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Reversion in Geometric Algebra is defined by the property $(AB)^\sim=\tilde B\tilde A$ (for any two multivectors $A$ and $B$), along with linearity, and that it leaves scalars and vectors unchanged. Equivalently, reversion is linear and multiplies any $k$-vector by $(-1)^{k(k-1)/2}$.

If $A\tilde A=c$ is a scalar, and $c\neq0$, then $c^{-1}\tilde A$ is the unique inverse of $A$, and it follows that $\tilde AA=c$ as well. (Proof below.)

Since the algebra is finite-dimensional, the powers of $A$ must be linearly dependent. Take a polynomial $p\neq0$ of lowest degree such that $p(A)=0$, and break off the constant term: $p(x)=xq(x)+p(0)$, so we have $Aq(A)=q(A)A=-p(0)$. Multiplying by $c^{-1}\tilde A$ on one side gives $q(A)=-p(0)c^{-1}\tilde A$. If $p(0)=0$ then $q$ contradicts minimality of $p$. So $p(0)\neq0$, and we find that $\tilde A=-p(0)^{-1}cq(A)$ is a polynomial in $A$, so it commutes with $A$. Thus $A\tilde A=\tilde AA$.

So what if $c=0$?

If the algebra is the basic Euclidean one (the dot product is positive-definite), then $A\tilde A=0$ implies $A=0$, since the scalar part $\langle A\tilde A\rangle_0=\lVert A\rVert^2$ is the sum of squares of components of $A$ (with respect to the canonical basis).

If it's a spacetime algebra (the dot product is indefinite), then there are bivectors $B=-\tilde B$ with $B\tilde B=-1$. The multivector $A=1+B$ has $A\tilde A=0$ even though $A\neq0$. But in this case $\tilde AA=0$ as well. Does that always happen?

mr_e_man
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  • Related question: "is every non-invertible multivector a left zero divisor?" I suspect non-invertibility implies left-or-right-zero-divisor-ness, in which case the related question is equivalent to "is every left or right sided zero divisor a two-sided zero divisor?" – Don Hatch Sep 27 '24 at 08:40
  • @DonHatch - Halfway through the above proof (let's forget about $A\tilde A=c$), if $p(0)\neq0$ then $A$ has a two-sided inverse $A^{-1}=-p(0)^{-1}q(A)$, and if $p(0)=0$ then we have both $Aq(A)=q(A)A=0$ but $q(A)\neq0$ (since $q(A)=0$ would contradict minimality of $p$), so $q(A)$ is a single multivector showing that $A$ is a two-sided zero-divisor. – mr_e_man Oct 08 '24 at 16:37

1 Answers1

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No.

Over $\mathbb R^{1,1}$ (that is, with $e_1^2=-e_2^2=1$), the multivector

$$A=1+e_1+e_2+e_1e_2$$

has $A\tilde A=0$, but $\tilde AA=4(e_1+e_2)\neq0$.

However, it feels a little unnatural to mix even and odd grades. So here's an even counter-example, over $\mathbb R^{4,1}$ (the space used for Conformal Geometric Algebra):

$$A=1+e_1e_2e_3(e_4+e_5)+e_4e_5.$$

This has $A\tilde A=0$, but $\tilde AA=4e_1e_2e_3(e_4+e_5)\neq0$.

mr_e_man
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