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I have seen two definitions for the norm in the Clifford algebra $\mathrm{Cℓ}_{p, q, r}$.

According to one of them $\Vert x\Vert = ⟨x. x^\dagger⟩_0$, where the dagger stands for the reversal of the order of all Clifford products. That is, if the basis is $e_1, e_2 \ldots$, then $ (e_1 . e_2 . e_3)^\dagger = e_3. e_2 . e_1$. The $⟨x. y⟩_0$ denotes the $0$ grade (i.e. scalar part) of the symmetric inner product $\frac{ x.y + y.x}{2}$.

The other definition is $\Vert x\Vert = x. \hat{x} $, where hat stands for Clifford conjugation, that is the composition of the main involution and the reversal.

If we take $\mathrm{Cℓ}_{0, 1, 0} \simeq \mathbf{C}$, then the Clifford conjugation coincides with the usual complex conjugation, while the reversal doesn't. That is, if we take $x=1+e_1$, then $\hat{x}=1 - e_1 = x^*$. It seems that, if we deal only with vector elements, the definitions are equivalent.

So the question is: which one is more natural definition? And which one holds also for bivectors and multivectors in general?

The literature is confusing.

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  • $\begingroup$ If we take instead $Cl_{2,0,0}^+\cong\mathbb C$, then reversion coincides with complex conjugation. $(1+e_1e_2)^\dagger=(1-e_1e_2)$ $\endgroup$
    – mr_e_man
    Commented Dec 6, 2019 at 5:27

2 Answers 2

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So the question is which one is more natural definition?

This is entirely a matter of opinion, but my vote would be with the one that matches complex conjugation. A better question would ask where each one is useful. That question would probably be best answered by paying attention to where they are used in the literature.

And which one holds also for bi-vectors and multivectors in general?

They are both defined for all elements of the algebra, and that includes all multivectors, so I'm not sure what you want.

The literature is confusing.

Yes, a lot of it is. Hopefully this will improve over time...

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  • $\begingroup$ Thanks for the answer. I am writing a computer algebra package so I was looking for definitions that are simple to code and yet general. My personal preference is also for the conjugation but I was reading the Leo Dorst's book there it is given with reverses. $\endgroup$
    – user48672
    Commented Feb 1, 2015 at 13:00
  • $\begingroup$ @user48672 What language are you writing the package in? $\endgroup$
    – rschwieb
    Commented Feb 1, 2015 at 13:23
  • $\begingroup$ The Maxima language. It is a CAS system. $\endgroup$
    – user48672
    Commented Feb 1, 2015 at 13:49
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    $\begingroup$ @user48672 Cool. There is a Python package for geometric algebra which might interest you (or maybe not, but it doesn't hurt.) $\endgroup$
    – rschwieb
    Commented Feb 1, 2015 at 15:25
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Since I see your question, I'd like to give some examples of uses of these norms. Generally speaking the naturalness of using one or the other really depends on the problem you are considering. I usually consider a little change in the first definition, and instead I take the "norms":

\begin{align} ||x||^{\dagger}=xx^{\dagger},\\ ||x||^{\hat{}}=x{\hat{x}}. \end{align}

We call them norms because both Clifford conjugation and reversion are antimorphisms, however they are not norms in the sense that they don't map every element in the Clifford algebra into a scalar. Consider for instance $a=\frac{\sqrt{2}}{2}({\bf{1}}+e_{1}e_{2}e_{3})$ in $\operatorname{Cl}_{3,0,0}$, then $\hat{a}=a$ and $||a||^{\hat{}}=a^2=e_{1}e_{2}e_{3}$ which is not a scalar. Of course you can always take only the scalar part of the norm.

There are some cases in which this quantities give a scalar as a result, and these are very interesting. Perhaps the simplest case is to consider a vector v, you obtain $||v||^{\hat{}}=-\Phi(v)\bf{1}$ and $||v||^{\dagger}=\Phi(v)\bf{1}$, where $\Phi$ is the quadratic form defined on the vector space.

You can also read for instance (https://arxiv.org/abs/0805.0311), and see that the Clifford conjugation norm, when applied to elements of the twisted Clifford group, returns a non-zero scalar. This is interesting since it implies that every element in this group can be "normalized" to $\pm{1}$, giving as a result the reduction of the twisted Clifford group to the Pin group.

When dealing with spinors each of the antimorhisms $^{\hat{}}$ and $^{\dagger}$ induces a norm of spinors (you can read this for instance in Pertti Lounesto's book, chapter 18, or in the book "An introduction to Clifford algebras and spinors" by J. Vaz and Roldão da Rocha, chapter 6). The construction takes some work, but loosely speaking to induce this norm, one has to pick an idempotent $f$, and if this idempotent is fullfills for instance $f^{\dagger}=f$, then the norm of spinors associated to reversion correspond to the norm $|| \ ||^{\dagger}$ which was defined on the whole algebra. The same happens for Clifford conjugation.

The classification of Clifford algebras states that every algebra is rather a matrix algebra over $\Bbb{R}$, $\Bbb{C}$, or $\Bbb{H}$ or direct sum of two identical copies of some of these algebras. When the Clifford algebra is a matrix algebra over $\Bbb{C}$ the construction of norms or inner products on spinor spaces is directly related to the complex structure of the matrix algebra. In this sense, in $\operatorname{Cl}_{3,0,0}\cong\operatorname{M}_2(\Bbb{C})$, because the pseudoscalar $I:=e_{1}e_{2}e_{3}$ acts as the imaginary unit, and because $I^{\dagger}=-I$, while $\widehat{I}=I$, the usual operation of complex conjugation and the corresponding complex norm of spinors in $\Bbb{C}^{2}$ is given by the reversion norm rather than the Clifford conjugation one. In this example, contrary to what happens in $\operatorname{Cl}_{0,1,0}$ the complex conjugation is given by reversion.

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