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Determinat formula in Einstein notation

  1. Jul 10, 2012 #1
    Hi all,

    I've been looking around at formulae for determinants (using them for tensor densities) and I just want to clarify that the expression below is correct (i.e. formulae are correct):
    [tex] |M| = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{1a_1}M_{2a_2} \ldots M_{na_n} = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{a_11}M_{a_22} \ldots M_{a_nn}[/tex]
    The reason I ask is that the second formulae lends itself to Einstein notation:
    [tex] |M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_1M^{a_2}{}_2 \ldots M^{a_n}{}_n[/tex]

    As an aside question, is this correct in the sense that there are unmatched indices on each side of the equation? I have found the following formula which seems to correct this:
    [tex] \epsilon_{b_1b_2 \ldots b_n}|M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_{b_1}M^{a_2}{}_{b_2} \ldots M^{a_n}{}_{b_n}[/tex]
    I think they are both correct...?

  2. jcsd
  3. Jul 10, 2012 #2

    Ben Niehoff

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    [tex]|M| = \frac{1}{n!} \varepsilon^{b_1 \ldots b_n} \varepsilon_{a_1 \ldots a_n} M^{a_1}{}_{b_1} \ldots M^{a_n}{}_{b_n}[/tex]
  4. Jul 10, 2012 #3
    OK, that's good. Putting [itex]\epsilon_{b_1 \ldots b_n}[/itex] on both sides gives my final formula.

    Also, according to http://en.wikipedia.org/wiki/Levi-Civita_symbol#Determinants does this not imply that my second formula is equivalent to yours?
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