State and prove necessary and sufficient conditions for a versor to be a blade.

Question

Any non-null blade in ${\mathbb G}^{p,q}$ can be expressed as the geometric product of orthogonal non-null vectors, so all blades are versors. But not all versors are blades. A simple example is $A = (e_1 - e_2)e_1 = 1 + e_1 e_2 = 1 + e_1 \wedge e_2$, where $\{e_1, e_2\}$ is a generating orthonormal basis for the algebra ${\mathbb G}^{2,0}$. $A$ is a versor but not a blade. Notice that $A$'s basis blade expansion has nonzero terms which differ in their grades: $1$ has grade zero, while $e_1 \wedge e_2$ has grade two.

I have a test for when a multivector is a versor, and I want to refine it to determine when a versor is a blade. On p. 533 of Geometric Algebra for Computer Science (GACS), an (incorrect) test is given for when a multivector will be a versor. GACS makes the further claim that a versor is a blade if and only if all terms in its expansion have the same grade, but it offers no proof of that claim.

Playing around with a few examples suggests that whether a versor $A = a_1 \cdots a_k$ will in addition be a blade has to do with whether the vector factors $a_1, \dots, a_k$ are orthogonal. Certainly orthogonality of all the vector factors implies that $A$ is a blade. But I don't see a general argument for non-orthogonal factors.

So my problem is to:

State and prove necessary and sufficient conditions for a versor to be a blade.

Answer 1

$ \newcommand\lcontr{\mathbin\rfloor} \newcommand\proj[1]{\langle#1\rangle} \newcommand\Lip{\mathrm{Lip}} \newcommand\maxgrade{\mathop{\mathrm{maxgrade}}} $

GACS makes the further claim that a versor is a blade if and only if all terms in its expansion have the same grade, but it offers no proof of that claim.

It is interesting to note that this is equivalent to the following: if $A$ is versor then $A^2$ is a scalar iff $A$ is a blade. The reverse direction is an obvious consequence of $A$ being a blade, and the forward direction follows from the above since in this case there is a scalar $\alpha=\pm1$ such that $$ \widetilde A = \alpha A \implies \alpha A\widetilde A = A^2. $$

However, we proceed with a proof of the theorem in the form GACS stated:

We proceed by induction on grade $k$. This is obvious when $k=0,1$, so suppose it holds true for $k$ and that $A$ is a versor of grade $k+1$. We can write $A = vA'$ with $v$ an invertible vector and $A'$ a $k$-versor. Then $$ A' = v^{-1}A = v^{-1}\wedge A + v^{-1}\lcontr A = v^{-1}\lcontr A $$ since $\proj{A'}_i = 0$ for all $i > k$. Thus $A'$ has grade $k$, and by the inductive hypothesis is a $k$-blade. But then $$ A = \proj A_{k+1} = \proj{vA'}_{k+1} = v\wedge A' $$ is a blade.

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Though we could easily replace $v^{-1}$ with just $v$ in the proof above, it still fails for arbitrary products of vectors since if $v^2=0$ then $A' \ne vA = 0$.

Next I give a proof that, as written, only works for versors in a space with definite signature, but is an interesting proof to comment on.

Let $A$ be a versor of grade $k$. Let $A = v_1\dotsb v_d$ (which has $\mathrm{span}\{v_1,\dotsc,v_d\}$ nondegenerate by definiteness). If $d < k$ then $A = 0$, which is impossible (and also a blade anyway). If $d = k$ then we're done since $$ A = \proj A_k = v_1\wedge\dotsb\wedge v_k. $$ Now suppose $d > k$. We have $$ 0 = \proj A_d = v_1\wedge\dotsb\wedge v_d $$ so $v_1,\dotsc,v_d$ are linearly dependent and $$ \dim(\mathrm{span}\{v_1,\dotsc,v_d\}) < d $$ By the Cartan-Dieudonne Theorem, $d$ is at most this dimension; but then $d < d$, a contradiction.

This last step is where the nondegeneracy is used, as Cartan-Dieudonne does not apply otherwise.

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Besides the application of Cartan-Dieudonne to get a bound on the number of vector factors in a versor, the above really uses nothing but the fact that $A$ is a product of vectors regardless of whether or not they are invertible.

As such, this connects to a problem mr_e_man posed: In a Clifford algebra over an n -dimensional non-degenerate space, is a product of any number of vectors a product of at most n vectors?. If this is true, then the above proof applies to arbitrary products of vectors where the vectors span a nondegenerate subspace, not just versors. Call this a nondegenerate Lipschitzian element. Hence, we can also conclude that mr_e_man's theorem is true $\implies$ the square of a nondegenerate Lipschitzian element being a scalar implies it is a blade. To be more clear:

• Let $\Lip(p,q)$ be the monoid generated by products of vectors and scalars in $\mathbb G^{p,q}$. There is a submonoid $\Lip_*(p,q)$ of nondegenerate Lipschitzian elements. Let $\deg A$ for $A \in \Lip(p,q)$ be the smallest $d$ such that $A$ is a product of $d$ vectors. Then the claim is $$ (\forall p,q.\forall A \in \Lip(p,q).\: \deg A \leq p+q) \implies (\forall p,q.\forall A\in\Lip_*(p,q).\: A^2\text{ is a scalar} \iff A\text{ is a blade}). $$

We do actually have a general bound of $2(p+q+r)$ even in the degenerate case $\mathbb G^{p+q+r}$. See Lipschitz monoids and Vahlen matrices (2005) by Jacques Helmstetter, Theorem 39. This is not good enough though because in the above proof we just get $d < 2d$.

Just to demonstrate that talking about this isn't completely vapid, an example of a nontrivial nondegenerate Lipschitzian element is $$ (e_1 + e_2)e_1 $$ where $e_1,e_2$ are orthogonal vectors with $e_1^2 = -e_2^2 = 1$. We could even have a product made entirely of null vectors, $$ (e_1 - e_3)(e_1 + e_2)(e_1 + e_3) $$ where $e_1^2 = -e_2^2 = -e_3^2 = 1$.