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In a recent question, I recalled the notion of differential operator, polyderivation, and principal symbol for a commutative algebra $A$ over some fixed commutative ring $k$. (I will not repeat those definitions here, because you can read them there.) In that question, I asked for sufficient conditions to assure that $A$ satisfied the following version of the "PBW theorem": whether the principal symbol map (which I called $s_n$) from ($n$th order) differential operators to ($n$th order) polyderivations is a surjection.

I thought I had an example to show that it is not always a surjection, but in the comments Vladimir Dotsenko pointed out that I was wrong. In fact, I have since been unable to come up with any counterexample to this "PBW theorem".

Because for some other part of my project I must restrict to the situation when $k\supseteq \mathbb Q$ (essentially because I only know how to define the commutator of polyderivations in that case), I will ask my question there:

Question: What's an explicit example of a commutative algebra $A$ over a commutative ring $k\supseteq \mathbb Q$ such that there exists a polyderivation $A^{\otimes n} \to A$ which is not the principal symbol of any differential operator.

If you can show that no such algebra exists, then I will gladly accept your answer both here and as an answer to my previous question.

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    $\begingroup$ It seems plausible that such an example could come from a ring $A$ whose ring of differential operators is not Noetherian. One standard such example is the coordinate ring of the cubic cone in $\mathbb{C}^3$. You can find more details here: math.tau.ac.il/~bernstei/Publication_list/publication_texts/… $\endgroup$
    – user91132
    Commented Nov 24, 2011 at 8:45

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Let me try to give a naive answer. Consider the following symmetric triderivation on $A=k[x]/x^2$: $$ x\partial_x\otimes\partial_x\otimes\partial_x:(x,x,x)\mapsto x $$ How could it be in the image of $s_3$, considering that any differential operator on $A$ is of order $\leq 2$?

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  • $\begingroup$ Yes, of course, I was being blind. $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 24, 2011 at 17:55

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