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:

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

  • 2
    $\begingroup$ 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 $$ \beta(v,u) = vAu, ~~~ A := \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} ~~.$$ $\endgroup$ Aug 26, 2019 at 14:17
  • $\begingroup$ In what context do you consider the product? $\endgroup$ Aug 26, 2019 at 14:20

1 Answer 1


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:


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


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


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