Let $B$ be a connected, solvable algebraic group (of dimension 2) acting on a projective variety $Y$ (of dimension $3$): for instance, let $B$ be a Borel subgroup of a reductive algebraic group $G$.

Question1: Is it always true that there is a $B$-invariant complete flag (i.e. codimension $1$, irreducible and $B$-invariant subvarieties), $Y \supset Y_1\supset Y_2 \supset Y_3$? For my purposes I can assume that there is a point in $Y$ with finite stabliser group and therefore can choose the orbit-closure of that point to be $Y_2$. How about $Y_1$? Is there always a point with $1$-dimensional stabiliser group of the $B$-action on the just described $Y_2$? Obviously there will always exist a $B$-fixed point $Y_3$, by Borel's fixed point theorem, if $Y_1$ is a projective variety on which $B$ acts.

Question2: Does this generalize to arbitrary dimension $n$ of $Y$ and any reductive algebraic group $G$ with a Borel subgroup $B$?

  • 1
    $\begingroup$ By the Borel fixed point theorem, every action of a solvable algebraic group on a proper $k$-scheme has a fixed point. For an action of $B$ on an irreducible, projective variety $Y$ of dimension $n$, there is an induced action of $B$ on each Hilbert scheme $\text{Hilb}_{Y/k}^{p(t)}$ of closed subschemes of $Y$ with Hilbert polynomial $p(t)= dt^{n}/n! + ...$ For appropriate choices of $p(t)$, this Hilbert scheme is nonempty, thus there exists a $B$-invariant closed subscheme. Any irreducible component of this subscheme of dimension $n-1$ gives $Y_1$. Now repeat with $Y_1$ in place of $Y$. $\endgroup$ Feb 1, 2017 at 13:28

2 Answers 2


Here is an argument which avoids the use of Hilbert schemes and works for any complete (irreducible) variety $Y$.

As in Jason's answer it suffices to construct a $B$-stable subset $Y_1$ of codimension $1$, provided that $\dim Y\ge1$. For this we consider two cases:

  1. $Y$ does not contain an open $B$-orbit. By Rosenlicht's theorem, there exists an open $B$-stable subset $Y_0\subseteq Y$ such that the orbit space $\pi:Y_0\to Z:=Y_0/B$ exists. Our assumption implies $\dim Z>0$. Therefore, $Z$ contains a subvariety $Z_1$ of codimension $1$. Now take for $Y_1$ the closure of $\pi^{-1}(Z_1)$ in $Y$.
  2. $Y$ does contain an open $B$-orbit $Y_0$. Then the above argument is not applicable since $Z$ is a point. Instead we use the well-known fact that homogeneous spaces for solvable groups are affine. Thus $Y_0$ is affine. But it not complete since $\dim Y_0\ge1$. Thus the complement $C$ of $Y_0$ in $Y$ is non-empty. We conclude with another well-known fact, namely that the complement of an affine open set is always pure of codimension one. So, take for $Y_1$ any component of $C$.

I am posting the comment as an answer. Let $Y$ be an irreducible, projective $k$-scheme of dimension $n$. Let $\mathcal{O}_Y(1)$ be an ample invertible sheaf. For some choice of degree $n-1$ numerical polynomial $p(t) = d t^{n-1}/(n-1)! + ...,$ the Hilbert scheme $\text{Hilb}_{Y/k}^{p(t)}$ is nonempty. The action of the solvable group $G$ on $Y$ induces an action of $B$ on $\text{Hilb}_{Y/k}^{p(t)}$. By the Borel fixed point theorem, there is a $B$-fixed point of this action of a solvable group on a projective $k$-scheme. For the corresponding $B$-invariant closed subscheme of $Y$, for any $(n-1)$-dimensional irreducible component $Y_1$ of the closed subscheme, $Y_1$ is a $B$-invariant, irreducible closed subscheme of $Y$ of dimension $n-1$. Now inductively apply the same argument to $Y_1$ to conclude the existence of a $B$-invariant flag of irreducible subvarieties.

Almost certainly this is in Hartshorne's thesis (and probably in other references as well). You could also just directly apply the Borel fixed point theorem to the action of $B$ on the flag Hilbert scheme (as developed in Sernesi's book).

Edit. The existence of $Y_1$ as above follows from Proposition 5.2, p. 44, of Hartshorne's article.

MR0213368 (35 #4232)
Hartshorne, Robin
Connectedness of the Hilbert scheme.
Inst. Hautes Études Sci. Publ. Math. No. 29 1966 5–48.


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.