What does a homogeneous space of a linear algebraic group know about the group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:38:46Z http://mathoverflow.net/feeds/question/56440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56440/what-does-a-homogeneous-space-of-a-linear-algebraic-group-know-about-the-group What does a homogeneous space of a linear algebraic group know about the group? Mikhail Borovoi 2011-02-23T19:42:08Z 2011-02-24T18:46:16Z <p>Let $X=G/H$, where $G$ is a connected linear algebraic group over the field $\mathbf{C}$ of complex numbers and $H\subset G$ is an algebraic subgroup. In general, we can write the algebraic variety $X$ as $X=G'/H'$ with $G'$ non-isomorphic to $G$. What can be said about $G'$ (and $H'$)?</p> <p>Of course, we can always take any surjective homomorphism $G'\to G$, then $X$ is naturally a homogeneous space of $G'$. Conversely, if the action of $G$ on $X$ is not effective, we can take a normal subgroup $N$ of $G$, contained in $H$ (say, the kernel of the action) and set $G'=G/N,\ H'=H/N$, then $X=G'/H'$. I am interested in less trivial examples.</p> <p>I would be happy to get a general answer, but will be also grateful for special cases, examples, comments, etc.</p> <p>I am obliged to Günter Harder for this nice title of my question.</p> http://mathoverflow.net/questions/56440/what-does-a-homogeneous-space-of-a-linear-algebraic-group-know-about-the-group/56445#56445 Answer by Dave Anderson for What does a homogeneous space of a linear algebraic group know about the group? Dave Anderson 2011-02-23T20:42:24Z 2011-02-23T20:42:24Z <p>For the special case where $X$ is projective, and reducing (by the "trivial" examples of surjections and normal quotients) to $G$ a semisimple group of adjoint type, with $H=P$ a parabolic, one version of the answer is given by Demazure in "Automorphismes et déformations des variétés de Borel", <em>Invent. Math.</em>, 1977. In a nutshell, the automorphism group of such an $X$ is the subgroup of $Aut(G)$ preserving the conjugacy class of $P$ --- unless the pair $(G,P)$ is one of several "exceptional" cases. (E.g., the $5$-dimensional quadric is both $G_2/P$ and $SO_7/P$.)</p> <p>In other words, for the non-exceptional cases, you can recover $G$ as the connected component of the identity in $Aut(X)$.</p> http://mathoverflow.net/questions/56440/what-does-a-homogeneous-space-of-a-linear-algebraic-group-know-about-the-group/56548#56548 Answer by Jim Humphreys for What does a homogeneous space of a linear algebraic group know about the group? Jim Humphreys 2011-02-24T18:46:16Z 2011-02-24T18:46:16Z <p>To add perspective to Dave's answer, it seems that homogeneous spaces realized as quotients of a connected reductive group <code>$G$</code> by parabolic subgroups are often natural and rich in structure besides having close connections with <code>$G$</code>. But I'm less confident about other types of homogeneous spaces. There are lots of arbitrary constructions, e.g., start with a product of connected 1-dimensional groups and factor out the product of one or more of them. At that extreme the homogeneous space has little connection with the given group. </p> <p>Even though factoring out a parabolic subgroup (especially a maximal one) tends to produce a relatively "small" variety, its automorphism group is often close to <code>$G$</code> (as seen in Demazure's paper). The only other type of maximal closed connected proper subgroup in <code>$G$</code> is itself reductive, as follows from Borel-Tits theory in arbitrary characteristic: see for instance Section 30.4 of my Springer text GTM 21. But then you get an <em>affine</em> homogeneous space. In fact, R.W. Richardson's classic theorem in Bull. London Math. Soc. 9 (1977) shows that <code>$G/H$</code> is affine iff the closed subgroup <code>$H$</code> of the reductive group <code>$G$</code> is itself reductive.</p> <p>All of which leads me to wonder how realistic it is in these affine cases to expect the homogeneous variety to know much about the given group? In other words, I'm uncertain about how broad the statement of the original question should be.</p>