A Lie algebra is an vector space $L$ equipped with a bilinear map called Lie bracket
\[
(x, y) \in L\times L \mapsto [x,y]\in L
\]
which satisfies the following properties:
\[
[x, x] = 0 \text{ for all } x\in L \qquad\qquad{(1)}
\]
\[
[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 \qquad {(2)}
\]
for all $x$, $y$, and $z \in L$.
Consider the set of screws over space $\cal E$ denoted $S({\cal E})$ and equipped with the bilinear map
\[
(\{{\cal U} \}, \{{\cal V} \} ) \in S({\cal E}) \times S({\cal E}) \mapsto \{{\cal U} \} \times \{{\cal V} \}
\]
corresponding to the vector field
\[
P \in {\cal E} \mapsto {\bf U}\times {\bf v}_P - {\bf V}\times {\bf u}_P
\]
We can show that $\{{\cal U} \} \times \{{\cal V} \}$ has the properties of a Lie bracket and that the set $S(\cal E)$
is a Lie algebra.
First we must show that $\{{\cal U} \} \times \{{\cal V} \}$ is a screw:
\[
{\bf U}\times {\bf v}_Q - {\bf V}\times {\bf u}_Q =
{\bf U}\times ( {\bf v}_P+ {\bf V}\times {\bf r}_{PQ}) - {\bf V}\times ({\bf u}_P+ {\bf U}\times {\bf r}_{PQ})
={\bf U}\times {\bf v}_P - {\bf V}\times {\bf u}_P
+({\bf U}\times{\bf V}) \times {\bf r}_{PQ}
\]
This shows that $\{{\cal U} \} \times \{{\cal V} \}$ is a screw of resultant ${\bf U}\times{\bf V}$.
Property (1) is obvious. Property (2) must be written as
\[
\{{\cal U} \} \times (\{{\cal V} \} \times \{\cal W \}) + \{{\cal V} \} \times (\{\cal W \} \times \{{\cal U} \})+\{\cal W \} \times (\{{\cal U} \} \times \{{\cal V} \}) = \{ 0 \}
\]
It follows from Jacobi's identity:
\[
({\bf U} \times {\bf V}) \times {\bf W} + ({\bf V} \times {\bf W}) \times {\bf U} + ({\bf W} \times {\bf U})
\times {\bf V} = {\bf 0}
\]
