Vector field with skew-symmetric Jacobian matrix

Question

Consider a vector field $X\in\mathfrak{X}(\mathbb{R}^n)$, and suppose to know that $JX(x) = \frac{\partial X(x)}{\partial x}\in\mathbb{R}^{n\times n}$ is symmetric for every $x$. Then it is well-known that there exists a potential function $I:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $\nabla I(x) = X(x)$.

I was wondering if a similar characterization of vector fields with a skew-symmetric jacobian matrix exists. Is there a generic form they must have on a simply-connected space like $\mathbb{R}^n$?

Playing around, I was thinking that one can proceed by decomposing the Jacobian matrix of a generic vector field in its symmetric and skew-symmetric part, to get $$ JX(x) = \nabla^2 I(x) + \frac{JX(x)-JX(x)^T}{2} $$ for some $I$. Now, is the skew-symmetric part relatable to the jacobian of a specifically structured vector field?

Another way to deal with the problem is through the analysis of the sensitivity matrix of the associated dynamical system, since defining $$ S(x(t)) = \frac{\partial x(t)}{\partial x_0} $$ one would get $$ \dot{x}(t) = X(x(t)),\quad x(0)=x_0 $$ $$ \dot{S}(t) = JX(x(t))S(t),\quad S(0)=I. $$ Thus a vector field $X$ is with skew-symmetric Jacobian if and only if its sensitivity matrix is orthogonal for every time. However I can not move forward, since I do not find classes of vector fields with this property.

Answer 1

In $\mathbb R^3$ the answer is quite simple when one knows where to look.

• A vector field that has an anti symmetric Jacobian is in the case of flat $\mathbb R^3$ nothing else than a Killing vector field that characterizes isometries of the metric $ds^2=dx^2+dy^2+dz^2\,.$ These isometries form the Lie group $ISO(3)$ which consists of translations and rotations. Therefore, the Killing vector fields are the Lie algebra $\mathfrak{iso}(3)$ of that group. That Lie algebra is six dimensional and is generated by $$ \underbrace{\partial_x\,,\partial_y\,,\partial_z\,,}_{3\;\text{translations}}\,\quad \underbrace{y\,\partial_x-x\,\partial_y\,,-z\,\partial_x+x\,\partial_z\,,z\,\partial_y-y\,\partial_z\,.}_{3\;\text{rotations}} $$ The vector fields $X$ with anti-symmetric Jacobian are precisely the linear combinations (with constant coefficients) of these six generating Killing vector fields.

Old version (outdated)

Sorry for repeating known stuff but I find your question interesting and want to put it into some context:

• In $\mathbb R^3$ the symmetry of the Jacobi matrix of $X$ is equivalent to $X$ being rotation free: $\nabla\times X\equiv 0\,.$ This is equivalent to the existence of a potential function $\Phi:\mathbb R^3\to\mathbb R$ such that $X=\nabla \Phi\,.$

• A vector field that is divergence free, $\nabla\cdot X\equiv 0\,,$ can be expressed as the rotation of a vector field $Y\,:$ $X=\nabla\times Y\,.$

• By the Helmholtz decomposition theorem any (smooth enough) vector field $X$ can be additively decomposed into a component that is rotation free and a component that is divergence free. Then the first to bullet points apply: $$ X=\nabla \Phi+\nabla\times Y\,. $$

• Assuming $X=\nabla \times Y$ the Jacobi matrix of $X$ is $$ \begin{pmatrix}\partial_1\partial_2Y_3-\partial_1\partial_3Y_2&\partial_2\partial_2Y_3-\partial_2\partial_3Y_2 &\partial_3\partial_2Y_3-\partial_3\partial_3Y_2\\[2mm] \partial_1\partial_3Y_1-\partial_1\partial_1Y_3&\partial_2\partial_3Y_1-\partial_2\partial_1Y_3&\partial_3\partial_3Y_1-\partial_3\partial_1Y_3\\[2mm] \partial_1\partial_1Y_2-\partial_1\partial_2Y_1&\partial_2\partial_1Y_2-\partial_2\partial_2Y_1&\partial_3\partial_1Y_2-\partial_3\partial_2Y_1&\end{pmatrix}\,.\tag{1} $$ This is in general not anti-symmetric.

• If we subtract from $X$ the component that has a symmetric Jacobi matrix (is rotation free) then, by the Helmholtz theorem, we must obtain a component that is divergence free and therefore have a Jacobi matrix of the form (1) which is not anti symmetric in general.

• If the Jacobi matrix of $X$ is anti-symmetric then its diagonal elements $\partial_iX_i$ must all vanish identically. In particular, their sum (divergence of $X$) must vanish. This means that anti-symmetry of the Jacobi matrix is a stronger condition than divergence freeness. Anti-symmetry implies the existence of $Y$ but is not necessary for it.

• Typical examples of vector fields $X$ that have anti-symmetric Jacobian are the three divergence free fields $$ \pmatrix{y\\-x\\0}\,,\quad \pmatrix{-z\\0\\x}\,,\quad \pmatrix{0\\z\\-y}\,. $$ As Helmholtz decompositons one can choose $\Phi\equiv 0$ and $Y$ being the corresponding one of $$ \frac12\pmatrix{0\\0\\x^2+y^2}\,,\quad \frac12\pmatrix{0\\x^2+z^2\\0}\,,\quad \frac12\pmatrix{y^2+z^2\\0\\0}\,. $$ This is however not a unique decomposition.