6
$\begingroup$

(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.

$\endgroup$
  • 2
    $\begingroup$ Since you are a geometer I suspect that you might be eventually interested in natural differential operators. There is a book: emis.de/monographs/KSM $\endgroup$ – Vít Tuček May 17 '17 at 12:20
  • $\begingroup$ Thanks for the reference, you are right in your suspicion:) $\endgroup$ – Asaf Shachar May 21 '17 at 15:51
7
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Can't you simplify this by applying your "the values of finitely many of the various derivatives are completely unconstrained" argument to $f$ instead of $g$? $\endgroup$ – Will Sawin May 18 '17 at 2:09
  • $\begingroup$ Oh, you're right of course. I was worried about having some sort of unusual cancellation in the nonlinear case but, yeah, that doesn't actually happen. $\endgroup$ – Terry Tao May 18 '17 at 5:01
  • $\begingroup$ I don't understand. What breaks down in this proof if we don't assume that $f_{xy} = f_{yx}$? Isn't "the values of finitely many of the various derivatives are completely unconstrained" precisely the statement the OP wants to prove here? $\endgroup$ – Vít Tuček May 20 '17 at 12:50
  • $\begingroup$ Given any finite set of coefficients $c_{i,j}$, one can create a function $f(x,y)$ with $\partial_x^i \partial_y^j f(0,0) = c_{i,j}$ for each $(i,j)$ in the finite set, simply by forming the finite Taylor series $\sum_{i,j} \frac{c_{i,j}}{i! j!} x^i y^j$. Of course, one cannot assign two different values to $f_{xy}$ and $f_{yx}$ by this procedure, as there is only one slot for both of these in the Taylor expansion. $\endgroup$ – Terry Tao May 20 '17 at 22:19
  • $\begingroup$ @TerryTao Right, so the answer to the question: "Are there any universal identities which are not consequences of the commutation of the mixed derivatives?" is "No, because universal identities would have to be true for polynomials and monomials are linearly independent." $\endgroup$ – Vít Tuček May 22 '17 at 12:59
5
$\begingroup$

Edit: after the edits of OP this seems to be an incomplete answer.

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

If by this ring you mean the subring of Maps($C^\infty(\mathbb{R}^n),C^\infty(\mathbb{R}^n)$) generated by partial derivatives (where I consider composition as multiplication), then the answer is, no. Indeed, the map from $\mathbb{Z}[v_1,\ldots,v_n]$ (the free commutative ring in $n$ generators, aka polynomials with integer coefficients), to Maps($C^\infty(\mathbb{R}^n),C^\infty(\mathbb{R}^n)$), sending each generator $v_i$ to the partial derivative operator $D_i$ is injective, since from

$\sum_{\alpha\in\mathbb{N}^n} c_\alpha D^\alpha (f)=0$ for all functions $f\in C^\infty(\mathbb{R}^n)$,

we can conclude that all coefficients $c_\alpha\in \mathbb{Z}$ are zero, by applying it to monomial functions $f(x_1,\ldots,x_n)=x^\alpha$. (Here $\alpha =(\alpha_1,\ldots,\alpha_n)$ is a multi-index and $x^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ etc.)

$\endgroup$
  • $\begingroup$ Thanks, you are right. After thinking a bit more, it seems the "right" question that I meant to ask is a little different: Choosing the ring to be the one you specified (which was a just interpretation of my question) , I see that we missed something: multiplication. I want to allow relations of the form of $f_xf_y=f_yf_x$ (trivially true) or $f_{xx}=f_yf_{xy}$ (clearly false). $\endgroup$ – Asaf Shachar May 16 '17 at 17:18
  • $\begingroup$ In fact, I am specifically interested whether the "cofactor identity" I mentioned in the question (divergence-free-rows) comes from the commutation of partial derivatives for $d>2$. (In $d>2$ multiplication is involved). So, we somehow need to consider all map $C^{\infty}(\mathbb{R}^n) \to C^{\infty}(\mathbb{R}^n)$ generated by the $D_i$ via composition, addition and multiplication... Can you think of the right algebraic setting for this? (Algebra? Module? this structure has $3$ operations...) $\endgroup$ – Asaf Shachar May 16 '17 at 17:18
  • $\begingroup$ I don't quite understand what you mean by multiplication vs composition. How does the multiplication of two partial derivatives act on a function? $\endgroup$ – Michael Bächtold May 17 '17 at 6:17
  • $\begingroup$ $D_x \times D_y$ should take $f$ to $f_x \cdot f_y$, $D_x \circ D_y$ should take $f$ to $f_{xy} $, $(D_x \circ D_x) \times D_y$ should take $f$ to $f_{xx} f_y$ etc. (Here $f$ is a scalar function, to extend the operations to $\mathbb{R}^d$-valued maps, jut act on each component separately. I tried to elaborate more on this in my edit... $\endgroup$ – Asaf Shachar May 17 '17 at 6:25
  • $\begingroup$ The "Cofactor lemma" is a good non-trivial example of the kind of "combined operations" I want to discuss. For $d>2$ the cofactor of $df$ involves multiplication of partial derivatives. In fact I am also allowing for the $i$-th component of output to depend on partial derivatives of all components of $f$. $\endgroup$ – Asaf Shachar May 17 '17 at 6:25
2
$\begingroup$

Consider the ring of $\mathcal{D}(n,k)$ of polynomial differential operators of $k$ smooth functions $f_1(x_1,\ldots,x_n), \ldots, f_k(x_1,\ldots, x_n)$ of $n$ variables. That is every element of $P \in \mathcal{D}(n,k)$ is a polynomials $P(x;f,\partial f, \partial^2 f, \ldots)$ in $f$ and finitely many of its partial derivatives, with coefficients being smooth functions of $x$. All of your examples fit into this category. Actually, $\mathcal{D}(n,k)$ is not only a ring, but also a differential ring; it has $n$ independent derivations commuting, $\partial_i \partial_j = \partial_j \partial_i$, and satisfying the Leibniz rule, $\partial_i(PQ) = (\partial_i P) Q + P (\partial_i Q)$. Also, it has the differential subring $\mathcal{D}(n,0) \subset \mathcal{D}(n,k)$ consisting of smooth functions of $n$ variables, with the usual action of partial derivatives on them.

Now, one can make the following statement:

The differential ring $\mathcal{D}(n,k) \cong \mathcal{D}(n,0)[f_1,\ldots, f_k]$ is a differential ring extension of $\mathcal{D}(n,0)$ freely generated by adjoining the generators $f_1, \ldots, f_k$.

Does this statement answer your question? The only identities used in the construction of the extension to $\mathcal{D}(n,k)$ from $\mathcal{D}(n,0)$ are those satisfied by all differential rings, namely the commutativity of the derivations (which just partial derivatives, in this case) and the Leibniz rule.

$\endgroup$
  • $\begingroup$ I think this statement may be what I am looking for, but I am not sure yet. Just a technical question: The generators $f_i$ are the partial derivations? (and the multiplication in your ring is pointwise multiplication, right?) $\endgroup$ – Asaf Shachar May 17 '17 at 11:47
  • $\begingroup$ Yes, the multiplication is the usual commutative pointwise multiplication. The generators are independent functions, for instance $f_1$ or any of its derivatives have no relation with $f_2$ or any of its derivatives. The derivatives are for instance $\partial_i f_1$ or $\partial_i\partial_j f_1$, where $i$ and $j$ may run from $1$ to $n$. You may want to look up the notion of a (partial) differential ring. $\endgroup$ – Igor Khavkine May 17 '17 at 11:58
  • $\begingroup$ @AsafShachar I think the right picture to have in mind here is to represent any smooth function $f$ by (left) multiplication operator $g \mapsto fg$. $\endgroup$ – Vít Tuček May 17 '17 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.