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Vector by bivector geometric product

  1. Feb 6, 2008 #1
    I am trying to teach myself and often get stuck.

    Right now I've come across a . B = 1/2(aB-Ba) where a is a vector and B is a bivector.

    what's confusing me is that it seems to require a change in the definition of geometric product as the sum of a symmetrical inner product and antisymmetrical wedge product, which implies that a . b = 1/2(ab + Ba).

    What have I misunderstood?


  2. jcsd
  3. Feb 28, 2008 #2
    An r-blade Ar can be factored into a product of anticommuting
    vectors a1,...,ar. The geometric product of an
    r-blade with an s-blade produces a set of multivectors having grades
    ranging from |r-s| to r+s and differing from each other by some
    multiple of 2. The multivector with the lowest grade is called
    Ar . As and the multivector with the highest
    grade is the wedge product. The geometric product of a vector (r=1)
    with an r-vector must therefore produce a dot product of grade r-1 and
    a wedge product of grade r+1. In the case of the product of a vector, a,
    and a bivector bc, the result will be a vector and a trivector. Now the
    trivector is the outer product of the anti-symmetric bivector with a
    vector and can always be written as the wedge product of the three vectors,
    a, b, c (You can always find two vectors b and c so that the bivector can
    be written as a wedge product.) Now exchange a with bc:
    a wedge b wedge c = -b wedge a wedge c = b wedge c wedge a. i.e. in order
    that the exchange of any two neighbouring vectors be antisymmetric, the
    exchange of the vector a with the bivector bc must be symmetric, so the
    sign must be +:

    a wedge B = 1/2(aB + Ba)

    Exchanging a and B on the rhs leaves the wedge product invariant, as we
    want. Now we said above that aB = a dot B + a wedge B, so we can
    substitute the above expression for a wedge B and solve for a dot B:

    a dot B = 1/2(aB - Ba).

    So where is the misunderstanding? Good question! I think you must be
    clear about the geometric product being more than just a sum of
    a dot product and wedge product. For two bivectors,

    [tex] AB = A \cdot B + A\times B + A\wedge B, [/tex]

    where the middle term is the commutator product (not the Gibbs cross
    product). Further, it is the antisymmetric nature of the wedge product
    of blades that must be maintained, so the sign that appears in the
    geometric product definition of a vector with an r-blade will be
    + if the r-blade is even and - if the r-blade is odd, but in all cases,
    the outer product of the vector with the blade must be antisymmetric
    because that is part of the definition of a blade. I hope this long-winded
    explanation is of some help.
    Last edited: Feb 29, 2008
  4. Feb 28, 2008 #3
    thank you very much. The part I didn't see was that in order to preserve the antisymmetrical nature of the wedge product the sign of reversing aB depends upon the grade of B.

    I think the text I'm using may not be clear enough for me.

    thanks again.
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