7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\nabla$ can be viewed as an unbounded linear differential operator between 2 function spaces. $\endgroup$ – Ranc Oct 12 '16 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 Oct 12 '16 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 Oct 12 '16 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 Oct 12 '16 at 11:28
1
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.