For what types of objects is the outer product defined in geometric algebra?

Question

I just started to learn geometric algebra from the "Geometric Algebra for Physicists" book. Authors first give definition of outer product of two vectors $a \wedge b$. Then they give geometric product in terms of inner and outer product: $ab = a \cdot b + a \wedge b$, but then they define the outer product through the geometric product: $a \wedge b = \frac{1}{2}(ab - ba)$.

As I understand, geometric product is defined for all elements in the algebra, so is outer product defined for all of them too?

For example, given $\mathcal{G}2$ algebra where elements are $a = a_0 + a_1 e_1 + a_2 e_2 + a_3 e_1 e_2$, what does the outer product of vector with bivector would mean? $a \wedge b = (a_1 e_1 + a_2 e_2) \wedge (b_3 e_1 e_2) = a_1 b_3 (e_1 \wedge (e_1 e_2)) + a_2 b_3 (e_2 \wedge (e_1 e_2))$ what do I do next?

Can I compute outer product of vector with the scalar? Like, $a_0 \wedge (b_1 e_1 + b_2 e_2)$. If it is defined, what does that mean?

Answer 1

The equation $ab = a\cdot b + a\wedge b$ is neither a definition of the geometric product nor even valid in general. When it is first introduced in Doran and Lasenby's Geometric Algebra for Physicists $a$ and $b$ are assumed to be vectors. Similarly, the equation $$ a\wedge b = \frac12(ab - ba) $$ is valid for vectors but is not valid in general. It is also not stated that this is a definition of the exterior product of vectors, merely that it can be if the geometric product is already defined.

You should consider reading Chapter 4 "Foundations of Geometric Algebra" if you have not already.

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One way to define the exterior product is via grade projections of the geometric product.

If $A_r$ is an $r$-vector and $B_s$ is an $s$-vector then $$ A_r\wedge B_s := \langle A_rB_s\rangle_{r+s} $$ where $\langle\cdot\rangle_{r+s}$ is the $(r+s)$-vector component (and is of course zero if there is no such component). Because every multivector is a sum of homogeneous multivectors, we simply extend this definition by linearity: if $n$ is the dimension of our base vector space and $$ A = \sum_{i=0}^nA_i,\quad B = \sum_{i=0}^nB_i $$ where $A_i, B_i$ are $i$-vectors, then $$ A\wedge B := \sum_{i=0}^n\sum_{j=0}^n\langle A_iB_j\rangle_{i+j}. $$

For your example, this looks like $$\begin{aligned} &(a_1 e_1 + a_2 e_2) \wedge (b_3 e_1 e_2) \\ &= a_1 b_3 e_1 \wedge (e_1 e_2) + a_2 b_3 e_2 \wedge (e_1 e_2) \\ &= a_1b_3\langle e_1e_1e_2\rangle_3 + a_2b_3\langle e_2e_1e_2\rangle_3 \end{aligned}$$ then $$ \langle e_1e_1e_2\rangle_3 = e_1^2\langle e_2\rangle_3 = 0, $$$$ \langle e_2e_1e_2\rangle_3 = -e_2^2\langle e_1\rangle_3 = 0 $$ so the whole expression is $0$. I've used the fact that $e_i^2$ is a scalar and that $e_ie_j = -e_je_i$ when $i\ne j$.

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Practically, what might help the most are the following facts:

1. The exterior product is associative and bilinear.
2. If $v_1,\dotsc,v_k$ are vectors then $$ V := v_1\wedge\dotsb\wedge v_k = 0 \iff \text{those vectors are linearly dependent} $$ and $V$ is always a $k$-vector (in fact, a $k$-blade). It follows that when $V \ne 0$ and $u$ is a vector then $$ u\wedge V = 0 \iff u \in \mathrm{span}\{v_1,\dotsc,v_k\}. $$

In fact, these properties completely characterize the exterior product. We can derive two more key facts:

3. If $v, w$ are vectors then $v\wedge v = 0$ and equivalently $v\wedge w = -w\wedge v$.
4. Let $v_1,\dotsc,v_s$ be linearly independent vectors and $w_1,\dotsc,w_r$ also be linearly independent vectors. Then there is a scalar $a$ such that $$ v_1\wedge\dotsb\wedge v_r = aw_1\wedge\dotsb w_s $$ if and only if $$ \mathrm{span}\{v_1,\dotsc,v_r\} = \mathrm{span}\{w_1,\dotsc,w_s\}. $$