(This is a cross-post from MSE).

Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth. The mixed partial derivatives commute: $f_{xy}=f_{yx}$. This identity is "universal" in the sense that it holds for any smooth map.

Question:

Are there any universal identities which are not consequences of the commutation of the mixed derivatives?

One way to phrase this is mentioned here:

Let $D_i$ be the differential operator which takes the partial derivative with respect to $x_i$. The symmetry can be written as an algebraic statement:

$$ D_i \circ D_j = D_j \circ D_i \tag{1}$$

So, the ring of differential operators with constant coefficients, generated by the $D_i$, is commutative.

(When we choose the domain for these operators to be the space of smooth maps $\mathbb{R}^d \to \mathbb{R}^d$, so we can compose operators).

Are there relations in this ring which are not consequences of the fundamental relation $(1)$?.

Edit:

Actually, this is not the "right" algebraic formulation which interests me: I want to be able to talk about relations of the form of $f_x f_y=f_yf_x$ (trivially true) or $f_{xx}=f_yf_{xy}$ (clearly false). Thus I need to add multiplication.

(Without multiplication, there are no additional relations as observed in this answer)

In particular, I am interested to know whether the "Cofactor Lemma" (divergence-free rows, see below) is a "consequence" of the commutation of mixed derivatives (for dimension $d>2$ this involves multiplication, as well as addition and composition).

So, we need to consider some algebraic structure $A$ which is a subset of the set of differential operators (which map $C^{\infty}(\mathbb{R}^d) \to C^{\infty}(\mathbb{R}^d)$), namely the set of operators which is "generated" by the $D_i$ via the $3$ operations - addition,composition,multiplication.

Do you have an idea to how formulate this algebraically?

The Cofactor Lemma:

Let $f:\mathbb{R}^d \to \mathbb{R}^d$ be smooth. Then the Cofactor of $df$ has divergence-free rows:

$$\sum_{j=1}^n \frac{\partial(Cof(Du))_{kj}}{\partial x_j} = 0, k=1,...,d.$$

In dimension $d=2$, it reduces to relation $(1)$:

Given $A= \begin{pmatrix} a & b \\\ c & d \end{pmatrix}$, $\operatorname{Cof}A= \begin{pmatrix} d & -c \\\ -b & a \end{pmatrix}$, so

$$ df= \begin{pmatrix} (f_1)_x & (f_1)_y \\\ (f_2)_x & (f_2)_y \end{pmatrix}, \operatorname{Cof}df= \begin{pmatrix} (f_2)_y & -(f_2)_x \\\ -(f_1)_y & (f_1)_x \end{pmatrix} .$$

We see that $\operatorname{div} (\operatorname{Cof}df)=0$ is equivalent to $(f_1)_{xy}=(f_1)_{yx},(f_2)_{xy}=(f_2)_{yx}$.

As stated above, for dimension $d>2$ we need multiplication to even phrase the question properly.