# Visualizing the geometric product?

The exterior product between blades has a relatively clear geometric interpretation: it gives the result of "extending" one factor along the other, with the direction pointing along the first factor and then the second. So the exterior product of two vectors can be imagined as the directed parallelogram between two arrows.

Is there a similar way to visualize the geometric product? Or must it just be considered algebraically?

• Well the geometric product of two vectors yields a scaled rotation (a rotor). This is easily seen by factoring the norm out of the product: $ab = \|a\|\|b\|(\cos(\theta) + i\sin(\theta))=\|a\|\|b\|e^{i\theta}$ where $i$ is a unit pseudoscalar for the subspace $\operatorname{span}(a,b)$. I don't know of a geometric interpretation for the geometric product of higher dimensional blades. – user137731 Nov 20 '15 at 0:03
• Of course parts of the geometric product have important meanings. For instance if $A$ is a $k$-blade and $B$ is a $j$-blade where $k\lt j$, then $\langle AB\rangle_{j-k}$ is the orthogonal complement of $\operatorname{proj}_BA$ in $B$. And of course $\langle AB\rangle_{j+k}$ is the $j+k$-blade representing $\operatorname{span}(A,B)$ if that span is $j+k$ dimensional. – user137731 Nov 20 '15 at 0:18
• @Bye_World Very useful. I'm mostly looking for a general technique, but the inner product as a downward projection is a good one. – Draconis Nov 20 '15 at 5:13
• I think the problem with finding a general geometric interpretation of the geometric product is that we don't have a general geometric interpretation of multivectors. We really only have geometric interpretations of blades and a few other convenient types of multivectors (rotors, paravectors, etc). But in general the geometric product of two blades will not be one of these special types of multivectors. – user137731 Nov 20 '15 at 14:52
• I think the key to developing a geometric understanding here would be to understand the geometric meaning of adding two multivectors with differently graded parts. Unless or until that happens, a general multivector will elude geometric intuition, whether it comes from a geometric product or somewhere else. – Muphrid Nov 20 '15 at 15:09