Idempotents in Rings of Differential Operators

Question

Differential Operators on General Commutative Rings

Let k be an algebraically closed field of characteristic zero, and let R be a commutative k-algebra. Then a (Grothendieck) differential operator on R is a k-linear endomorphism $\delta$ of R, with the property that there is some $n\in \mathbb{N}$ such that for any $r_0,r_1...r_n\in R$, the iterated commutator vanishes: $$ [...[[\delta,r_0],r_1]...,r_n]=0$$ Let the smallest such $n$ be the order of $\delta$.

The set of all differential operators is then a subring of $End_k(R)$, which has an ascending filtration given by the order, and with $D_0(R)=R$. If $R=k[x_1,...x_r]$, then $D(R)$ will be polynomial differential operators (in the calculus sense) in r-variables. More generally, if R is the ring of regular functions on a smooth affine variety, then $D(R)$ is the usual ring of differential operators generated by multiplication operators and directional derivatives.

However, if $Spec(R)$ is not smooth, then $D(R)$ does not have an obvious geometric interpretation. For example, if $R=k[x]/x^n$, then all k-linear endomorphisms of R are differential operators, and so $$D(k[x]/x^n)=Mat_n(k)$$

Idempotents

For both research reasons and curiosity, I am interested in idempotent elements in $D(R)$, for R a general commutative ring. An idempotent is an element $\delta\in D(R)$ such that $\delta^2=\delta$. Idempotents in a commutative ring $R$ correspond to projections onto disconnected components of $Spec(R)$, but $D(R)$ is not commutative. If the base ring $R$ does have idempotents, then they will also be idempotents under the inclusion $R\subset D(R)$.

However, there can be idempotents of higher order. Consider the example from before, of $R=k[x]/x^n$. Here, $D(R)=Mat_n(k)$, and there are many idempotents in $Mat_n(k)$, even though $R$ here has none. As an explicit example, take $k[x]/x^2$, and consider the endomorphism which sends 1 to 0 and x to itself. This can be realized by the differential operator $x\partial_x$ (which has a well-defined action on $k[x]/x^2$), and it squares to itself. In general, I believe that $R$ must have nilpotent elements if $D(R)$ will have idempotents of positive order (since the symbol needs to square to zero).

My general question is, what is known about general idempotent elements in $D(R)$? Has anyone seriously looked at them? Do they correspond to something geometric? Is there a condition one can put on a subspace decomposition $V\oplus W=R$ such that the projection onto $V$ which kills $W$ is a differential operator for the algebra structure on $R$?

Answer 1

Here's a cute partial result, classifying idempotents of order 1.

Let $\delta$ be an idempotent differential operator of order 1. Then there is a unique decomposition $R\simeq A\oplus M$, with $A$ a subring and $M$ a square-zero ideal, and an element $m\in M$, such that $$ \delta = \epsilon + m - (1-2\epsilon) \pi_M $$ where $\epsilon$ is an idempotent in $R$ and $\pi_M$ is the projection onto $M$ with kernel $A$ (which is a derivation). Note that if $R$ is an integral domain, then $\epsilon$ is $1$ or $0$.

As a consequence, when $R$ is an integral domain, the decomposition of $R$ corresponding to $\delta$ and $1-\delta$ is $R=A'\oplus M$, where $A'$ is a shear translation of $A$ given by $a\rightarrow a+m$ (for some fixed $m$).