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Anyone familiar with Peano' (or Grassman) algebra?

  1. Mar 7, 2004 #1

    enigma

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    I'm an engineer, not a mathematician...

    the professor has assigned several proof questions, and I'm having difficulty answering them...

    (This may need to be moved to homework help, but the topic is unusual, so I thought I'd get better response here)

    Terminology: v is a join operation, ^ is a meet operation D[] is a bracket operation

    Example:

    Show that when:

    [tex] A = a_1 \vee a_2, B = b_1 \vee b_2, \ and\ n = 2 [/tex]

    then

    [tex] A \wedge B = -D[a_2 , b_1 , b_2] a_1 + D[a_1 , b_1 , b_2] a_2 [/tex]
    [tex]= D[a_1 , a_2 , b_2] b_1 - D[a_1 , a_2 , b_1] b_2 [/tex]

    If the meet is zero, then assuming that the sets [itex] (a_1, a_2) [/itex] and [itex] (b_1, b_2) [/itex] are both independent, the four brackets must be zero. Show that, in this case, the two subspaces [itex] A_s[/itex] and [itex] B_s [/itex] are the same.

    Now, proving the first part is fairly simple: just run though the definition of the meet.

    I'm having difficulty with the second part. I thought that the definition of the meet is the intersection of the two subspaces. If the subspaces are the equal, then wouldn't the meet be either As or Bs?
     
    Last edited: Mar 7, 2004
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  3. Mar 7, 2004 #2

    enigma

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    Answering my own question somewhat:

    OK. If the sets are independant, and [itex] \alpha a_1 + \beta a_2 = 0[/itex] , then [itex] \alpha \ and \ \beta[/itex] need to be zero because independancy implies [itex]a_1 \ and \ a_2[/itex] are not equal.

    If [itex] D[a_i, b_i, b_j] \ and \ D[a_i, a_j, b_i] [/itex] are zero for any combination of i and j, that means that the two subspaces A and B are linearly dependant for any combination of bases. This must mean that they are the same subspace.

    Is that correct?
     
  4. Mar 7, 2004 #3

    Janitor

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    Pardon my ignorance; I have barely ever done any reading on Grassman algebra. I probably won't be able to help you. But what is the definition of n?
     
  5. Mar 7, 2004 #4

    enigma

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    n is the number of dimensions in addition to the projective plane: PA[Rn+1], where PA represents peano algebra.

    so, the dimensions of a two dimensional projective space is PA[R2+1] corresponding to e0 e1 and e2 as the three coordinates, with e1 being the x direction, e2 being the y direction and e0 being the perspective direction.

    *I hope I'm not botching up my terminology here...*
     
  6. Mar 7, 2004 #5

    Janitor

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    Ummm

    I'll get back to you in a few weeks on this.
     
  7. Mar 28, 2004 #6

    Janitor

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    You must know the answer by now! If you have time, could you sketch it out for us? If you're too busy, that's understandable.
     
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