The conjugate of an element $x$ in a Clifford Algebra $Cl_n$ is defined by $\bar{x} = \alpha(x^t)$, where
- $(x)^t$ reverses the order of all the vectors in each term ("reversion")
- $\alpha(x)$ applies a minus sign to a term if it has an odd number of vectors in the clifford product ("grade involution").
For example, the conjugate of $x=e_1 e_2 + e_1 e_3 e_4$ would be $\bar{x}=e_2 e_1 -e_4 e_3 e_1 $.
The norm is defined as $N(x)=x\bar{x}$, which helps us compute inverses as $x^{-1} = \frac{\bar{x}}{N(x)}$.
My question is, why are the conjugate and norm defined this way?
I have read in section 4.5 of Lounesto's book "Clifford Algebras and Spinors" that, at least for the case of $Cl_3$, the norm $N(x)=x\bar{x}$ is always in the center of $Cl_3$ (meaning it will commute with all elements in $Cl_3$), so division is the same from both the left and the right.
Is this true in general for all $Cl_n$? If so, how can we prove it?
My attempt at a proof so far:
Take a Clifford algebra $Cl_n$ over a field $F$ with non-degenerate quadratic form $Q:Cl_n \rightarrow F$.
Take $u\in Cl_n$ to be of the form $u=e_{i_1} e_{i_2} ... e_{i_k}$ and take $w \neq u \in Cl_n$ to be of the form $w=e_{i_1} e_{i_2} ... e_{i_h}$
I must show that the following norm is in the center of $Cl_n$
$N(u+w) = (u+w)(\bar{u} + \bar{w})= u\bar{u} + w\bar{w} + u\bar{w} + w\bar{u}$
The parts $u\bar{u} + w\bar{w}$ are obviously in the center, since they are scalars. So it's left to show that $u\bar{w} + w\bar{u}$ is also in the center.
I know that for even dimension $n$, only the scalars are in the center, and for odd dimension $n$, the center includes only the scalars and the "top" element $e_{i_1} e_{i_2} ... e_{i_n}$, and linear combinations of those.
I also know the rule for sign changes on a reversal is $u^t=(e_{i_1} e_{i_2} ... e_{i_k})^t = (e_{i_k} ... e_{i_2} e_{i_1}) = u(-1)^{k(k-1)/2}$
For even dimension $n$ it seems I must show that $u\bar{w} + w\bar{u} = 0$, but I am stumped about how to do this.
Consider, in $Cl_4$ with the positive-definite form,
$$x=e_1e_2e_3+2e_4$$
$$xx=3$$
$$\overline x=e_1e_2e_3-2e_4$$
$$x\overline x=-5-4e_1e_2e_3e_4$$
$$\overline xx=-5+4e_1e_2e_3e_4$$
$$(x\overline x)^{-1}=\tfrac19(\overline xx)$$
$$(x\overline x)\overline x=\overline x(\overline xx)=-5e_1e_2e_3+10e_4-4e_4+8e_1e_2e_3=3e_1e_2e_3+6e_4$$
$$\overline x(x\overline x)=(\overline xx)\overline x=-5e_1e_2e_3+10e_4+4e_4-8e_1e_2e_3=-13e_1e_2e_3+14e_4$$
There are several problems here: (1) $x\overline x$ is not in the centre. (2) $x\overline x\neq\overline xx$, so the definition of the norm is "asymmetric" in some sense. (3) $x\overline x$ does not commute with $\overline x$, so the division is undefined.
But we can divide on one side, to get
$$\overline x(x\overline x)^{-1}=\tfrac19\overline x(\overline xx)=\tfrac13e_1e_2e_3+\tfrac23e_4=\tfrac13x=x^{-1}$$
This formula works in general, because $(x\overline x)^{-1}=\overline x^{-1}x^{-1}=\overline{(x^{-1})}x^{-1}$ exists whenever $x^{-1}$ exists, and $x\big(\overline x(x\overline x)^{-1}\big)=(x\overline x)(x\overline x)^{-1}=1$. And $(x\overline x)^{-1}$ should be easier to calculate than $x^{-1}$ itself, because $x\overline x$ is in a smaller space. (Being its own conjugate, it only has terms of grade $0$ and $3\text{ mod }4$).
For that matter, $xx^t$ should work the same way, only having terms of grade $0$ and $1\text{ mod }4$.
Generalizing, if we can find any invertible $y$ such that $xy$ is easy to invert, then
$$x^{-1}=y(xy)^{-1}$$
"Why is the Clifford conjugate defined the way it is?" One reason is that it commutes with these other operations: $\overline{(x^t)}=(\overline x)^t,\;\overline{\alpha(x)}=\alpha(\overline x),\;\overline{(x^{-1})}=(\overline x)^{-1}$. Another reason is that it separates the space into two subspaces, in a basis-independent way. The same can be said for reversion and grade involution. I don't know Lounesto's reasons. Indeed, I think this norm is a bad definition; at least, it shouldn't be called a "norm", as it's not a scalar.