Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric algebra

  1. Dec 10, 2003 #1
    can anyone explain how commutators act on tri-vectors (in orthonormal conditions)?

    on bi-vectors i know that it ends up to be a bivector again,
    but with tri-vectors it vanishes if its lineraly dependent.
    what about the case if its not linearly dependent,
    does that mean it remains a tri-vector?

    how does a vector transform under a transformation generated by exponentiation of a trivector ?

    a transformation is a rotation or reflection,
    but who can explain the exponentiation?
  2. jcsd
  3. Dec 11, 2003 #2
  4. Dec 12, 2003 #3
    so using exp(ix)=cos(x)+ isin(x), multiplying it by the vector, this will result in a rotation, correct?

    i have found examples for bi-vectors, but how does this change with tri-vectors?

    for a bivector:


    exp(Ix)=cos(x)+I sin(x)
    exp(Kx)=cos(hx)+K sin(hx)

    cos (hx)=0.5(exp(x)+exp(-x)
    sin(hx) = 0.5(exp(x)-exp(-x)



    is this a lorentz transformation?

    also, i read that complex numbers represent vectors as points with the transformation...?
  5. Dec 12, 2003 #4
    Yes, R+iI <-> (R,I).
  6. Dec 12, 2003 #5
    Ah, now I think I understand what you mean.
    Of course, you can't usually define the exponentiation of a vector. But you can define the eponentiation of a linear operator (matrix).
    Like this: Let A be a matrix, then
    \exp{(A)}=\sum_{k=0}^\infty \frac{A^k}{k!}.
    Let's say a transformation can be written in the form
    x' = \exp{(A)} \cdot x.
    Now, if A = 1 + Gt with some parameter t, then G is called the generator of this transformation. For rotations, t is the angle.
  7. Dec 12, 2003 #6
    if the commutator of a bi-vector [A,B] is found by AB-BA, is the commutator of a tri-vector then ABC-BCA-CAB?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook