= Solution
For a surface with unit normal $\mathbf n$, <traction> is $t_i=\tau_{ij}n_j$. Thus $\tau_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is $\mathbf e_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile <normal stress> positive in this convention.
The symmetric <strain> has six independent components and so does the symmetric <stress>. A general <linear map> between these two six-dimensional spaces therefore has
$$
\boxed{6\times6=36\text{ independent constitutive coefficients}.}
$$
Equivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the <strain> indices have no observable effect and are omitted from this representative.
A further distinction is important: if a <strain energy density> exists, $C_{ijkl}=\partial^2W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot7/2=21$ independent coefficients. Thus \b[36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption], as in the variational setting later in the paper. These are the <minor and major symmetries of linear elasticity>.
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