Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's have symbol $\delta_{ijkl}$. Is it equal to 1 for $i = j = k = l$ and $0$ in the other cases?

share|cite|improve this question

1 Answer 1

It depends on the context. $\delta_{ijkl}$ could be defined as a generalization of Kronecker delta (then it is -as you wrote- equal to 1 if $i=j=k=\ell$ and 0 otherwise), but it could also denote the Levi-Civita symbol. In 4 dimensions it's defined as

$$\delta_{ijkl} := \frac{(i-j) \cdot (i-k) \cdot (i-l) \cdot (j-k) \cdot (j-l) \cdot (k-l)}{12}$$

where $i,j,k,l \in \{1,2,3,4\}$. If you want to generalize it you can use permutations, so

$$\delta_{ijkl} = \begin{cases} 1 & (i,j,k,l) \, \, \text{is an even permutation} \\ -1 & (i,j,k,l) \, \, \text{is an odd permutation} \\ 0 & \text{otherwise} \end{cases}$$

share|cite|improve this answer
I met this Kroneker delta when read about a 4-rank stiffness tensor written for cube-symmetric body. It means that it's invariant under rotations at pi/2 angle. It's components can be representated as $$ a_{ijkl} = b\delta_{ij}\delta_{kl} + b(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) + c\delta_{ijkl}, $$ where the last summand is invariant under rotations only at angle $\pi k/2$ round an x-,y-,z- axis. I can't prove it without determine a $\delta_{ijkl}$. – John Taylor Dec 19 '12 at 11:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.