Restricted universal enveloping algebra of Abelian p-Lie algebra

Question: Let $p$ be a prime. Let $k$ be a commutative ring such that $p=0$ in $k$.

Let $\mathfrak g$ be an abelian $p$-restricted Lie algebra over $k$. In other words, let $\mathfrak g$ be a $k$-module along with a $\mathbb Z$-linear map ${}^{[p]}:\mathfrak g\to\mathfrak g$ (written postfix) that satisfies $\left(\lambda v\right)^{[p]}=\lambda^p v^{[p]}$ for all $\lambda\in k$ and $v\in\mathfrak g$.

Let $U^{[p]}\left(\mathfrak g\right)$ be the restricted universal enveloping algebra of $\mathfrak g$. In other words, let $U^{[p]}\left(\mathfrak g\right)$ be the factor algebra of the symmetric algebra of $\mathfrak g$ modulo the ideal generated by elements of the form $x^p-x^{[p]}$ with $x\in\mathfrak g$. Note that $U^{[p]}\left(\mathfrak g\right)$ is not a graded algebra, but a filtered one.

Does the canonical projection $\otimes \mathfrak g\to\mathrm{Sym}\mathfrak g \to U^{[p]}\left(\mathfrak g\right)$ (where $\otimes \mathfrak g$ means the tensor algebra of $\mathfrak g$) split canonically?

Motivation: If this holds, then it is an analogue of the fact that over a ring $k$ in which $1$, $2$, $3$, ... are invertible (e. g., a field of characteristic $0$), the projection from the tensor algebra to the symmetric algebra of a module splits canonically (the splitting is the symmetrizer).

• The splitting of $\mathfrak{g} \to U^{[p]}(\mathfrak{g})$ is discussed in Friedlander and Parshall Rational actions associated to the adjoint representation at the end of $\S$1; they give examples where no splitting exists.
– M T
May 11, 2015 at 16:07

I don't think so. Consider the case which should be the most difficult to split canonically, the case when the $$p$$'th power map is zero. The automorphism group is then equal to the linear automorphism group of $$\mathfrak g$$ and I assume further that $$k$$ is an infinite field and $$\mathfrak g$$ a finite dimensional vector space. I then interpret "canonical" as saying in particular that a canonical splitting respects the action of the automorphism group, the general linear group of $$\mathfrak g$$. As the relations in the restricted enveloping algebra are of the form $$x^p=0$$ everything is graded and the grading can be read off from how scalar multiplication acts. As $$k$$ is infinite this means that a canonical splitting must be homogeneous. This implies that the map from $$\mathrm{Sym}^p\mathfrak g$$ onto the degree $$p$$ part of $$U^{[p]}(\mathfrak g)$$ must split equivariantly. However, the kernel of this map is the image of $$\mathfrak g^{(p)}$$ given by the $$p$$'th power map in the symmetric algebra and it is well-known that that inclusion does not split as a map of representations of the general linear group.