# Bilinear form and cross product in hyperbolic geometry

I'm reading Patrick J. Ryan's Euclidean and non-Euclidean geometry, page 152. There is a bilinear form defined by $b\left( {x,y} \right) = {x_1}{y_1} + {x_2}{y_2} - {x_3}{y_3}$ on ${\mathbb{R}^3}$ and a norm defined by $\left| v \right| = {\left| {b\left( {v,v} \right)} \right|^{\frac{1}{2}}}$. Vectors $v,w$ are orthogonal if $b\left( {v,w} \right) = 0$. He uses standard definitions for a normal vector and orthonormal set.

Vector is space-like if $b\left( {v,v} \right) > 0$, time-like if $b\left( {v,v} \right) < 0$ and light-like if $b\left( {v,v} \right) = 0$. Let $\left\{ {{e_1},{e_2},{e_3}} \right\}$ be an orthonormal basis. We want to show that two of those are space-like and one is time-like. It's easy to see that at least one is time-like and at least one is space-like, let those be ${e_1},{e_3}$.

Now, $\left( {{e_1} \times {e_3}} \right) \times {e_2} = 0$ so ${e_2}$ is a multiple of ${{e_1} \times {e_3}}$ and obviously $- b\left( {{e_1},{e_1}} \right)b\left( {{e_3},{e_3}} \right) = 1$.

However, it is claimed that $b\left( {{e_1} \times {e_3},{e_1} \times {e_3}} \right) = - b\left( {{e_1},{e_1}} \right)b\left( {{e_3},{e_3}} \right)$.

My question: How do I see that $b\left( {{e_1} \times {e_3},{e_1} \times {e_3}} \right) = - b\left( {{e_1},{e_1}} \right)b\left( {{e_3},{e_3}} \right)$. I've tried direct algebraic manipulation, the statement is equivalent to $- 2{x_2}{y_3}{x_3}{y_2} - 2{x_3}{y_1}{x_1}{y_3} + 2{x_1}{y_2}{x_2}{y_1} = - x_1^2y_1^2 - x_2^2y_2^2 - x_3^2y_3^2$ for $\left| {b\left( {x,x} \right)} \right| = \left| {b\left( {y,y} \right)} \right| = 1$. Is there an easier way to see this holds?

EDIT: I've just realized by skipping to the text ahead that the cross product with respect to the bilinear form $b$ is not the same as the standard cross product on ${\mathbb{R}^3}$. How would, in general, one find a cross product with respect to the given bilinear form?

I've just realized by skipping to the text ahead that the cross product with respect to the bilinear form b is not the same as the standard cross product on R3. How would, in general, one find a cross product with respect to the given bilinear form?

It's not too hard to answer this, provided one doesn't mind a little abstract linear algebra.

Given a right-handed orthonormal basis $(e_1, e_2, e_3)$ of $\mathbb{R}^3$, we can form the dual basis $(\theta^1, \theta^2, \theta^3)$, and then by construction $\epsilon := \theta^1 \wedge \theta^2 \wedge \theta^3$ is a volume form for the bilinear form $b$, and $\epsilon$ is independent of the basis chosen (if we instead use a left-handed basis, we get the negative of that volume form). We can regard $\epsilon$ as a map $E: \mathbb{R}^3 \times \mathbb{R}^3 \to (\mathbb{R}^3)^*$, namely, $$(X, Y) \mapsto \epsilon(X, Y, \,\cdot\,)$$ that is, the map that sends the pair $(X, Y)$ of vectors in $\mathbb{R}^3$ to the linear functional $Z \mapsto \epsilon(X, Y, Z)$. Now, we can regard the bilinear form $b$ as an isomorphism $B: \mathbb{R}^3 \to (\mathbb{R}^3)^*$, namely, $$X \mapsto B(X, \,\cdot\,),$$ that is, the map that sends the vector $X$ to the functional $Z \mapsto B(X, Z)$. Then, the composition $B^{-1} \circ E$ is a map $\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$, and we can show using these definitions that map this is a cross product compatible with the bilinear form $b$.

Note that this construction works both for the usual cross product and for the cross product you mention (sometimes called a "split" cross product).

Remark There's another natural way to view your (split) cross product, by the way. We can identify $\mathbb{R}^3$ with the space of tracefree $2 \times 2$ matrices (which in this context is sometimes called the algebra of "split quaternions"), namely, those of the form $$A = \begin{pmatrix} x & y\\ z & -x\end{pmatrix},$$ and this space has a natural nondegenerate quadratic form, namely $$Q(A) := - \det A,$$ and polarizing this quadratic form gives a bilinear form $b$. This form is compatible with the cross product $$A \times B := -\text{tf}(AB),$$ where $\text{tf}$ denotes the tracefree part; in terms of components, this is $$\begin{pmatrix} x & y\\ z & -x\end{pmatrix} \times \begin{pmatrix} x' & y'\\ z' & -x'\end{pmatrix} = \begin{pmatrix} -yz' + zy' & - xy' + yx'\\ - zx' + xz' & -(-yz' + zy')\end{pmatrix}.$$

Found it on page 137. For given vectors ${u,v}$ and a bilinear form $b$ there exists a unique vector $w$ such that for all $z$ we have

$b\left( {w,z} \right) = \sqrt {\det B} \det \left( {z,u,v} \right)$, where $B$ is a matrix representation of $b$ in any base.

Then we say that $w = u{ \times _b}v$.

For the given bilinear form from the question, we get $u{ \times _b}v = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{ - {\mathbf{k}}} \\ {{u_1}}&{{u_2}}&{{u_3}} \\ {{v_1}}&{{v_2}}&{{v_3}} \end{array}} \right|$.