Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?

The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.


This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form

$$1+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}^2/2 + \dots = \begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} $$ is smooth, irreducible, and closed under $n$th powers since $$\begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}^n =\begin{pmatrix} 1 & na & n^2 ab/2 \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{pmatrix} ,$$ but isn't a subgroup.

Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.

  • $\begingroup$ Thank you! But I think it's true for torus. Is it true for any commutative groups? $\endgroup$ – sawdada Sep 15 '19 at 19:38
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    $\begingroup$ @sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^{2p-1}$ will provide counterexamples. $\endgroup$ – Will Sawin Sep 15 '19 at 20:23
  • $\begingroup$ @sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties. $\endgroup$ – Will Sawin Sep 15 '19 at 22:49
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    $\begingroup$ @sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X \to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption. $\endgroup$ – Will Sawin Sep 16 '19 at 0:55
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    $\begingroup$ @sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.) $\endgroup$ – Will Sawin Sep 16 '19 at 1:03

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