Given a vector field $F=(f_1,f_2,f_3)$, we can associate it to a differential 2-form $\omega = f_1 dy\wedge dz+f_2 dx\wedge dz+f_3dx\wedge dy$. Taking the exterior derivative is sort of an equivalent of taking the divergence of $F$. If we have a scalar field $f= f(x_1, \ldots, x_n)$ we can see it as a 0-form, and the exterior derivative would be like calculating $\nabla f$ and we can define something similar for the curl.

All of that was using the standard basis, but what if I want to use a change of coordinates, say spherical, for instance. How would these things above translate?

My attempt so far consisted in using the pullback. I tried it on polar coordinates for the gradient of $f$:

$$ \phi(r,\varphi) = (r\cos\varphi, r\sin\varphi)\\ \phi^{\ast}(f) = f\circ\phi \\ d\phi^{\ast}(f) = \frac{\partial f}{\partial r}dr+\frac{\partial f}{\partial \varphi} d\varphi $$

As you can see it's missing the $1/r$ in the second term. So I don't think I'm doing it right. Any hints? Thanks.

EDIT: So after asking my professor he said to use the Jacobian. Using this hint I got the expression for the gradient in generalized coordinates as follows:

Let $\phi(x,y,z) = (u,v,w)$ be our new coordinates. Then we have the Jacobian to be:

$$D\phi(\vec{x}) = \begin{pmatrix}\frac{\partial u}{\partial x} && \frac{\partial u}{\partial y} && \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} && \frac{\partial v}{\partial y} && \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} && \frac{\partial w}{\partial y} && \frac{\partial w}{\partial z}\end{pmatrix}$$

Now, given $i,j,k$ the usual basis, we can find a new basis by using:

$$D\phi(\vec{x})(i), D\phi(\vec{x})(j), D\phi(\vec{x})(k)$$

And we can normalise it, which gives the new basis vectors:

$$e_u = D\phi(\vec{x})(i)/||D\phi(\vec{x})(i)||, e_v = D\phi(\vec{x})(j)/||D\phi(\vec{x})(j)||, e_w = D\phi(\vec{x})(k)/||D\phi(\vec{x})(k)||$$

Then $\nabla f = \frac{\partial f}{\partial u}e_u+\frac{\partial f}{\partial v}e_v + \frac{\partial f}{\partial w}e_w $. This gives the correct expressions for the gradient. How can I generalize this for divergence and curl?


1 Answer 1


Let $M$ be a psuedo-Riemannian $n$-dimensional orientable manifold, $TM$ its tangent bundle. In order to define divergence of a vector field $X=X^ie_i$ (with $e_i$ a basis for $TM$), $\mathrm{div}X$ a bit more work is required than what you realized. Let $g_{ij}$ be the metric. First two notations:

  1. Flat $\flat$: The exterior derivative $d$ acting on forms and not vector fields, one needs a way to transform a vector field to a form. This is done via $X^\flat=g_{ij}X^ie^j=X_ie^i$ (where $e^j$ is the dual basis). Now we can use operator $d$.
  2. Hodge dual $\star$: Let $\omega=e^1\wedge e^2\wedge\cdots \wedge e^n$, which is known as the volume form. We define the operator $\star$ such that given any two $k$-forms $\alpha, \beta$ one has $\alpha\wedge \star \beta = \langle \alpha, \beta\rangle \omega$. Explicitley if $\sigma$ is a permutation of $\{1, \cdots, n\}$, then $$ \star (e^{\sigma_1}\wedge \cdots \wedge e^{\sigma_k})= \mathrm{sgn}\:\sigma \:e^{\sigma_{k+1}}\wedge \cdots \wedge e^{\sigma_n} $$

If $M=\mathbb{R}^3$, $e_i=\partial_i$ and $g_{ij}=\delta_{ij}$ then for a vector field $X=f_i\partial_i$ you have $X^\flat=f_i dx^i$ and $\star X^\flat=f_1dx_2\wedge dx_3 + f_2dx_3\wedge dx_1 + f_3 dx_1\wedge dx_2$. Taking the exterior derivative, we have $$ d(\star X^\flat) = \mathrm{div} X \:\omega $$ Therefore $\boxed{\mathrm{div} X = \star d\star X^\flat}$. This is in fact how divergence is defined in terms of differential forms in general. You can check that this definition works just as well for other coordinate systems. In fact, as you can see, this definition is coordinate free.

If you define $\sharp$ as the inverse of $\flat$, then curl of $X$ (in a 3d manifold) can be defined as $\boxed{\mathrm{curl}\: X=(\star dX^\flat)^\sharp}$.


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