TITLE: Why does $\nabla$ behave like a member of $\mathbb{R}^3$ (Euclidean vector in 3d space) in many cases?
QUESTION [10 upvotes]: Is there a reason why $\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$, behaves like a member of $\mathbb{R}^3$ (Euclidean vector in 3d space) in so many cases:


Dot products and cross products of $\vec{\nabla}$ with multivariable vector function like $\vec{\nabla} \cdot \mathbf{\vec{f}}$ and $\vec{\nabla} \times \mathbf{\vec{f}}$ or with scalar functions (gradient) are vectors (i.e. independent of particular choice of co-ordinates).




Properties like the Lagrange's formula that are valid for vectors have direct analogues involving the $\vec{
\nabla}$ operator.




We can write pseudo-determinants for curls just like we could with components of vectors.




Most of its vector properties hold if $\vec{\nabla}$ is replaced by any other vector.


There will be separate proofs for all of these properties. But is there any underlying reason that works for all such properties? Is there a fundamental reason why $\vec{\nabla}$ is $\mathbb{R}^3$ vector-ish?

REPLY [2 votes]: $$\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$$ behaves like a vector because $$\vec{\nabla}f $$  is a vector.
The vector space in which it is defined is not the usual $ \mathbb R^n$
It is a linear operator which operates on functions and the outcome is a vector whose components are partial derivatives of the function.