# Lagrange's Identity in $\Bbb L^3$.

I'm working in $\Bbb L^3 = (\Bbb R^3, \langle \cdot, \cdot \rangle_L)$, with $\langle {\bf x},{\bf y}\rangle_L = x_1y_1+x_2y_2-x_3y_3$. Let $\|\cdot\|_L = \sqrt{|\langle \cdot, \cdot \rangle_L|}$ be the "norm" and $\times_L$ be the vector product (it is the euclidean product reflected on the plane $z = 0$). I'm studying Lagrange's Identity: $$\|{\bf x}\times_L {\bf y}\|_L^2 = \|{\bf x}\|_L^2\|{\bf y}\|_L^2 - \langle {\bf x},{\bf y}\rangle_L^2.$$

I have proved that if ${\bf x},{\bf y} \in \Bbb L^3$ span a spacelike plane, then it is true. It is also true if they are lightlike and parallel. If ${\bf x}$ is spacelike, ${\bf y}$ is lightlike and ${\bf x}\times_L{\bf y}$ is also lightlike, then the identity is still true. There is only one case that I'm struggling to kill here: when both ${\bf x}$ and ${\bf y}$ are spacelike, with ${\bf x}\times_L{\bf y}$ lightlike. In every other case is false.

I don't know how exactly to go at it, and I'm having trouble finding a counter-example. If we had the situation $\langle {\bf x},{\bf y}\rangle_L = 0$ possible, then it would be easy, but this can't happen, otherwise the plane spanned would be spacelike and not lightlike. Can someone give me a hand? Thanks.

Build a clifford algebra from this space. Let $\epsilon = e_1 e_2 e_3$ with $e_3 e_3 = -1$ and $e_1 e_1 = e_2 e_2 = +1$. Then see that multiplication with $\epsilon$ produces the following:

$$\epsilon e_1 e_2 = -e_3, \quad \epsilon e_2 e_3 = e_1, \quad \epsilon e_3 e_1 = e_2$$

This allows us to write the Lorentzian cross product as

$$a \times b = \epsilon (a \wedge b)$$

Now consider the product

$$\langle abba \rangle_0 = (a \cdot b)^2 - (a \wedge b)^2 = a^2 b^2$$

See that $\epsilon^2 = +1$, so we can substitute $a \wedge b = \epsilon(a \times b)$ and write this as

$$(a \cdot b)^2 - (a \times b)^2 = a^2 b^2$$

This identity is always true; note the sign difference from the identity you're considering. This is the correct Lorentzian analogue of the Lagrange identity, as the sign on $(a \times b)^2$ depends on the sign of $\epsilon^2$--in Euclidean 3d space, $\epsilon^2 = -1$ and you get the "usual" Lagrange identity.

Edit: retract this section. While I admit I have not directly answered your question, it's my hope that I've illustrated why the usual Lagrange identity doesn't hold, but that it does have an analogue in Lorentzian space that does hold.

Indeed, in the case that $a\times b$ is lightlike, then this Lorentzian analogue suggests

$$(a \cdot b)^2 - (a \times b)^2 = (a \cdot b)^2 - 0 = a^2 b^2$$

And we can see that this is true for your example, as well.

• Hi, thanks for taking the time to answer. I'm not familiar at all with clifford algebras.. Can you point me some reference for it? I would like to understand completely what you wrote. About your last remark, the situation is possible: take for example ${\bf x} = (1,1,1)$ and ${\bf y} = (1,2,1)$. The cross product is $(-1,0,-1)$, which is lightlike. – Ivo Terek Feb 9 '15 at 0:17
• I would suggest some work by David Hestenes, or perhaps by Chris Doran and Anthony Lasenby. Alan Macdonald also has a good, undergraduate level pair of books for the algebra and its associated calculus, making frequent contact with traditional matrix notations or with vector calculus as appropriate. - The bottom line is that clifford algebra introduces an associative, but not necessarily commutative, product of vectors. For a given vector $a$, $aa = a \cdot a$, and if $u,v$ are orthogonal, then $uv=-vu$. All other properties follow. – Muphrid Feb 9 '15 at 4:10
• I'll look into it. I'll leave the question open for two more days, just in case someone comes up with a solution for the specific case, then I'll accept the answer. You've made your point about the identity having an analogue, it was helpful. Thanks again. :) – Ivo Terek Feb 9 '15 at 9:17
• I also added a section on using the analogous identity to help prove the case you were interested in. – Muphrid Feb 9 '15 at 15:31