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In a geometric algebra, is every non-invertible element a left zero divisor, with respect to the geometric product?

Background:

This answer apparently gives (for dimensions <= 5) an explicit formula for:

  • the (necessarily two-sided) inverse of given element M, if it's invertible
  • if M is non-invertible, a right zero divisor showing that M is a left zero divisor (proving M is non-invertible, since it's a basic fact, in any ring, that a zero divisor is non-invertible).

But closer inspection reveals that it doesn't really do that, in the non-invertible case-- in particular, for dimension 4, sometimes the given formula shows that M is a right zero divisor (thus still proving that M is non-invertible), but without any evidence that M is a left zero divisor. See the commments on that answer for more details.

This leaves me wondering: given a non-invertible multivector M, is it always possible to show that M is a left zero divisor at all?

More generally, which of the following statements are true in every geometric algebra? And if they are not all true in all geometric algebras, what are the implications among them?

  • (A) Every non-invertible element is a left or right zero divisor (true up to dimension 5, by the formulas given in that answer)
  • (B) Every non-invertible element is a left zero divisor (equivalently: every non-invertible element is a right zero divisor)
  • (C) Every left or right zero divisor is a two-sided zero divisor (where two-sided zero divisor is defined in wikipedia here-- note that the nonzero $x$ such that $a x = 0$ may be different from the nonzero $y$ such that $y a = 0$)
  • (D) Every non-invertible element is a two-sided zero divisor

As an example where it's not clear whether a right zero divisor $M$ is a left zero divisor, here is a right zero divisor $M$ in GA(4) along with its clifford conjugate ${\overline M}$ which happens to be the corresponding left zero divisor: $$ M = 1 + e1 - e2 - e3 - e4 - e12 - e13 - e23 - e14 - e24 - e34 - e123 - e124 - e134 - e234 - e1234 \\ {\overline M} = 1 - e1 + e2 + e3 + e4 + e12 + e13 + e23 + e14 + e24 + e34 - e123 - e124 - e134 - e234 - e1234 \\ {\overline M} M = 0 \\ M {\overline M} = -8*e234 - 8*e1234 \neq 0 $$ So this shows $M$ is a right zero divisor, but I don't know whether it's a left zero divisor at all. I've tried various products of various kinds of conjugations of $M$, but haven't found a right zero-complement for it.

I have been unable to find any real references or evidence one way or the other.

Don Hatch
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  • If you're only interested in real, nondegenerate Clifford algebras then you'll probably want to use the fact that every such Clifford algebra is either a matrix algebra over $\mathbb R,\mathbb C,\mathbb H$ (in even dimension) or is a direct sum of two copies of one such algebra (in odd dimension). See https://en.m.wikipedia.org/wiki/Classification_of_Clifford_algebras – Nicholas Todoroff Sep 27 '24 at 17:55
  • I may have made a mess of this. My motivation is that I want to write a procedure to compute the inverse of an element, in an experimental geometric algebra library I'm writing. The linked answer gives a nice way to do it, in dimensions <= 5, as long as the element is invertible. And, at first glance, it also looks like it gives a nice way to calculate a right zero-complement, in the case that the element is not invertible, which would be a simple certificate of non-invertibility. So I thought I'd try to write a procedure that always returns one of those two things. Not so simple, it seems. – Don Hatch Sep 27 '24 at 20:17

2 Answers2

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If "in geometric algebra" just means you are working with the Clifford algebra for a finite dimensional vector space, then such an algebra is finite dimensional and hence left and right Artinian. This does not depend on the field being used or any degeneracy of the bilinear form, but only the fact the generating vector space is finite dimensional.

It is known that in an Artinian ring, every element is either a unit or a zero divisor. (It is also known that one-sided invertible elements are two-sided invertible in an Artinian ring, if you are worried about that. One-sided invertibility or zero divisibility is two-sided in an Artinian ring.)

rschwieb
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(A) For any finite dimensional algebra, $x$ is (right) invertible iff (left) multiplication by $x$ is bijective, iff it is injective. So if $x$ is not (right) invertible then (left) multiplication is not injective, so $xy = xz$ and thus $x(y-z) = 0$ for some $y\ne z$.

(B) In Clifford algebras, an element is left invertible iff it is right invertible (as rchwieb says). By the above argument, every non-invertible element has a left zero complement and a right zero complement.

(C) This is false. Let $v$ be a null vector and let $w$ be a vector such that $w\cdot v \ne 0$. Then $v(vw) = 0$ but $$(vw)v = v(2v\cdot w - vw) = 2(v\cdot w)v \ne 0.$$

I am unsure about (D).

Nicholas Todoroff
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  • I'm having trouble following your argument against (C). First of all I don't follow why $(v w) v = v(2 v \cdot w - v w)$, but I'm not sure it matters. It looks to me like you're giving an example in which there's a right-zero-complement of v (namely vw) which isn't a left-zero-complement of v (that's easy), but that doesn't prove v doesn't have some other left-zero-complement, right? – Don Hatch Sep 27 '24 at 23:48
  • For (B) you say a non-invertible element "has a left zero divisor and a right zero divisor"... that's not meaningful wording, is it? I think you must mean "has a right zero-complement and a left zero-complement)" or equivalently "is a left zero divisor and a right zero divisor"? – Don Hatch Sep 28 '24 at 00:01
  • Also I'm having trouble making sense of your argument for (A). It looks like you're saying something of the form "for any element in any ring, [I] implies [II]. so if not [I] then not [II]" which isn't valid reasoning (it looks to me like it's's the logical fallacy called "denying the antecedent"?) – Don Hatch Sep 28 '24 at 00:07
  • @DonHatch Addressing (C): $v$ is a left zero complement of $vw$: $v(vw) = v^2w = 0$. But it is not a right zero complement of $vw$. The geometric product is associative $(vw)v = v(wv)$ and the identity $2w\cdot v = wv + vw$ is the defining identity of a Clifford algebra. – Nicholas Todoroff Sep 28 '24 at 00:11
  • Addressing (B): whatever you want to call it. If $AB = 0$ then $A$ is a left divisor of $0$ and $B$ is a right divisor of $0$, so that is where my language came from. – Nicholas Todoroff Sep 28 '24 at 00:13
  • Addressing (C): I should perhaps restrict myself to finite dimensional vector space algebras rather than rings; in this case injectivity of endomorphisms is equivalent to bijectivity, so not right invertible $\iff$ left multiplication is not bijective $\iff$ left multiplication is not injective. – Nicholas Todoroff Sep 28 '24 at 00:17
  • Thanks for explaining where $(vw)v = v(2 v \cdot w - v w)$ came from (and you might want to consider putting that brief explanation in your answer; there may be readers other than me for whom it's not obvious, I don't know). In any case, again, yes we agree that you've shown a case where $a b = 0$ but $b a \neq 0$. But that's not a counterexample to (C). – Don Hatch Sep 28 '24 at 00:25
  • @DonHatch Then explain what you mean by (C) in notation, because it very much is a counterexample by my interpretation. Also, I just noticed that you literally used the terminology "has a left/right zero divisor" in your question, so why is it not ok if I'm the one to use it? – Nicholas Todoroff Sep 28 '24 at 00:29
  • Let us continue this discussion in chat. – Don Hatch Sep 28 '24 at 00:29