Restricted universal enveloping algebra of Abelian p-Lie algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:25:50Z http://mathoverflow.net/feeds/question/66925 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66925/restricted-universal-enveloping-algebra-of-abelian-p-lie-algebra Restricted universal enveloping algebra of Abelian p-Lie algebra darij grinberg 2011-06-04T23:49:19Z 2012-08-10T06:33:55Z <p><strong>Question:</strong> Let $p$ be a prime. Let $k$ be a commutative ring such that $p=0$ in $k$.</p> <p>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 $v\in\mathfrak g$.</p> <p>Let $U^{[p]}\left(\mathfrak g\right)$ be the <a href="http://en.wikipedia.org/wiki/Restricted_Lie_algebra#Restricted_universal_enveloping_algebra" rel="nofollow">restricted universal enveloping algebra</a> 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.</p> <p>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?</p> <p><strong>Motivation:</strong> 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).</p> http://mathoverflow.net/questions/66925/restricted-universal-enveloping-algebra-of-abelian-p-lie-algebra/66934#66934 Answer by Torsten Ekedahl for Restricted universal enveloping algebra of Abelian p-Lie algebra Torsten Ekedahl 2011-06-05T04:42:58Z 2011-06-05T04:42:58Z <p>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.</p>