# Proof - epsilon permutation and metric tensor relation

1. Jun 23, 2010

### symmetric

1. The problem statement, all variables and given/known data

$$\mbox{Prove that}\,g^{ij} \epsilon_{ipt}\epsilon_{jrs}\,=\, g_{pr}g_{ts}\,-\,g_{ps}g_{tr}$$
Notation :
$$e_{ijk}\,=\,e^{ijk}\,=\,\left\{\begin{array}{cc}1,&\mbox{ if ijk is even permutation of integers 123...n }\\-1, & \mbox{if ijk is odd permutation of integers 123...n}\\0&\mbox{in all other cases} \end{array}\right$$

$$\epsilon_{ijk}\,=\,\sqrt{g}e_{ijk}$$
$$\epsilon^{ijk}\,=\,\frac{1}{\sqrt{g}}e^{ijk}$$

$$\mbox{where}\,g\, = | g_{ij}| \mbox{ value of determinant formed by metric components of space }$$

2. Relevant equations

3. The attempt at a solution

$$\epsilon_{ipt}\epsilon_{jrs}\,=\,ge_{ipt}e_{jrs}$$

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\,g^{ij}ge_{ipt}e_{jrs}\,=\,g^{ij}\,g\left| \begin{array}{ccc}\delta_{ij}&\delta_{ir}&\delta_{is}\\ \delta_{pj}&\delta_{pr}&\delta_{ps}\\ \delta_{tj}&\delta_{tr}&\delta_{ts}\end{array}\right|$$

$$= g^{ij}\,\left| \begin{array}{ccc}g_{ij}&g_{ir}&g_{is}\\ g_{pj}&g_{pr}&g_{ps}\\ g_{tj}&g_{tr}&g_{ts}\end{array}\right|$$

$$= g^{ij}g_{ij} ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, g^{ij}g_{ir} ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, g^{ij}g_{is} ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )$$

$$= ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )$$

$$= g_{pr}g_{ts} \, - \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts} \, + \, g_{ps}g_{tr} \, + \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts}$$

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts}$$

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, - ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} )$$

Why am I getting unexpected -ve sign ?

2. Jun 25, 2010

### symmetric

Got the correction ! The modified solution is as follows -

From above solution continuing up to step -

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\, g^{ij}g_{ij} ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, g^{ij}g_{ir} ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, g^{ij}g_{is} ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )$$

then -

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\,=\delta_j^j( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )$$

$$= 3 ( g_{pr}g_{ts} \, - \, g_{ps}g_{tr} ) \, - \, \delta_r^j ( g_{pj}g_{ts} \, - \, g_{ps}g_{tj} ) \, + \, \delta_s^j ( g_{pj}g_{tr} \, - \, g_{pr}g_{tj} )$$

$$= 3( g_{pr}g_{ts} ) \, - \,3( g_{ps}g_{tr} ) \, - \, g_{pr}g_{ts} \, + \, g_{ps}g_{tr} \, + \, g_{ps}g_{tr} \, - \, g_{pr}g_{ts}$$

$$g^{ij}\epsilon_{ipt}\epsilon_{jrs}\, = \, g_{pr}g_{ts} \, - \, g_{ps}g_{tr}$$

which is same as required !