4

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?

paco gil
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2 Answers2

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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.

amnesiac
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1

This paper by D. S. Shirokov (2021) claims to give a basis-free formula using only basic operations and various involutions on multivectors. Preprint.

Nicholas Todoroff
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  • (I haven't read this in detail.) In Lemmas 5 and 6, they divide by $3$ to get formulas for the inverse. I wonder if anything really goes wrong when the field has characteristic $3$. Consider this for comparison: If $B$ is a bivector, the expression $B\wedge B\wedge B/6$ seems to be definable even in characteristic $3$. – mr_e_man Apr 11 '23 at 03:46
  • @mr_e_man I don't think the $1/3$ is all that important, the real question is whether or not $N(U) = 0 \iff U = 0$ (or some appropriate "scaling" of $N(U)$ that is definable) when you change the characteristic. For your $B\wedge B\wedge B/6$ idea though, perhaps related is the definition of exterior exponential for arbitrary modules given in section 4.5 of Quadratic Mappings and Clifford Algebras by Helmstetter and Micali. – Nicholas Todoroff Apr 14 '23 at 17:14
  • To be a little clearer about my last comment, we would define $${"}B\wedge B\wedge B/6,{"} = \langle\exp B\rangle_6$$ or more generally $${"}B^{\wedge k}!/k!,{"} = \langle\exp B\rangle_{2k}.$$ – Nicholas Todoroff Apr 15 '23 at 00:48