Symmetry and Antisymmetry

Relevant parts to questions...

• A rank-2 tensor is symmetric if $T_{ij}=T_{ji}$, antisymmetric if $T_{ij}=-T_{ji}$.
• Any tensor splits uniquely as symmetric + antisymmetric parts: $T_{ij}=\frac{1}{2}(T_{ij}+T_{ji})+\frac{1}{2}(T_{ij}-T_{ji})$.
• Symmetry is frame-independent under orthogonal transformations.
• Symmetry is defined for indices at the same level (both upper, or both lower).

A second-rank tensor $T_{ij}$ is:

• Symmetric::$T_{ij}=T_{ji}$.
• Antisymmetric (skew-symmetric)::$T_{ij}=-T_{ji}$ (which forces $T_{ii}=0$, no sum).

Examples from earlier notes: the Kronecker Delta ${\delta }_{ij}={\delta }_{ji}$ is symmetric; the Alternating Tensor ${ϵ}_{ijk}=-{ϵ}_{jik}$ is antisymmetric in any two indices.

Frame Independence

Symmetry Lemma

If $A_{ij}$ is symmetric in one Cartesian frame, it is symmetric in every Cartesian frame.

Proof. If $A_{ij}=A_{ji}$, then under a rotation:

$$A_{ij}^{\prime }=L_{ik}{L}_{jm}{A}_{km}=L_{ik}{L}_{jm}{A}_{mk}=L_{jm}{L}_{ik}{A}_{mk}=A_{ji}^{\prime }$$

The same argument with a sign flip proves antisymmetry is also frame-independent. ✓

Symmetric/Antisymmetric Decomposition

Decomposition Theorem

Any rank-2 tensor $T_{ij}$ decomposes uniquely as

$T_{ik}=S_{ik}+A_{ik}$, where

$S_{ik}=\frac{1}{2}(T_{ik}+T_{ki})$ is the symmetric part,

$A_{ik}=\frac{1}{2}(T_{ik}-T_{ki})$ is the antisymmetric part.

The decomposition is immediate from $S_{ik}+A_{ik}=T_{ik}$, and uniqueness follows because if $T=S_1+A_1=S_2+A_2$, then $S_1-S_2=A_2-A_1$ is simultaneously symmetric and antisymmetric, hence zero.

In Generalised Coordinates

Symmetry lives at fixed index positions

In a generalised coordinate system, symmetry or antisymmetry is defined only for indices at the same level:

• $A_{ik}^{\cdot \cdot \ell }$ is symmetric in $i,k$ if $A_{ik}^{\cdot \cdot \ell }=A_{ki}^{\cdot \cdot \ell }$ (both lower).
• $B_{\cdot \cdot \ell }^{ik}$ is antisymmetric in $i,k$ if $B_{\cdot \cdot \ell }^{ik}=-B_{\cdot \cdot \ell }^{ki}$ (both upper).

You cannot compare symmetry between one upper and one lower index - the concept requires the same index position.

Raising or lowering a single index via the Metric Tensor can break symmetry, since it changes which metric components appear.

Applications

1. Alternating-tensor contraction kills the symmetric part::for any $T_{jk}$, ${ϵ}_{ijk}{T}_{jk}$ is sensitive only to the antisymmetric part (the symmetric part contracts to zero by the antisymmetry of $ϵ$). So ${ϵ}_{ijk}{T}_{jk}=0$ forces $T_{jk}$ to be symmetric.
2. Moment of inertia / stress tensors in physics are symmetric - a physical, frame-independent statement.
3. Curl and angular velocity correspond to the antisymmetric part of $\frac{\partial u_i}{\partial x_j}$.

${ϵ}_{ijk}{T}_{jk}=0⟹T_{ij}$ symmetric

Fix $i=1$: ${ϵ}_{1jk}{T}_{jk}={ϵ}_{123}{T}_{23}+{ϵ}_{132}{T}_{32}=T_{23}-T_{32}=0⟹T_{23}=T_{32}$.

Repeating for $i=2$ and $i=3$ gives $T_{12}=T_{21}$ and $T_{13}=T_{31}$, so $T_{ij}=T_{ji}$. $\boxed{T_{ij}=T_{ji}}$ ✓

Symmetric ↔ $+$ swap, antisymmetric ↔ $-$ swap. Any tensor is the sum of the two.