what does $(\nabla_x + \nabla_\xi)^n$ mean? (In the context, $x,\xi\in \mathbb{R}^d$). In my notes, the author states this should be a tensor, but what exactly does he mean. Is this clear notation?
My operator acts on a function $f:\,\mathbb{R}^d\times\mathbb{R}^d \to \mathbb{R}$.
So in one dimensions, this is $(\partial_x + \partial_\xi)^n f$.
Note, the dimension is $d$ and in general $d\neq n$.
On
It seems to me that it means an operator defines using covariant derivatives, which we can write recursively, as in: $$(\nabla_x + \nabla_\xi)^nV = (\nabla_x+\nabla_\xi)(\nabla_x+\nabla_\xi)^{n-1}V$$ for $n > 1$, and $$(\nabla_x+\nabla_\xi)V = \nabla_x V + \nabla_\xi V,$$ where $\nabla_x V = (x[v_1], \ldots, x[v_d])$ and $\nabla_\xi V = (\xi [v_1], \ldots, \xi[v_d])$, where $V = (v_1,\ldots, v_d)$ is a vector field in $\Bbb R^d$. If we had more context, maybe I could speculate something better..
I also saw the notation $\nabla_\xi f$ to designate the directional derivative $\frac{\partial f}{\partial \xi}$, but I am not sure if this is standard. The above remarks would still apply in this case, with this interpretation.
It is meant that $n$ is dimension of the tensor. This means that $$\left((\nabla _x )^n\right)_{i_1,\dots i_n} =\partial_{x_{i_1}}\dots \partial_{x_{i_n}}$$ where $i_j \in \{1,\dots d\}$ for $j\in\{1,\dots, n\}$.