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Basis of bivector

  1. Jan 20, 2014 #1
    What is the basis of a bivector?

    For example (see the attachment and http://en.wikipedia.org/wiki/Bivector#Axial_vectors first):
    [tex]e_{11}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix}[/tex]
    or
    [tex]e_{11}=\begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}[/tex]
    or ##e_{11}## is equal to what?

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2014 #2

    tiny-tim

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    Hi Jhenrique! :smile:
    The basis is ##\mathbf{e}_i\wedge \mathbf{e}_j##, for all i ≠ j :wink:

    (wikipedia writes that as ##e_{ij}##, which i find confusing :redface:)

    For example, the electromagnetic 4-vector (E;B) is:​

    ##E_x\mathbf{i}\wedge\mathbf{t}+E_y\mathbf{j}\wedge\mathbf{t}+ E_z\mathbf{k}\wedge\mathbf{t} +## ##B_x\mathbf{j}\wedge\mathbf{k}+ B_y\mathbf{k}\wedge\mathbf{i}+ B_z\mathbf{i}\wedge\mathbf{j}##
     
  4. Jan 21, 2014 #3
    I asked what is e11 in terms of matrix...
     
  5. Jan 21, 2014 #4

    tiny-tim

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    e11 doesn't exist :confused:

    e1 ##\wedge## e1 = 0​
     
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