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Is there a reason why $\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$, behaves like a vector in so many cases:

  1. 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).

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

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

  4. 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 so-vectorish?

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    $\begingroup$ could you share with us your definition of vector? In the most common mathematical usage of vector (ie an element of a vector space), $\nabla$ is trivially a vector. $\endgroup$ – Jonathan Mar 18 '18 at 12:59
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    $\begingroup$ The set of linear operators on some vector space is again a vector space, so if you want to you can consider a linear operator to be a vector (at least you can in standard mathematical usage). $\endgroup$ – Jonathan Mar 18 '18 at 13:08
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    $\begingroup$ @Jonathan But there are so many cases where we use $\nabla$ as if $\nabla \in \mathbb{R}^n$. Why are we able to do that, when clearly $\nabla \not \in \mathbb{R}^n$? $\endgroup$ – Truth-seek Mar 18 '18 at 13:11
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    $\begingroup$ The question is "why does $\nabla$ behave like a vector in $\mathbb{R}^{3}$?" I personally don't see why people are having difficulties understanding the statement of the question. $\endgroup$ – Will R Mar 18 '18 at 13:16
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    $\begingroup$ I think just changing "vector" to "element of $\mathbb{R}^{3}$" ought to be a big help. What I don't get is that you've put specific things you would like explained and peoples' responses have nothing to do with them, e.g., the fact that $\nabla$ is a linear operator does not imply that it should admit product formalism-like expressions. $\endgroup$ – Will R Mar 18 '18 at 13:24
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The gradient operator acts on a scalar differentiable function $f(\vec x)$, where $\vec x \in \mathbb R^n$, and returns a vector:

$$\text{grad} \ f(\vec x) = \nabla f(\vec x) \equiv \sum_{i=1}^n \frac{\partial f (\vec x)}{\partial x_i} \vec e_i $$

where $\{\vec e_i \dots\vec e_n\}$ is an orthogonal basis of $\mathbb R^n$.

The divergence operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^n$, and returns a scalar function:

$$\text{div} \ \vec F(\vec x) = \nabla \cdot \vec F(\vec x) \equiv \sum_{i=1}^n \frac{\partial F_i (\vec x)}{\partial x_i} $$

The curl operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^3$, and returns a vector field:

$$\text{curl} \ \vec F(\vec x) = \nabla \times \vec F(\vec x) = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat i+\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat j+ \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat k$$

where $\hat i, \hat j, \hat k$ are the unit vectors of the three Cartesian axes.

Notice that, unlike the gradient and divergence, the curl operator does not generalize simply in $n$ dimensions. Also, the notation $\nabla \times \vec F$ is only a mnemonic device useful when we work in cartesian coordinates: in other coordinate systems, applying $\nabla \times \vec F$ will hold the wrong result.

We should probably also mention the laplacian operator, which is the divergence of the gradient:

$$\nabla^2 f(\vec x) \equiv \text{div} \ (\text{grad} \ f(\vec x)) = \nabla \cdot (\nabla f(\vec x))$$


So, to sum up, $\nabla$ is just a useful notation that is used in the representation of three different vector operators. It turns out that we can often formally manipulate $\nabla$ as if it were a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless.

To see this, just consider one of the fundamental properties of vector spaces: if $v,w$ are elements of the vector space $V$, then $v+w$ is also an element of $V$.

Let's consider the vector space $\mathbb R^n$: what meaning should we give to an expression such as

$$\nabla + \vec x \ ?$$

the answer is: no meaning at all, because $\nabla$ is not a vector.

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    $\begingroup$ @Von Neumann. It turns out that we can often formally manipulate $\nabla$ as if it were a vector. Yeah, sir. Exactly. I want an intuition for why we are able to do that $\endgroup$ – Truth-seek Mar 18 '18 at 13:07
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    $\begingroup$ @Truth-seek Ahh ok then. I misunderstood-ish the question. Then I'll think and edit. $\endgroup$ – Von Neumann Mar 18 '18 at 13:08
  • $\begingroup$ @VonNeumann Good answer. I fully agree your statements "It turns out that we can often formally manipulate $\nabla$ as if it were a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless." $\endgroup$ – ChoF Mar 18 '18 at 13:40
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    $\begingroup$ @ChoF That statement is incorrect since $\nabla$ is an element of the Cartesian cube of the tangent space, for example. Therefore, is is a proper vector in a vector space, and it is not meaningless. $\endgroup$ – blueInk Mar 18 '18 at 13:44
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    $\begingroup$ @blueInk I agree. But it is a language of geometry, not a calculus. $\endgroup$ – ChoF Mar 18 '18 at 13:54
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$$\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$$ 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.

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