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In geometric (Clifford) algebra, both $k$-vectors and inhomogeneous multivectors may have inverses, which are unique (if they exist). I want to prove the following statement.

Let $A = ⟨A⟩_k$ be a $k$-vector. Is its inverse $A^{-1} = ⟨A^{-1}⟩_k$ necessarily a $k$-vector?

We can prove that $A^{-1}$ is of grade $k$ if we assume it is homogeneous. For example, assuming $A^{-1} = ⟨A^{-1}⟩_m$, then from $$ AA^{-1} = ⟨A⟩_k ⟨A^{-1}⟩_m = ⟨⟨A⟩_k ⟨A^{-1}⟩_m⟩_{|k-m|} + ⟨⟨A⟩_k ⟨A^{-1}⟩_m⟩_{|k-m| + 2} + \cdots + ⟨⟨A⟩_k ⟨A^{-1}⟩_m⟩_{k + m} = 1 $$ we must have that the lowest-grade part $⟨⟨A⟩_k ⟨A^{-1}⟩_m⟩_{|k-m|} = 1$ is one, requiring $k = m$. Thus $A^{-1} = ⟨A^{-1}⟩_k$ is also a $k$-vector.

In an effort to restrict the possible grades of $A^{-1}$ given that $A = ⟨A⟩_k$ is a $k$-vector, I have made the following observations (without assuming $A^{-1}$ is homogeneous):

  1. If $k$ is even/odd, then so is $A^{-1}$. This follows because the even-subalgebra is closed, and hence for $AA^{-1} = 1$ we have either $(\text{even})(\text{even}) = (\text{even})$ or $(\text{odd})(\text{odd}) = (\text{even})$.

  2. If $\tilde{A} = (-1)^{k(k - 1)/2} A$ is the reverse of $A$, then $\tilde{A^{-1}} = (-1)^{k(k - 1)/2}A^{-1}$ follows from $\widetilde{AA^{-1}} = \tilde{A^{-1}}\tilde{A} = 1$. This implies that $A^{-1}$ may only contain parts of grade $\{ m \in \mathbb{Z} \mid (-1)^{k(k - 1)/2} = (-1)^{m(m - 1)/2}\}$.

  3. Combining the two points above, we conclude that if $A = ⟨A⟩_k$ then $A^{-1}$ may only contain parts of grade $k + 4\mathbb{Z}$.

  4. In view of the last point, and the fact that (pseudo)scalars and (pseudo)vectors always square to a scalar (and hence have an inverse of the same grade), we can conclude that $A^{-1}$ must be a $k$-vector in algebras of dimension $\le 7$.

But in $8$ dimensions, I can't eliminate the possibility that a $2$-vector has an inverse with parts of grade $6$, and vice versa.

Is there a more direct way to show this, or is there a counter example?

Jollywatt
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  • Warning: My logic in points 3. and 4. about it holding true for dimensions ≤ 7 is wrong… see the answer! – Jollywatt Jan 03 '24 at 06:06

1 Answers1

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The inverse of a $k$-vector is not necessarily a $k$-vector !

To my astonishment, I have found a counterexample.

Consider the $6$ dimensional bivector $$ a ≔ _{12} + _{34} + _{56} $$ whose dual (with $I ≔ _{123456}$) is $$ Ia = aI = -(_{3456} + _{1256} + _{1234}) $$ and whose square is $$ a^2 = -3 + 2(_{3456} + _{1256} + _{1234}) = -3 - 2aI $$ and whose inverse is $$ b ≔ -\frac{a + 2I}{3} $$ as verified by $$ ab = - \frac{a^2 + 2aI}{3} = \frac{(3 + 2aI) - 2aI}{3} = 1 $$ $\square$

Jollywatt
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  • $aI \ne Ia$, in even dimensions it is $aI = -Ia$. I'm not sure if that matters for your counterexample or not, I'll have to study it more carefully later. Something I'm also not sure of is whether or not right inverse $\implies$ left inverse since $\mathrm{Cl}_6(\mathbb R) \cong M_3(\mathbb H)$. Can you pinpoint anywhere where my proof of the converse is incorrect? It seems too simple to be incorrect. – Nicholas Todoroff Jan 03 '24 at 06:25
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    Sorry, the very first thing I said was wrong; $a$ and $I$ commute because $a$ is a bivector, not a vector. – Nicholas Todoroff Jan 03 '24 at 06:54
  • It’s definitely the case that $AA^{-1} = 1 ⟺ A^{-1}A = 1$ in geometric algebra… but I can’t prove it other than by saying “GAs are like a matrix algebras, and they have two-sided inverses, thus so do GAs”.

    I.e., any GA over a $d$-dimensional vector space over $ℝ$ has a linear representation in $2^d × 2^d$ matrices, and we know $AA^{-1} = I ⟺ A^{-1}A = I$ for matrices over $ℝ$…

     – Jollywatt Jan 03 '24 at 09:26
  • Yeah, I think this counterexample is correct. – Nicholas Todoroff Jan 03 '24 at 14:55
  • The proof contained here should apply to a one-sided inverse of $A$ and shows that it is necessarily a two-sided inverse; the only requirement seems to be that our algebra is finite dimensional. – Nicholas Todoroff Jan 03 '24 at 15:03
  • Just for fun: we can find that $$b = \frac{-1}9a^3 -\frac{10}9a$$ and hence $$0 = a^4 + 10a^2 + 9.$$ – Nicholas Todoroff Jan 03 '24 at 15:47