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.

    Reply
    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)}.

      Reply
      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

        Reply

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