Let $x\in \Omega\subseteq \mathbb{R}^n \to V(x)\in \mathbb{R}^n$ a smooth vectorial field defined on a open set. It is well known that in Cartesian coordinates the differential is represented by the Jacobian matrix. I would like to know what characterization holds for tensor fields.
Let $x\in \Omega\subseteq \mathbb{R}^n \to A(x)\in \mathcal{Lin}(\mathbb{R}^n, \mathbb{R}^n)$ a smooth map from an open subset in to the space of linear applications (i.e. a second order tensor field). I racall that the differential in $x$ is the operator $L(x)$ such that $$ A(x+h)- A(x) =L(x)[h] + o(||h||) $$ as $h\to 0$ (so in our case is a third order tensor). Is it possible to characterize it? At least in Cartesian coordinates?
Yes, if you fix a common basis in the domain and codomain of $ A:\mathbb{R}^n\to\mathbb{R}^n $, then you get a map from $ \mathbb{R}^n $ to the smooth manifold $ M(n,\mathbb{R}^n) $, which can be seen as $ \mathbb{R}^{n^2} $:$$ T:\mathbb{R}^n\to M(n,\mathbb{R})\cong\mathbb{R}^{n^2},x\mapsto(A_i^j) .$$ Then the standard definition of differential of maps between smooth manifolds is given as:$$ dT=\frac{\partial A_i^j}{\partial x^k}dx^k\otimes dx^i\otimes \frac{\partial}{\partial x^j} .$$ If you are not familiar with $ dx^k\otimes dx^j\otimes \frac{\partial}{\partial x^i} $, you can think that $ dT $ is a $ (1,2)- $tensor, whose components read as $ (dT)_{ki}^j=\frac{\partial A_i^j}{\partial x^k} $.