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Permutations and Transpositions

  1. Mar 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Attached are some screen shots of portion of the textbook I'm currently working through:
    Capture.PNG Capture'.PNG

    2. Relevant equations


    3. The attempt at a solution

    My first question, why exactly can't ##\Delta## contains ##x_p - x_q## only once (note, switched from ##i,j## to ##p,q##)? As you can see, the author didn't give many very details concerning this. Clearly ##\Delta## can also be written ##\Delta = \prod_{(i,j) \in S} (x_i - x_j)##, where ##S = \{(i,j) ~|~ 1 \le i < j \le n \}##. Since sets don't contain duplicates of elements, ##S## won't contain any pair ##(p,q)## twice, implying that ##x_p - x_q## won't appear in ##\Delta## more than once. Would this be the reason, that ##S## cannot contain duplicates? Seems to be a rather unremarkable reason, but if it gets job done...

    Next, I am trying to prove that ##\sigma(\Delta)## contains either ##x_p - x_q## or ##x_q - x_p##, but not both. For simplicity, let ##g = \sigma^{-1}##. Suppose that ##\sigma (\Delta)## contains both factors. Then ##\sigma (\Delta) = (x_p - x_q)(x_q - x_p) \prod_{(i,j) \in S \setminus \{(p,q),(q,p)\}}##, and therefore

    $$g (\sigma(\Delta)) = (x_{g(p))} - x_{g(q)})(x_{g(q)} - x_{g(p)}) \prod (x_{g(i)} - x_{g(j)})$$

    $$\Delta = - (x_{g(p))} - x_{g(q)}) (x_{g(p))} - x_{g(q)}) \prod (x_{g(i)} - x_{g(j)})$$,

    showing that ##\Delta## contains ##(x_{g(p))} - x_{g(q)})## twice, contradicting what we showed above.

    I know: it isn't great. For one thing, the RHS could be ##- \Delta##, so that is one flaw in the argument. I hope someone can help. For all DF's verbosity, it doesn't really clearly spell out the details very well, which is why I don't like DF very much, although it has massive number of problems.
     
  2. jcsd
  3. Mar 13, 2017 #2

    haruspex

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    I presume you mean, why not more than once.
    It's because of the definition of the product. It is taken over the pairs (i,j) for which i<j. Therefore for a given pair of indices with i<j, the pair (i,j) occurs exactly once and the pair(j,i) does not occur at all.

    For the same result after permuting the indices, it seems reasonably obvious to me, so I would be happy with the text as it stands. But if you feel it needs to be proved, your proof looks ok.
     
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