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The reverse of a multivector $M$ will be denoted by ${\tilde M}$. The scalar part of $M$ is indicated by $\langle M \rangle$.

Let $A$ be a multivector in a geometric algebra ${\mathbb G}^{p,q}$ generated by a real scalar product space ${\mathbb R}^{p,q}$. I have several examples where ${\tilde A}A$ and $A{\tilde A}$ are equal nonzero scalars. The most important are when $A$ is a versor, i.e., is a geometric product $A = a_k \cdots a_1$ of multiple nonnull vectors $a_1,\dots,a_k$. However there do exist examples where $A$ is not a versor.

Note that if ${\tilde A}A$ and $A{\tilde A}$ are both scalars, they must be equal, for ${\tilde A}A = \langle {\tilde A}A \rangle = \langle A{\tilde A} \rangle = A{\tilde A}.$

My question is this: Will ${\tilde A}A$ be a scalar whenever $A{\tilde A}$ is a nonzero scalar? If the answer is “no”, please provide a counterexample.

That $A{\tilde A}$ be a nonzero scalar is necessary. Otherwise $A = 1 + e_1 + e_2 + e_1 e_2$, where $\{e_1, e_2\}$ is the standard orthonormal basis for ${\mathbb R}^{1,1}$, provides a counterexample; for then $A{\tilde A} = 0$ but ${\tilde A}A = 4 e_1 + 4 e_2$.

My question is asked in the context of determining when a multivector $A$ must be a versor. I have necessary and sufficient conditions for such, but I am trying to determine whether ${\tilde A}A$ a scalar needs to be part of the conditions if one already knows that $A{\tilde A}$ is a nonzero scalar.

(By the way, neither ChatGPT nor Copilot seem to be able to answer the question. They make mistake after mistake, acknowledging an error when one is explained to them, but then repeatedly make the same mistake in their revised answers. Those programs have a long way to go before much credence should be put into their answers about mathematics.)

Greg Grunberg
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  • $A\widetilde A$ being a nonzero scalar implies that $A$ is invertible, as proven here by mr_e_man. Thus it implies that $\widetilde AA$ is the same scalar. More generally, if $AB$ is a nonzero scalar then $A,B$ are invertible and $BA$ is the same scalar. – Nicholas Todoroff Mar 08 '25 at 18:19

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Given that $A{\tilde A}$ is a nonzero scalar, the product ${\tilde A}A$ will not only be a scalar, but will necessarily equal $A{\tilde A}$.

Proof: $A$ is an element in a finite-dimensional linear space ${\mathbb G}^{p,q}$, so there is some smallest natural number $k$ such that the $k+1$ elements $1=A^0, A=A^1, A^2, \dots, A^k$ are linearly dependent. That is to say, there exist scalars $c_0, c_1, \dots, c_k$, not all zero and $c_k \neq 0$ in particular, such that $$c_0 + c_1 A + c_2 A^2 + \ldots + c_k A^k = 0.$$ Right multiply both sides of this equation by $\displaystyle \frac {\tilde A}{A{\tilde A}}$, which is a right inverse of $A$, to obtain $$\frac {c_0}{A{\tilde A}} {\tilde A} + c_1 + c_2 A + \ldots + c_k A^{k-1} =0.$$ From this we see that $c_0 \ne 0$, for otherwise we would have a relationship of linear dependence among the $k$ elements $1, A, A^2, \dots, A^{k-1}$, in violation of our choice of $k$.

Divide both sides of the last displayed equation by the nonzero scalar $\displaystyle \frac{c_0}{A{\tilde A}}$, and then move all terms except the first from the left side to the right. The result, $${\tilde A} = b_1 + b_2 A + \ldots + b_k A^{k-1}$$ where each $b_i = \displaystyle -\frac{c_i}{c_0}A{\tilde A}$ is a scalar, shows that ${\tilde A}$ is a polynomial in $A$. As such, it commutes with A: $${\tilde A}A = A{\tilde A}.$$

Greg Grunberg
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