@mr_e_man

@NicholasTodoroff - I'm not following your calculation. How does the second displayed line relate to the first? Where is $\sqrt\epsilon$ coming from?

Yes, I've wondered about continuity. It may be possible to find a (rational?) formula for the decomposition into $\leq n$ vectors; of course it wouldn't be unique, and it wouldn't work in all cases, but it may have a limit as the multivector approaches non-invertibility. The classical Cartan-Dieudonne Theorem only says that a decomposition exists; it doesn't give a formula. But maybe we can look at the proof and get something stronger.

I've also been thinking about characterizing products of vectors in a different way. For $P$ to be such a product, it's necessary but not sufficient that $P\tilde P=\tilde PP\in\mathbb R$. It's also necessary that $Pv\tilde P\in V$ for all $v\in V$. Is this sufficient (given non-degeneracy of $V$)?

The second display line is an intermediate calculation. I should reference An algorithm for the Cartan-Dieudonné theorem on generalized scalar product spaces by M.A. Rodríguez-Andrade, G. Aragón-González, J.L. Aragón, Luis Verde-Star, though the idea is simple.

Let $$ V = e_1(e_1+e_2)(e_1+e_3)e_5(e_1+e_5)e_3 $$ be the versor we wish to reduce. Perturb this to $$ V_\epsilon = e_1(e_1+e_2)(e_1+e_3 + \epsilon(e_1-e_3))e_5(e_1+e_5)e_3. $$ Denote the action of a versor $W$ on $X$ by $W[X] = \hat WXW^{-1}$.

We proceed by setting $r_1 = V_\epsilon[e_1] - e_1$ this difference is nonzero and $r_1 = 1$ otherwise, then $V_{\epsilon,1} = r_1V_\epsilon$. By construction $V_{\epsilon,1}[e_1] = e_1$. Then $r_2 = V_{\epsilon,1}[e_2] - e_2$ or $r_2 = 1$ and $V_{\epsilon,2} = r_2V_{\epsilon,1}$. It's not too hard to show that $V_{\epsilon,2}[e_1] = e_1$, and by construction $V_{\epsilon,2}[e_2] = e_2$.

Continuing in this way, the action of $V_{\epsilon,5} = r_5r_4r_3r_2r_1V_\epsilon$ will be the identity, so $r_1r_2r_3r_4r_5 \propto V_\epsilon$. If one of the $r_i$ is null then of course we need to modify this algorithm; that is possible, but does not occur in this specific case. Normalizing $r_1r_2r_3r_4r_5$ properly involves scaling by $\sqrt{|V_\epsilon\widetilde V_\epsilon|} = 4\sqrt\epsilon$, and this factor is of course crucial when taking $\epsilon \to 0$.

In my calculation (which clearly had various sign errors but I got "close enough") I got $r_4 = 1$ and normalized $r_3, r_5$ as $$ r_3 = \frac1{2\sqrt\epsilon}[(1-\epsilon)e_3-(1+\epsilon)e_5],\quad r_5 = \frac1{2\epsilon}[(1-\epsilon^2)e_3-(1+\epsilon^2)e_5] = \frac12[1+\epsilon+(1-\epsilon)e_3e_5]. $$ These are the only two $r_i$ I got that depend on $\epsilon$.

The point I was making was that I didn't see how to calculate $\lim_{\epsilon\to0}\sqrt\epsilon r_3r_5$ without first multiplying out $r_3r_5$, and issues like this seem like a roadblock to using this method in general.

@mr_e_man

There is really good stuff in Lipschitz monoids and Vahlen matrices by Jacques Helmstetter, and I think looking at his other work with be worthwhile too. I have only skimmed this so far and haven't delved into any of the arguments.

He defines the Lipschitz monoid of any Clifford algebra as the monoid generated by scalars, vectors, and elements of the form $1+xy$ where $x, y$ are vectors. Except in the the case of the exterior algebra or some exceptional cases over the field $\mathbb F_2$, the Lipschitz monoid is the monoid generated by all vectors.

He has an invariance theorem: let $\beta$ be a bilinear form and $\mathrm{Lip}(\beta)$ the Lipschitz monoid of $x \mapsto \beta(x,x)$. Given $\beta$, we can canonically identify ${\bigwedge}V = \mathrm{Cl}(0)$ with $\mathrm{Cl}(\beta)$. In doing so, the theorem states that $\mathrm{Lip}(\beta) = \mathrm{Lip}(0)$ as sets.

This is Section 4.

In Section 6, Theorem 38 states the following: Consider the algebra $A = \mathrm{Cl}(q)\mathbin{\hat\otimes}\mathrm{Cl}(-q)$, i.e. the graded tensor product algebra. An element $a \in \mathrm{Cl}(q)$ is in the Lipschitz group iff $a\otimes\widetilde a$ is a scalar in $A$, i.e. is a multiple of $1\otimes 1$. Note that $A \cong \mathrm{Cl}(q\oplus(-q))$ with the direct sum done orthogonally.

I just noticed, in my last display equation above you can ignore the last $=$ and everything after it, I just messed up when I copy-and-pasted.

"...is in the Lipschitz group..." should say Lipschitz monoid.

Theorem 39 bounds arbitrary products of vectors to $2n$ factors! Specifically: let $$\mathrm{RKer}(a) = \{x \in V \;:\; ax = 0\}$$ be the right kernel of $a$ in the Lipschitz monoid. This is a totally isotropic subspace of $V$ since $x^2 \ne 0$ would imply $a = 0$. Then for any basis $x_1,\dotsc,x_r$ of $\mathrm{RKer}(a)$ there is a $g$ in the Lipschitz group such that $a = gx_1\dotsb x_r$. We can make a similar statement about the left kernel $\mathrm{LKer}(a)$.

The argument looks good. The core idea is to consider a product $g_0x_1g_1x_2g_2\dotsb x_rg_r$ where $g_i$ is in the Lipschitz group and $x_i$ are isotropic, and then define its "complexity" by $(r, s)$ where $s$ is the number of $g_i \ne 1$. These complexities are totally ordered lexicographically, and the argument proceeds by assuming we have a minimal complexity.

We then use the "moving" trick $g_ix_{i+1}g_{i+1} = g_ig_{i+1}(g_{i+1}^{-1}xg_{i+1})$ to show that minimality implies all $g_i = 1$ for $i > 0$. Then $x_ix_{i+1} = (x_i + x_{i+1})x_{i+1}$ so if they are non-orthogonal then $x_i + x_{i+1}$ is invertible, contradicting minimality.