# Deriving geometric product of basis vectors from outer product

I was exploring geometric algebra in general, when I came across these two formulas relating the inner and outer products of two objects with their geometric products:

\begin{align} \mathbf{u} \cdot \mathbf{v} &= \frac{1}{2}(\mathbf{uv} + \mathbf{vu}) \\ \mathbf{u} \wedge \mathbf{v} &= \frac{1}{2}(\mathbf{uv} - \mathbf{vu}) \\ \end{align}

I also learned that the geometric products of orthogonal vectors are anticommutative (which means that such of basis vectors of an orthogonal basis are too):

\begin{align} \mathbf{\hat{e}_1}\mathbf{\hat{e}_2} &= -\mathbf{\hat{e}_2}\mathbf{\hat{e}_1} \\ \mathbf{\hat{e}_1}\mathbf{\hat{e}_3} &= -\mathbf{\hat{e}_3}\mathbf{\hat{e}_1} \\ \mathbf{\hat{e}_2}\mathbf{\hat{e}_3} &= -\mathbf{\hat{e}_3}\mathbf{\hat{e}_2} \\ \end{align}

As part of my exploration, I tried to derive $$\mathbf{\hat{e}_1}\wedge\mathbf{\hat{e}_2}\wedge\mathbf{\hat{e}_3} = \mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\mathbf{\hat{e}_3}$$ by working forwards using the definition of the outer product. However, I ended up proving that $$\mathbf{\hat{e}_1}\wedge\mathbf{\hat{e}_2}\wedge\mathbf{\hat{e}_3} = 0$$ ...

\begin{align} &\ \mathbf{\hat{e}_1}\wedge\mathbf{\hat{e}_2}\wedge\mathbf{\hat{e}_3} \\ &= \frac{1}{2}(\mathbf{\hat{e}_1}\mathbf{\hat{e}_2} - \mathbf{\hat{e}_2}\mathbf{\hat{e}_1})\wedge\mathbf{\hat{e}_3} \\ &= \frac{1}{2}(\mathbf{\hat{e}_1}\mathbf{\hat{e}_2} + \mathbf{\hat{e}_1}\mathbf{\hat{e}_2})\wedge\mathbf{\hat{e}_3} \\ &= \mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\wedge\mathbf{\hat{e}_3} \\ &= \frac{1}{2}(\mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\mathbf{\hat{e}_3} - \mathbf{\hat{e}_3}\mathbf{\hat{e}_1}\mathbf{\hat{e}_2}) \\ &= \frac{1}{2}(\mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\mathbf{\hat{e}_3} + \mathbf{\hat{e}_1}\mathbf{\hat{e}_3}\mathbf{\hat{e}_2}) \\ &= \frac{1}{2}(\mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\mathbf{\hat{e}_3} - \mathbf{\hat{e}_1}\mathbf{\hat{e}_2}\mathbf{\hat{e}_3}) \\ &= 0\ \text{(subtraction cancels to zero)} \\ \end{align}

Where am I going wrong?

• A small correction: basis vectors are not perpendicular by definition; this is an extra property of a basis and is called orthogonality, but in general, an arbitrary basis need not be orthogonal. So, to be precise, you should say that products of orthogonal vectors are anticommutative, and in particular this holds for the elements of an orthogonal basis. Jun 28, 2021 at 17:14
• @JackozeeHakkiuz Edited, thanks. Jun 28, 2021 at 18:20

The mistake is the fourth line passing to the fifth.

You are applying your identity to $$u=e_1e_2$$ and $$v=e_3$$ but $$e_1e_2$$ is not a vector.

The identity applies only to vectors $$u,v$$.

The relevant identity in your case is that

$$a\wedge b=\frac12(ab-ba)$$

The correct extension of that is the generalized symmetrized product

$$\bigwedge_{i=1}^n a_i=\frac{1}{n!}\sum_{\sigma\in Sym(n)}\prod_{i=1}^n a_{\sigma(i)}$$

In your case, $$e_1\wedge e_2\wedge e_3\\ =\frac16(e_1e_2e_3+e_2e_3e_1+e_3e_1e_2 \\ -e_1e_3e_2-e_2e_1e_3-e_3e_2e_1)\\ =\frac16(6e_1e_2e_3)=e_1e_2e_3$$

• What identity does apply? Or what can/should I apply to get the right answer? (Or is that beyond the scope of the question - should I ask a new one for that?) Jun 28, 2021 at 11:07
• @KenHilton I think it is reasonable to answer in this question. I’ll write something up when I can. Jun 28, 2021 at 11:31
• Is there any way it can be represented recursively using outer products of only pairs? Programmatically, it's easier to evaluate $(e_1 \wedge e_2) \wedge e_3$ than to build $e_1 \wedge e_2 \wedge e_3$ and lazily evaluate it. (Thank you for this representation, though, in any case! I haven't seen it before and it'll definitely come in useful.) Jun 28, 2021 at 12:55
• @KenHilton In principle it should be possible but I don't see how, immediately. It seems to boil down to deciding how the geometric product interacts with the wedge product. Jun 28, 2021 at 14:51

A more fundamental approach is to use the grade selection operators and the contraction axiom $$\mathbf{x}^2 = \mathbf{x} \cdot \mathbf{x}$$ to define wedge products, to demonstrate the anticommutitivity property, and to prove the symmetric product identity for the dot product.

The anticommutive property follows from the contraction and distribution axioms. If $$\mathbf{a} \cdot \mathbf{b} = 0$$, then we have \begin{aligned} 0 &= \left( { \mathbf{a} + \mathbf{b} } \right)^2 - \left( { \mathbf{a} + \mathbf{b} } \right) \cdot \left( { \mathbf{a} + \mathbf{b} } \right) \\ &= \mathbf{a}^2 + \mathbf{b}^2 + \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} - \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} \\ &= \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a},\end{aligned} so for any orthogonal vectors $$\mathbf{a}, \mathbf{b}$$, we have $$\mathbf{a} \mathbf{b} = -\mathbf{b} \mathbf{a}$$

Now supposed that we have an Euclidean orthonormal basis $$\left\{ {\mathbf{e}_1, \mathbf{e}_2, \cdots} \right\}$$. Let $$\mathbf{u} = \sum_k \mathbf{e}_k u_k$$ and $$\mathbf{v} = \sum_k \mathbf{e}_k v_k$$, then \begin{aligned} \mathbf{u} \mathbf{v} &= \sum_{j,k} \mathbf{e}_j \mathbf{e}_k u_j v_k \\ &= \sum_{j = k} \mathbf{e}_j \mathbf{e}_k u_j v_k + \sum_{j \ne k} \mathbf{e}_j \mathbf{e}_k u_j v_k \\ &= \sum_{j} (\mathbf{e}_j \cdot \mathbf{e}_j) u_j v_j + \sum_{j < k} \mathbf{e}_j \mathbf{e}_k \left( { u_j v_k - u_k v_j } \right).\end{aligned} The scalar part of this sum is completely symmetric, whereas the bivector portion of this sum is completely antisymmetric (this is a general statement, and can also be shown easily for non-Eucidean bases). We must then have $$\left\langle{{\mathbf{u} \mathbf{v}}}\right\rangle = \frac{1}{{2}} \left( { \mathbf{u} \mathbf{v} + \mathbf{v} \mathbf{u} } \right),$$ and $${\left\langle{{\mathbf{u} \mathbf{v}}}\right\rangle}_{2} = \frac{1}{{2}} \left( { \mathbf{u} \mathbf{v} - \mathbf{v} \mathbf{u} } \right)$$ It is also clear that the scalar portion of this coordinate expansion is the dot product of the two vectors, which means $$\mathbf{u} \cdot \mathbf{v} = \frac{1}{{2}} \left( { \mathbf{u} \mathbf{v} + \mathbf{v} \mathbf{u} } \right),$$ Now, we define the wedge of two vectors as $$\mathbf{u} \wedge \mathbf{v} = {\left\langle{{ \mathbf{u} \mathbf{v} }}\right\rangle}_{2},$$ from which we see $$\mathbf{u} \wedge \mathbf{v} = \frac{1}{{2}} \left( { \mathbf{u} \mathbf{v} - \mathbf{v} \mathbf{u} } \right).$$

We define the wedge of three vectors as $$\mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w} = {\left\langle{{ \mathbf{u} \mathbf{v} \mathbf{w} }}\right\rangle}_{3}.$$ More generally, we define the dot and wedge of a vector $$\mathbf{u}$$ and a k-blade (the wedge of k vectors) $$V_k$$ as $$\mathbf{u} \cdot V_k = {\left\langle{{ \mathbf{u} V_k }}\right\rangle}_{{k-1}},$$ and $$\mathbf{u} \wedge V_k = {\left\langle{{ \mathbf{u} V_k }}\right\rangle}_{{k+1}}.$$

and (still) more generally for a j-blade $$U_j$$ and a k-blade $$V_k$$ $$U_j \cdot V_k = {\left\langle{{ U_j V_k }}\right\rangle}_{\left\lvert{k-j}\right\rvert},$$ and $$U_j \wedge V_k = {\left\langle{{ U_j V_k }}\right\rangle}_{{k+j}}.$$

• (How) Do your last two definitions extend to the dot and wedge of two k-blades? Jun 28, 2021 at 16:31
• answered above. Jun 28, 2021 at 16:59
• Thanks, that clarified everything I needed. Jun 28, 2021 at 17:10