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The underlying variety of a linear elgebraic group (say, over an algebraically closed field) is affine, so doesn't have nontrivial (infinitesimal) deformations. I'm curious to know whether it's possible to deform the group structure on a fixed variety (that admits at least a structure of an algebraic group).

Edit: I should have added, though when I wrote the question I mistakenly didn't expect reductivity made any difference, that I was mostly interested in reductive groups (in which case the comment of user54268 shows my question was not that naive after all..).

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    $\begingroup$ The group law on a unipotent algebraic group deforms non-trivially in general. For instance by scaling the symplectic form you get a one parameter deformation of the three dimennsional Heisenberg group (upper triangular $3\times 3$ matrices with ones on the diagonal) to the additive group of a $3$ dimensional vector space. $\endgroup$
    – Tony Pantev
    Commented Aug 28, 2014 at 1:48
  • $\begingroup$ @Tony Pantev Huh. I guess I just wrote down this. For some reason your comment didn't show up while I was writing my answer... $\endgroup$
    – Stephen
    Commented Aug 28, 2014 at 1:57
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    $\begingroup$ If $G$ and $H$ are smooth affine group schemes with connected reductive fibers over an artin local ring $R$ (with arbitrary residue field) and their geometric special fibers are isomorphic then the functor ${\rm{Isom}}(G,H)$ is represented by a disjoint union of smooth affine $R$-schemes, as can be checked over a finite etale cover that splits $G$ and $H$ (and then using etale descent, which is avoids some effectivity problems since Spec($R$) is a fat point). Thus, an isomorphism of special fibers (if one exists) always lifts to an $R$-isomorphism. $\endgroup$
    – user54268
    Commented Aug 28, 2014 at 2:00
  • $\begingroup$ @user54268 So, you are saying that reductive groups are rigid, yes? $\endgroup$
    – Stephen
    Commented Aug 28, 2014 at 2:04
  • $\begingroup$ @StephenGriffeth: Yes, though I'd prefer to say "infinitesimally rigid" (and be explicit about the connectivity assumption; hmm, perhaps fibral connectivity can be dropped, but I haven't thought it through). The proof rests on quite a bit of the theory of reductive group schemes in SGA3. $\endgroup$
    – user54268
    Commented Aug 28, 2014 at 3:43

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A most excellent example of "non-rigidity" beyond the reductive case (depending on how loose one wants to be about the meaning of "rigidity") is given in 5.2--5.10 of Exp. XIX of SGA3: a smooth affine group scheme $G$ over $k[t]$ for any field $k$ of characteristic 0 such that $G|_{t \ne 0}$ is a form of ${\rm{PGL}}_2$ (reductive!) but the fiber $G_0$ is solvable with 2 geometric connected components. By definition, $G$ is the automorphism scheme of the Lie algebra over $k[t]$ whose underlying $k[t]$-module is free on a basis $\{X,Y,H\}$ that satisfy $$[H,X]=X,\,\,\,[H,Y]=-Y,\,\,\,[X,Y]=2tH.$$ (It is clear that $G|_{t\ne 0}$ becomes ${\rm{Aut}}_{\mathfrak{sl}_2} = {\rm{PGL}}_2$ over the degree-2 finite etale cover defined by $\sqrt{t}$, but proving that $G$ is $k[t]$-smooth requires some cleverness.)

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  • $\begingroup$ Nice! It's never a bad idea to read SGA. Do you know if it is possible to see this deformation as a commutative deformation living within a larger family also containing quantum $\mathfrak{sl}_2$? $\endgroup$
    – Stephen
    Commented Aug 28, 2014 at 14:43
  • $\begingroup$ @Stephen Griffeth: I know less than nothing about quantum-math, so I have nothing meaningful to say about your question; sorry. $\endgroup$
    – user27920
    Commented Aug 29, 2014 at 16:25
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    $\begingroup$ I don't really know the general definition of "reductive", but is this $G$ reductive "over $k[t]$"? There are many examples of reductive groups degenerating to solvable groups, of which this is a particularly nice one. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 2, 2014 at 22:45
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How about this: for each $t \in F$, define a group structure on $F^3$ by $$(a,b,c) \cdot (d,e,f)=(a+d,b+e,c+f+tae).$$ When $t=0$ this is just $F^3$ with coordinate-wise addition, while for $t \neq 0$ it is $3$ by $3$ unipotent matrices.

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