# What is the parity inversion of antisymmetric tensor

1. Mar 5, 2015

### genxium

First by antisymmetric tensor I mean the "totally antisymmetric tensor" like this:

$\epsilon^{\alpha\beta\gamma\delta} = \left\{ \begin{array}{clcl} +1 \;\; \text{when superscripts form an even permutation of 1,2,3,4} \\ -1 \;\; \text{when superscripts form an odd permutation of 1,2,3,4} \\ 0 \;\; otherwise \end{array} \right.$

I'm ok with that 3-vectors, 4-vectors are invariant under parity inversion. However I'm confused by WHAT IS THE PARITY INVERSION of antisymmetric tensor? There's NO COORDINATE in it.

Any help is appreciated :)

2. Mar 5, 2015

### Orodruin

Staff Emeritus
The answer is basically already in the link you posted. It relates the components in the new system to the determinant of the coordinate transformation. What happens with the determinant if you make a parity flip?

3. Mar 5, 2015

### genxium

Hi @Orodruin, do you mean that to verify whether parity inversion changes the antisymmetric tensor I can perform sth like:

$\left\{ \begin{array}{cl} \frac{\partial x'}{\partial x}=-1 \\ \frac{\partial y'}{\partial y}=-1 \\ \frac{\partial z'}{\partial z}=-1 \end{array} \right.$ and $J(parity \; inversion) = \left[ \begin{array}{clclcl} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} & \frac{\partial x'}{\partial z} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} & \frac{\partial y'}{\partial z} \\ \frac{\partial z'}{\partial x} & \frac{\partial z'}{\partial y} & \frac{\partial z'}{\partial z} \end{array} \right]$

Thus $det(J) = -1$ implies that antisymmetric tensor is inverted after the parity inversion?