Let $X$ be a reasonable smooth scheme over some base $S$. The tangent sheaf $T\_X$ is a Lie algebroid, locally free as a $\mathcal{O}\_X$ module, and its Universal Enveloping Lie Algebroid $\mathfrak{U}(T\_X)$ is the sheaf of differential operators on $X$.


However, when $X$ is not necessarily smooth, for any open $U \subset X$, we could still apply the Universal Enveloping Lie Algebroid construction to $T\_X(U)$ and get a presheaf. Is it a sheaf when X is affine? Or the question is: does the Universal Enveloping Lie Algebroid construction commute with localization?


Thank you.



About the construction of the Universal Enveloping Lie Algebroid
Suppose that $X$ is affine.


A natural way to prove that localisation commutes with the universal enveloping Lie algebroid construction is via Rinehart's Theorem 3.1 in "Differential forms on general commutative algebras." Transactions of the American Mathematical Society (1963): 195-222. This says that if the global object of the Lie algebroid, a $(\mathcal{O}\_S(X),\mathcal{O}\_X(X))$-Lie algebra $L$, is a projective $\mathcal{O}\_X(X)$-module then the associated graded ring of $U(L)$ with respect to its natural filtration is isomorphic to the symmetric algebra $\mathrm{Sym}(L)$. Thus in this case one can use the fact that the symmetric algebra construction commutes with localisation to deduce the same for the universal enveloping Lie algebroid. 


The condition that $L$, in your case $T\_X(X)$, is a projective $\mathcal{O}\_X(X)$-module is necessary for this approach. I don't know what you could do in general. I imagine that it might be possible to find examples where it is false but I don't know of any. 


