Dear MathOverFlow: As many of you know, the Lie algebra of the group of $3 \times 3$ orthogonal matrices (with determinant one) is isomorphic to $\Bbb{R}^3$ endowed with the cross-product $a \times b$; furthermore the Killing form in this case is the usual dot-product $a \cdot b$. Linear algebra teaches us that these two operations on $\Bbb{R}^3$ are related by an elegant $2 \times 2$ determinantal identity --- a special case of what is known as a {\it contraction formula} --- namely:

$$ \big( a \times b \big) \cdot \big( c \times d \big) \ = \ \text{det} \, \begin{pmatrix} a \cdot c & a \cdot d \\ b \cdot c & b \cdot d \end{pmatrix} $$

My question is: If $\frak{g}$ is the Lie Algebra (say of a compact group), is there some 'contraction identity' for

$$ \big[ a , b \big] \, \cdot \, \big[ c , d \big] $$

where $a,b,c,d \in \frak{g}$ and where $\cdot$ denotes the Killing form ?

Kind regards for the consideration of this question.

  • 1
    $\begingroup$ A Killing form is not that deadly... Your title somehow misses a capitalized "K" ;) $\endgroup$ – Loïc Teyssier Oct 10 '13 at 16:07
  • $\begingroup$ I was aiming for a more poetical reading of the title ... as in 'identity is killing form' ;) $\endgroup$ – A. Leverkuhn Oct 14 '13 at 8:48

The usual generalisation of the contraction identity for the cross product is the Binet–Cauchy identity involving the dot and the wedge product (a cross-product exists only in dimension $3$ and $7$, but in dimension $7$ it does not satisfy the Jacobi identity): $$ (a\wedge b)\cdot (c\wedge d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c) $$ for vectors $a,b,c,d$. If one searches for contraction identities for Lie algebras then one comes across a "contraction theory". This means that a given Lie algebra is "contracted" to another one of the same dimension, usually more abelian than the original one. This has been studied a lot in mathematics and physics, see for example http://arxiv.org/abs/math-ph/0608018.


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