The reflection – Geometric Algebra

Reflection of a vector

If we want to reflect the vector $\pmb{a}$ with respect to the plane identified by the normal vector unit $\pmb{\hat{n}}$ , we can break it down into the components $\pmb{a} = a_T + a_N$ and the reflected $\pmb{b}$ vector will keep the tangent component, reversing the normal one:

$\pmb{b} = a_T - a_N = \pmb{a} - 2a_N$

where the normal component to the plane is given by the dot product of $\pmb{a}$ over $\pmb{\hat{n}}$ :

$\pmb{b} = \pmb{a} - 2 (a \cdot n) n$

expressing the scalar product as a semisum of the symmetrical part of the geometric product:

$\begin{equation} \pmb {b} = a - 2 \frac {an + na} {2} n = a - an ^ 2 -nan = -nan$

Reflection of a bivector

Now suppose we reflect a bivector B, formed for example by $a \wedge b$ with respect to a plane identified by the normal $n$ . We do this by reflecting $\pmb{a}$ and $\pmb{b}$ :

$B '= \pmb{a}' \wedge \pmb{b} '= (-nan) \wedge (-nbn)$

remembering the expression $a \wedge b = 1/2 (ab-ba)$ we get:

$B '= \pmb{a}' \wedge \pmb {b} '= (-nan) \wedge (-nbn) = 1/2 (nannbn - nbnnan) =$
$= 1 / 2 n (ab-ba) n = n (\pmb{a} \wedge \pmb {b}) = nBn$

So be careful: the transformation formula is the same, but in the case of a bivector there is no minus sign. This, however, explains the difference between vectors and bivectors in the parity transformation, which consists precisely in specular reflection.

In general, for the reflection of a multivector $A$ of degree $j$ on a multivector $B$ of degree $k$ we will have:

$A '= (-1) ^ {j (k + 1)} BAB ^ {- 1}$

For example, trivectors transform by reflection in the same way as vectors, i.e. they change sign (chirality).