Suppose one day you run into the following algebraic group, defined on $\mathbb{Z}$-algebras $R$ by

$$G(R) = \left\{ \begin{pmatrix}

a_{11} & a_{12} & a_{13} & a_{14} \\

a_{12} & a_{11} & -a_{14} & -a_{13} \\

a_{13} & -a_{14} & a_{11} & -a_{12} \\

a_{14} & -a_{13} & -a_{12} & a_{11}

\end{pmatrix} \in\mathrm{GL}_4(R) \right\}$$

Can you figure out what this group is? I actually did run into this group one day, but luckily I discovered its true identity. Today we'll see one way to do so.

One thing to try to identify unknown algebraic groups in practice is to check out their matrix multiplication. If $A = (a_{ij}), B = (b_{ij})$ have the above form then $C = AB$ has first row

$$

c_{11} = a_{11} b_{11} + a_{12} b_{12} + a_{13} b_{13} + a_{14} b_{14} \\c_{12}= a_{12} b_{11} + a_{11} b_{12} – a_{14} b_{13} – a_{13} b_{14} \\c_{13}= a_{13} b_{11} – a_{14} b_{12} + a_{11} b_{13} – a_{12} b_{14} \\c_{14}= a_{14} b_{11} – a_{13} b_{12} – a_{12} b_{13} + a_{11} b_{14}

$$

(We only need to specify the first row to get the whole matrix in this group). Can you tell what the group is yet? At first, I couldn't, so let's try something else. One can check now that this group is actually commutative, and this would be enough to determine what it is given the context of where it came from, but let's assume we don't know that. Instead, let's take the determinant:

$$

\mathrm{det}(A) = (a_{11} + a_{12} + a_{13} – a_{14})(a_{11} + a_{12} – a_{13} + a_{14})\\\times(a_{11} – a_{12} + a_{13} + a_{14})(a_{11} – a_{12} – a_{13} – a_{14})$$

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