# For which algebras does \{Differential Operators\} satisfy a PBW-like theorem?

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.

• And just two years ago I was stuck on mathoverflow.net/questions/3477 . Nov 20, 2011 at 19:45
• 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. Nov 20, 2011 at 20:58
• 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. Nov 20, 2011 at 21:01
• @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. Nov 20, 2011 at 23:27

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.
• 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! Nov 24, 2011 at 11:08
• 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). Nov 24, 2011 at 13:31