# I Outer product in geometric algebra

1. Jul 30, 2017

### Silviu

Hello! I am reading so very introductory stuff on geometric algebra and at a point the author says that, as a rule for calculation geometric products, we have that $e_{12..n}=e_1\wedge e_2 \wedge ...\wedge e_n = e_1e_2...e_n$, with $e_i$ the orthonormal basis of an n-dimensional space, and I am not sure I understand this. As far as I understood, the wedge product of 2 vectors, is just the cross product and we have $e_1e_2=e_1 \cdot e_2 + e_1 \wedge e_2=e_1 \wedge e_2$, which makes sense as the dot product of 2 perpendicular vectors is 0. But for 3 vectors we would have $e_1e_2e_3=(e_1 \wedge e_2)e_3=(e_1 \wedge e_2) \cdot e_3 + (e_1 \wedge e_2)\wedge e_3 = e_1 \wedge e_2 \wedge e_3$, which is not true in euclidian space. So can someone explain to me how does the rule the author mentioned, works? Thank you!

2. Jul 30, 2017

### FactChecker

(e1∧e2) is not the cross product. It is not a vector in the direction of e3. It is an oriented element of area in the plane of e1 and e2. The vector e3 is orthogonal to that plane. So (e1∧e2)⋅e3 = 0. Visualizing (e1∧e2) as the cross product vector is a habit that will cause trouble.

3. Jul 30, 2017

### Silviu

Ok, this makes more sense. But what is the direction of $e_1 \wedge e_2$, if it is not perpendicular to the plane?

4. Jul 30, 2017

### FactChecker

It is not a vector with a direction. It is an area element in the e1, e2 plane with a clockwise or counter-clockwise orientation.

EDIT: More precisely, e1∧e2 is not in the familiar R3 vector space with basis e1, e2, and e3. It is an oriented area vector in the vector space of oriented areas with basis e1∧e2, e2∧e3, and e3∧e1.

Last edited: Jul 30, 2017
5. Jul 30, 2017

### Staff: Mentor

A smooth function $f$ is a vector, as it can be viewed as an element of the vector space $\mathcal{C}^\infty(\mathbb{R})$. In which direction does $f$ point? With $v \wedge w$ it is similar. It is a vector as an element of a Graßmann algebra, but without a basis, no coordinates and without coordinates no direction. If $\{u,v,w\}$ are linear independent vectors in, say $\mathbb{R}^3$, then $\{1,u,v,w,u\wedge v, v\wedge w, w\wedge u, u\wedge v \wedge w\}$ are also linear independent in $\bigoplus_n \bigwedge^n(\mathbb{R}^3)$.

Last edited: Jul 30, 2017