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(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 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.

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 $f:\mathbb{R}^d \to \mathbb{R}^d$).

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.

(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.

(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.

(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 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.

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 $f:\mathbb{R}^d \to \mathbb{R}^d$).

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.

.(This is a cross-post from MSE)

Consider the following well-known fact:

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be smooth. Then the mixed partial derivatives commute: $f_{xy}=f_{yx}$. This identity is "universal" in the sense that it holds for any smooth map, no matter what are its properties or any special features it may or may not posses (similarly for maps $\mathbb{R}^d \to \mathbb{R}$).

Question:

Are there any universal identities (satisfied by all smooth maps) 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:

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

From this relation it follows that 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 $\mathbb{R}^d \to \mathbb{R}^k$).

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

It's not always trivial to see explicitly if a relation comes from the symmetry:

For example, there is the following universal result:

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

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

For the case $d=2$, It indeed 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 $(f_1)_{xy}=(f_1)_{yx},(f_2)_{xy}=(f_2)_{yx}$.

I guess this also hold for $d>2$ but this seems less transparent.

(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.