General form of an isotropic fourth rank tensor According to [1], a representation of the most general isotropic tensor of rank 4 is
$$ T_{ijkl} = \lambda \delta_{ij}\delta_{kl} + \mu(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) + \nu(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) $$
How is it derived? 
I'm trying to understand the Navier-Stokes equation [2] and is crucial to understand that tensor representation. 
 A: A general derivation can be found in 'On Isotropic Cartesian Tensors' by Hodge in 1961 or here which is based on the mentioned reference. Another 'derivation' based on the individual components is here
Since the derivation is well documented I only outline the necessary steps.
Below we follow the one given by Hodge.
You start from the definition of isotropic tensors and require invariance under infinitesimal rotations to arrive at the condition
$$
\epsilon_{mis}a_{sjkl} 
+ \epsilon_{mjs}a_{iskl} 
+ \epsilon_{mks}a_{ijsl} 
+ \epsilon_{mls}a_{ijks} = 0
$$
Which you multiply by $\epsilon_{mit}, \epsilon_{mjt}, \epsilon_{mkt}, \epsilon_{mlt}$ and set $t=i, t=j, t=k, t=l$ respectively to obtain 4 equations.
This equations you 'simplify' by realizing that the 4th order isotropic tensors with two internal indices contracted are actually 2nd order isotropic tensors, which are proportional to the Kronecker delta, i.e. you can replace terms like $a_{iikl} = \lambda \delta_{kl}$. This results in equations like
$$
2 a_{ijkl} + a_{jikl} + a_{kjil} + a_{ljki} = 
\lambda\delta_{ij}\delta{kl}
+ \mu\delta_{ik}\delta_{jl}
+ \nu\delta_{il}\delta_{jk} \\
\vdots
$$
From this 4 equations you can deduce relations like
$$
a_{ijkl} = a_{jilk} = a_{klij} = \ldots
$$
by subtracting the summed pairs from each other.
Using the relations you found in this way you can rewrite the former 4 equations to yield
$$
2 a_{ijkl} + a_{ijlk} + a_{ikjl} + a_{ilkj} = 
\lambda\delta_{ij}\delta{kl}
+ \mu\delta_{ik}\delta_{jl}
+ \nu\delta_{il}\delta_{jk} \,.
$$
Here you obtain two more equations via cyclic permutation of $j,k,l$ while leaving $i$ untouched.
The resulting 3 equations are then summed followed by a symmetry argument to give
$$
a_{ijkl} + a_{iklj} + a_{iljk}
= a_{ijkl} + a_{ikjl} + a_{ilkj} 
= \frac{1}{5} (\lambda + \mu + \nu) (
\delta_{ij}\delta{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}
)\,.
$$
Substituting this int he equation before while defining
$$
    \alpha = \frac{4 \lambda - \mu     - \nu}{10}, \qquad \qquad
    \beta  = \frac{4 \mu     - \lambda - \nu}{10}, \qquad \qquad
    \gamma = \frac{4 \nu     - \lambda - \mu}{10}
$$
You get the general form of a isotropic tensor of 4th rank
$$
a_{ijkl} = \alpha \delta_{ij}\delta_{kl} 
+  \beta \delta_{ik}\delta_{jl} 
+  \gamma \delta_{il}\delta_{jk}
$$
You can easily find a relation that yields the form you stated with the one given here.
