Calculating the inverse of a multivector

Question

Given a multivector, what is the easiest way to compute its inverse? To take a concrete example, consider a bivector $ B = e_1(e_2 + e_3) $. To compute $ B^{-1} $, I can use the dual of $ B $: $$ B = e_1e_2e_3e_3 + e_1e_2e_2e_3 = I(e_3-e_2) = Ib $$ $$ BB^{-1} = 1 = Ib B^{-1} $$ $$ B^{-1} = -b^{-1}I = -\frac{b}{b^2}I$$ But this won't work for a bivector in 4-dimensions for example. Is there a more general/easier way?

Answer 1

Not sure where you got the idea that inverses should involve duality. Usually this is done merely through reversion. Let $B^\dagger$ denote the reverse of $B$. Then the inverse is

$$B^{-1} = \frac{B^\dagger}{B B^\dagger}$$

For a bivector, $B^\dagger = -B$. I believe this works for any object that can be written as a geometric product of vectors (i.e. that can be factored; which is why it works for rotors and spinors), but don't quote me on that. Of course in mixed signature spaces, anything that has a null factor is not invertible.

Answer 2

With respect to the geometric product, if B is a non-null versor (could be a mixed signature GA), the inverse of B will be:

$$B^{-1} = \frac{B^\dagger}{B B^\dagger}$$

For a null-versor, the inverse with respect to the geometric product does not exist.

Answer 3

If your multivector is a product of vectors ($x = x_1x_2\cdots{}x_n$) then the invers is given by

\begin{align} x^{-1} & = \left(x_1x_2\cdots{}x_n\right)^{-1}\\ & = x_n^{-1}\cdots{}x_1^{-1}. \end{align}

Since none null vectors are always invertible ($a^{-1} = \frac{a}{\Vert{}a\Vert^2}$) you can compute your inverse from that.