# What is the Clifford/geometric product in terms of the inner and exterior product

The geometric product is defined generally as the algebra on the space of multivectors $\bigwedge \Lambda(V)$

$$ab = a \cdot b + a \wedge b$$

with $\cdot$ a product on the vector space $V$ and $\wedge$ the exterior product. But how do those definitions extend to general multivectors? That is, for the multivectors

\begin{eqnarray} A_i \in \Lambda^i V,\ A &=& \sum_{i=0}^n a_i A_i\\ B_i \in \Lambda^i V,\ B &=& \sum_{i=0}^n b_i B_i \end{eqnarray}

What are the products $A \cdot B$ and $A \wedge B$? For the exterior product I'm guessing it will simply be

$$A \wedge B = (\sum_{i=0}^n a_i A_i) \wedge (\sum_{j=0}^n b_j B_j)$$

since that is a well defined quantity, but in the case of the inner product, it seems a bit harder. From some books it seems to be something of the form of the sum over every combination of the basis contracting with each other, for instance,

$$e_1 \cdot (e_2 \wedge e_3) = \frac 12 \left[(e_1 \cdot e_2) e_3 + (e_1 \cdot e_3) e_2 \right]$$

which maps $n$-blades to $(n-1)$-blades, with the appropriate generalization for inhomogeneous multivectors and arbitrary $n$-blades, although I'm not sure how the inner product of two $2$-blades work, either, or what happens in the case of $0$-forms, is it simply the product in this case? Is this the correct product, and in particular, is there a definition of it that is independant of the choice of basis?

The geometric product is defined generally as the algebra on the space of multivectors $\bigwedge \Lambda(V)$ $ab = a \cdot b + a \wedge b$

I'm not sure what you're reading, and that's not surprising given the disparate descriptions available, but I don't think I have ever seen the geometric product defined in terms of this identity. In most places, you see the geometric product defined on the product of chosen orthonormal basis vectors, extended linearly. Then $\wedge$ is defined and it is shown that $ab = a \cdot b + a \wedge b$ follows.

I don't think a single description of the geometric product on the entire algebra in terms of $\cdot$ and $\wedge$ exists. This is mainly because of what you mention next:

What are the products $A \cdot B$ and $A \wedge B$?

The dot and wedge products are not defined for general elements (there are many useful extensions). Describing everything in terms of $\cdot$ and $\wedge$ would require that.

If we viewed the geometric algebra as the quotient of the tensor algebra $T(V)$ by the ideal generated by $v\otimes v-v\cdot v$, this would amount to asking for a general form of the product of two arbitrary elements in terms of this quotient, and that seems too messy to have a simple answer.

The lesson should be that a lot of geometric algebra hype gets one's hopes up in terms of simple explanations and tidy expressions. While this is true in a lot of places, it doesn't seem to be true in this situation. Not in the "basic" geometric algebra at least. A lot of advances have been made by embedding geometric structures in larger versions of the algebra, so perhaps there is something more appealing there.

• One source I found (arxiv.org/pdf/1205.5935.pdf) defined the inner product in term of the Clifford product as $$A \cdot B = \sum_{r,s} \langle A_r B_s \rangle_{s-r}$$though it's not quite clear what this means from the point of view of the components – Slereah Jan 15 '18 at 15:47
• @Slereah That's among the list of alterantive extensions listed in the link I gave. – rschwieb Jan 15 '18 at 16:15