The volume of the parallelepiped formed by those vectors.
#3
JakeD
14
0
Thank you.
Jake
#4
chogg
129
10
No, this is just wrong! The volume of the parallelepiped would be all outer products, e_1 \wedge e_2 \wedge e_3.
(e_1 \wedge e_2) is a bivector. So you're asking: what is the inner product of a vector with a bivector? That has a clear geometrical interpretation. For a vector a and bivector B, a \cdot B does the following:
Project a onto the plane defined by B
Rotate 90 degrees in the "sense" of B
Dilate by the magnitude of B
Note that this uses all three defining characteristics of the bivector B:
Attitude (basically the angle the plane makes in space)
Orientation (clockwise vs. counterclockwise)
Magnitude (i.e. area)
With the inner product used by Hestenes et al, you also have
a \cdot B = - B \cdot a
which let's you answer your question.
By the way, in 3D, your construction is equivalent to the "double cross product" (not the "vector triple product"):
<br />
(e_1 \wedge e_2) \cdot e_3 = - (e_1 \times e_2) \times e_3<br />
Note how the GA version (described above) is much more intuitive and easy to visualize -- the VA version (double cross product) will give you carpal tunnel from all those applications of the right-hand rule!
#5
JakeD
14
0
Thanks chogg for your details answer; I later noticed indeed that his answer is wrong.
GAViewer also demonstrates this very nicely.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?