# Vector product between orthogonal terms?

Let $$\mathbf{v}$$ and $$\mathbf{u}$$ be vectors:

$$\mathbf{v}:=x_v \sigma_x+ y_v \sigma_x+ z_v \sigma_v\\ \mathbf{u}:=x_u \sigma_x+ y_u \sigma_x+ z_u \sigma_v$$

What standard notation, if any, gives the following product:

$$\mathbf{v}?\mathbf{u}=x_v(y_u+z_u)+y_v(x_u+z_u)+z_v(x_u+y_u)$$

• I am not sure if there exists a standard notation for this particular product. Something to note though is that it is a symmetric (indefinite) bilinear form: $\beta: V \times V \rightarrow k$ given by $$\beta(v,u) = vAu, ~~~ A := \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} ~~.$$ Aug 26, 2019 at 14:17
• In what context do you consider the product? Aug 26, 2019 at 14:20

This is an ordinary dot product; more precisely, a symmetric bilinear form with signature $$(1,2)$$. Consider a standard basis $$\{e_1,e_2,e_3\}$$ for such a form:

$$e_1\cdot e_1=1,\quad e_2\cdot e_2=-1,\quad e_3\cdot e_3=-1,\\ e_1\cdot e_2=e_1\cdot e_3=e_2\cdot e_3=0$$

The null cone, the set of all vectors squaring to $$0$$, has $$e_1$$ as its axis. Now define a set of vectors $$\{\sigma_k\}$$ equally spaced around the cone:

$$\sigma_1=\frac{\sqrt2}{\sqrt3}\big(e_1+e_2\big),\quad\sigma_2=\frac{\sqrt2}{\sqrt3}\Big(e_1-\frac12e_2+\frac{\sqrt3}{2}e_3\Big),\quad\sigma_3=\frac{\sqrt2}{\sqrt3}\Big(e_1-\frac12e_2-\frac{\sqrt3}{2}e_3\Big)$$

These are linearly independent, so $$\{\sigma_k\}$$ can be used as a basis instead of $$\{e_k\}$$. You can calculate $$\sigma_1\cdot\sigma_1=\sigma_2\cdot\sigma_2=\sigma_3\cdot\sigma_3=0$$ , and $$\sigma_1\cdot\sigma_2=\sigma_1\cdot\sigma_3=\sigma_2\cdot\sigma_3=1$$ . So the dot product of any two vectors is

$$(u_1\sigma_1+u_2\sigma_2+u_3\sigma_3)\cdot(v_1\sigma_1+v_2\sigma_2+v_3\sigma_3)=u_1v_2+u_1v_3+u_2v_1+u_2v_3+u_3v_1+u_3v_2$$

which agrees with your formula, up to differences in notation.