Proof? Kronecker delta is the only isotropic second rank tensor

[mSSM]

It is pretty straight forward to prove that the Kronecker delta \delta_{ij} is an isotropic tensor, i.e. rotationally invariant.

But how can I show that it is indeed the only isotropic second order tensor? I.e., such that for any isotropic second order tensor T_{ij} we can write
<br /> T_{ij} = \lambda \delta_{ij}<br />

 

[Fredrik]

Not sure what the answer is, or even what the question means. So this may or may not be useful.

What does "rotationally invariant" mean here? Is it something like $T_{ij}=R_{ik}T_{kl}R_{lj}$? This can be written as a matrix equation: $T=RTR^T$. If this is supposed to hold for all R, you can try many different choices of R. Each choice gives you a little more information about the components of T. Perhaps you can find a bunch of rotations that give you enough conditions on T to determine all its components.

 

[mSSM]

Rotationally invariant in this context means that
<br /> T_{ij} = T&#039;_{ij} = R_{ip} R_{jq} T_{pq}<br />

For example, for the Kronecker delta this is pretty straight forward to show:
<br /> \delta_{ij} = R_{ip} R_{jq} \delta_{pq} = R_{ip}R_{jp} = \delta_{iq}\text{,}<br />
where we have used that R_{ij} are orthogonal matrices, i.e. \mathbf{R}\mathbf{R}^T = \mathbf{1}.

 

[dextercioby]

Do you know Schur's lemma ?

 

[Fredrik]

Rotationally invariant in this context means that
<br /> T_{ij} = T&#039;_{ij} = R_{ip} R_{jq} T_{pq}<br />

Ah, yes, that's what I was thinking, but I put two indices in the wrong order. What I wrote corresponds to the matrix equation $T=RTR$. What you wrote corresponds to $T=RTR^T$. My idea should still work, but I suspected that there would be a fancier way, and it looks like dextercioby has given you one.

 

[TrickyDicky]

It is pretty straight forward to prove that the Kronecker delta \delta_{ij} is an isotropic tensor, i.e. rotationally invariant.

But how can I show that it is indeed the only isotropic second order tensor? I.e., such that for any isotropic second order tensor T_{ij} we can write
<br /> T_{ij} = \lambda \delta_{ij}<br />

It is probably worth noting that the uniqueness only applies to (cartesian) tensors in Euclidean space.

 

[lpetrich]

If a tensor T is rotationally invariant, that means that for every rotation R, that T = R^T.T.R. Note that R^T = R^-1.

Since pure rotations form a Lie group, we can use its Lie algebra: R = 1 + ε*L for small ε. Since L^T = -L, that gives us commutator [L,T] = 0.

A rotation-algebra generator has form (L_ab)_ij = δ_aiδ_bj - δ_ajδ_bi to within some multiplicative factor. Its commutator with T is

[L_ab,T]_ij = δ_aiT_bj - δ_biT_aj - δ_bjT_ia + δ_ajT_ib

If a,b,i,j are all different, then this expression is zero. But let's try ab = 12 and ij = 13 for definiteness. Then it equals T₂₃. Thus, for number of dimensions >= 3, all off-diagonal T is zero.

Let's turn to ab = 12 and ij = 11. We get T₂₁ + T₁₂ = 0, or T_ij ~ ε_ij, the antisymmetric symbol.

Turning to ab = 12 and ij = 12, we get T₂₂ - T₁₁ = 0

We thus find two possible invariants: T ~ δ for all numbers of dimensions and also T ~ ε for two dimensions.

-

Alternately, we can contract the commutator on b and j, giving δ_aiTr(T) - T_ai - (n-1)*T_ia for n dimensions.

Adding (ai) and (ia) gives 2*δ_aiTr(T) - n*(T_ai + T_ia)

Thus, the symmetric part of T is proportional to δ.

Subtracting instead gives (n-2)*(T_ia - T_ai)

Thus, the antisymmetric part of T vanishes, except for 2 dimensions, where it is proportional to ε.