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First the definition of the inner and outer product in terms of grade projection:

$A_{r} \cdot B_{s} = \langle A_{r}B_{s} \rangle_{|r-s|}$

$A_{r} \wedge B_{s} = \langle A_{r}B_{s} \rangle_{r+s}$

So the inner and outer products are defined in terms of the geometric product (terms in the angle brackets) as far as I understand. Now let's calculate an example with two bivectors:

$A_{2}= 6e_{1}e_{2}$

$B_{2}= e_{2}e_{3}$

$A_{2}B_{2} = 6e_{1}e_{3}$

Ok that's easy to calculate, but what exactly is this term? If you evaluate the calculation with the grade projection you get:

$A_{2} \cdot B_{2} = \langle A_{2}B_{2} \rangle_{0}$

$A_{2} \wedge B_{2} = \langle A_{2}B_{2} \rangle_{4}$

So this should mean that the inner and outer product of the bivectors is zero, because there are no such terms in the result of the geometric product between the bivectors. Is that right? But what is the term $6e_{1}e_{3}$ ?

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The product in geometric (Clifford) algebra is the geometric product. A Clifford algebra $Cl_{p,q}(\Bbb R)$ is an algebra over $\Bbb R$ with respect to the geometric product.

So then you might ask some questions: (1) What are the inner and outer products? (2) Do they contain all of the information of the geometric product? (3) If not, why are they even important?

Let's go through these questions. You've already the answer to $(1)$ in your question (at least for two blades). The inner and outer products (I say the inner product, but I really should say an inner product because there is more than one common inner product in use in GA, and in fact I typically use a different one than you) are simply defined as certain grade projections of the geometric product -- the outer product being the largest possible grade projection and the inner product being the smallest possible.

(2) But that doesnt mean that the inner and outer products of two multivectors $A$ and $B$ contains all of the information of the geometric product. The geometric product can very easily contain other graded elements than strictly the highest and lowest ones possible given the grades of the $A$ and $B$. The only way you can know for sure (based solely on the grades of $A$ and $B$) that the inner and outer products give you all of the information on the geometric product is if the highest grade of either $A$ or $B$ (you only need one of them to have this property) is $1$.

(3) Given that, why do we even care about the inner and outer products? Because they both have geometric meaning. Let $A$ be a $k$-blade representing a $k$-dimensional subspace $\mathcal A \subseteq\Bbb R^n$ and $B$ be a $j$-blade representing a $j$-dimensional subspace $\mathcal B\subseteq\Bbb R^n$. Then if $\mathcal A \oplus \mathcal B = \mathcal C$ is a $(k+j)$-dimensional subspace then $A\wedge B$ is a $(k+j)$-blade representing $\mathcal C$. Likewise if $\mathcal A$ is a subset of $\mathcal B$ then $A\cdot B$ is a $(j-k)$-blade representing the orthogonal complement of $\mathcal A$ in $\mathcal B$ (a similar statement can be made if $\mathcal B$ is a subset of $\mathcal A$ instead).


Looking at your example we see that $A_2 \cdot B_2 = 0$ therefore neither of the two subspaces represented by $A_2$ and $B_2$ is a subset of the other. Moreover $A_2\wedge B_2 = 0$ and thus the subspace $\mathcal A_2 + \mathcal B_2 = \mathcal C$ is not $(2+2)$-dimensional (it is in fact $3$-dimensional). Therefore we can immediately see that $\mathcal A_2$ and $\mathcal B_2$ are two subspaces, neither of which is a subset of the other, but also which are not entirely in the orthogonal complement of the other, either. So each "shares some but not all" vectors which the other.

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  • $\begingroup$ Hi thanks. Ok I got it. The inner product gives the lowest grade and the outer product the highest grade but there can also be grades in between those. That makes sense because $A= \langle A \rangle_{0}+ \langle A \rangle_{2}$ is only a special case. After some tryouts I realized that the term $6e1e3$ is actually the result of the commutator product of which I read that this product often involves the product of bivectors, so it makes sense. I'm not fully understanding any of this but I'm getting there. Thx for your answer. $\endgroup$ – JonnyPython Mar 8 '16 at 15:41
  • $\begingroup$ *special case when you have 1-vectors. $\endgroup$ – JonnyPython Mar 8 '16 at 15:47
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Well, yes, with these definitions, the inner and outer products are both zero. In my opinion, all those inner/outer products on Clifford algebras are basically ill-behaved (this does not mean that they are meaningless or useless, but they mainly serve computational purposes).

The thing is that the Clifford algebra is not a $\mathbb{Z}$-graded algebra : it is a $\mathbb{Z}$-filtered algebra and a $\mathbb{Z}/2\mathbb{Z}$-graded algebra, but only a $\mathbb{Z}$-graded vector space. So interactions between the product and the $\mathbb{Z}$-grading are bound to be imperfect.

In particular, you can't expect to extract all the information of the so-called "geometric product" using your inner and outer product.

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  • $\begingroup$ I don't fully understand your answer right now, but I have a vague feeling of what it means for example that the geometric product only picks out certain terms or something. But thx for your answer, I'll come back and read it if I learned more about the subject. $\endgroup$ – JonnyPython Mar 8 '16 at 15:56

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