13
$\begingroup$

Let $S$ be a scheme and let $X := \mathrm{Spec}(\mathscr{O}_S[t, t^{-1}])$ be the underlying $S$-scheme of the $S$-group scheme $(\mathbb{G}_m)_S$. Is there only one structure of a commutative $S$-group scheme on $X$ for which $t = 1$ is the identity section?

$\endgroup$
26
$\begingroup$

Yes if $S$ is reduced, and no otherwise. The case of a field is classical (ultimately because $k[t,1/t]^{\times} = k^{\times}\cdot t^{\mathbf{Z}}$ for fields $k$), and I assume you are familiar with that. So in general if $S$ is reduced then by passing to the case of affine $S$ and then noetherian $S$ (by the usual limit business) we have $S = {\rm{Spec}}(R)$ with just finitely many generic points. At each generic fiber the desired equality of hypothetical group laws is known by the settled case of fields, so it holds globally by reducedness of the base.

In the non-reduced case the presence of "extra" units creates non-homomorphic automorphisms of the pointed scheme. Say $S = {\rm{Spec}}(R)$ and $r \in R$ is a nonzero nilpotent element satisfying $r^2=0$. Since $r+t \in R[t,1/t]^{\times}$ (with inverse $(1-r/t)/t = 1/t - r/t^2$), we get an automorphism of the pointed scheme $({\rm{GL}}_1,1)$ via $t \mapsto (1-r)(r+t)=r+(1-r)t$. Transferring the usual group law through this automorphism provides another group law.

But maybe you are not asking the question you intend. There are very strong results in SGA3 concerning the infinitesimal deformation theory of relative tori and rigidity results thereof. What is your motivation for the question posed?

$\endgroup$
  • $\begingroup$ Thank you. I am not familiar with the field case, so it would help if you could add further details on that. My motivation is that Deligne and Rapoport seem to casually use this fact in the sentence that breaks between pages 39 and 40 of their article, and I don't understand the justification for their claim there. The base there need not be reduced, but perhaps they are taking advantage of the extra information known about the action on $\mathbb{P}^1$?.. $\endgroup$ – Lisa S. Apr 20 '15 at 15:24
  • $\begingroup$ Nevermind about the field case: I think it follows from the classification of connected $1$-dimensional linear algebraic groups over an algebraically closed field (there are only $\mathbb{G}_m$ and $\mathbb{G}_a$) and a small calculation with $k$-scheme automorphisms of $\mathbb{G}_m$. If you could comment on Deligne-Rapoport though, that would help. $\endgroup$ – Lisa S. Apr 20 '15 at 16:07
  • $\begingroup$ @LisaS.: By requiring the $\mathbf{G}_m$-action on itself to extend to an action on $\mathbf{P}^1$ preserving the $S$-points $0$ and $\infty$, only the usual group law works (i.e., the intervention of "extra units" is ruled out). The key thing is that distinct $S$-points of $\mathbf{P}^1_S$ can have the same open complement when nilpotents are involved, so when $S$ may be non-reduced one has to specify those sections (which come from the generalized elliptic curve structure). $\endgroup$ – grghxy Apr 21 '15 at 4:37
  • $\begingroup$ Could I see it like this? The action of the purported exotic $(\mathbf{G}_m)_S$ on $\mathbf{P}^1_S$ defines a homomorphism towards the subgroup of $\mathrm{Aut}(\mathbf{P}^1_S) = (\mathrm{PGL}_2)_S$ that fixes $0, \infty \in \mathbb{P}^1(S)$, i.e., towards the "diagonal" torus. But this homomorphism measures exotic translation on $\mathbf{G}_m$ itself (because the exotic $\mathbf{G}_m$ is required to have the usual identity section), so therefore must be an isomorphism. $\endgroup$ – Lisa S. Apr 21 '15 at 5:54
  • 1
    $\begingroup$ @LisaS.: I'm not sure what "measures exotic translation" means, but it sounds like you now understand an answer. $\endgroup$ – grghxy Apr 22 '15 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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