# Outer product in geometric algebra

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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!

## Answers and Replies

Science Advisor
Homework Helper
Gold Member
(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.

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

Science Advisor
Homework Helper
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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:
Mentor
2022 Award
Ok, this makes more sense. But what is the direction of ##e_1 \wedge e_2##, if it is not perpendicular to the plane?
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:
hjaramil
(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.
Is there a formal proof of (e1∧e2)⋅e3 = 0 ? Thanks

Mentor
2022 Award
Is there a formal proof of (e1∧e2)⋅e3 = 0 ? Thanks
The Graßmann algebra ##\bigwedge^m( V)## normally doesn't have an inner product. However, if ##V## has, as in the case of a real vector space, there can be defined one. So your question comes down to: "How is an inner product on a Graßmann algebra defined, if the vector space has one?", i.e. how to interpret the dot in the equation above. The definition goes
$$\langle (u_1\wedge \ldots \wedge u_m)\; , \;(v_1\wedge \ldots \wedge v_m)\rangle = \det \begin{bmatrix}\langle u_1,v_1\rangle&\ldots& \langle u_1,v_m\rangle \\ \vdots & \ldots &\vdots \\ \langle u_m,v_1\rangle&\ldots& \langle u_m,v_m\rangle \end{bmatrix}$$
which shows that only products of equal grade are defined. To expand this on the entire Graßmann algebra, subspaces of different grade are defined to be orthongonal. It is in accordance to the definition, because such a determinant would also be zero.

steenis
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!

It would be very helpful if you mention who the author is and which book you are reading. It helps in understanding the context.

Math Amateur
hjaramil
It would be very helpful if you mention who the author is and which book you are reading. It helps in understanding the context.
I believe I answered my own question. There are several definitions of inner products. Here is one.
The Graßmann algebra ##\bigwedge^m( V)## normally doesn't have an inner product. However, if ##V## has, as in the case of a real vector space, there can be defined one. So your question comes down to: "How is an inner product on a Graßmann algebra defined, if the vector space has one?", i.e. how to interpret the dot in the equation above. The definition goes
$$\langle (u_1\wedge \ldots \wedge u_m)\; , \;(v_1\wedge \ldots \wedge v_m)\rangle = \det \begin{bmatrix}\langle u_1,v_1\rangle&\ldots& \langle u_1,v_m\rangle \\ \vdots & \ldots &\vdots \\ \langle u_m,v_1\rangle&\ldots& \langle u_m,v_m\rangle \end{bmatrix}$$
which shows that only products of equal grade are defined. To expand this on the entire Graßmann algebra, subspaces of different grade are defined to be orthongonal. It is in accordance to the definition, because such a determinant would also be zero.

I believe I answered my own question. There are several definitions of inner prodcut. Here is one (right contraction) if $$A_j$$ is a j-vector and $$B_k$$ is a k-vector then
$$A_j \cdot B_k= \langle AB \rangle_{j-k}$$. Then $$(e_1 \wedge e_2) \cdot e_3 = \langle e_1 e_2 e_3 \rangle_{2-1} = 0$$.