Lie order - Knowlpedia

Let $R$ be a Noetherian integral domain with fraction field $K$ , and let $\mathfrak g$ be a finite-dimensional $K$ -Lie algebra. A Lie $R$ -order in $\mathfrak g$ is a full $R$ -lattice $L\subseteq\mathfrak g$ satisfying $[L,L]\subseteq L$ .

Thus the Lie bracket restricts to $L$ , and $K\otimes_R L\cong\mathfrak g$ as Lie algebras. Such an $L$ is also called an integral Lie form or Lie lattice; the word “order” here refers to the Lie operation.

The rank-three example

For $K=\mathbb Q$ , put

e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.

Then $\mathfrak{sl}_2(\mathbb Z)=\mathbb Ze\oplus\mathbb Zh\oplus\mathbb Zf$ is a Lie order because

[h,e]=2e,\qquad[h,f]=-2f,\qquad[e,f]=h.

This is the basic Chevalley basis example.

Changing the base ring

The same basis gives a Lie $\mathbb Z[i]$ -order $\mathfrak{sl}_2(\mathbb Z[i])$ in $\mathfrak{sl}_2(\mathbb Q(i))$ . Under restriction of scalars it is also a rank-six Lie $\mathbb Z$ -order, with basis $e,ie,h,ih,f,if$ .

Neither trace-zero lattice is an associative order under matrix multiplication: $ef=\operatorname{diag}(1,0)$ is not trace zero. The corresponding associative examples are the full matrix rings.

References

The displayed matrix products directly verify bracket closure and failure of associative closure.
For the general semisimple construction and its literature, see Chevalley basis and Chevalley lattice and integral model.