MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Question. 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?

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.

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 $\mathbb G_m$?

share|cite|improve this question
@Piotr: If you are in Bonn, come to my office at the Max-Planck-Institut für Mathematik, Bonn, on Monday Feb 21, or on Tuesday, and we can discuss all this. – Mikhail Borovoi Feb 15 '11 at 8:08
up vote 4 down vote accepted

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 idem 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.

share|cite|improve this answer

Colliot-Thélène's paper Résolutions flasques des groupes linéaires connexes, J. für die reine und angewandte Mathematik (Crelle) 618 (2008) 77--133,, contains the following theorem:

Theorem 5.6. 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 $S$.

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.

See also, 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.

share|cite|improve this answer

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 T are classified by homomorphisms from the character group of T 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.

If I recall correctly this is discussed nicely in Serre's Groupes Algébriques et Corps de Classes.

share|cite|improve this answer

I think the first place where this problem has been discussed is: S. Eilenberg, S. MacLane Cohomology theory in abstract groups. II. Group extensions with a non-abelian kernel, Ann. of Math., 48 (1947), 326-341.

share|cite|improve this answer

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).

See for more details.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.