An operator in clifford algebra

In my research I found it useful to define a following operator ($^{*}$) in Clifford algebra:
If $a \in G_{p,q}$, (where $p$ is number of elements that square to $1$, and $q$ is the number of elements that square to $-1$) that is $a=\sum_{i=1}^{2^{p+q}}a^{i}\bar G_{p,q}[i]$, where $\bar G_{p,q}[i]$ denotes $i$'th basis of clifford algebra $G_{p,q}$ ,then $$a^{*}=\sum_{i=1}^{2^{p+q}}a_i\big(\bar G_{p,q}[i]\big)^{3}$$

This operator has the property that $Re(a^{*}a)=\sum_{i=1}^{2^{p+q}}a_i^2$, where $Re(a)$ is defined as $Re(a)=a_0$.
I searched on the internet, but I could not find any mention of this operator. I wonder if I can somehow represent this operator in terms of the known operators(the ones I know are reverse, conjugate and grade involution, however they are all blind to the exact values of p,q and only depend on their sum) so that I don't need to introduce new notation. Or maybe someone has already given a name to this operator but I can not find it? What is a good name for this operator?

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Did you intend $\overline{G}_{p,q}[i]$ to be the basis arising from an orthonormal basis? – rschwieb Jan 2 '13 at 20:12
@rschwieb yes, that is what i meant – Sunny88 Jan 3 '13 at 6:14

Suppose you begin with an orthonormal basis (orthonormal in the sense that the elements are orthogonal and square to either 1 or -1) for $V$.

Then (if I understand your notation correctly), the standard basis elements square to $\pm1$, so $(\overline{G}_{p,q}[i])^3=\pm[\overline{G}_{p,q}[i]]$ where the sign depends on the grade of the basis element and the mixture of elements with square 1 and square -1 appearing there.

The effect on the basis for $V$ is that the elements squaring to -1 are mirrored in the opposite direction, but the elements squaring to 1 are unaffected. It's an involution of some kind. In the quaternions, for example, that map is just the usual conjugate. Unfortunately, I have no ideas how the involutions of general Clifford algebras behave :(

If I had to make up a name for an involution that reversed the basis elements with square -1 this way, I would call it a "time-like reflection".

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How does cubing an element give it back up to sign? – Mariano Suárez-Alvarez Jan 2 '13 at 20:33
@MarianoSuárez-Alvarez If I'm reading his notation correctly, that's what would happen for orthonormal basis elements. – rschwieb Jan 2 '13 at 20:50
Oooo. I see. ${}{}$ – Mariano Suárez-Alvarez Jan 2 '13 at 20:53
I found out that this is called just "conjugate" and denoted by $^{\dagger}$, at least in this book: "Geometric algebra with applications in engineering" by Christian Perwass. – Sunny88 Jan 17 '13 at 11:09