Lie algebra actions on schemes - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T20:51:51Zhttp://mathoverflow.net/feeds/question/37588http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/37588/lie-algebra-actions-on-schemesLie algebra actions on schemesDaniel Larsson2010-09-03T08:57:24Z2010-09-04T05:51:58Z
<p>Let us assume first of all that we are in the affine case (we can worry about globalization later) and that we have $X$ affine over $S$, where $S$ is some unspecified scheme (but in practice probably the spectrum of a field), with $X=\mathrm{Spec}(A)$ (thus $A$ is an $\mathcal{O}_S$-algebra). We are emphatically <em>not</em> assuming $X$ to be smooth over $S$. </p>
<p>Assume that we are given a map $\mathfrak{g}\to\mathrm{Der}_S(A)$ of Lie algebras and that we are viewing $\mathfrak{g}$ as a Lie-sub-algebra of $\mathrm{Der}_S(A)$. </p>
<p>In analogy with the differential-geometric case we can interpret this as a distribution on $X$ and so we can ask: what are the integral subschemes of this distribution? Specifically, is there, through every point, a <em>unique</em> integral subscheme? And even more importantly, what can go wrong in the singular points and can we "integrate" this action to an analogue of a Lie groupoid?</p>
<p>I'm seriously betting the answer to most of the above is a resounding "NO!" but I'm curious to know <em>what</em> can go wrong and what is <em>known</em> to go wrong? In short: what is known concerning this? Can one form something like "$X/\mathfrak{g}$"? </p>
<p>Finally, let me iterate that I'm not assuming $X$ to be $S$-smooth. </p>
http://mathoverflow.net/questions/37588/lie-algebra-actions-on-schemes/37695#37695Answer by Victor Protsak for Lie algebra actions on schemesVictor Protsak2010-09-04T05:51:58Z2010-09-04T05:51:58Z<p>I don't think that there is any hope for "integral subschemes" in the usual sense even in the nonsingular case, because it is exceptionally rare for a polynomial vector field to have algebraic trajectories. For example, consider an "irrational line on a torus" given by the vector field</p>
<p>$$z\frac{\partial}{\partial z}-\lambda w\frac{\partial}{\partial w}$$</p>
<p>with irrational $\lambda.$ Its integral curves have local equations $z^\lambda w=C,$ hence these integral curves are not algebraic. </p>