Is a vector field irrotational iff its Jacobian is symmetric?

Question

Let $F: \mathbb R^n \to \mathbb R^n$ be a smooth vector field. I conjecture that $F$ is irrotational iff its Jacobian is a symmetric matrix.

In two and three dimensions, this seems clear from Green and Stokes theorem. I conjecture this is true for any $n$ dimensions.

Is this true? If so, it raises a few additional questions:

1. We normally use curl to describe a vector field. Can we use curl, or at least its magnitude, to describe matrices? That is, can we use the magnitude of curl to measure how much a matrix rotates, or how non symmetric it is? Can we use the components of curl to tell us the nature of this asymmetry / rotation?

2. Would it be correct to say that for any two distinct dimensions $i, j$, that the measure of asymmetry $A_{ij} - A_{ji}$ is a measure of how much the matrix rotates a vector in the $i,j$ plane?

3. There are many well known types of matrices with interesting properties (e.g. orthogonal). If the Jacobian of $F$ is an "interesting" matrix, what does it tell us about $F$? For example, if $J(F)$ is orthogonal?

Answer 1

Part of the question here is to make sense of what irrotational means in $\Bbb R^n$ for $n \geq 4$. One way to approach a definition is as follows.

Recall that a smooth vector field ${\bf X} = (P, Q, R)$ with domain $\Bbb R^3$ is irrotational, i.e., has curl ${\bf 0}$, if and only if it is conservative, that is, if and only if $${\bf X} = \operatorname{grad} \varphi$$ for some (smooth) function $\varphi : \Bbb R^3 \to \Bbb R$, equivalently, $$\varphi_x = P, \qquad \varphi_y = Q, \qquad \varphi_z = R .$$ Indeed, since $\varphi$ is smooth, Clairaut's Theorem implies that the components $P, Q, R$ of $\bf X$ satistfy the differential condition $$Q_x = \varphi_{yx} = \varphi_{xy} = P_y$$ and its analogues for other pairs of independent variables, $P_z = R_x$ and $R_y = Q_z$. But if these $3$ conditions hold, then $$\operatorname{curl} {\bf X} = (R_y - Q_z, P_z - R_x, Q_x - P_y) = {\bf 0} .$$

In this case, the Jacobian of $\bf X$, regarded as a map $\Bbb R^3 \to \Bbb R^3$ is simply the (automatically symmetric) Hessian of the potential function $\varphi$: $$\operatorname{Jac} ({\bf X}) = \pmatrix{ P_x & P_y & P_z \\ Q_x & Q_y & Q_z \\ R_x & R_y & R_z } = \pmatrix{ \varphi_{xx} & \varphi_{xy} & \varphi_{xz} \\ \varphi_{xy} & \varphi_{yy} & \varphi_{yz} \\ \varphi_{xz} & \varphi_{yz} & \varphi_{zz}}$$

In $n$ dimensions the situation is essentially the same: If a vector field ${\bf X} = (X^1, \ldots, X^n)$ with domain $\Bbb R^n$ is conservative, say, with potential $\varphi$, then for all indices $i, j$ we have $$\partial_{x_j} X^i = \partial_{x_j} \partial_{x_i} \phi = \partial_{x_i} \partial_{x_j} \phi = \partial_{x_i} X^j,$$ a total of $\frac{1}{2} n (n - 1)$ nontrivial conditions. So it's natural to define the $n$-dimensional analogue of curl to be a map $\alpha$ that takes a vector field $\bf X$ to to the collection $\partial_{x_i} X^j - \partial_{x^j} X^i$ of functions, where $1 \leq i < j \leq n$, and we can declare ${\bf X}$ to be irrotational if $\alpha({\bf X}) = (0, \ldots, 0)$. Again it turns out that vanishing of $\alpha({\bf X})$ is not only necessary but also sufficient to guarantee that ${\bf X}$ is a gradient. As in the $3$-dimensional case, the conditions $\partial_{x_i} X^j = \partial_{x_j} X^i$ are exactly that the Jacobian $\operatorname{Jac} ({\bf X})$ of $\bf X$ (at each point) is symmetric (hence equal to the Hessian of the potential), giving the desired equivalence of conditions: $$\boxed{\textrm{$\bf X$ is irrotational} \Leftrightarrow \textrm{$\operatorname{Jac}({\bf X})$ is symmetric}} .$$

There are two more systematic ways to organize the above construction:

(1) Regard $\alpha$ as the map taking a vector field ${\bf X}$ to the skew-symmetrization of its Jacobian: $$\operatorname{Alt} \operatorname{Jac} ({\bf X}) = \frac12 [\operatorname{Jac} ({\bf X}) - (\operatorname{Jac} ({\bf X}))^\top].$$ This viewpoint makes the desired equivalence of conditions more or less immediate.

(2) Regard the map $\alpha$ as taking a vector field ${\bf X} = \sum_i X^i {\bf e}_i$ to the bivector field, $$\alpha({\bf X}) := \sum_{i < j} (\partial_{x_i} X^j - \partial_{x^j} X_i) \, {\bf e}_i \wedge {\bf e}_j .$$ From this point of view $\alpha$ fits into a natural sequence of operators that can be identified with the exterior derivative on Euclidean space.

Example In dimension $3$, there is a canonical identification between vectors and bivectors on $\Bbb R^3$, namely, $$ {\bf i} \leftrightarrow {\bf j} \wedge {\bf k}, \qquad {\bf j} \leftrightarrow {\bf k} \wedge {\bf i}, \qquad {\bf k} \leftrightarrow {\bf i} \wedge {\bf j} , $$ and under this identification we can think of $\alpha$ as a map that takes vector fields to vector fields, and this map is exactly the usual curl. Also using a similar identification for $\operatorname{div}$ lets us regard the sequence of operators mentioned in (2) with the usual sequence $\operatorname{grad}, \operatorname{curl}, \operatorname{div}$ on $\Bbb R^3$.

Jacobians satisfying matrix conditions

It turns out that the symmetric part $$\operatorname{Sym} \operatorname{Jac} {\bf X} = \frac12 [\operatorname{Jac} ({\bf X}) + (\operatorname{Jac} ({\bf X}))^\top]$$ of the Jacobian of ${\bf X}$ measures exactly how the metric structure of $\Bbb R^n$ distorts infinitesimally when we let points "flow" along ${\bf X}$ (more precisely, we can identify that symmetric part with the Lie derivative $\mathcal{L}_{\bf X} \bar g$ of the flat metric $\bar g$ on $\Bbb R^n$ by $\bf X$.) So, a vector field whose Jacobian (at every point) has zero symmetric part---equivalently, whose Jacobian is skew-symmetric---is exactly one under whose flow the metric structure is preserved. Such vector fields are called Killing (vector) fields.

Similarly, the trace $\operatorname{tr} \operatorname{Jac} {\bf X}$ of the Jacobian of ${\bf X}$ measures how volume distorts infinitesimally when we let points flow "along" ${\bf X}$, and so a vector field that has tracefree Jacobian (at each point) is exactly one under which volume is preserved.

Remark In the above presentation I specifically treated the case that the domain of the vector field is all of $\Bbb R^n$. If that's not the case, most of the above still holds with one important exception: It's still always true that if $\bf X$ is conservative then it satisfies $\alpha(\bf X) = {\bf 0}$ (in dimension $3$, $\operatorname{curl} {\bf X} = {\bf 0}$), but the converse need only hold when the domain of $\bf X$ is simply connected, or, roughly speaking, has no "holes" of a certain type. This observation can be turned around to use $\alpha$ to detect holes in a set and is the beginning of de Rham cohomology.

Answer 2

Let $F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth vector field. I conjecture that is irrotational iff its Jacobian is a symmetric matrix. [...] I conjecture this is true for any dimensions. Is this true?

To answer you question, we have to extend the concept of a vector field being irrotational (curl-free) to higher dimensions. This requires a more generalized form of curl that fits into the context of differential forms and exterior derivatives. The notion of a symmetric Jacobian implies that the mixed partial derivatives of the vector field components are equal (Schwarz's theorem), which aligns with the field being irrotational in a generalised sense. Let’s look a this in more detail.

In the language of differential forms, a vector field $F$ being irrotational in $\mathbb{R}^n$ means that its associated differential 1-form $\omega$ has a vanishing exterior derivative, i.e., $d\omega = 0$. In other words, the irrotational vector fields correspond to the closed 1-forms Note that d^2 = 0 implies that any exact form is closed, so any conservative vector field is irrotational. Conversely, we then have that closed 1-forms are exact if U is simply connected.

The 1-form $\omega$ is related to $F$ through a musical isomorphism (\flat or \sharp) between the tangent and cotangent bundles, typically depending on the metric of the space, e.g. the Riemannian metric.

Let $F = (F_1, F_2, \ldots, F_n)$. The associated differential 1-form $\omega$ is given by

$$\omega = F_1 dx_1 + F_2 dx_2 + \ldots + F_n dx_n$$

The exterior derivative $d\omega$ of this 1-form in $ \mathbb{R}^n$ is a 2-form, given by

$$ d\omega = \sum_{i < j} \left(\frac{\partial F_j}{\partial x_i} - \frac{\partial F_i}{\partial x_j}\right) dx_i \wedge dx_j$$

For F to be irrotational ($d\omega = 0)$, each coefficient in this sum must vanish, meaning we must have

$$\frac{\partial F_j}{\partial x_i} = \frac{\partial F_i}{\partial x_j} \quad \forall i, j $$

Great!

Now this directly translates to the symmetry of the Jacobian matrix of $F$, where each off-diagonal element $\frac{\partial F_j}{\partial x_i}$ is equal to its transpose counterpart $\frac{\partial F_i}{\partial x_j}$. Hopefully this answers any doubts you had with the formulation of your conjecture to higher dimensions.

We normally use curl to describe a vector field. Can we use curl, or at least its magnitude, to describe matrices? That is, can we use the magnitude of curl to measure how much a matrix rotates, or how non symmetric it is? Can we use the components of curl to tell us the nature of this asymmetry / rotation?

Traditionally, curl is a vector operation defined for vector fields, not matrices. Though I guess the idea behind curl (measuring local rotation) can be conceptually extended to matrices. Say you have a matrix representing a linear transformation, while you can't directly apply the curl operation, you can interpret the antisymmetric part of a matrix as an indicator of rotational properties, where the antisymmetric part is given by $J-J^\text{T}$. You could think of the magnitude of this antisymmetric part as an analogue to the magnitude of curl for a vector field, indicating the degree of rotation/asymmetry? Not too sure what you’re asking here.

Would it be correct to say that for any two distinct dimensions i,j, that the measure of asymmetry $A_\text{ij}−A_\text{ji}$ is a measure of how much the matrix rotates a vector in the i,j plane?

Indeed, the elements $A_\text{ij}−A_\text{ji}$ represent the antisymmetric part of the Jacobian matrix. In the language of differential forms, these components correspond to the coefficients of the exterior derivative of the associated 1-form of the vector field. More specifically, they appear in the 2-form obtained by taking the exterior derivative, reflecting the twisting or rotational properties of the vector field in higher dimensions.

There are many well known types of matrices with interesting properties (e.g. orthogonal). If the Jacobian of F is an "interesting" matrix, what does it tell us about F? For example, if $J(F)$ is orthogonal?

If J(F) is orthogonal, it means that the linear approximation of F at each point preserves lengths and angles. In physical terms, this implies that the transformation F locally behaves like a rotation/reflection. Also, orthogonal transformations are isometries, meaning that F preserves the Euclidean structure of the space.

For more examples of so-called interesting matrices:

• Symmetric Jacobian $\rightarrow$ F resembles a scaling transformation locally, stretching or compressing along orthogonal axes.
• Skew-Symmetric Jacobian $\rightarrow$ F is locally “pure” rotation (e.g.: in 3 dimensions, a skew-symmetric matrix can be associated with an angular velocity vector).
• Diagonal Jacobian $\rightarrow$ F acts independently on each coordinate axis, scaling each axis separately.