Let $C$ be a smooth projective curve over an algebraically closed field $k$.

The tangent lie algeborid $\mathcal{T}_C$ of $C$ is just sheaf of vector fields on $C$ equipped with the usual lie bracket (commutator of derivations).

Definition: We'll say a lie algebroid $\mathcal{E}$ over a regular ring $R$ split iff it is free as a module over $R$ and there exists a basis for $\mathcal{E}$ consisting of mutually commuting sections (commuting for the bracket on $\mathcal{E}$). Is there already a name for this property in the literature?

For instance the tangent sheaf of a curve is locally split when it is locally spanned by mutually commuting vector fields.

Clearly the tangent lie algebroid of $\mathbb{A}^1$ is split and hence the tangent lie algebroid of $\mathbb{P}^1$ is locally split in the zariski topology.

Since every smooth variety is etale locally isomorphic to affine space we get that (at least for smooth varieties) the tangent lie algebroid is always locally split in the etale topology.

My hunch is that for the Zariski topology this fails miserably however i'm unsure about how to come up with an example. Hence the question:

Is $\mathbb{P}^1$ the only smooth projective curve with a locally split (in the zariski topology) tangent lie algebroid? If it is, how can this be shown? If not is there some kind of classification of those with this property?

  • $\begingroup$ The Zariski-local structure of smooth morphisms as in Will Sawin's answer (i.e., etale onto an affine space) is absolutely crucial for developing the theory of smooth morphisms and the robustness of the etale topology (e.g., proving many theorems in etale cohomology), being one of several equivalent characterizations (the proof of equivalences among which is far from elementary). That semistable singularities don't generally admit a Zariski neighborhood etale onto the model situation (only an etale neighborhood, by Artin approximation) makes them more subtle to study rigorously. $\endgroup$
    – nfdc23
    Apr 23, 2017 at 13:54

1 Answer 1


If $f: X \to \mathbb A^n$ is an etale map, then we can pull back vector fields on $\mathbb A^n$ to vector fields on $X$. This pullback operation is a lie algebroid homomorphism. Hence if we pull back a commuting basis of vector fields from $\mathbb A^n$, we obtain a commuting basis on $X$.

Every smooth $n$-dimensional variety Zariski-locally admits an etale map to $\mathbb A^n$: If you choose a basis for the maximal ideal of the local ring of a point, these functions will extend to some Zariski neighborhood of the point, defining a map to $\mathbb A^n$ on that neighborhood, which is etale at that point.

So every smooth variety has a locally split tangent Lie algebroid.

  • 1
    $\begingroup$ If $k\ne \overline{k}$ then this recipe to make an etale map to an affine space Zariski-locally around each closed point wouldn't work at closed points whose residue field is not separable over $k$ (e.g., if $p={\rm{char}}(k)>0$ and $C={\rm{Spec}}(k[t])$ then for $a\in k -k^p$ and $P=\{t^p=a\}$, a $k(P)$-basis of $m_P/m_P^2$ is given by $\{t^p-a\}$, for which the associated $k$-map $C\to\mathbf{A}^1_k$ is not etale at $P$). So it is more "relative" to use $f_1,\dots,f_n\in O_X(U)$ for which $\{df_1,\dots,df_n\}$ is a frame of the vector bundle $\Omega^1_{U/k}$, as exist for $U$'s covering $X$. $\endgroup$
    – nfdc23
    Apr 23, 2017 at 13:46

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