# Tensors: symmetric and anti-symmetric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.6.

Tensors, like matrices, can be symmetric or anti-symmetric. Since a tensor can have a rank higher than 2, however, a single tensor can have more than one symmetry. For a rank-2 tensor ${T^{ij}}$, it is symmetric if ${T^{ij}=T^{ji}}$ and anti-symmetric if ${T^{ij}=-T^{ji}}$. In matrix terminology, a symmetric rank-2 tensor is equal to its transpose, and an anti-symmetric rank-2 tensor is equal to the negative of its transpose.

A higher rank tensor can be symmetric or anti-symmetric in any pair of its indices, provided both indices are either upper or lower. For example, ${F_{\;\;\;\; lm}^{ijk}}$ can be symmetric or anti-symmetric in any pair selected from ${i,j,k}$ or in the pair ${l,m}$, but not in one upper and one lower index. Thus if ${F_{\;\;\;\; lm}^{ijk}=F_{\;\;\;\; lm}^{kji}}$ and ${F_{\;\;\;\; lm}^{ijk}=-F_{\;\;\;\; ml}^{ijk}}$, then ${F_{\;\;\;\; lm}^{ijk}}$ is symmetric in ${i}$ and ${k}$, and anti-symmetric in ${l}$ and ${m}$.

Returning to rank-2 tensors, we can show that the symmetry property is an invariant:

 $\displaystyle F^{\prime ij}$ $\displaystyle =$ $\displaystyle \frac{\partial x^{\prime i}}{\partial x^{k}}\frac{\partial x^{\prime j}}{\partial x^{l}}F^{kl}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial x^{\prime i}}{\partial x^{k}}\frac{\partial x^{\prime j}}{\partial x^{l}}F^{lk}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F^{\prime ji} \ \ \ \ \ (3)$

If a tensor is symmetric in a pair of upper indices, then if both indices are lowered, the resulting tensor is also symmetric in the two lower indices:

 $\displaystyle F_{ij}$ $\displaystyle =$ $\displaystyle g_{ik}g_{jl}F^{kl}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ik}g_{jl}F^{lk}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F_{ji} \ \ \ \ \ (6)$

Similarly, the anti-symmetric property persists through lowering of indices. If ${T^{ij}=-T^{ji}}$

 $\displaystyle T_{ij}$ $\displaystyle =$ $\displaystyle g_{ik}g_{jl}T^{kl}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -g_{ik}g_{jl}T^{lk}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -T_{ji} \ \ \ \ \ (9)$

If ${T^{ij}=-T^{ji}}$ then all diagonal elements must be zero, since ${T^{ii}=-T^{ii}}$ has only zero as a solution. Also, the trace is

 $\displaystyle T_{\;\; i}^{i}$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -g_{ij}T^{ji}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -g_{ji}T^{ji}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -T_{\;\; i}^{i} \ \ \ \ \ (13)$

In line 3, we used ${g_{ij}=g_{ji}}$, since in terms of the basis vectors, ${g_{ij}=\mathbf{e}_{i}\cdot\mathbf{e}_{j}}$, and thus the metric tensor is symmetric. Thus the trace is also zero for an anti-symmetric tensor.

A rank 2 symmetric tensor in ${n}$ dimensions has all the diagonal elements and the upper (or lower) triangular set of elements as independent components, so the total number of independent elements is ${1+2+\ldots+n=\frac{1}{2}n\left(n+1\right)}$. An anti-symmetric tensor has zeroes on the diagonal, so it has ${\frac{1}{2}n(n+1)-n=\frac{1}{2}n(n-1)}$ independent elements.

## 3 thoughts on “Tensors: symmetric and anti-symmetric”

1. Serkan

I think something wrong about last two summation. If there is n element in matrix, that matrix has n number of elements not n(n+1)/2. It might be about something about tensor; I would like to know.

1. growescience

An ${n\times n}$ matrix has ${n^{2}}$ elements, but if the matrix is symmetric, then all elements above the diagonal are the same as those below the diagonal. Thus the number of independent elements is 1 in the first row, 2 in the second row and so on, up to ${n}$ in the ${n}$th row. The total number of independent elements is thus as stated. I’ve just used the standard formula for the sum of integers from 1 to ${n}$, which is ${\frac{1}{2}n\left(n+1\right)}$.

1. Serkan

I thought matrix likes

1 2 3 4
5 6 7 8
……. n

but what you are saying is
independent elements
1 1
1 1 2
1 1 1 3
1 1 1 1 4

1 1 1 . ……. 1 n