Symmetric tensor

A symmetric tensor is unchanged by every permutation of its slots. A symmetric second-rank tensor is the rank-two case; symmetry for higher rank requires invariance under every such slot exchange.

For $v=(a,b,0)$ with $r=\sqrt{a^2+b^2}>0$, the symmetric tensor $S=v{v}^T+c{r}^2{e}_3{e}_3^T$ has principal axeswith eigenvalues $r^2,0,c{r}^2$, respectively. This follows directly from the outer product: $v{v}^{T}v=r^{2}v$, while it annihilates vectors perpendicular to $v$. The orthogonal matrix with rows $e_r^T,e_{\theta }^T,e_3^T$ gives $MS{M}^T=\mathrm{diag}(r^2,0,c{r}^2)$.

If $a,b$ depend on position, this is a pointwise eigenframe, not necessarily one fixed Cartesian change of axes throughout space. For $v=0$, the displayed tensor vanishes and every orthonormal basis is a principal basis. Subtracting a scalar multiple of the identity leaves its eigenvectors unchanged and shifts every eigenvalue by the same amount; this is a simple case of coaxial symmetric tensors.

A symmetric tensor invariant under rotations about a fixed nonzero vector $y$ has the form $\alpha I+\beta y{y}^T$. Choose the third coordinate axis parallel to $y$: rotation by a half-turn makes the mixed axial-transverse entries vanish, and all transverse rotations make the transverse block a scalar multiple of the identity. Its two transverse eigenvalues are $\alpha$ and its axial eigenvalue is $\alpha +\beta |y{|}^2$. The trace and the quadratic contraction $y^{T}Ty$ determine these two coefficients. Full isotropy holds exactly when $\beta =0$.

For $|y|=a$ and $n>-1$, this surface integral is finite: near $x=y$ its integrand has size $O(|x-y{|}^{2n})$ and the two-dimensional area element supplies one further power of distance. Rotational covariance and symmetry make it an axisymmetric symmetric rank-two tensor. Put $K=\pi (2a{)}^{2n+2}$. Polar integration about the $y$ axis gives $\mathrm{tr}T=K/(n+1)$ and $y^{T}Ty=a^{2}K/(n+2)$. Solving $3\alpha +\beta a^2=\mathrm{tr}T$ and $\alpha +\beta a^2=y^{T}Ty/a^2$ gives $\alpha =K/[2(n+1)(n+2)]$ and $\beta =\alpha (2n+1)/a^2$. Thus the weighted chord tensor becomes an isotropic tensor exactly at $n=-1/2$.

Over a field of characteristic zero, degree-$n$ symmetric tensors are spanned by pure powers $a^{\otimes n}$. Inclusion-exclusion expands ${\sum }_{J\subseteq [n]}(-1{)}^{n-|J|}({\sum }_{j\in J}{a}_j{)}^{\otimes n}$ into the sum of all permutation orderings of $a_1\otimes \cdots \otimes a_n$. Every surviving term uses each of the $n$ labels once. This converts an invariant-tensor spanning problem into pure tensor powers.

A second-rank tensor is symmetric when $T_{ij}=T_{ji}$. Every second-rank tensor has the unique decompositioninto symmetric and antisymmetric parts.