Is the tangent space functor from PD formal groups to Lie algebras an equivalence? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:55:31Z http://mathoverflow.net/feeds/question/8661 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8661/is-the-tangent-space-functor-from-pd-formal-groups-to-lie-algebras-an-equivalence Is the tangent space functor from PD formal groups to Lie algebras an equivalence? S. Carnahan 2009-12-12T08:37:27Z 2010-06-21T19:16:39Z <p>The previous version of this question was rather badly broken, and I hope this version makes some sense.</p> <p>There have been a few questions on MathOverflow about how much representation-theoretic information is lost when passing from a Lie group to its Lie algebra, e.g., away from the semisimple case, Lie algebras have many more representations. In the algebraic setting, there is an intermediate construction between an algebraic group and its Lie algebra, given by the formal group. One completes the algebraic group along the identity to get a formal scheme equipped with a group law, and one can pass from there to the tangent space to get the Lie algebra. In characteristic zero, the tangent space functor is an equivalence of categories from formal groups to Lie algebras, but in positive characteristic, formal groups form an honest intermediate category since the tangent space can lose a lot of information. For example, there is only one isomorphism class of one-dimensional Lie algebra, but one-dimensional formal groups have a rich arithmetic theory, with a moduli space stratified by positive integer heights. The completions at the identity of the additive group and the multiplicative group have very distinct formal group structures, and one way to explain the lack of isomorphism is by the presence of denominators in the usual logarithm and exponential power series.</p> <p>It seems to me that in positive characteristic, there could be an intermediate construction between formal groups and Lie algebras, given by passing to PD rings and replacing the coordinate ring of the formal group with the divided power envelope of the identity section. If I'm not mistaken, this construction yields a group object in PD formal schemes.</p> <p>Here is a bit of explanation for the uninitiated (see Berthelot-Ogus for more): PD rings are triples $(A,I,\gamma)$, where $A$ is a commutative ring, $I$ is an ideal, and <code>$\gamma = \{ \gamma_n: I \to A \}_{n \geq 0}$</code> is a system of <a href="http://en.wikipedia.org/wiki/Divided_power_structure" rel="nofollow">divided power operations</a>. I think they arose when Grothendieck tried to get De Rham cohomology to give the expected answers for proper varieties in characteristic $p$, since the naïve definition tended to yield infinite dimensional spaces. There is a forgetful functor $(A,I,\gamma) \mapsto (A,I)$ from PD rings to ring-ideal pairs, and it has a left adjoint, called the divided power envelope. In characteristic zero, $\gamma$ is canonically given as $\gamma_n(x) = x^n/n!$, so both functors are equivalences in that case. The notion of PD ring can be sheafified and localizations have canonical PD structures, so one has notions of PD scheme and PD formal scheme.</p> <blockquote> <p><strong>Question:</strong> Do PD formal groups contain any more information than the underlying Lie algebra?</p> </blockquote> <p>I have a suspicion that the answer is "no" and the answer to the title question is "yes". Vague word-association suggests that the divided power structure is exactly what one needs to get a formal logarithm, but maybe there is a more fundamental obstruction.</p> <p>I was originally motivated by the question of how Gelfand-Kazhdan formal geometry would differ in charateristic $p$ if I switched between ordinary and PD structures (cf. <a href="http://mathoverflow.net/questions/6057/formal-geometry" rel="nofollow">David Jordan's question</a>). Unfortunately, I was laboring under some misconceptions about formal completions, and I'm still a bit confused about the precise structure of the automorphism group of the completion (PD or ordinary) of a smooth variety at a point in characteristic $p$.</p> http://mathoverflow.net/questions/8661/is-the-tangent-space-functor-from-pd-formal-groups-to-lie-algebras-an-equivalence/8895#8895 Answer by Ben Webster for Is the tangent space functor from PD formal groups to Lie algebras an equivalence? Ben Webster 2009-12-14T18:34:35Z 2009-12-14T18:34:35Z <p>Have you looked at <a href="http://front.math.ucdavis.edu/0501.5247" rel="nofollow">Fedosov quantization in positive characteristic</a>? That has a page or so on formal geometry in characteristic p.</p> http://mathoverflow.net/questions/8661/is-the-tangent-space-functor-from-pd-formal-groups-to-lie-algebras-an-equivalence/28979#28979 Answer by Torsten Ekedahl for Is the tangent space functor from PD formal groups to Lie algebras an equivalence? Torsten Ekedahl 2010-06-21T19:16:39Z 2010-06-21T19:16:39Z <p>I think that this <em>MR0277590 (43 #3323) André, M. Hopf algebras with divided powers. J. Algebra 18 1971 19--50</em> may be relevant. It says that a graded commutative divided power Hopf algebra is the co-enveloping algebra of a graded Lie algebra. I think that the grading could be replaced by a condition of completion instead which would give your correspondence.</p>