2
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – M T
    May 11, 2015 at 16:07

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.