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

Question

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?

Answer 1

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.

Answer 2

$ \newcommand\grade[1]{\langle#1\rangle} \newcommand\lcontr{{\rfloor}} \newcommand\rcontr{{\lfloor}} $

The Exterior Product

Given multivectors $A = \sum_{i=0}^n A_i$ and $B = \sum_{i=0}^n B_i$ where $A_i, B_i$ are $i$-vectors, then since $\wedge$ is bilinear $$ A\wedge B = \sum_{i=0}^n\sum_{j=0}^n A_i\wedge B_j. $$ To define $\wedge$ in terms of the geometric product, typically we'd go with $$ A_i\wedge B_j = \grade{A_iB_j}_{i+j} $$ where $\grade{}_k$ is the projection onto the grade $k$ part; then you extend by linearity, meaning $$ A\wedge B = \sum_{i=0}^n\sum_{j=0}^n\grade{A_iB_j}_{i+j}. $$ If $a_1, a_2,\dotsc, a_k$ are vectors, then we can also define their exterior product as the completely antisymmetrized sum: $$ a_1\wedge a_2\wedge\cdots\wedge a_k = \frac1{k!}\sum_{\sigma\in\mathrm{Perm}(k)}\mathrm{sgn}(\sigma)a_{\sigma(1)}a_{\sigma(2)}\cdots a_{\sigma(k)}, $$ where the summation is over all permutations of $\{1,\dotsc,k\}$. You can probably extend this to blades, but I don't know how off the top of my head.

The Inner Product

Rather than trying to work with some sort of symmetric dot product, it is much better to work with the left ($\lcontr$) and right ($\rcontr$) contractions. See here. These are most naturally defined as the operations adjoint to the exterior product: $$ \grade{(A\wedge B)C}_0 = \grade{A(B\lcontr C)}_0,\quad \grade{A(B\wedge V)}_0 = \grade{(A\rcontr B)C}_0 $$ for arbitrary multivectors $A, B, C$. There are also the universal identities $$ (A\wedge B)\lcontr C = A\lcontr(B\lcontr C),\quad A\rcontr(B\wedge C) = (A\rcontr B)\rcontr C. $$ As you may have gotten a sense for, the left and right contractions have the same identities but reversed; as such, I'll just focus on the left contraction from now on.

On vectors, we simply have $a\lcontr b = a\cdot b$. Contraction by a vector is a (graded) derivation: $$ a\lcontr(B\wedge C) = (a\lcontr B)\wedge C + \hat B\wedge(a\lcontr C) $$ where $\hat B$ is the grade involution (or main involution) of $B$.

In general, we can find that $$ A_i\lcontr B_j = \begin{cases} 0 &\text{if } j - i < 0 \\ \grade{A_iB_j}_{j-i} &\text{if } j - i \geq 0. \end{cases} $$ Extending by linearity to general multivectors $A = \sum_{i=0}^nA_i$ and $B = \sum_{i=0}^nB_i$, $$ A\lcontr B = \sum_{i=0}^n\sum_{j=0}^nA_i\lcontr B_j = \sum_{i=0}^n\sum_{j=0}^n\grade{A_iB_j}_{j-i}, $$ with the convention that $\grade{\cdots}_{j-i} = 0$ if $j - i < 0$.

On blades $A_i, B_j$, the contraction has a geometric interpretation: $A_i\lcontr B_j = 0$ if and only if $i > j$ or there is a vector in $A_i$ completely perpendicular $B_j$. When it is not zero, $A_i\lcontr B_j$ is a subblade of $B_j$ completely perpendicular to $A_i$.

The Geometric Product

There is not a simple way (I know of) to express the geometric product of arbitrary multivectors in terms of contraction and exterior product. That being said, for $a$ a vector and $B$ an arbitrary multivector, we do have $$ aB = a\lcontr + a\wedge B,\quad Ba = B\rcontr a + B\wedge a. $$