Compute if a multivector inverse exists in Clifford Algebra

clifford-algebrasgeometric-algebras

Suppose we have a 4 dimension positive signature clifford algebra. In Calculating the inverse of a multivector and Inverse of a general nonfactorizable multivector, the inverse of a multivector is presented as a solution when vectors/bivectors are present

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

but the above is not true for any multivector. For example, how to know if

$(1+e_{1234})^{-1}$

exists and how to compute it?

Best Answer

Naively speaking, the existence of inverses will depend on the signature $(p,q)$ of the quadratic space $\mathbb{R}^{p,q}=(\mathbb{R}^{p+q},g)$, in which for an orthonormal basis $\{e_i\}_{i=1}^{n=p+q}$ and $v=\sum v^ie_i$ we have $$g(v,v)=(v^1)^2+(v^2)^2+\cdots+(v^p)^2-(v^{p+1})^2-\cdots-(v^{p+q})^2.$$

For your example, notice that if $(e_{1234})^2=-1$, then $$(1+e_{1234})\frac{1}{2}(1-e_{1234})=1,$$ which means that $(1+e_{1234})^{-1}=\frac{1}{2}(1-e_{1234})$.

Now, if $(e_{1234})^2=1$, then there is no inverse for $(1+e_{1234})$, which is due to the fact that $x\overline{x}=0$. More specifically, one can derive conditions for which there are inverses for the cases where $p+q=n\leq 5$. The discovery of new faster methods for higher dimensions, which do not depend on the signature is a problem still under development as of today. As an example, we could cite this.

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