For which algebras does \{Differential Operators\} satisfy a PBW-like theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:51:45Z http://mathoverflow.net/feeds/question/81446 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81446/for-which-algebras-does-differential-operators-satisfy-a-pbw-like-theorem For which algebras does \{Differential Operators\} satisfy a PBW-like theorem? Theo Johnson-Freyd 2011-11-20T19:45:08Z 2011-11-20T23:30:38Z <p>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:</p> <p><strong>Definition (differential operator):</strong> 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 <em>$n$th order differential operator</em> if $s_{n+1}D = 0$.</p> <p><strong>Examples:</strong> $s_0D = D(1) \in A$. $s_1D(a) = D(a) - aD(1)$. $s_2D(a,b) = D(ab) - aD(b) - bD(a) + abD(1)$.</p> <p><strong>Remark:</strong> $s_nD$ 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 <em>symmetric polyderivation</em>. It deserves the be called the <em>principal symbol</em> of $D$. It measures the failure of $D$ to be an $(n-1)$th order differential operator.</p> <blockquote> <p><strong>Question:</strong> 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 $n$?</p> </blockquote> <p><strong>Examples:</strong> 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. <strike>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.</strike> <strong>Edit:</strong> I don't know a characteristic-$0$ example for which PBW theorem fails, but I don't expect it to always hold.</p> <p><strong>Remark:</strong> 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.</p> http://mathoverflow.net/questions/81446/for-which-algebras-does-differential-operators-satisfy-a-pbw-like-theorem/81450#81450 Answer by DamienC for For which algebras does \{Differential Operators\} satisfy a PBW-like theorem? DamienC 2011-11-20T20:43:19Z 2011-11-20T20:48:25Z <p>I think a sufficient condition is: </p> <blockquote> <p>the $A$-module $Der_k(A)$ of $k$-derivations of $A$ is projective.</p> </blockquote> <p>A hint for the proof you may find in an old paper of G.S. Rinehart: <a href="http://www.ams.org/journals/tran/1963-108-02/S0002-9947-1963-0154906-3/S0002-9947-1963-0154906-3.pdf" rel="nofollow">Differential Forms on General Commutative Algebras</a> (just have to notice that $(A,Der_k(A))$ is an example of a $(k,A)$-Lie algebra - today known as <em>Lie-Rinehart algebra</em>, or <em>Lie algebroid</em>). Especially: you might want to have a look at Theorem 3.1 and its proof. </p>