Determinat formula in Einstein notation

1. Jul 10, 2012

ianhoolihan

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):
$$|M| = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{1a_1}M_{2a_2} \ldots M_{na_n} = \sum^n_{a_1,a_2, \ldots ,a_n = 1} \epsilon_{a_1a_2 \ldots a_n} M_{a_11}M_{a_22} \ldots M_{a_nn}$$
The reason I ask is that the second formulae lends itself to Einstein notation:
$$|M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_1M^{a_2}{}_2 \ldots M^{a_n}{}_n$$

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:
$$\epsilon_{b_1b_2 \ldots b_n}|M| = \epsilon_{a_1a_2 \ldots a_n} M^{a_1}{}_{b_1}M^{a_2}{}_{b_2} \ldots M^{a_n}{}_{b_n}$$
I think they are both correct...?

Cheers

2. Jul 10, 2012

Ben Niehoff

Try

$$|M| = \frac{1}{n!} \varepsilon^{b_1 \ldots b_n} \varepsilon_{a_1 \ldots a_n} M^{a_1}{}_{b_1} \ldots M^{a_n}{}_{b_n}$$

3. Jul 10, 2012

ianhoolihan

OK, that's good. Putting $\epsilon_{b_1 \ldots b_n}$ 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?