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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).

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up vote 4 down vote accepted

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 respect 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.

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Nice one! I see that the well-known fact you used in your last sentence was detailed in your answer at mathoverflow.net/questions/15336 . –  darij grinberg Jun 5 '11 at 8:55
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