# Coordinate free definition of $\nabla$ operator

There are a number of posts on this site asking similar questions and some of them have been answered (to my taste) at least partially but none give a complete answer that I am satisfied with. See links at the bottom of this question for a small selection of posts asking related (or even the same) questions.

My question is as follows. The following is often written down:

$$\nabla = \frac{\partial}{\partial_x} \hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z} \hat{z}$$

Some people will call this an operator, some will call it a vector, some will call it a vector operator, and some will adamantly claim that it is not properly anything at all and you shouldn't call it any of these things and you should just treat it as a "notational convenience".

One can then go on to use this "vector operator" to calculate things like $\nabla f$, $\nabla \cdot \vec{F}$ or $\nabla \times\vec{F}$ where the operator is treated notationally as if it were a vector.

First I want to take issue with the final claim that it is purely a notational convenience. I think it is more than just a notational convenience for the following reason. It is possible, by following certain transformation rules, to express $\nabla$ in different coordinate systems, for example cylindrical or spherical. That might be fine, but there is a FURTHER point which makes me think $\nabla$ must be more than a notational convenience. if you express $\nabla$ in different coordinates you can then calculate something like $\nabla \cdot \vec{F}$ in the new coordinates and get the right answer. An answer which you could have arrived at by explicitly converting the cartesian expression for $\nabla \cdot \vec{F}$ into the new coordinate system. In other words, the $\nabla$ allows you to actually skip a step of calculation you would have had to do otherwise. This is evidence that the symbol carries some sort of mathematical structure to it which should be able to be captured in an independent definition.

To that end I'm interested in a coordinate free definition of this symbol. The definition I gave above relies on using the usual Cartesian coordinates above. I have searched but haven't been able to find a coordinate free definition of the $\nabla$ symbol. Can one exist? In particular, I am interested in such a formula so that it is algebraically evident how one should calculate the components of $\nabla$ in any given coordinate system.

Is there a coordinate free definition of $\nabla$?

I am aware of a few complications with this endeavor that I'll just list here:

1) If this is to be some kind of vector or some kind of operator then it is not clear what space it should live in. For example, it is an object which can take a function $f$ and map it to a vector space. But at the same time it is an object which can be fed as an argument to a dot product together with a vector (form a different space) and return a scalar.

2) If I put on my differential geometry hat it becomes a very weird object. In differential geometry I come to think of vectors as actually being things like $\frac{\partial}{\partial x}$ and that $\vec{x}$ notation is eschewed. However the $\nabla$ symbol above contains both of these sitting next to each other. it's like a vector of vectors.. The idea of two vectors sitting next to eachother made me think it might be some kind of rank 2 contravariant tensor but I think that may have been a stretch.

3) I am aware that the cross product and curl operator are only defined in 3 dimensions so it does not need to be pointed out that that limits the possibility of defining such an operator for arbitrary dimension. I am happy to say we are working in 3 dimension.

4) I understand that the idea of divergence and curl depends on the presence of a metric for a space. Ok, that is fine. We can work in a space that has a metric defined on it.

5) Maybe the metric needs to be flat? Even that is fine as long as we can work in coordinate systems such as cylindrical or spherical where the metric is still flat but no longer has a trivial component representation. I am happy to restrict analysis to $\mathbb{R}^3$ if that is necessary.

6) Finally if such a definition truly cannot be formulated then could you at least answer why I can calculate BOTH $\nabla f$ and $\nabla \cdot \vec{F}$ by either 1) computing $\nabla f$ or $\nabla \cdot \vec{F}$ in xyz coordinates, then convert everything to spherical or 2) compute $\nabla$ in xyz coordinates, covert to spherical, then calculate $\nabla f$ and $\nabla \cdot \vec{F}$ and get the same answer in both cases? It just seems slightly too powerful/structured to be JUST a notational convenience.

Here are a few other related questions:

Is there a general formula for the del operator $\nabla$ in different coordinate systems?

Can $\nabla$ be called a "vector" in any meaningful way?

Coordinate transformation on del operator

• From what it looks like, Christoph gives a good answer here. Apologies if you’ve already seen that before though. If I recall correctly, there’s a nice explicit using the musical isomorphisms and exterior derivative, but I can’t remember it off the top of my head.
– user261214
Jul 21, 2018 at 6:22
• Thank you for the link. Christoph's answer is quite good and is definitely in the vein of what I am looking for. The problem is that it only address the use of $\nabla$ in its function in taking the gradient. It is a satisfactory coordinate independent definition of $\nabla f$ but it isn't a definition for $\nabla$ itself. A definition which can be used to calculated both divergences and gradients for multiple choices of coordinates. Jul 21, 2018 at 6:27
• Ahh, I see. It turns out that div, grad, and curl can be expressed as compositions of the musical isomorphism, the exterior derivative, and the Hodge star operator. I’ll have to see if I can find a diagram, there’s one I’m thinking of that shows it quite nicely.
– user261214
Jul 21, 2018 at 6:30
• I have found this. en.wikipedia.org/wiki/…. Unfortunately exterior derivatives and the hodge star are a little out of my understanding paygrade at the moment. I am trying to learn more to understand this machinery now. However, my suspicion is that, if you restrict yourself to $\mathbb{R}^n$ then that machinery isn't necessary to answer the question. Perhaps simplifying that notation accordingly would reveal the answer I'm looking for.. Jul 21, 2018 at 6:34
• Ah, I follow. Do you want to move this to a chat room to streamline things?
– user261214
Jul 21, 2018 at 6:36

Looking at the coordinate-free expressions that you've found in Wikipedia, it is easy to convince yourself that $\mathrm{grad}$, $\mathrm{curl}$, and $\mathrm{div}$ are instances of the exterior derivative in disguise: just think of the musical isomorphisms and the Hodge star as of means of identification. In the same article one can find a coordinate-free formula, that can be taken as the definition of the exterior derivative. This observation essentially closes the question.

The classical vector calculus deals with $\mathbb{R}^3$, which possesses some specific or exceptional structures, in particular, it has a canonical (Euclidean) coordinate system, the Euclidean metric, and the cross product, that all are extensively used in theory and calculations. If you want tor restict yourself to this case, then I doubt that it is ever possible to find a pure coordinate-free way of expressing the quantities under consideration (i.e. $\nabla f$, $\nabla \cdot \vec{F}$, and $\nabla \times\vec{F}$), as the space $\mathbb{R}^3$ itself is defined by explicitly presenting a single coordinate chart! In other words, you are forced to deal with coordinates and the dimension-related tricks in order to handle these quantities.

Coming back to the expressions in Wikipedia, notice that they use the Hodge star, but we have not received yet any convincing answer on how to give a coordinate-free definition for it. This doubles my pessimism, but I can be wrong and overlook something important.

Nevertheless, I find that this question and the other attempts to answer it are very insightful. For further discussion I suggest to look at the references below.

The best picture that shows that $\mathrm{grad}$, $\mathrm{curl}$, and $\mathrm{div}$ are closely related is given in [1], where they are combined into the de Rham complex. This text is perhaps too advanced, but a diligent undergraduate should be able to follow the first two paragraphs there, and the details can be recovered from [2] and [3].

References:

1. M.G. Eastwood, A complex from linear elasticity, http://calvino.polito.it/~salamon/seminar/srni99.pdf
2. W.G. Faris,Vector fields and differential forms, September 25, 2008, http://math.arizona.edu/~faris/methodsweb/manifold.pdf
3. E.H.Goins, T.M. Washington, A Tasty Combination: Multivariable Calculus and Differential Forms, https://arxiv.org/abs/0910.0047
• I was also suspicious that the restriction to $\mathbb{R}^3$ would put "too much structure" onto the problem to allow a clean coordinate free expression so I see what you mean by that. The links are very nice. I especially appreciate the expressions for div grad and curl for arbitrary coordinate systems in the second reference. I've seen those expressions before but they are very nicely derived from the more general machinery of differential forms. It seems to be the case that these formulas are the most general that can be given for div grad and curl in $\mathbb{R}^3$ Jul 22, 2018 at 20:30
• I think this may close out the question of a search for a coordinate free expression of these different operations by at least giving expressions for arbitrary coordinate frames. However, there is still a question remaining as to what I see as the unexplained effectiveness of $\nabla$ as an operator. That is: why can manipulations on the $\nabla$ operator (supposedly purely a notational convenience) be used to find the answer to expressions for div grad and curl in arbitrary coordinate systems in $\mathbb{R}^3$. Jul 22, 2018 at 20:35
• I will wait a little while to accept an answer to this question. If nothing more shows up on the effectiveness of the $\nabla$ symbol I may open a new question asking more specifically about that using some of what I have learned in this post. Jul 22, 2018 at 20:35

Let $V$ be an $n$-dimensional real vector space equipped with inner product $\langle.,.\rangle$ and $f\colon V\to\mathbb R$ a differentiable function. Then $D_pf$, the differential of $f$ in $p$, is a linear form $V\to\mathbb R$. It’s well known that in the presence of an inner product there exists a unique vector, we call it $\nabla^{\langle.,.\rangle}_pf$, which represent that linear form, that is, for all $v\in V$ we have that $$D_pf(v)=\langle \nabla^{\langle.,.\rangle}_pf,v\rangle.$$ In case that the inner product is the usual dot product we simply denote that vector as $\nabla_pf$.

Quite coordinate-free, isn’t it?

• This notation seems to capture how the gradient can interact with a vector to give a directional derivative but again, I don't see how it generalizes to also work for divergence, for example. Jul 21, 2018 at 17:11

Here are my two cents. The definitions below are independent of coordinates. Given a scalar function $F$, define $\nabla F$ as the function such that the line integral is equal to the net change

$$\int_{\vec{a}}^{\vec{b}} \nabla F \cdot d\vec{s} = F(\vec{b}) - F(\vec{a})$$

Given a vector function $\vec{F}$, define $\nabla \cdot \vec{F}$ as the function such that the volume integral is equal to the flux

$$\iiint_V \nabla \cdot \vec{F} \; dV = \iint_S \vec{F} \cdot d\vec{S}$$

Given a vector function $\vec{F}$, define $\nabla \times \vec{F}$ as the function such that the surface integral is equal to the circulation

$$\iint_S \nabla \times \vec{F} \cdot d\vec{S} = \int \vec{F} \cdot d\vec{s}$$

I'm using hindsight a bit, because there's no reason for me to expect that such functions exist satisfying the integral definitions, or if they are unique. That aside, the purpose of a derivative (in physical application) is either to be integrated, or serve in a linear approximation. The above definitions highlight this practical use. Anyways, instead of the 3 "large-scale" definitions above (which I have no reason to suspect that such functions exist or are unique), I could have given 3 infinitesimal definitions (limit definitions). The divergence can be defined as the limit

$$\nabla \cdot \vec{F} = \lim_{Vol\to 0} \frac{\iint_{\text{surface}} \vec{F} \cdot d\vec{S}}{\text{Vol}}$$

That is, the divergence is the flux through the surface of an infinitesimal volume divided by the volume. An instantaneous flux per unit volume. I don't know how you would carry out this limit for the same reason I don't know how you would carry out the limit, mass density $= \lim_{V\to0} m(V)/V$ because I have no idea how I would write down $m(V)$ analytically. I know what's going on, but it would be hard to carry out the operation. A better definition of density is $\int \rho dV = m$ (still we have the uniqueness issue on $\rho$ - then maybe the more precise definition would be the limit definition - or you could extend the integral definition to any volume (a part, a whole, whatever) which might make $\rho$ unique). Anyways, You could also give a limit definition to the curl (circulation per unit area) and gradient (height per unit length), however they are a bit trickier because of the dot product (I have to worry about directions and whatnot in a precise definition). If you want something a little more explicit (in terms of how you carry out the operation), it is possible is write each operation in a coordinate-free way. For instance, following Appendix A of Griffiths' book Introduction to Electrodynamics, you can define the divergence as

$$\nabla \cdot \vec{F} = \frac{1}{fgh}\Big[\frac{\partial}{\partial u}(ghF_u) + \frac{\partial}{\partial v}(fhF_v) + \frac{\partial}{\partial w}(fgF_w)\Big]$$

where $u,v,w$ indicate some coordinate space (such as cartesian $x,y,z$ or spherical $r,\theta, \phi$). In cartesian, $f = g = h = 1$. In spherical, $f = 1$, $g = r$, and $h = r\sin\theta$. He does this for the curl and gradient as well. To be general, he used some type of parameterization but I haven't studied it fully. Using integrals to define derivatives might be analogous to using addition to define subtraction (if you allow negative numbers). Anyways, hope this helps

• Thanks for this nice answer. Unfortunately it looks like three independent definitions for $\nabla f$ $\nabla \cdot \vec{F}$ and $\nabla \times \vec{F}$. As such it doesn't explain why the calculational power of the notation for $\nabla$ I introduced in the beginning of my question. I have come across these definitions and it is useful to have them here. The reference to the Griffith's appendix is nice. I may spend some more time looking at that. Jul 21, 2018 at 18:25