I used to think that $\nabla$ (or $\vec \nabla$) was just some fancy notation to represent some differential operators ($\nabla f \equiv \text{grad} \ f$, $\nabla \cdot \vec v \equiv \text{div} \ \vec v$, $\nabla \times \vec v \equiv \text{curl} \ \vec v$), which is particularly convenient because it happens to behave like a vector in algebraic manipulations.

However, I've read in some posts on this site (like this answer or this answer) that seem to suggest that there could in fact be a way to consider $\nabla$ as a vector in a formally meaningful way.

For example, quoting from this answer:

There are at least two layers of ideas here. First, as you say, the "dual space" $V^*$ to a real vector space is (by definition) the collection of linear maps/functionals $V\rightarrow \mathbb R$, with or without picking a basis. Nowadays, $V^*$ would more often be called simply the "dual space", rather than "covectors".

Next, the notion of "tangent space" to a smooth manifold, such as $\mathbb R^n$ itself, at a point, is (intuitively) the vector space of directional derivative operators (of smooth functions) at that point. So, on $\mathbb R^n$, at $0$ (or at any point, actually), $\{\partial/\partial x_1, \ldots, \partial/\partial x_n\}$ forms a basis for that vector space of directional-derivative operators.

It thus seems to me that there could be a meaningful, rigorous way to interpret $\nabla$ as an element of some vector space, maybe the dual space of an appropriate vector space of functions. Is this line of reasoning correct?

PS I am a physicist, not a mathematician, and I only have a very basic background on functional analysis and differential geometry.

  • $\begingroup$ $\nabla$ can be viewed as an unbounded linear differential operator between 2 function spaces. $\endgroup$
    – Ranc
    Commented Oct 12, 2016 at 10:53
  • 1
    $\begingroup$ There are certainly operator spaces and C$^*$-algebras whose elements are operators on some Hilbert space. But I'm not aware (though that doesn't mean much) of a space that includes unbounded operators. $\endgroup$
    – user137731
    Commented Oct 12, 2016 at 11:12
  • $\begingroup$ @Bye_World , An example may not have to be of mathematical intereset. Consider the space of infinitely differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$. This is a vector space and the gradient is well defined for every element (No norm introduced). $\endgroup$
    – Ranc
    Commented Oct 12, 2016 at 11:14
  • $\begingroup$ Try to make a change of coordinates, for instance in dimension 2 to see if it behave like a vector. For instance if $X= ax, Y= by+x, Z=z$, compute the $\nabla$ operator in the coordinates system $X,Y,Z$. $\endgroup$
    – Thomas
    Commented Oct 12, 2016 at 11:28

3 Answers 3


In differential geometry, it is common to identify vectors (or "vector fields") and "derivations" (acting on the space of smooth functions). For instance, in $3$-dimensional space $\mathbb{R}^3$, the differential operator $$ \frac{\partial}{\partial x} + \frac{\partial}{\partial z} $$ is the same thing as the (constant) vector field $$ \overrightarrow{V} = (1, 0, 1).$$ This is because vector fields "act" on functions simply by declaring that $\overrightarrow{V}(f) := df(\overrightarrow{V})$ (what you might want to write $\overrightarrow{\nabla} f \cdot \overrightarrow{V}$). So, it is perfectly ok to write $\overrightarrow{V} = \frac{\partial}{\partial x} + \frac{\partial}{\partial z}$.

I wrote this because I thought you would be interested to know, but it does not really answer the question about $\nabla$. The symbol $\nabla$ is called a covariant derivative, and it is not a vector field (even though it kind of looks like one when using it on functions). This covariant derivative is another kind of differential operator, in general it is only well-defined when you have a Riemannian metric (which is the case on $\mathbb{R}^3$, the natural Euclidean metric). Contrary to the differential of a function, notions such as the gradient of a function or the divergence of a vector field require a metric. In $\mathbb{R}^3$, there is a last ingredient needed to give its full power to the notation $\nabla$: the cross product (which technically identifies vectors to bi-vectors, allowing to define the rotational as a vector field). These special features of $\mathbb{R}^3$ is the reason why the magical notations $$\overrightarrow{\nabla} f = \text{grad} \ f \qquad \overrightarrow{\nabla} \cdot \vec V = \text{div} \ \vec V \qquad \overrightarrow{\nabla} \times \vec V = \text{curl} \ \vec V$$ only work in $\mathbb{R}^3$, or I should say, only have partial generalizations to higher dimensions or more general spaces.

In conclusion, I would say that yes $\nabla$ is a real mathematical object that it is possible to define properly (it is more than just a notation), but no it is not quite right to say that it is a vector, and finally the formulas that you know involving it don't work 100% the same in general (that being said, it allows to do many other things).


If you have a function $\mathbb{R}^n \rightarrow \mathbb{R}$ then consider it's directional derivative $D_{v}f=(\nabla f,v)$. In differential geometry we have the notion of a pushforward differential between the tangent spaces.

Commonly we consider the vector $v$, which is the direction in which we take this derivative in, as a member of the domain of $f$, but in fact thinking about this in terms of the pushforward is much better.

In this case the vector $\nabla f $ (given $f$) is a 1 row matrix(Jacobian) acting on the tangent space of the domain at some point which in the $\mathbb{R}^n$ case happens to be $\mathbb{R}^{n}$ itself. In this way $v$ is a member of the tanget space and $ (\nabla f ,v)$ becomes $(\nabla f)^{T} v$

If we have an arbitrary manifold I dont think we can take the directional as defined in $\mathbb{R}^{n}$ unless this vector is in the tangent space of the manifold. Hence I think the the outline above is the only reasonable way to think about it.


$\nabla$ can (in particular) be viewed as a linear map $\nabla: \Gamma(TM) \to \mathcal{T}_{1}^1(M)$, where $\mathcal{T}_{1}^1(TM)$ denotes the bundle of $(1, 1)$ tensors (or equivalently, endomorphisms of $\Gamma(TM)$). Therefore it is an element of the vector space $\operatorname{Hom}(\Gamma(TM),\mathcal{T}_{1}^1(M) )$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .