Solid Mechanics -$\nabla\times\nabla\times\varepsilon = 0$ - having trouble with Einstein notation Note: this is not an assignment for a grade, just me trying to improve my solid mechanics.
The task at hand is to show that the compatibility condition $$\nabla\times\nabla\times\varepsilon = 0$$ implies $$\varepsilon_{ij,kl} + \varepsilon_{kl,ij} - \varepsilon_{il,jk} - \varepsilon_{jk,il}=0$$
The hint is to "multiply" by two additional alternating tensors. I'm kind of stuck thinking about $\varepsilon$ being the matrix
$$\varepsilon = \frac{1}{2}(\nabla\textbf{u}+\nabla\textbf{u}^T)$$
So how do I even define the curl? Is the curl applied row-wise? I tried drawing out the resulting matrices but I'm not sure how to write them in the notation required to compare to the second equation above. 
Is there any advice on how to approach this? Any questions y'all could give to simplify the steps and I'll try to answer from there? Many thanks.
 A: The important thing about the curl is that it is the anti-symmetric part of the derivative.
In Einstein notation, $\varepsilon$ would be given by $$\varepsilon_{ij} = \frac{1}{2} ( u_{i,j} + u_{j,i}).$$
Also in Einstein notation, we pick out the anti-symmetric part with the Levi-Civita symbol $\epsilon_{ijk}$. Thus the curl is $$(\nabla\times\varepsilon)_{ij} = \epsilon_{iln} \varepsilon_{jl,n} = \frac 1 2 \epsilon_{iln} (u_{j,ln} + u_{l,jn}) = \frac{1}{2} \epsilon_{iln} u_{l,nj} %= \frac{1}{2}(\nabla \times u)_{i,j}.$$
Note that since $\varepsilon_{ij}$ is symmetric, it doesn't matter if its first or second index is contracted with the Levi-Civita, and the last equality is symmetry of partial derivatives. We will only need the first equality, though.
Now, we can apply the curl again, $$(\nabla\times(\nabla\times\varepsilon)\big)_{ij} = \epsilon_{imk}\epsilon_{jln} \varepsilon_{ml,nk}.$$
Now the trick is to note that if $\epsilon_{ijk} T_{ij\ldots} = 0$ for all $k$, then in fact $T_{ij\ldots} - T_{ji\ldots} = 0$. Here $\ldots$ stand for additional indices carried by the tensor $T$. Applying it once, $$0 = \epsilon_{jln} (\varepsilon_{ml,nk} - \varepsilon_{kl,nm})$$
and twice,
$$0 = \varepsilon_{ml,nk} - \varepsilon_{mn,lk} - \varepsilon_{kl,nm} + \varepsilon_{kn,lm}.$$
This is the identity you're looking for, modulo relabeling indices and using that $\epsilon_{ml,nk}$ is symmetric in in $ml$ and $nk$.
