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According to Wikipedia, a Quaternion can be represent in 48 different $4\times4$ matrix forms. Finding this number can be proved by using permutation?

$$\binom 4 2 \cdot4=\frac{4!}{(4-2)!}\cdot4=48$$

As in each column, there are $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$ different permutations. Is this correct?

For example, two of those 48 representations are mentioned below:

\begin{align} & \begin{bmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix} \\[10pt] = {} & a \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} +b \begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix} +c \begin{bmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{bmatrix} +d \begin{bmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{bmatrix} \end{align}

OR \begin{align} & \begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix} \\[10pt] = {} & a\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}+ b\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix}+ c\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}+ d\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} \end{align}

But my main question is how they can be found? Obviously, the reason is because of Hamilton's famous equation:

$$\mathbf{i}^2 =\mathbf {j}^2=\mathbf {k}^2 = \mathbf {i} \mathbf {j} \mathbf {k} =-1$$

But how?

Michael Hardy
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David
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    This is false. There are uncountably many, and they are all related by change of coordinates. Said another way, there are uncountably many different homomorphisms $\mathbb{H} \to M_4(\mathbb{R})$, and they are all conjugate wrt the conjugation action of $GL_4(\mathbb{R})$. – Qiaochu Yuan Nov 27 '17 at 00:54
  • Note that $\displaystyle \binom 4 2$ can be coded as \binom 4 2, so you don't need the much more complicated code that you wrote. (If more than one object is in the position of either the $4$ or the $2$, such as a two-digit number, then one also needs {curly braces}, so $\displaystyle \binom{42}{24}$ is coded as \binom{42}{24}.) – Michael Hardy Nov 27 '17 at 01:11
  • @QiaochuYuan, could you explain more? I cannot get your point. – David Nov 27 '17 at 01:15
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    Suppose $f(q) : \mathbb{H} \to M_4(\mathbb{R})$ is a representation of the quaternions by $4 \times 4$ matrices. Pick any real invertible $4 \times 4$ matrix $X$. Then $X f(q) X^{-1}$ is another such representation. Furthermore, any two such representations are related by such a matrix $X$. There are some particularly nice choices of $X$ given by the signed permutation matrices, but there are $2^4 \cdot 4! = 192$ of these. – Qiaochu Yuan Nov 27 '17 at 01:17
  • Okay, I guess $-I$ acts trivially, so signed permutation matrices give at most $96$ possibilities, and if there's another element of order $2$ acting trivially on the image of $\mathbb{H}$ (which seems likely) that would cut it down to $48$. – Qiaochu Yuan Nov 27 '17 at 01:23
  • The reference from Wikipedia has this as its abstract: “We establish that there are a total of $48$ distinct ordered sets of three $4\times4$ (skew-symmetric) signed permutation matrices which will serve as the basis of an algebra of quaternions.” [http://www.sciencedirect.com/science/article/pii/S0024379502005359?via%3Dihub] – Steve Kass Nov 27 '17 at 01:41

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