Reduction of a blade pseudovector quotient?

In summary, the triplex product of two blades can be reduced to the dot product of the individual blades.
  • #1
Peeter
305
3
For two blades:

[tex]
A \in {\bigwedge}^r
[/tex]
[tex]
B \in {\bigwedge}^{k-r}
[/tex]

and a corresponding pseudovector for the wedge product of the two:

[tex]
I \in {\bigwedge}^k
[/tex]

intuition tells me that the following scalar quotient:

[tex]
\frac{A \wedge B}{I}
[/tex]

can be reduced to

[tex]
A \cdot \left(B \frac{1}{I}\right) = A \cdot \left(B \cdot \frac{1}{I}\right)
[/tex]

Presuming it this is correct, I curious if it is clear to anybody how to prove this.

Motivation for the question comes from computing the coordinates of a multivector using a basis generated from not necessarily orthonormal vectors (ie: calculation of the reciprocal frame vector/bivector/...). My Doran/Lasenby text has an approach that I have trouble following (I understand the lead up and the end result but not the middle), so I was trying to puzzle it out on my own in a more natural fashion. Such a grade reduction is required as an intermediate step.
 
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  • #2
Wow --- it's been a long time since I've tried to look at the GA stuff. From recollection, isn't there something about projecting and rejecting vectors and multivectors from each other? Your desired step looks something like one of the identities there of that comes from that.
 
  • #3
There's a vector blade reduction identity for r< s:

[tex]
(a \wedge A_r) \cdot A_s = a \cdot (A_r \cdot A_s)
[/tex]

If I try to prove this for myself I need to use:

[tex]
a \wedge A_r = a A_r - a \cdot A_r
[/tex]

So that I can first express this "triple" product as a direct product (with an observation that the dot product term doesn't contribute) :

[tex]
(a \wedge A_r) \cdot A_s = {\langle a A_r A_s \rangle}_{\lvert s - (r+1) \rvert}
[/tex]

Reduction follows by expanding this by first taking the products of [tex]A_r[/tex] and [tex]A_s[/tex] and looking at the grades of the various terms.

But without that first wedge = product - dot result I didn't see how to do this for arbitrary grade blades, but after writing this down I think I now see how to approach it (not much different from above).
 

1. What is a blade pseudovector quotient?

A blade pseudovector quotient is a mathematical concept used in geometric algebra to represent a geometric object in terms of its orientation and magnitude.

2. How is the reduction of a blade pseudovector quotient useful?

The reduction of a blade pseudovector quotient allows for simplification and manipulation of geometric objects, making them easier to work with in calculations and applications.

3. What is the process for reducing a blade pseudovector quotient?

The process involves finding a simpler representation of the geometric object by eliminating unnecessary elements and simplifying the remaining components using mathematical operations.

4. Can the reduction of a blade pseudovector quotient be applied to any geometric object?

Yes, the reduction process can be applied to any geometric object represented using a blade pseudovector quotient in geometric algebra.

5. Are there any limitations to the reduction of a blade pseudovector quotient?

While the reduction process can simplify many geometric objects, it may not always result in the most concise representation and may not be applicable to all types of geometric objects.

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