(This is a cross-post from [MSE](https://math.stackexchange.com/questions/2276367/universal-differential-identities)).


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](https://mathoverflow.net/questions/269907/universal-differential-identities/269909#269909)).


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](https://en.wikipedia.org/wiki/Adjugate_matrix) 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. 



"Universal" differential identities
Let's work in two dimensions for notational simplicity. We claim that there is no non-trivial polynomial identity of the form
$$ P( f, f\_x, f\_y, f\_{xx}, f\_{xy}, \dots ) = 0$$
relating some finite number of derivatives of $f$, for any smooth $f$. Suppose for contradiction that this is the case. Taking a functional derivative (i.e. replacing $f$ by $f+ \varepsilon g$ for arbitrary smooth $g$, and then differentiating at $\varepsilon=0$, one obtains a linearised (in $g$) identity of the form
$$ P\_f(f,f\_x,\dots) g + P\_{f\_x}(f,f\_x,\dots) g\_x + \dots = 0.$$
Now, at any point in space, the values of finitely many of the various derivatives $g, g\_x, g\_y, \dots$ are completely unconstrained (as can be seen by taking $g$ to be a finite Taylor series around that point). So the only way the above identity can hold for all $f$ and $g$ is if one has
$$ P\_f(f,f\_x,\dots) = 0$$
$$ P\_{f\_x}(f,f\_x,\dots) = 0$$
etc.. By induction on the degree of $P$, this can only happen if $P\_f, P\_{f\_x}, \dots$ vanish identically, so that $P$ is constant, hence zero, so the identity is trivial.


