Vectors in Three Dimensional Space with Cross Product forms Lie Algebra
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Theorem
Let $S$ be the set of vectors in $3$ dimensional Euclidean space.
Let $\times$ denote the vector cross product on $S$.
Then $\struct {S, \times}$ is a Lie algebra.
Proof
By definition of Lie algebra, it suffices to prove two properties:
- $(1): \forall a \in S: a \times a = 0$
- $(2): \forall a, b, c \in S: a \times \paren {b \times c} + b \times \paren {c \times a} + c \times \paren {a \times b} = 0$
Proof of $(1)$
Cross Product of Vector with Itself is Zero
$\Box$
Proof of $(2)$
Vector Cross Product satisfies Jacobi Identity
$\Box$
Both properties hold, and the result follows.
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem