbf k\)
\(\ds \mathbf j \mathbf k = - \mathbf k \mathbf j\) \(=\) \(\ds \mathbf i\)
\(\ds \mathbf k \mathbf i = - \mathbf i \mathbf k\) \(=\) \(\ds \mathbf j\)
\(\ds \mathbf i^2 = \mathbf j^2 = \mathbf k^2 = \mathbf i \mathbf j \mathbf k\) \(=\) \(\ds - \mathbf 1\)


Now consider the quaternions $\Bbb H$ defined as ordered pairs $\tuple {x, y}$ where $x, y \in \C$ are complex numbers, on which the operation of multiplication is defined as follows:


Let $w = a_1 + b_1 i, x = c_1 + d_1 i, y = a_2 + b_2 i, z = c_2 + d_2 i$ be complex numbers.

Then $\tuple {w, x} \tuple {y, z}$ is defined as:

$\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$

where $\overline w$ and $\overline x$ are the complex conjugates of $w$ and $x$ respectively.


These two definitions are equivalent.


Proof

First we identify the following:

\(\text {(1)}: \quad\) \(\ds \mathbf 1\) \(=\) \(\ds \tuple {1, 0}\)
\(\text {(2)}: \quad\) \(\ds \mathbf i\) \(=\) \(\ds \tuple {i, 0}\)
\(\text {(3)}: \quad\) \(\ds \mathbf j\) \(=\) \(\ds \tuple {0, i}\)
\(\text {(4)}: \quad\) \(\ds \mathbf k\) \(=\) \(\ds \tuple {0, 1}\)

We can see that:

\(\ds \mathbf 1^2\) \(=\) \(\ds \tuple {1, 0} \tuple {1, 0}\)
\(\ds \) \(=\) \(\ds \tuple {1 \times 1 - 0 \times 0, 1 \times 0 + 0 \times 1}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {1, 0}\)
\(\ds \) \(=\) \(\ds \mathbf 1\)


\(\ds \mathbf i^2\) \(=\) \(\ds \tuple {i, 0} \tuple {i, 0}\)
\(\ds \) \(=\) \(\ds \tuple {i \times i - 0 \times 0, \paren {-i} \times 0 + 0 \times i}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {-1, 0}\)
\(\ds \) \(=\) \(\ds -\mathbf 1\)
\(\ds \mathbf j^2\) \(=\) \(\ds \tuple {0, i} \tuple {0, i}\)
\(\ds \) \(=\) \(\ds \tuple {0 \times 0 - i \times \paren {-i}, 0 \times i + i \times 0}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {-1, 0}\)
\(\ds \) \(=\) \(\ds -\mathbf 1\)
\(\ds \mathbf k^2\) \(=\) \(\ds \tuple {0, 1} \tuple {0, 1}\)
\(\ds \) \(=\) \(\ds \tuple {0 \times 0 - 1 \times 1, 0 \times 1 + 1 \times 0}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {-1, 0}\)
\(\ds \) \(=\) \(\ds -\mathbf 1\)
\(\ds \mathbf i \mathbf j\) \(=\) \(\ds \tuple {i, 0} \tuple {0, i}\)
\(\ds \) \(=\) \(\ds \tuple {i \times 0 - i \times 0, \paren {-i} \times i + 0 \times 0}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {0, 1}\)
\(\ds \) \(=\) \(\ds \mathbf k\)
\(\ds \mathbf j \mathbf k\) \(=\) \(\ds \tuple {0, i} \tuple {0, 1}\)
\(\ds \) \(=\) \(\ds \tuple {0 \times 0 - 1 \times \paren {-i}, 0 \times 1 + i \times 0}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {i, 0}\)
\(\ds \) \(=\) \(\ds \mathbf i\)
\(\ds \mathbf k \mathbf i\) \(=\) \(\ds \tuple {0, 1} \tuple {i, 0}\)
\(\ds \) \(=\) \(\ds \tuple {0 \times i - 0 \times 1, 0 \times 0 + 1 \times i}\) from $\tuple {w, x} \tuple {y, z} := \tuple {w y - z \overline x, \overline w z + x y}$
\(\ds \) \(=\) \(\ds \tuple {0, i}\)
\(\ds \) \(=\) \(\ds \mathbf j\)


Let:

$w = a_1 + b_1 i$
$x = d_1 + c_1 i$
$y = a_2 + b_2 i$
$z = d_2 + c_2 i$

and so:

$\overline w = a_1 - b_1 i$
$\overline x = d_1 - c_1 i$

Then substituting for $(1)$ to $(4)$ above, we have:

\(\ds w\) \(=\) \(\ds a_1 \mathbf 1 + b_1 \mathbf i\)
\(\ds x\) \(=\) \(\ds d_1 \mathbf k + c_1 \mathbf j\)
\(\ds y\) \(=\) \(\ds a_2 \mathbf 1 + b_2 \mathbf i\)
\(\ds z\) \(=\) \(\ds d_2 \mathbf k + c_2 \mathbf j\)


Notice the way $\mathbf j$ and $\mathbf k$ are configured. See that they are what appears to be in the wrong order.


We can then demonstrate the equivalence by showing that:

\(\ds \tuple {w, x} \tuple {y, z} \ \ \) \(\ds =\) \(\) \(\ds (\)
\(\ds \) \(\) \(\ds \paren {a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2} + \paren {a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2} i\)
\(\ds \) \(,\) \(\ds \paren {a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2} + \paren {a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2} i\)
\(\ds \) \(\) \(\ds )\)

which is equivalent to:

\(\ds \mathbf x_1 \mathbf x_2 \ \ \) \(\ds =\) \(\) \(\ds \paren {a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2} \mathbf 1\)
\(\ds \) \(+\) \(\ds \paren {a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2} \mathbf i\)
\(\ds \) \(+\) \(\ds \paren {a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2} \mathbf j\)
\(\ds \) \(+\) \(\ds \paren {a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2} \mathbf k\)

where:

$x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
$x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$

in accordance with Quaternion Multiplication.


Thus:

\(\ds w y\) \(=\) \(\ds \paren {a_1 a_2 - b_1 b_2} + \paren {a_1 b_2 + b_1 a_2} i\)
\(\ds z \overline x\) \(=\) \(\ds \paren {d_1 d_2 + c_1 c_2} + \paren {d_1 c_2 - c_1 d_2} i\)
\(\ds \overline w z\) \(=\) \(\ds \paren {a_1 d_2 + b_1 c_2} + \paren {a_1 c_2 - b_1 d_2} i\)
\(\ds x y\) \(=\) \(\ds \paren {d_1 a_2 - c_1 b_2} + \paren {d_1 b_2 + c_1 a_2} i\)

So:

\(\ds w y - z \overline x\) \(=\) \(\ds \paren {a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2} + \paren {a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2} i\)
\(\ds \overline w z + x y\) \(=\) \(\ds \paren {a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2} + \paren {a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2} i\)

as required.

$\blacksquare$


Sources

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