# Curl: invariant under change of basis or not?

I wondered how the curl$$\text{rot}\mathbf{F}=\left( \begin{array}{ccc}\partial_y F_3-\partial_z F_2 \\ \partial_z F_1-\partial_x F_3 \\ \partial_x F_2-\partial_y F_1 \end{array} \right)$$of a vector field $\mathbf{F}=(F_1,F_2,F_3)$ changes when the basis $\mathbb{R}^3$ is changed. I would have thought that it is invariant, because of the intuitive and physical interpretation of the curl and because, if I correctly understand other quantities typical of vector fields, like divergence, are (if I am not wrong $\text{div}\mathbf{F}$ is the trace of the Jacobian matrix $J_{\mathbf{F}}$ of $\mathbf{F}$, and the trace of $E J_{\mathbf{F}} E^{\text{T}}$ - see below - is the same of $J_{\mathbf{F}}$), but I have got serious problem to prove it.

Trial:

If I am not wrong, if $E\in\text{O}(3)$ is the basis change matrix, and if we define the function $\mathbf{G}$ as $\mathbf{y}\mapsto E\mathbf{F}(^t E \mathbf{y})$, invariancy is equivalent to $$E\text{rot}\mathbf{F}=\text{rot}\mathbf{G}$$please correct me if I am wrong. The Jacobian matrix $J_{\mathbf{G}}(\mathbf{y})$ of $\mathbf{G}$ in $\mathbf{y}$ should be:$$J_{\mathbf{G}}(\mathbf{y})=E J_{\mathbf{F}}(\mathbf{x}) E^{\text{T}}$$where $E^{\text{T}}$ is the transpose, and inverse, matrix of $E$.

I have used such an identity to find the expression of $\text{rot}\mathbf{G}(\mathbf{y})$ in terms of the components of $J_{\mathbf{F}}(\mathbf{x})$ and $E$, but my calculations do not give me the expected result: for example, in the first component of the expression of $\text{rot}\mathbf{G}$ calculated in such a way , the coefficient of $\partial_x F_2$ is $(e_{21}e_{32}-e_{22}e_{31})$, while, in the first component of $E\text{rot}\mathbf{F}$, the coefficient of $\partial_x F_2$ is $e_{13}$...

Is the curl invariant under a change of orthogonal basis and, if it is, how can it be correctly proved? Thank you so much for any answer!

• Your wording is somewhat imprecise. You refer to a "change of basis" throughout most of this question, but then you say the coordinates would change. These concepts are different: do you mean a change of coordinates or not? Or do you mean merely (and only) a change of basis? Commented Jul 3, 2015 at 16:03
• @Muphrid Thank you for the remark! I mean: I chose a different basis, the same basis, for both the domain and the codomain, represented by the matrix $E$, therefore the coordinates of $\mathbf{x}$ and $\mathbf{F}(\mathbf{x})$ become $E\mathbf{x}$ and $E\mathbf{F}(\mathbf{x})$. If a quantity, whose coordinates are $\mathbf{v}$ in the original basis, is invariant under the change of basis, its coordinates become $E\mathbf{v}$. Where am I wrong? Commented Jul 4, 2015 at 8:36
• Ok, let's impose a distinction between coordinates (which are the arguments for a scalar or vector field) and components (which are individual scalar functions that, along with a basis, describe a vector field). - With that in mind, you could have a transformation change only the components (e.g. a local rotation) without the coordinates, or you could have a transformation that changes the coordinates and the components (because any map that changes the coordinates also defines a new coordinate basis). So which do you want? change coordinates and evaluate wrt the new coord basis? Or not? Commented Jul 4, 2015 at 14:56
• I think what might be true is this: $\nabla \times \mathbf F$ is invariant under actions of $SO(3)$, but changes sign under the action of a $Q \in O(3)$, $Q \notin SO(3)$, i.e., $\det Q = -1$. Need to think more on this; will post if I get somewhere worth posting. Cheers! Commented Jul 15, 2015 at 17:52
• Sorry I haven't written anything yet; I've been thinking about this but have nothing definitive as of yet. Will keep you posted. Cheers! Commented Jul 19, 2015 at 22:57

Update

In the first version of my answer (reprinted below) I tried to give a "geometrical" or "physical" explanation of the curl vector, but in fact did not address your question, which is the following: How do the coordinates of the curl vector ${\bf c}$ transform when we replace the given orthonormal coordinates $(x_i)$ by an equally oriented orthonormal coordinates $(\bar x_l)\,$?

Denote the transform matrix from the old to the new coordinates by $T=[t_{il}]\in SO(3)$. The columns of $T$ then are the old coordinates of the new basis vectors $\bar{\bf e}_l$, and the rows of $T$ are the new coordinates of the old basis vectors ${\bf e}_i$. In your setup the coordinates of the points ${\bf x}$ and those of the force field ${\bf F}$ transform in the same way according to formulas like $$x_i=\sum_l t_{il}\bar x_l,\quad \bar F_l=\sum_i t_{il} F_i\ .$$ Now in any orthonormal system the coordinates of the curl vector are given by $$c_i={\partial F_{i-1}\over\partial x_{i+1}}-{\partial F_{i+1}\over\partial x_{i-1}}\ .$$ In order to determine the new coordinates $\bar c_l$ of ${\bf c}$ we have to compute partial derivatives of the $\bar F_k$ using the chain rule: $${\partial \bar F_k\over\partial \bar x_l}=\sum_i t_{ik}\left(\sum_j{\partial F_i\over\partial x_j}\>{\partial x_j\over\partial\bar x_l}\right)=\sum_{i, \>j}{\partial F_i\over\partial x_j}\>t_{ik}t_{jl}\ .$$ We therefore obtain $$\bar c_l={\partial \bar F_{l-1}\over\partial\bar x_{l+1}}-{\partial \bar F_{l+1}\over\partial \bar x_{l-1}}=\sum_{i, \>j}{\partial F_i\over\partial x_j}\>(t_{i,l-1}t_{j,l+1}-t_{i,l+1}t_{j,l-1})\ .\tag{1}$$ Now $$t_{i,l-1}t_{j,l+1}-t_{i,l+1}t_{j,l-1}=({\bf e}_j\times {\bf e}_i)_l\ ,$$ which is $=0$ when $i=j$. This allows to write $(1)$ in the form $$\bar c_l=\sum_i {\partial F_{i-1}\over\partial x_{i+1}}({\bf e}_{i+1}\times {\bf e}_{i-1})_l+ \sum_i {\partial F_{i+1}\over\partial x_{i-1}}({\bf e}_{i-1}\times {\bf e}_{i+1})_l\ ,$$ or $$\bar c_l=\sum_i \left({\partial F_{i-1}\over\partial x_{i+1}}- {\partial F_{i+1}\over\partial x_{i-1}}\right)({\bf e}_i)_l=\sum_i c_i\>t_{il}\ .$$ This shows that the coordinates of the curl vector ${\bf c}$ transform in the same way as the coordinates of the vectors ${\bf x}$ and ${\bf F}$.

First version:

Given a force field ${\bf F}$ on a domain $\Omega\subset{\mathbb R}^3$ its curl measures the "local nonconservativity" of ${\bf F}$. Consider a point ${\bf p}\in\Omega$ and a small parallelogram $P$ of diameter $|P|$ with center ${\bf p}$, and spanned by two vectors ${\bf X}$ and ${\bf Y}$. Calculating the integral of ${\bf F}$ around $\partial P$ (which would be $=0$ when ${\bf F}$ is conservative) gives the following: $$\int_{\partial P}{\bf F}\cdot d{\bf x}= d{\bf F}({\bf p}).{\bf X}\cdot {\bf Y}- d{\bf F}({\bf p}).{\bf Y}\cdot {\bf X}+o\bigl(|P|^2\bigr)\ .$$ Here the main part of the right hand side is a skew bilinear function $R$ of the vectors ${\bf X}$ and ${\bf Y}$ spanning $P$: $$\int_{\partial P}{\bf F}\cdot d{\bf x}=R({\bf X},{\bf Y})+o\bigl(|P|^2\bigr)\ .$$ This setup makes sense in euclidean spaces of any dimension. Now in dimension $3$, such a skew bilinear function $R$ can be represented by a vector as follows: There is a well defined vector ${\bf c}$, called the curl of ${\bf F}$ at ${\bf p}$, such that $$\int_{\partial P}{\bf F}\cdot d{\bf x}={\bf c}\cdot({\bf X}\times{\bf Y})+o\bigl(|P|^2\bigr)\ .$$ My answer to your question would therefore be the following: As long as you stick to (linear) orthonormal coordinates of positive orientation the usual formula for ${\rm curl\,}{\bf F}$ is valid; but as soon as you introduce more general coordinates you would have to explain more carefully what you mean by the curl in the context.

• ...where the sums are intended for the indices from 1 to 3 and where $0\equiv 3$ and $4\equiv 1$ as if $i,l\in\mathbb{Z}/3\mathbb{Z}$. What a beautiful and complete answer, to be tasted as a rare wine. I find the use of the vector product very elegant: I had not thought about it. You have saved me from two months of headaches. I heartily thank you. If the new basis has the opposite orientation, i.e. if $T\in O(3)\setminus SO(3)$, what changes? $t_{i-1,l-1}t_{i+1,l+1}-t_{i-1,l+1}t_{i+1,l-1}$ still is the $l$-th coordinate of $\mathbf{e}_{i+1}\times\mathbf{e}_{i-1}$ [continues] Commented Sep 8, 2015 at 22:43
• (where I understand that you have called $\mathbf{e}_k$ the $k$-th vector of the basis, say $\mathcal{B}$, with respect to which $x$ are the coordinates of the argument) with respect to the basis (say $\bar{\mathcal{B}}$) where the argument's coordinates are $\bar{x}$ and $t_{il}$ still is the $l$-th coordinate of $\mathbf{e}_i$, with respect to $\bar{\mathcal{B}}$, but I think that, if $\bar{\mathcal{B}}$ has the opposite orientation to $\mathcal{B}$, then $\mathbf{e}_{i+1}\times\mathbf{e}_{i-1}=-\mathbf{e}_i$, [continues] Commented Sep 8, 2015 at 22:44
• and therefore $\text{rot}F(x)=-T\text{ rot}\bar{F}(\bar{x})$. Am I right? I $\infty$-ly thank you again! Commented Sep 8, 2015 at 22:45

My guess was that this might work only for infinitesimal rotations.

And there seems to be a question (link) and an answer for this kind of problem: (link)

• I've edited the original post to include some reflections about a conceptual error I committed. Thank you for the answer and the links! Forgive me: I don't understand how that answer is related to a change of basis: not sure whether my low level of mathematical knowledge is sufficient to relate the two issues... Thank you a lot again!!! Commented Jul 1, 2015 at 7:27
• Your change of basis through an orthogonal transformation corresponds to a rotation and/or reflection, e.g. a 30 degrees angle and requires invariance for this finite amount. I believe you will only have invariance for an infinitesimal angle.
– mvw
Commented Jul 1, 2015 at 7:37