New answers tagged clifford-algebras
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To reveal my deeper motivations, I see Clifford Algebra as an intriguing puzzle that exposes the fundamental principles behind regular algebra, revealing it to be an essentially spatial logic, a logic of geometrical concepts interacting in spatial ways. And that, in turn, reveals the essential principles of the human mind, and how it makes sense of reality. ...
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When you say $c = ab$ is a bivector, you're taking $a \perp b$ implicitly. The geometric product will in general have both terms: $ab = a \cdot b + a \wedge b$, remember? A scalar and a bivector.
It is, admittedly, pretty hard to visualize what that is geometrically. Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and ...
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The quaternion procedure is probably the simplest, easiest to implement, and most computationally economical way to go.
In practice you would likely be doing all of this in a computer anyhow, and computing the product of two quaternions (in the big scheme of things) is not much harder than two real numbers, or two complex numbers. I think the multiplication ...
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I do not believe there is without passing through some alternate representation (quaternion, matrix, ...). This is one of the known disadvantages of axis-angle compared to the others, while an advantage is the triviality of inversion (simply negate the angle or the axis).
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Look at the following link: Axis–angle representation
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If the k-vector is simple it can be written as a (geometric)product of k vectors.
$A_r=a_1...a_r$
The proof is by induction. For a vector you know
$a^{\dagger}a=|a|^2$
which is scalar by definition, so you know this to be true for r=1
suppose it is true for r
$A_r^{\dagger}A_r=|A_r|^2$
it follows that
$A_{r+1}^{\dagger}A_{r+1}$
...
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part 1)
The differential forms approach is indeed very powerful, what Hestenes points out in his "From Clifford Algebra to Geometric Calculus" is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book you will find an alternative. The starting point (as was pointed out above) is the notion of a ...
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In the context of Clifford/geometric algebras (which we have been talking about recently), I can give you a crash course on how a special case of a real geometric algebra is built. Keep in mind that what I'm describing is a special case, and there are more possibilities.
Here is a brief introduction to why Clifford algebras for finite dimensional vector ...
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You may be intrested in Hurwitz' theorem.
A "simpler" proof that the dimension is always a power of two goes like this:
Theorem. If there is a continuous odd map $S^{n-1}\times S^{n-1}\to S^{n-1}$, then $n$ is a power of $2$.
Applying this theorem to the map $(x,y)\mapsto \frac{xy}{\Vert xy\Vert}$ (no zero divisors!) gives the desired result baout ...
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