Principal bundles over groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:52:49Z http://mathoverflow.net/feeds/question/55408 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55408/principal-bundles-over-groups Principal bundles over groups Piotr Achinger 2011-02-14T13:29:22Z 2011-02-14T22:45:54Z <p>If we have an extension of groups (say algebraic groups or group schemes) $1\to F\to P\to G\to 1$, then $P$ is a principal $F$-bundle over $G$ (is it locally trivial?). How about going in the opposite direction?</p> <blockquote> <p><strong>Question.</strong> Let $F$ and $G$ be groups. $P$ be a principal $F$-bundle over $G$. When does $P$ carry a structure of a group such that $1\to F\to P\to G\to 1$ is an extension? How to classify all such structures?</p> </blockquote> <p>Sounds connected with (some kind of) group cohomology of $G$ with coefficients in $F$, but I can't figure out what exactly this group is. Does the obstruction lie in some $H^2$, which is only a pointed set, and if it vanishes then the structures are classified by the corresponding $H^1$? The fact that the principal bundle itself corresponds to an element of $H^1(G, F)$ confuses me a bit.</p> <p>For example, if $G$ is an abelian variety and $L$ a line bundle on it, then we have the corresponding principal $F=\mathbb G_m$-bundle $P_L$. When does it give an extension of $G$ by $\mathbb G_m$?</p> http://mathoverflow.net/questions/55408/principal-bundles-over-groups/55413#55413 Answer by Dan Petersen for Principal bundles over groups Dan Petersen 2011-02-14T14:29:48Z 2011-02-14T14:35:36Z <p>I can at least answer your last question: a line bundle on an AV gives an extension by $\mathbb G_m$ if and only if it has degree zero. (So I guess there is an obstruction in a particularly simple $H^2$!) This is usually stated as an isomorphism $A^\vee = \operatorname{Pic}^0 A \cong \mathrm{Ext}(A,\mathbb{G}_m)$. More generally, extensions of an abelian variety by a torus <em>T</em> are classified by homomorphisms from the character group of <em>T</em> to $A^\vee$, and when $T = \mathbb{G}_m$ such a homomorphism is just determined by the element of $A^\vee$ that a generator is sent to.</p> <p>If I recall correctly this is discussed nicely in Serre's Groupes Algébriques et Corps de Classes.</p> http://mathoverflow.net/questions/55408/principal-bundles-over-groups/55414#55414 Answer by Mikhail Borovoi for Principal bundles over groups Mikhail Borovoi 2011-02-14T14:49:56Z 2011-02-14T18:04:25Z <p>Colliot-Thélène's paper Résolutions flasques des groupes linéaires connexes, J. für die reine und angewandte Mathematik (Crelle) 618 (2008) 77--133, <a href="http://www.math.u-psud.fr/~colliot/resolflsq_211107.pdf" rel="nofollow">http://www.math.u-psud.fr/~colliot/resolflsq_211107.pdf</a>, contains the following theorem:</p> <blockquote> <p><strong>Theorem 5.6.</strong> Let $k$ be a field and $G$ be a connected linear algebraic $k$-group, assumed reductive when $\mathrm{char}(k)>0$. Let $S$ be a smooth $k$-group of multiplicative type. Let $p\colon Y\to G$ be a torsor over $G$ under $S$, whose fiber is trivial over the neutral element $e_G\in G(k)$. Then there exists a structure of algebraic $k$-group on $Y$ such that $p\colon Y\to G$ is a homomorphism of algebraic $k$-groups with central kernel $S$.</p> </blockquote> <p>Corollary 5.7 says that that $\mathrm{Ext}_{k-\mathrm{gr}}(G,S)$ is in a canonical bijection with $\mathrm{Ker}[H^1(G,S)\to H^1(k,S)]$. Here $H^1(G,S)$ means étale cohomology.</p> <p>See also <a href="http://arxiv.org/abs/0912.0408" rel="nofollow">http://arxiv.org/abs/0912.0408</a>, Lemma 2.13, for a relative version of this theorem in characteristic 0 in the case $S=\mathbb{G}_m$. We consider a torsor $Y\to X$ under a connected linear $k$-group $G$ over a smooth $k$-variety $X$, and a torsor $Z\to Y$ under $\mathbb{G}_m$. We prove that $Z\to X$ has a structure of a torsor under some $k$-group $G_1$, a central extension of $G$ by $\mathbb{G}_m$. The class of such an extension $G_1$ is uniquely determined.</p> http://mathoverflow.net/questions/55408/principal-bundles-over-groups/55419#55419 Answer by Torsten Ekedahl for Principal bundles over groups Torsten Ekedahl 2011-02-14T15:40:01Z 2011-02-14T15:40:01Z <p>There is the Grothendieck theory of bitorsors (see e.g. SGA 7, Exp VII) which gives an abstract answer to this question. The key point is that when you have an actual group extension you do not just have a torsor but a bitorsor, i.e., $P$ is an $F$-torsor in two ways given by left and right multiplication. The group structure on $P$ can then be formulated in terms of maps of bitorsors (see <i>idem</i> for details). The two specific criteria mentioned by Dan and Mikhail do not follow formally from this description but they nevertheless fit very well into this framework.</p> http://mathoverflow.net/questions/55408/principal-bundles-over-groups/55427#55427 Answer by pasquale zito for Principal bundles over groups pasquale zito 2011-02-14T17:18:28Z 2011-02-14T17:18:28Z <p>I think the first place where this problem has been discussed is: <em>S. Eilenberg, S. MacLane</em> <strong>Cohomology theory in abstract groups. II. Group extensions with a non-abelian kernel</strong>, Ann. of Math., 48 (1947), 326-341.</p> http://mathoverflow.net/questions/55408/principal-bundles-over-groups/55465#55465 Answer by David Roberts for Principal bundles over groups David Roberts 2011-02-14T22:45:54Z 2011-02-14T22:45:54Z <p>The 'some $H^2$' which classifies group extensions you are thinking of is actually cohomology $H^1$ with values in the crossed module $F \to Aut(F)$ (or the 2-group corresponding to it). </p> <p>See <a href="http://ncatlab.org/nlab/show/nonabelian+group+cohomology" rel="nofollow">http://ncatlab.org/nlab/show/nonabelian+group+cohomology</a> for more details.</p>