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Expression with levi-civita symbol

  1. Feb 17, 2010 #1
    Hello,
    during a calculation I got the following term:

    [tex]\varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n})[/tex]

    where [tex]\varepsilon[/tex] is the levi-civita symbol and [tex]a_i^j[/tex] are real numbers.

    Is it possible to simplify that expression?
     
    Last edited: Feb 17, 2010
  2. jcsd
  3. Feb 17, 2010 #2

    tiny-tim

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    Hi mnb96! :wink:

    I think you can write it [tex]{a_{[i_1 \ldots i_n]}}^{[j_1 \ldots j_n]}[/tex] :smile:

    (or maybe that's out by a factor of 1/n!2 ? :redface:)
     
  4. Feb 17, 2010 #3
    ?? is that the solution ??

    ...I was wondering if that expression might perhaps be equivalent to the determinant of something. Any ideas?
     
  5. Feb 17, 2010 #4

    tiny-tim

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    waa! I didn't notice that you changed the question a minute before I posted my answer. :cry:
     
  6. Feb 17, 2010 #5
    ...sorry :) :P
    I realized of the mistake later, after re-reading my post :(
     
  7. Feb 18, 2010 #6
    I thought about the following: since we have that,

    [tex]
    \varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} = det(A)
    [/tex]

    then,

    [tex]
    \varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) =
    [/tex]


    [tex]
    =\sum_{k=1}^{n!}sgn(\sigma_k(n))\cdot \varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} =
    [/tex]


    [tex]
    =\sum_{k=1}^{n!}sgn(\sigma_k(n))\cdot det(A)
    [/tex]

    where [tex]\sigma_k(n)[/tex] is the k-th permutation on the sequence [itex](1,\ldots,n)[/itex].
    Is that correct?
    Isn't that supposed to be zero?!
     
  8. Feb 18, 2010 #7

    tiny-tim

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    Hi mnb9! :smile:

    (just got up :zzz: …)

    I think it's clearer (and better, because it's accurate :wink:) to write it …

    [tex]
    \varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} = det((a_1,\cdots a_n))
    [/tex]

    then you get

    [tex]
    \varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) =
    [/tex]


    [tex]
    \varepsilon^{i_1 \ldots i_n}det((a_i_1,\cdots a_i_n))
    [/tex]


    which is … ? :smile:
     
  9. Feb 18, 2010 #8
    if it is true that given (1,...,n) there are always as many odd-permutations as even-permutations, then we get...zero! :) right?

    BTW, I have some troubles with your notation [tex]det((a_1,\cdots a_n))[/tex]. What are those [tex]a_i[/tex] ?
    I assume they are vectors.
    I also assume those [tex]a_{i1}[/tex] and [tex]a_{in}[/tex] are instead the vectors [tex]\mathbf{a}_{i_1}[/tex] and [tex]\mathbf{a}_{i_n}[/tex]
    Did I get it right?

    Finally, do you know a good source in which I can find useful exercises and tricks about manipulations with this notation?
     
    Last edited: Feb 18, 2010
  10. Feb 18, 2010 #9

    tiny-tim

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    Hi mnb96! :smile:
    I don't think so … doesn't interchanging two indices multiply both the epsilon and the determinant by -1 ?
    Yup! :biggrin:

    (I couldn't be bothered to write it properly :rolleyes:)
    Sorry, no idea. :redface:
     
  11. Feb 18, 2010 #10
    Damn! that's true, whenever we have an odd permutation the determinant should change its sign, but also the epsilon will be -1, so we finally have (crossing fingers):


    [tex]
    \varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) = n!\cdot det((\mathbf{a}_1, \ldots, \mathbf{a}_n))
    [/tex]

    ...Oh, BTW is it true that "given (1,...,n) there are always as many odd-permutations as even-permutations"? :redface:
     
  12. Feb 18, 2010 #11

    tiny-tim

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    (btw, i think we usually use small letter for "btw" :wink:)

    Only for n > 1 …

    try starting the proof with "if there exists an odd permutation π, then … " :smile:
     
  13. Feb 18, 2010 #12
    Ah...why?
    if n=1 we have:

    [tex]
    \varepsilon^{i_1}\varepsilon_{j_1}a_{i_1}^{j_1} = \varepsilon^{1}\varepsilon_{1}a_{1}^{1} = a_{1}^{1} = det(\mathbf{a}) = 1!det(\mathbf{a}) = n!\cdot det(\mathbf{a})
    [/tex]

    I assumed that permuting (1) zero times, means even permutation so that [tex]\varepsilon^{1}=+1[/tex]
     
  14. Feb 18, 2010 #13

    tiny-tim

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    That's a different question. :redface:

    (and there are no odd permuatations for n = 1 :wink:)
     
  15. Feb 18, 2010 #14
    Uhm...I am just wondering if we can just define the levi-civita symbol being equal to 1 in case the sequence has just one element, but that's not terribly important in any case :smile:

    This reinforces the fact that the levi-civita symbol cannot be -1 for n=1, but there is no repetition of indices, so it should not even be 0.
     
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