Given the total derivative $Df$ of a (sufficiently) smooth function $f:\mathbb{R}^n \to \mathbb{R}^n$, the trace of the total derivative matrix corresponds to the divergence of $f$ (considered as a vector field $\mathbf{F}$), i.e. $\operatorname{trace}(Df) = \nabla \bullet \mathbf{F}$. (For context, I've asked a question about this result for the divergence on this website before.)
Question: Given a sufficiently smooth function $f: \mathbb{R}^n \to \mathbb{R}^n$, what operation on matrices applied to its total derivative $Df$ corresponds to the (generalized) curl $\nabla \times f$?
In other words, what matrix operation is to "curl" as "trace" is to "divergence"?
Clarification: By "curl" I mean the general version, which is a "bivector field", I don't mean the vector field notion specific to $\mathbb{R}^n$ for $n=3$. However an answer that addresses at least that special case could still be accepted (provided it is a very strong hint for how to address the general $n$ case).
Also an answer addressing the "2D curl" from Green's theorem could also still be accepted (with the same caveat).
I was thinking that maybe something related to the minors of the total derivative considered as a matrix might be relevant, but I wasn't able to make that work. It also seems like the exterior derivative should be related to the antisymmetrization of the total derivative in some way (cf. this other previous question I've asked about that).
This other question is definitely related, but the given answer does not address this question: it neither gives a specific function of matrices for the curl, nor proves that no such function can exist. (It also seemingly ignores the case when $n\not=3$ and the fact that the curl still exists as a bivector field then.)
Best Answer
$ \newcommand\dd{\mathrm D} \newcommand\Tr{\mathop{\mathrm{Tr}}} $Let $V$ be our (real) $n$-dimensional vector space. Something that must be noted is that divergence—being the trace of the differential—is an intrinsic quantity, but curl is not in the following sense: given any nondegenerate symmetric bilinear form $(v, w) \mapsto v\cdot w$ (henceforth, a metric), we can define a basis-independent symbol $\nabla$ in such a way that $v\cdot\nabla$ is the $v$-directional derivative and $\nabla\cdot f(x)$ is always the one and only divergence. The curl $\nabla\wedge f(x)$, however, is always dependent on the choice of metric. (Since you mention curl as being a bivector field, I am taking this expression as the definition of curl.)
Taking it on faith for the moment that $v\cdot\nabla f(x) = \dd f_x(v)$, what we want to do is extract a linear map $V \to V$ from the curl somehow. As a tensor, the curl acts on $V^{*}\wedge V^*$ in the following way $$ \langle\nabla\wedge f(x),\, \omega\wedge\eta\rangle = \omega(\nabla)\eta(f(x)) - \eta(\nabla)\omega(f(x)). $$ Using the raising operator $\sharp : V^{*} \to V$, we can write this as $$ (\omega^\sharp\cdot\nabla)\eta(f(x)) - (\eta^\sharp\cdot\nabla)\omega(f(x)) = \eta(\dd f_x(\omega^\sharp)) - \omega(\dd f_x(\eta^\sharp)). $$ In index notation, this reads $$ \eta_i\partial_jf^ig^{jk}\omega_k - \omega_k\partial_jf^kg^{ji}\eta_i = (\partial_jf^ig^{jk} - \partial_jf^kg^{ji})\omega_k\eta_i. \tag{$*$} $$ Here we see the metric ($g$) dependence very explicitly. To get a matrix, we want to raise one of $\omega$ or $\eta$; choosing $\eta$, we write $$ (\partial_jf^ig^{jk} - \partial_jf^kg^{ji})g_{il}\omega_k\eta^l = (\partial_jf^ig^{jk}g_{il} - \partial_jf^k\delta^j_l)\omega_k\eta^l = (g^{jk}\partial_jf^ig_{il} - \partial_lf^k)\omega_k\eta^l. $$ The first term is exactly the metric adjoint $\overline{\dd f_x}$ of the differential, defined for any $T : V \to V$ by $$ v\cdot T(w) = \overline T(v)\cdot w $$ for all $v, w \in V$. When $V = \mathbb R^n$ and the metric is the standard inner product, the adjoint is precisely the transpose of $T$ when expressed in the standard basis.
So we have that the following map $V \to V$ is equivalent to the curl of $f$ at $x$: $$ \overline{\dd f_x} - \dd f_x. $$ In this sense the curl is the "antisymmetric part" of $\dd f_x$.
Returning to the index notation expression, let's forcibly raise $\omega$ as well: $$ (g^{jk}\partial_jf^ig_{il} - \partial_lf^k)\omega_k\eta^l = (g^{jk}\partial_jf^ig_{il} - \partial_lf^k)g_{km}\omega^m\eta^l = (\delta^j_m\partial_jf^ig_{il} - \partial_lf^kg_{km})\omega^m\eta^l = (\partial_mf^ig_{il} - \partial_lf^kg_{km})\omega^m\eta^l. $$ But now we can absorb the metric into $f$! We get $$ (\partial_mf_l - \partial_lf_m)\omega^m\eta^l. $$ This is a new operation which takes in a (not necessarily linear) function $f : V \to V^{*}$—in other words, a one-form—and outputs a two-form. We can realize the parenthized expression as $$ (\dd f_x)^{*} - \dd f_x $$ where by $({-})^{*}$ we mean the dual of a linear transformation; because $\dd f_x : V \to V^{*}$ it's dual is also $V \to V^{*}$ and the above expression makes sense. This is the exterior derivative.
What exactly is $\nabla$, and how can we see the metric (in)dependence of the curl (divergence)? How can we see more clearly the relationship between the curl and the exterior derivative?
Let $L : V^*\times V \to W$ be linear in the first argument. We define a symbol $\Delta$ which is analogous to $\nabla$ but covector like instead of vector like; we will write its argument on the left instead of the right so that $(v)\Delta$ is a scalar-like differential operator defined by $$ (v)\Delta f(x) = \dd f_x(v) $$ for all $v \in V$.
We want to define $L(\Delta; x)$ with $\Delta$ differentiating the $x$-dependence. To this end, let $v \in V$ and $\omega\in V^*$. We consider the following manipulations reasonable: $$ L(\omega\,(v)\Delta; x) = (v)\Delta L(\omega; x) = \dd[L(\omega; {-})]_x(v). $$ Contracting $\omega$ and $v$, which we write as $\Tr_{\omega(v)}$, we define $$ L(\Delta; x) = \Tr_{\omega(v)}\dd[L(\omega; {-})]_x(v). $$ Call this the derivative of $L$. We already know of two derivatives:
Let's write out the last one. $$ \Delta\wedge\Psi(x) = \Tr_{\omega(v)}\dd[\omega\wedge\Psi ]_x(v) = \Tr_{\omega(v)}\omega\wedge\dd\Psi_x(v). $$ Now using index notation with a basis $e_i$ with dual $e^i$, $$ \Delta\wedge\Psi(x) = e^i\wedge\partial_i\Psi, $$ which is the exterior derivative.
Now, given a metric we have the musical isomorphism $\sharp : V^* \cong V$, and we simply define $$ \nabla = \Delta^\sharp. $$ Now it's clear why the divergence of $f : V \to V$ is metric-invariant: $$ \nabla\cdot f(x) = \Delta^\sharp\cdot f(x) = (f(\dot x))\dot\Delta. $$ In a sense, the metricity of $\nabla$ and the metricity of $\cdot$ "cancel out". We also see why the curl is metric dependent $$ \nabla\wedge f(x) = \Delta^\sharp\wedge f(x) $$ and the relationship with the exterior derivative is readily apparent.
Finally, we easily derive that $$ v\cdot\nabla f(x) = v\cdot\Delta^\sharp f(x) = (v)\Delta f(x) = \dd f_x(v). $$