Let $k$ be a commutative ring, $A$ a commutative $k$-algebra, and for some other part of why I'm asking this question I only care about the case when $k \supseteq \mathbb Q$. Recall the following notion, I think originally due to Grothendieck:

Definition (differential operator): Let $D : A\to A$ be $k$-linear. Define $s_nD : A^{\otimes n} \to A$ by: $$ s_nD(a_1,\dots,a_n) = \sum_{I \subseteq \lbrace 1,\dots,n\rbrace} (-1)^{|I|}\; \left( \prod_{i\not\in I} a_i\right) \;D\left(\prod_{i\in I}a_i\right) $$ One says that $D$ is an $n$th order differential operator if $s_{n+1}D = 0$.

Examples: $s_0D = D(1) \in A$. $s_1D(a) = D(a) - aD(1)$. $s_2D(a,b) = D(ab) - aD(b) - bD(a) + abD(1)$.

Remark: $s_nD$ is symmetric in the $a_i$s. If $D$ is an $n$th order differential operator, then $s_nD(-,a_2,\dots,a_n)$ is a derivation, for $a_2,\dots,a_n$ fixed. Thus if $D$ is an $n$th order differential operator, $s_nD$ is a symmetric polyderivation. It deserves the be called the principal symbol of $D$. It measures the failure of $D$ to be an $(n-1)$th order differential operator.

Question: For which algebras $A$ (i.e. what are natural, checkable conditions) does the following "PBW theorem" hold: $$ s_n: \lbrace n\text{th order differential operators}\rbrace \to \lbrace \text{symmetric }n\text{-polyderivations} \rbrace $$ surjective for every $n$?

Examples: This PBW theorem holds for $A = k[x_1,\dots,x_n]$ and $A = k [\\![ x_1,\dots,x_n ]\\!]$ and $A = \mathscr C^\infty(M)$ where $M$ is a smooth manifold. This PBW theorem failes for $A = k[x]/x^2$, as the space of symmetric biderivations is one-dimensional spanned by $x \frac{\partial}{\partial x}\otimes \frac{\partial}{\partial x}$, whereas every second-order differential operator is also a first-order differential operator. Edit: I don't know a characteristic-$0$ example for which PBW theorem fails, but I don't expect it to always hold.

Remark: I would expect that algebras $A$ for which the PBW theorem holds are those for which $\operatorname{spec}(A)$ is "smooth" in the appropriate sense, but I don't know if this is "smooth" in other usual senses of the word.

  • $\begingroup$ And just two years ago I was stuck on mathoverflow.net/questions/3477 . $\endgroup$ Nov 20, 2011 at 19:45
  • $\begingroup$ For finitely generated algebras, it seems very likely that the answer is if and only if $\operatorname{Spec}(A)$ is regular (i.e. has regular local rings). This is a local question, so you just have to prove PBW holds for a local ring if and only if it's regular. $\endgroup$
    – Ben Webster
    Nov 20, 2011 at 20:58
  • $\begingroup$ I am a bit confused with what you say about k[x]/x^2. It appears to me that d/dx is a second order differential operator on k[x]/x^2 (it's not of order 1, put a=b=x in the definition of s_2) whose symbol is precisely the one you give. $\endgroup$ Nov 20, 2011 at 21:01
  • $\begingroup$ @Vladimir: Do you mean the operator $x \mapsto 1$, $1\mapsto 0$? I refuse to call this operator $\frac{\partial}{\partial x}$, because it is not a derivation! But of course you are right, this fails to be a derivation by the map that sends $x\otimes x \mapsto 0 - x - x + 0 = -2x$, and so $-\frac12$ of this operator does the trick. $\endgroup$ Nov 20, 2011 at 23:27

1 Answer 1


I think a sufficient condition is:

the $A$-module $Der_k(A)$ of $k$-derivations of $A$ is projective.

A hint for the proof you may find in an old paper of G.S. Rinehart: Differential Forms on General Commutative Algebras (just have to notice that $(A,Der_k(A))$ is an example of a $(k,A)$-Lie algebra - today known as Lie-Rinehart algebra, or Lie algebroid). Especially: you might want to have a look at Theorem 3.1 and its proof.

  • $\begingroup$ Great, so projectivity is certainly enough, and probably in the general case this is the best possible result. Do you happen to know if this PBW theorem holds more generally when we're in characteristic zero? $\endgroup$ Nov 22, 2011 at 19:10
  • 1
    $\begingroup$ I think that you need a bit more in positive characteristic. The problem is that in Rinehart's language the symbol belongs to $I^{k-1}/I^k$ where $I\in A\otimes A$ is the kernel of the multiplication map, not a polyderivation. I believe that to use theorem 3.1 in our case, you also need to ask something along the lines that both $S^m(Der(A))$ and $SDer^m(A)$ (the latter being symmetric $m$-polyderivations) are projective [or maybe take the representing objects and talk about 1-forms and m-forms?], and outside char=0 this requires to assume more! $\endgroup$ Nov 24, 2011 at 11:08
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    $\begingroup$ Agreed! In order to go from Rinehart's result to Theo's problem one needs that $S^m_A(Der(A))=SDer^m(A)$ (even in characteristic zero this might not old when $Der(A)$ is not projective). In characteristic zero, evene in the smooth situation there is an issue: differential operators are not given by the universal envelopping algebra of $(A,Der_k(A))$, but rather by the restricted version (notice that $(A,Der_k(A))$ is a restricted Lie algebroid). $\endgroup$
    – DamienC
    Nov 24, 2011 at 13:31

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