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Let $k$ be a field of characteristic $p > 0$ (algebraically closed, if you want; that doesn't make a difference). Consider a finite type $k$-group scheme $E$ that is a (central) extension of $\alpha_p$ by $\mathbb{G}_m$. Is $E$ necessarily commutative?

Edit: $E$ is an extension of $A$ by $B$ if it fits into a short exact sequence (which is part of the data of an extension) $$ 1 \rightarrow B \rightarrow E \rightarrow A \rightarrow 1, $$ and the extension is central if $B$ is in the center of $E$. So in the case at hand I am looking at central extensions $$ 1 \rightarrow \mathbb{G}_m \rightarrow E \rightarrow \alpha_p \rightarrow 1. $$

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    $\begingroup$ Could you please specify in the question which way the extension goes? The terminology "extension of A by B" is somewhat ambiguous. $\endgroup$
    – Matthias Wendt
    Commented Oct 10, 2014 at 18:23
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    $\begingroup$ Actually, I know of no serious source that uses that phrase to mean an extension of the form $0 \rightarrow A \rightarrow E \rightarrow B \rightarrow 0$, but I would love to be shown an example. $\endgroup$
    – R.P.
    Commented Oct 10, 2014 at 19:04
  • $\begingroup$ OK, I've clarified this. $\endgroup$
    – Question Mark
    Commented Oct 10, 2014 at 19:36
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    $\begingroup$ What is $\alpha_p$? The additive group $\mathbb{G}_a$ or the finite group scheme with $\alpha_p(R)=\{a\in R|a^p=0\}$. If the latter, there are no nontrivial extensions, see Oort 1966, LNM 15. $\endgroup$
    – anon
    Commented Oct 10, 2014 at 20:13
  • $\begingroup$ The latter (I thought this was standard notation), so it is a finite group scheme. In other words, $\alpha_p$ is the Frobenius kernel of $\mathbb{G}_a$. Could you give a more precise reference within LNM 15? $\endgroup$
    – Question Mark
    Commented Oct 10, 2014 at 20:40

1 Answer 1

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So, bilinear maps $\alpha_p \times \alpha_p \rightarrow \mathbf{G}_m$ are classified by maps from $\alpha_p$ to itself (since $\alpha_p$ is Cartier self-dual). The collection of such maps is a $1$-dimensional vector space $V$ over $k$. The group $\mathbf{Z}/ 2 \mathbf{Z}$ acts on the vector space $V$ by "swapping'' the two factors of $\alpha_p$. This action is trivial (this follows from the fact that you can write down the Cartier self-duality on $\alpha_p$ in a symmetric way). It follows that if $p \neq 2$, there are no nonzero skew-symmetric bilinear maps from $\alpha_p$ to $\mathbf{G}_m$, so any central extension of $\alpha_p$ by $\mathbf{G}_m$ must be commutative.

If $p = 2$, then the symmetric bilinear maps $\alpha_2 \times \alpha_2 \rightarrow \mathbf{G}_m$ are also skew-symmetric, so this argument does not apply. It is still true that there are no noncommutative central extensions of $\alpha_2$ by $\mathbf{G}_m$, but you have to work a little bit harder to show this. You can find an argument in Example 3.2.7 of my paper with Mike Hopkins "Ambidexterity in K(n)-Local Stable Homotopy Theory" (we need this fact to discuss "alternating powers" of 1-dimensional p-divisible groups, which arise naturally from some calculations in algebraic topology).

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  • $\begingroup$ In the p=2 reference you cite, a cocycle computation is performed. To say that the central extension comes from a cocycle requires the map from E to alpha_p to have a section in the category of schemes. Why is this the case? $\endgroup$
    – Peter McNamara
    Commented Oct 11, 2014 at 2:31
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    $\begingroup$ Any line bundle over the spectrum of a local ring is trivial. $\endgroup$
    – Jacob Lurie
    Commented Oct 11, 2014 at 4:22
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    $\begingroup$ To prove the extension $E$ is commutative it is enough to consider $k$ that is algebraically closed, and in particular perfect, so Theorem 6.1.1(B) in Expose XVII of SGA3 is a suitable reference for the splitting (with a bit more generality on the kernel term; see Example 6.4(b),(c) there for why that generality makes the perfectness assumption on $k$ unavoidable). The key idea is to split the exact sequence of Frobenius kernels via computation with $p$-Lie algebras (a powerful tool in the non-commutative infinitesimal case). $\endgroup$
    – user27920
    Commented Oct 11, 2014 at 4:56

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