Question: Given a twice totally differentiable (not necessarily $C^2$) function $f: \mathbb{R}^m \to \mathbb{R}^n$, do its $n$ Hessian matrices correspond to the exterior derivatives of its $n$ gradient 1-forms? What about the skew-symmetrizations of the Hessians?
Background: This previous question appears to be related. The "Edit 1" of the accepted answer says "I think the exterior derivative is just the antisymmetric part of the Jacobian", but I haven't found anything definitive answering this question.
This unanswered question (which I admittedly I do not fully understand) also appears to be related.
I understand that, given a totally differentiable function $f: \mathbb{R}^m \to \mathbb{R}^n$, each of the $n$ rows of the $n \times m$ Jacobian (total derivative) matrix corresponds to the gradient 1-form of the associated differentiable scalar function $\mathbb{R}^m \to \mathbb{R}$.
Likewise, given a twice totally differentiable function $f: \mathbb{R}^m \to \mathbb{R}^n$, the second total derivative corresponds to a $n \times m \times m$ 3rd order tensor, for which each of the $n$ slices that are $m \times m$ matrices correspond to the Hessian of the associated differentiable scalar function $\mathbb{R}^m \to \mathbb{R}$. Each $m \times m$ matrix slice can be thought of either as a linear function $\mathbb{R}^m \otimes \mathbb{R}^m \to \mathbb{R}$ or equivalently (via the universal property of the tensor product) as a bilinear function $\mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}$.
The inputs to that bilinear funciton would presumably be the two directions in which one wants to take directional derivatives in order to get a "second directional derivative". (Analogous to how the input to the gradient 1-form is the direction in which one wants to take a directional derivative).
Unlike arbitrary linear functions $\mathbb{R}^m \to \mathbb{R}$ that (vacuously) are always skew-symmetric (aka alternating) and hence are one-forms, an arbitrary bilinear function $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ will not necessarily be skew-symmetric.
This suggests that the slices of the $n \times m \times m$ second derivative tensor are not themselves the second exterior derivative (i.e. exterior derivative of corresponding gradient), which would be an alternating bilinear form.
But then what would correspond to the exterior derivatives of the gradient 1-forms? The skew-symmetrization of these Hessian slices? (Cf. these Wikipedia articles [1][2][3] for definition of skew-symmetrization.) That seems kind of too good to be true?
On the other hand, we know that if $f$ is $C^2$ (twice continuously differentiable or equivalently has continuous $\mathbb{R}^m \to \mathbb{R}$ second partial derivatives) by the Schwarz-Clairaut theorem that these Hessian matrices are symmetric matrices. Hence the anti-symmetrization of their corresponding bilinear functions would be identically zero.
That would appear to be consistent with the result that $d^2 = 0$ for exterior derivatives. But it hardly proves that the second exterior derivative is the anti-symmetrization of the Hessian.
Follow-up (bonus) question: If the second exterior derivative is the anti-symmetrization of the Hessian, then for twice totally differentiable functions whose Hessians are not symmetric (hence are necessarily not twice continuously differentiable functions), is the identity $d^2 = 0$ false? I.e. does the identity $d^2 = 0$ implicitly invoke the assumptions of the Schwarz-Clairaut theorem, i.e. continuous second partial derivatives?
