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:
\(\displaystyle \mathbf i \mathbf j = - \mathbf j \mathbf i\) \(=\) \(\displaystyle \mathbf k\)
\(\displaystyle \mathbf j \mathbf k = - \mathbf k \mathbf j\) \(=\) \(\displaystyle \mathbf i\)
\(\displaystyle \mathbf k \mathbf i = - \mathbf i \mathbf k\) \(=\) \(\displaystyle \mathbf j\)
\(\displaystyle \mathbf i^2 = \mathbf j^2 = \mathbf k^2 = \mathbf i \mathbf j \mathbf k\) \(=\) \(\displaystyle - \mathbf 1\)


Now consider the quaternions $\Bbb H$ defined as ordered pairs $\left({x, y}\right)$ 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 $\left({w, x}\right) \left({y, z}\right)$ is defined as:

$\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$

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:

\(\text {(1)}: \quad\) \(\displaystyle \mathbf 1\) \(=\) \(\displaystyle \left({1, 0}\right)\)
\(\text {(2)}: \quad\) \(\displaystyle \mathbf i\) \(=\) \(\displaystyle \left({i, 0}\right)\)
\(\text {(3)}: \quad\) \(\displaystyle \mathbf j\) \(=\) \(\displaystyle \left({0, i}\right)\)
\(\text {(4)}: \quad\) \(\displaystyle \mathbf k\) \(=\) \(\displaystyle \left({0, 1}\right)\)

We can see that:

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


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


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:

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


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:

\(\displaystyle \left({w, x}\right) \left({y, z}\right) \ \ \) \(\displaystyle =\) \(\) \(\displaystyle (\)
\(\displaystyle \) \(\) \(\displaystyle \left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) + \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i\)
\(\displaystyle \) \(,\) \(\displaystyle \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) + \left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) i\)
\(\displaystyle \) \(\) \(\displaystyle )\)

which is equivalent to:

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

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:

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

So:

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

as required.

$\blacksquare$


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