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I am studying from the book "Geometric Algebra for Physicists: by Chris Doran and Anthony Lasenby." In the book they define a map $\dagger : \mathcal G \to \mathcal G$, where $\mathcal G$ is a geometric algebra, by $(a_1\ldots a_r)^\dagger=a_r\ldots a_1$ where $a_1,\ldots,a_r$ are vectors. My question is if $$a_1\ldots a_r=b_1 \ldots b_s$$ where $r$ is not necessarily equal to $s$ how does one know that $(a_1\ldots a_r)^\dagger=(b_1\ldots b_s)^\dagger$?

Jason Rodriques
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  • It's not difficult to prove that these these multivectors each have inverses constructed from their reverses: if $A = a_1 \ldots A_r$, then $A^{-1} = A^\dagger/(A A^\dagger)$. The inverses must be identical: $A^{-1} = B^{-1}$. Proving the reverses are identical is harder; that this point, all I've been able to show is that the reverses must be scalar multiples of each other. – Muphrid Feb 14 '14 at 06:35
  • It's trivial in the standard basis. The reversion just changes the sign of certain grades. It doesn't matter that you can write the same multivector in two different ways via the geometric product of some vectors. The components in the standard basis will be identical and therefore the reversions will be identical. – Andrey Sokolov Feb 18 '14 at 05:22
  • I guess you don't even need the standard basis. The formula for reversion in terms of the grades should be sufficient to convince you that it is well-defined. – Andrey Sokolov Feb 18 '14 at 05:59
  • To Andrey Sokolov: – Jason Rodriques Feb 20 '14 at 03:43
  • To Andrey Sokolov:

    Given an $r$-blade I understand that reversion changes the sign depending $r$. But as reversion is defined in "Geometric Algebra for Physicists" where the product of vectors considered may not necessarily anti-commute, it appears that the issue of well-defined is skirted when defining reversion?

     – Jason Rodriques Feb 20 '14 at 03:51
  • I am not sure that $A^{-1} = A^\dagger/(A A^\dagger)$ holds in all cases. I can give a counterexample in the quaternion algebra$\mathcal{Cl}_{0,2,0}$. – user48672 Feb 01 '15 at 11:53

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Such a map exists and is unique by the universal property of Clifford algebras : let $\mathcal{C}=\mathcal{C}(V,Q)$ be the Clifford algebra on $V$ where $Q$ is a quadratic form on the $\Bbbk$-vector space $V$. By definition, $\mathcal{C}$ is generated as a $\Bbbk$-algebra by the vectors of $V$ and the relations $vv+Q(v)=0$ (conventions differ). It is universal in that for any $\Bbbk$-algebra $\mathcal A$ and any map of $\Bbbk$-vector spaces $\phi:V\to\mathcal{A}$ such that for all $v\in V$, $\phi(v)^2+Q(v)=0$, there exits a unique extension of $\tilde\phi$ to $\mathcal{C}\to\mathcal{A}$ that is a morphism of $\Bbbk$-algebras: $$\begin{matrix} V&\xrightarrow{\phi}&\mathcal A\\ \downarrow&\nearrow_{\tilde\phi}\\ \mathcal C \end{matrix}$$ You then simply apply the universal property to the $\Bbbk$-algebra $\mathcal A=\mathcal C^{\text{op}}$ and the canonical inclusion $V\to\mathcal A=\mathcal C$ (the equality is that of vector spaces). By definition, this map reverses the order of multiplication, that is $$\tilde\phi(e_1\cdots e_n)=\phi(e_1)\times^{\text{op}}\cdots\times^{\text{op}}\phi(e_n)=e_1\times^{\text{op}}\cdots\times^{\text{op}}e_n=e_n\times\cdots\times e_1$$ (multiplication in the two central terms is that of $\mathcal C^{\text{op}}$, and the right hand side is the translation in terms of products in $\mathcal C$.)

Olivier Bégassat
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  • This is precisely the construction I used to formalize the reversion operator in the Lean theorem prover, as clifford_algebra.reverse (although I didn't find this post until after). Note that to arrive back at an endomorphism of k-modules $\mathcal{C} \to \mathcal{C}$, I had to compose your algebra morphism from $\mathcal{C} \to \mathcal{C}^\mathrm{op}$ with the canonical automorphism of k-modules $\mathcal{C} ^\mathrm{op} \to \mathcal{C}$ – Eric Sep 03 '21 at 09:29