**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).

Rational actions associated to the adjoint representationat the end of $\S$1; they give examples where no splitting exists. $\endgroup$ – M T May 11 '15 at 16:07