In the category of groups, it is elementary that all central extensions of a cyclic group are abelian. Is the same true, in the category of (finite?) group schemes over a field $k$, for central extensions of the group $\mu_n$ of $n$th roots of unity?

  • $\begingroup$ Unless I'm missing some subtlety, this should be true by using the functor of points. That is, you will get an exact sequence $1\rightarrow A(S) \rightarrow B(S) \rightarrow C(S)$ for every $k$-scheme $S$. Now, any subgroup of a cyclic group is cyclic, so what you want should follow from the usual fact. $\endgroup$ – Minhyong Kim Aug 23 '11 at 1:12
  • $\begingroup$ @Minhyong: It seems that $\mu_n(S)$ is not cyclic for general $S$. I don't quite follow your argument, so I don't know if this is a problem. $\endgroup$ – S. Carnahan Aug 23 '11 at 2:51
  • $\begingroup$ Scott: You're probably right. The argument is OK when the characteristic of $k$ doesn't divide $n$, but I suppose all the fun is when it does. I'll think about it some more. $\endgroup$ – Minhyong Kim Aug 23 '11 at 3:10
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    $\begingroup$ In the additive case there are interesting central extensions of $G_a$ over a finite field, the "fake Heisenberg groups" whose representation theory is a motivating example in the works of Boyarchenko and Drinfeld. $\endgroup$ – David Ben-Zvi Aug 23 '11 at 3:40
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    $\begingroup$ Heh. I see my kneejerk response was very naive. I will leave the comment up anyways, so others can learn from my silliness. $\endgroup$ – Minhyong Kim Aug 23 '11 at 6:06

If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$ with $A$ abelian, then we get a commutator mapping $\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an non-abelian extension to exist there must be a non-zero map $\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first the case when $n=p$, the characteristic of the field $k$ (which we may assume is algebraically closed). A non-zero map $\Lambda^2A\rightarrow B$ would give a non-zero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$ or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is isomorphic to $\mathrm{Hom}(D(B),D(A))$, where $D(-)$ denotes the Cartier dual. However $D(\mu_p)=\mathbb Z/p$ so when $A=\mu_p$ we get that $\mathrm{Hom}(D(B),D(A))=\mathbb Z/p$ and there is only the zero map from $A=\mu_p$ into it. In the other case $D(A)=\alpha_{p^m}$ and $\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$ is zero. If instead $n=p^k$, the argument is the same. The case when $n=\ell^k$ is even simpler so in all cases all possible commutator maps are zero and the extension is commutative.

(When $A=B=\mathbb G_a$ then there are candidates for commutator maps and in fact $(a,b)(a',b')=(a+a',b+b'+a^pa')$ gives a non-commutative central extension which I imagine is the fake Heisenberg group.)

Addendum: A general comment is that it is more convenient to work with sheaves (in the fppf topology say) as that means that we essentially can pretend that we work with set-theoretic groups. It is however also necessary if we want to see the commutator map as a map $\Lambda^2A\to B$ as the sheaf $\Lambda^2A$ (of $A$ considered as an abelian sheaf) is in general not reprsentable. The $\langle-,-\rangle\colon\Lambda^2A\to B$ view point is convenient as it allows us to do what one usually does when having a pairing: We get for instance a map $A\to\mathrm{Hom}(A,B)$ given by $a\mapsto (a'\mapsto \langle a,a'\rangle)$ just from the fact that $\langle-,-\rangle$ is biadditive.

I have implicitly assumed that $B$ is of finite type (as I claim that its Frobenius kernel is finite) even though it may not be necessary (a limit argument anyone?).

  • $\begingroup$ "Now, as sheaves Hom(A,B) is isomorphic to Hom(D(A),D(B))". Shouldn't that be Hom(D(B),D(A))? $\endgroup$ – Kevin Buzzard Aug 23 '11 at 8:05
  • $\begingroup$ It certainly should (fixed). $\endgroup$ – Torsten Ekedahl Aug 23 '11 at 8:21
  • $\begingroup$ [There's still one Hom(D(A),D(B)) left.] $\endgroup$ – Kevin Buzzard Aug 23 '11 at 20:11
  • $\begingroup$ Hmmm... I don't know what $\Lambda^2$ of a group scheme is, but I suppose I don't have to, as I certainly agree that the commutator will determine a morphism $A \to \mbox{Hom}(A,B)$. Also, this proof seems to assume $B$ is finite? For otherwise, even after Frobenius, couldn't $B$ be something humongous? Anyway, that's the only case I care about. Many thanks!! $\endgroup$ – Michael Thaddeus Aug 23 '11 at 23:23
  • $\begingroup$ @Kevin: Sorry for the sloppy first editing. @Michael: I have added some comments which hopefully clarifies. $\endgroup$ – Torsten Ekedahl Aug 24 '11 at 4:29

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