In continuum mechanics, why is the stress vector $T=\sigma\cdot n$ not a covector? In continuum mechanics, the stress vector (see Cauchy stress tensor) $T=\sigma\cdot n$ is the surface density of a force. Forces are covectors, since they map a displacement vector to a scalar energy. Then, why is $T$ not a covector too?
Edit: I talking about the stress vector, defined for a given unit normal vector $n$ by $\sigma\cdot n$, not to be confused with the stress (second-order) tensor $\sigma$.
 A: When people study continuum mechanics they usually do so at first in $\mathbb{R}^3$ where we have usually implied the usual metric tensor $(g_{ij}) = \operatorname{diag}(1,1,1)$ and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert between vector fields and one-forms by means of raising and lowering indices.
So if $M$ is your space and $(x,U)$ a coordinate system, if $X$ is a vector field, which on $U$ can be written in coordinates as
$$X = X^i\dfrac{\partial}{\partial x^i},$$
then $g$ allows you to build the one-form equivalent to it by setting $\omega = g(X,\cdot)$, that is, in $U$ we can write
$$\omega(Y) = g(X,Y) \Longrightarrow \omega = g_{ij}X^idx^j,$$
where $g_{ij}$ are the components of $g$ on the coordinate system $(x,U)$, that is, functions that allows us to write $g = g_{ij}dx^i\otimes dx^j$.
Now the stress tensor you speak off is usually defined as a linear map that takes vectors into vectors: it is capable of taking one normal and giving back one force. Now linear maps on a vector space $V$ may be identified with the tensor product $V\otimes V^{\ast}$ and so linear maps and tensors of type $(1,1)$ are the same.
In that setting it is best to think about the stress tensor as this $(1,1)$ tensor $\sigma$ which on $(x,U)$ can be written
$$\sigma = \sigma^{i}_{j} \dfrac{\partial}{\partial x^i}\otimes dx^j.$$
Now in the same way such a tensor can map vectors to vectors it can map covectors to covectors. In that way, if you consider force as a covector, $\sigma$ can map it to. Now because you have a metric tensor, those operations can all be "matched" using the musical isomorphism. More importantly, when $g$ is the usual metric tensor of $\mathbb{R}^3$ you see no difference at all.
