# Most effective way to calculate the products of Geometric Algebra

I'm somewhat confused about the different products in GA. It appears to me that the most general and most effective way is just to calculate everything with the geometric product, and then apply the projection operator to single out the wanted products. Of course I could use the formulas for the inner and outer products separately which are also defined in terms of the geometric product , but I don't like the formulas and they are also not that general. For example I could calculate $aB$ where $a$ is a one-vector and $B$ is a two-vector as:

$a \cdot B= \frac{1}{2}(aB-Ba)$

$a \wedge B = \frac{1}{2}(aB+Ba)$

Then I have to think about two different formulas where I also have to keep track of the signs in the anti-commutator and commutator terms (which is reversed for the product of a 1-vector and a 2-vector), but instead of memorising all the formulas I could just think of the projection operator and the geometric product with:

$a \cdot B = \langle aB \rangle_{1}$

$a \wedge B = \langle aB \rangle_{3}$

Which is a lot easier to memorise. Does this make sense? Further, is it right to think about the grades as the components of a multivector? So to say that I project out the components of a multivector with the grade operator?

• $a\cdot B$ should equal $\langle aB\rangle_1$. But I'm guessing that was just a typo.
– user137731
Mar 17, 2016 at 14:18
• Yes, thank you. Mar 17, 2016 at 18:59
• well, that how it's defined in MacDonald's "Linear and Geometric Algebra". The problem arise when you are dealing with close subalgebras, like the even subalgebra of even grade $\mathrm{Cl}_+(\Bbb R^n)$, because $\langle AB\rangle_{r+s}\neq A\wedge B$. Apr 15, 2023 at 0:19

Yes, the grade selection is a perfectly fine way to define the dot and wedge products. This is, for example, exactly how Doran and Lasenby define the general dot and wedge product operations for grade r,s blades $$A_r, B_s$$ respectively (4.43)
$$A_r \wedge B_s = \langle A_r B_s \rangle_{r + s}$$
$$A_r \cdot B_s = \langle A_r B_s \rangle_{\left\lvert {r - s} \right\rvert}$$