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Silviu

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Silviu

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- #3

Silviu

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Ok, this makes more sense. But what is the direction of ##e_1 \wedge e_2##, if it is not perpendicular to the plane?(e_{1}∧e_{2}) is not the cross product. It is not a vector in the direction of e_{3}. It is an oriented element of area in the plane of e_{1}and e_{2}. The vector e_{3}is orthogonal to that plane. So (e_{1}∧e_{2})⋅e_{3}= 0.

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It is not a vector with a direction. It is an area element in the e_{1}, e_{2} plane with a clockwise or counter-clockwise orientation.

EDIT: More precisely, e_{1}∧e_{2} is not in the familiar R^{3} vector space with basis e_{1}, e_{2}, and e_{3}. It is an oriented area vector in the vector space of oriented areas with basis e_{1}∧e_{2}, e_{2}∧e_{3}, and e_{3}∧e_{1}.

EDIT: More precisely, e

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

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hjaramil

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Is there a formal proof of (e_{1}∧e_{2}) is not the cross product. It is not a vector in the direction of e_{3}. It is an oriented element of area in the plane of e_{1}and e_{2}. The vector e_{3}is orthogonal to that plane. So (e_{1}∧e_{2})⋅e_{3}= 0. Visualizing (e_{1}∧e_{2}) as the cross product vector is a habit that will cause trouble.

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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 goesIs there a formal proof of (e_{1}∧e_{2})⋅e_{3}= 0 ? Thanks

$$

\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,

- #8

steenis

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

- #9

hjaramil

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I believe I answered my own question. There are several definitions of inner products. Here is one.It would be very helpful if you mention who the author is and which book you are reading. It helps in understanding the context.

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 $$(e_1 \wedge e_2) \cdot e_3 = \langle e_1 e_2 e_3 \rangle_{2-1} = 0$$.

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