Timeline for Is every $p$-group the $\mathbb{F}_p$-points of a unipotent group
Current License: CC BY-SA 4.0
26 events
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| S Mar 28, 2019 at 19:24 | history | bounty ended | CommunityBot | ||
| S Mar 28, 2019 at 19:24 | history | notice removed | CommunityBot | ||
| Mar 27, 2019 at 23:46 | comment | added | YCor | Do you require $G$ to be connected? | |
| Mar 24, 2019 at 5:43 | comment | added | Peter McNamara | @DrorSpeiser the Zariski closure of a finite set is itself (so not unipotent). I don't see any reason why we should expect any given embedding of D_8 in a unipotent group to come from the F_2-points of a connected subgroup. | |
| Mar 23, 2019 at 22:06 | comment | added | Dror Speiser | Take $a=1+E_{12}+E_{34}+E_{56}$, $b=1+E_{23}+E_{45}+E_{67}\in Mat_{7\times7}(\mathbb{F}_2)$, where $E_{ij}$ has a unique nonzero element at row $i$ and column $j$ equal to 1. Set $c=ab$. Let $V$ be $\overline{\mathbb{F}}_2^7$, and let $D_{16}\cong<x, y|x^2=y^8=xyxy=1>$ act on $V$ by sending $x$ to $a$ and $y$ to $c$. With this action $V$ is an indecomposable module for $D_{16}$. Let $\overline{D_{16}}$ be the Zariski closure of $<a, c>$ in $GL(V)$, which is unipotent. Does it satisfy $\overline{D_{16}}(\mathbb{F}_2)=<a,c>\cong D_{16}$? | |
| Mar 20, 2019 at 19:01 | comment | added | David E Speyer | @DrorSpeiser I decided to put the bounty on my own version, rather than the duplicate because I'd rather edit my own question and because it has a bunch of useful comments. If I had (two years ago) seen Gjergji's link to the duplicate before a bunch of those comments showed up, I probably would have self-closed, but it doesn't seem like a big deal either way. Of course, if someone posts a good answer, I'll link it on that question as well to make sure searchers find it. | |
| Mar 20, 2019 at 17:26 | comment | added | Dror Speiser | Given that this is a duplicate of a question six years earlier, how is this not closed? | |
| S Mar 20, 2019 at 16:31 | history | bounty started | David E Speyer | ||
| S Mar 20, 2019 at 16:31 | history | notice added | David E Speyer | Draw attention | |
| Mar 20, 2019 at 16:30 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Feb 23, 2017 at 10:42 | comment | added | David E Speyer | Okay, I'll take a look at Abhyankar's conjecture and see what that does for me. In the mean time, I'll record that the formula was meant to be $(x_1, y_1, z_1) \ast (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2 +(x_1 y_2)^p-x_1 y_2)$, sorry for the incorrect subscripts in the previous one. | |
| Feb 23, 2017 at 9:25 | comment | added | Jason Starr | I am specifically suggesting to use the faithful representations of $p$-groups arising from Abhyankar's conjecture. If we realize $\Gamma$ as the Galois group of a finite etale extension $\mathbb{F}_q[t] \subset R$, then for every subgroup $\Lambda \subset \Gamma$, then $R$ is a finite etale extension of the $\Lambda$-invariant subring $S$, with Galois group $\Lambda$. This is a "complete reducibility" result that is missing for arbitrary representations. I suggest we projectively complete these rings at $\infty$, and quotient by the maximal ideal of $\infty$ in $\mathbb{F}_q[t^{-1}]$. | |
| Feb 23, 2017 at 2:11 | comment | added | David E Speyer | Suppose there were such a $G$, and let $Z$ be the center $\{ (0,0,z) \}$ of $U$. Case 1: $G \supset Z$. Then $G(\mathbb{F}_p) \supset \{ (0,0,z) : z \in \mathbb{F}_p \}$, which is not in $\Gamma$. Case 2: $G \cap Z$ is finite. In that case, the projection $G \to U/Z$ is surjective. But any subvariety of $U$ which surjects onto $U/Z$ generates $U$, contradicting that $G$ is two dimensional. | |
| Feb 23, 2017 at 2:08 | comment | added | David E Speyer | I think there is a fatal flaw in @JasonStarr's approach. It seems that Jason's method would show that, if $U$ is a unipotent group, and $\Gamma$ is a subgroup of $U(\mathbb{F}_p)$, then there is a unipotent subgroup $G$ of $U$ with $G(\mathbb{F}_p) = \Gamma$. This isn't true. Let $U$ be the group structure on $\mathbb{A}^3$ given by $(x_1,y_1,z_1) \ast (x_2,y_2,z_2) = (x_1+x_2, y_1+y_2, z_1+z_2+x_1^p y_1^p - x_1 y_1)$ and let $\Gamma = \{ (x,y,0) : x,y \in \mathbb{F}_p \} \subset U(\mathbb{F}_p)$. Then $\Gamma$ is not $G(\mathbb{F}_p)$ for any $2$-dimensional subgroup of $U$. (continued) | |
| Feb 21, 2017 at 19:29 | comment | added | Jason Starr | @DavidSpeyer. I should not have called this the unique smallest subgroup. I am saying that for a vector space $V$ and a finite subgroup $Z$ of the group of $\mathbb{F}_p$-points of the scheme $\text{Spec}\ \text{Sym}_{\mathbb{F}_p}(V)$, the kernel of the restriction map $V\to \prod_{z\in X} \kappa(z)$ is an $\mathbb{F}_p$-linear subspace of $V$ generating an ideal whose associated closed subscheme is a closed subgroup scheme whose associated group of $\mathbb{F}_p$-points equal to $Z$. | |
| Feb 21, 2017 at 19:11 | comment | added | David E Speyer | @JasonStarr It seems like you are saying that, given an connected abelian group $H$ (in your setting, $Z(U)$), and a discrete subgroup $B$, there is a smallest connected group $A$ containing $B$. But that isn't true: consider $H = \mathbb{G}_a^2$ and $B = \{ (x,0) : x \in \mathbb{F}_p \}$. Then $B$ is contained in $\mathbb{G}_a$ embedded in $H$ by either $t \mapsto (t,0)$ or $t \mapsto (t,t^p-t)$. What am I missing? | |
| Feb 21, 2017 at 16:33 | comment | added | nfdc23 | @JasonStarr: I was wondering about that connectedness issue too but thought I might be overlooking something. | |
| Feb 21, 2017 at 16:21 | comment | added | Jason Starr | @nfdc23. You are correct. I realize now the problem with the approach I suggest: why should the centralizer of $Z(\Gamma)$ in $U$, much less the center of the centralizer, be a connected group? | |
| Feb 21, 2017 at 16:17 | comment | added | nfdc23 | @JasonStarr: In view of your inductive suggestion it seems that what you call "$n$" (from SL$_n$) is not "$n$" as in the statement of the question, just some unknown integer, in which case such $\rho$ is provided by the regular representation of $\Gamma$ on itself over $\mathbf{F}_p$ (that gives $\rho$ into ${\rm{GL}}_N(\mathbf{F}_p)$, necessarily landing inside SL$_N({\mathbf{F}}_p)$ and even conjugating into $U_N(\mathbf{F}_p)$ since the latter is a $p$-Sylow subgroup). | |
| Feb 21, 2017 at 13:58 | comment | added | Jason Starr | . . . So there is a unique minimal, connected subgroup $A\subset Z(U)$ such that $A(\mathbb{F}_p)$ equals $Z(\Gamma)$. Now replace $U$ by $U/A$, and use induction on the rank. | |
| Feb 21, 2017 at 13:56 | comment | added | Jason Starr | Here is how I would try to prove this. There exists a faithful homomorphism, $\rho:\Gamma\to \textbf{SL}_n(\mathbb{F}_p)$. This follows, for instance, from the solution of the Abhyankar conjecture by Raynaud and Harbater. By counting $\mathbb{F}_p$-points of the flag variety of $\textbf{SL}_n$, there exists a Borel subgroup $B$ that contains the image of $\rho$. Thus, there exists a faithful homomorphism from $\Gamma$ to a unipotent group. Replace that unipotent group by the centralizer $U$ of $Z(\Gamma)$. Thus $Z(U)(\mathbb{F}_p)$ is an additive group that contains $Z(\Gamma)$ . . . | |
| Feb 21, 2017 at 8:13 | comment | added | Marty | A step towards an inductive proof: if $G$ is a unipotent $k$-group, and $c \in Z^2(G(k), Z / p Z)$ is a 2-cocycle (for trivial action), does there exist a polynomial 2-cocycle $\gamma \in Z^2(G, G_a)$ such that $\gamma$ gives $c$ upon taking $k$-points? I want to attack this one inductively with inflation-restriction with respect to $G$ and $[G,G]$. But no time for more than a sketchy comment. | |
| Feb 21, 2017 at 5:09 | comment | added | Gjergji Zaimi | This was asked a while ago, as well, but there was no answer. mathoverflow.net/questions/69397/… | |
| Feb 21, 2017 at 4:10 | history | edited | YCor |
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| Feb 21, 2017 at 3:55 | comment | added | nfdc23 | If $U$ is smooth connected unipotent of dimension $n$ over $k=\mathbf{F}_p$ then $\#U(k)=p^n$ (use composition series with successive quotients $\mathbf{G}_a$). Thus, if $\Gamma=G(k)$ for a unipotent $k$-group $G$ of dimension $n$ then by passing to $G_{\rm{red}}$ so that $G$ is smooth we see that if $G$ is not connected then $G^0(k)$ of size $p^n$ must exhaust $G(k)$. In other words, we lose nothing by assuming $G$ is smooth and connected. So it is the same as asking for smooth connected unipotent $k$-groups $U$ if there is any constraint on $U(k)$ beyond its order being $p^n$. Good puzzler. | |
| Feb 21, 2017 at 3:26 | history | asked | David E Speyer | CC BY-SA 3.0 |