(This is a cross-post from MSE).

Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth. The mixed 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?

More explicitly, 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 not universal). 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 differential operators (which map $C^{\infty}(\mathbb{R}^d) \to C^{\infty}(\mathbb{R}^d)$), that is "generated" by the $D_i$ via the **$3$** operations - addition, composition, and multiplication*.

(I am not sure if there is a term for such an "algebraic creature", $A$ is a ring w.r.t both operations $(+,\cdot)$ and $(+,\circ)$, but these two "multiplicative" operations have relations, namely $$(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h).$$

Does such a structure have a name?

*By *multiplication* (as opposed to *composition*) of operators I mean the following:

$$D_x \times D_y(f)=f_x \cdot f_y \, , \, D_x∘D_y(f)=f_{xy} \, , \, (D_x \circ D_x) \times D_y(f)=f_{xx}f_y$$ etc. (Here $f$ is a scalar function, to extend the operations to $\mathbb{R}^d$-valued maps, jut act on each component separately. I am also allowing for the $i$-th component of output to depend on partial derivatives of *all* components of $f:\mathbb{R}^d \to \mathbb{R}^d$).

*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.