I read article about stress tensor on wiki and find following formula at the top:
$$ \mathbf{T}^{(\mathbf n)}= \mathbf n \cdot\boldsymbol{\sigma}\quad \text{or} \quad T_j^{(n)}= \sigma_{ij}n_i. $$
As far I understand $T^{(n)}$ is covariant (row; "horizontal") vector - we can find it explicit definition few lines above this paragraph on that wiki aritcle which looks as follows:
$$ \left[{\begin{matrix} T^{(\mathbf n)}_1 & T^{(\mathbf n)}_2 & T^{(\mathbf n)}_3\end{matrix}}\right]=\left[{\begin{matrix} n_1 & n_2 & n_3 \end{matrix}}\right]\cdot \left[{\begin{matrix} \sigma _{11} & \sigma _{12} & \sigma _{13} \\ \sigma _{21} & \sigma _{22} & \sigma _{23} \\ \sigma _{31} & \sigma _{32} & \sigma _{33} \\ \end{matrix}}\right] $$
As we see to calculate it using matrix multiplication we need to assume that normal vector $\mathbf n$ is row vector (for column vector left-multiplication is not allowed).
As far I know typical vector is contravariant (column; "vertical") vector .
I wonder: what is the 'default' form of normal vector in math - is it column vector or it is row vector?
$$ \mathbf n =\qquad [n_1\ n_2\ n_3]\qquad or \qquad \begin{bmatrix} n_1 \\ n_2 \\ n_3 \\ \end{bmatrix} \qquad ? $$
Second question: Are there some advantages of using row vector?