# Idempotents in Rings of Differential Operators

## 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 $$R$$?

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 $m$).