Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: algebraic-geometry, group-theory · Tags:

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})$$

I actually found this using the Sage command


Though I am sure since this matrix has such a nice form it should be easy to come up with a slick way to do this calculation. Anyways, now that we have the determinant, which looks suspiciously nice, we could define a function $G(R)\to \mathbb{G}_m^4(R)$ by

$$A\mapsto \bigl( \left(a_{11} + a_{12} + a_{13} – a_{14}\right), \left(a_{11} + a_{12} – a_{13} + a_{14}\right),\\ \left(a_{11} – a_{12} + a_{13} + a_{14}\right),\left(a_{11} – a_{12} – a_{13} – a_{14}\right) \bigr)$$

Because the determinant is a homomorphism of algebraic groups, this map is a homomorphism too, and from here it’s easy to construct an inverse homomorphism of algebraic groups. Hence we have $G\cong\mathbb{G}_{m,\mathbb{Z}}^4$. Of course, there are other ways…

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