Let $G$ be a connected reductive group over an algebraically closed field $k$ of characteristic 0. Fix a Borel subgroup $B$ and a maximal torus $T \subset B$. Let $P \subset G$ be a parabolic subgroup containing $B$, and let $L \subset P$ be a Levi subgroup containing $T$.

Suppose given an action of $L$ on $\mathbb{A}^n_k$ such that (1) the commutator subgroup $[L,L]$ fixes $\mathbb{A}^n_k$, and (2) $\mathbb{A}^n_k$ is a toric variety for some quotient $T'$ of the torus $L/[L,L]$. In other words, $T' \cong \mathbb{G}_m^n$, and under this isomorphism, $T'$ acts on $\mathbb{A}^n_k$ in the natural way. The origin $0 \in \mathbb{A}^n_k$ is fixed by $T'$, and there are "many" one-parameter subgroups $\lambda': \mathbb{G}_m \to T'$ such that $\lim_{t \to 0} \lambda(t) \cdot z = 0$ for all $z \in \mathbb{A}^n_k$. On the other hand, the composition $T \to L \to T'$ of the inclusion map and the quotient map allows us to associate to any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ a one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ (given by the composition of $\lambda$ with the map $T \to T'$).

My question is this: does there exist some choice of $\lambda: \mathbb{G}_m \to T$ such that (1) the corresponding one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ contracts all of $\mathbb{A}^n_k$ to $0$, and (2) $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ (equivalently, $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$ of $G$ with respect to $T$ and the base given by $B$).

In the case where $G = P = \mathrm{SL}_n$, $B = L$ is the subgroup of upper triangular matrices, and $T$ is the subgroup of diagonal matrices, everything works out very nicely. In this case, $T = T'$ and $\lambda(t) = \mathrm{diag}(t^{m_1},\dots,t^{m_n})$ for some $m_i \in \mathbb{Z}$, and the condition that $\lim_{t \to 0} \lambda(t) z = 0$ for all $z \in \mathbb{A}^n_k$ is the statement that $m_i > 0$ for all $i$. The positive roots of $G$ with respect to $T$ are the maps $T \to \mathbb{G}_m$ sending $\mathrm{diag}(t_1,\dots,t_n) \mapsto t_i/t_j$ for $i < j$, so we may pick for $\lambda$ any choice of $m_i > 0$ such that $m_i > m_j$ for $i < j$.

In general, I think if we embed $G$ in $\mathrm{SL}_n$ and pick a basis where $T$ and $T'$ are diagonalized, we might be able to do something similar. I'm not convinced that this works, though. Essentially, the condition that $\lambda$ contracts all of $\mathbb{A}^n_k$ to a point says that $\langle \lambda, e_i \rangle > 0$ for all $1 \leq i \leq r$, where the characters $e_i: T \to \mathbb{G}_m$ factor through $T'$ and form a certain basis for the character group $\mathcal{X}(T')$. It's not clear to me that these hyperplanes will in general intersect all the hyperplanes $\langle \lambda,\alpha \rangle > 0$ for $\alpha$ a positive root. I'm not particularly comfortable with arguments about roots of reductive groups and one-parameter subgroups, so I would really appreciate any suggestions you may be able to give me!

  • $\begingroup$ This (somewhat convoluted) setup comes from a situation involving the local structure theorem on spherical varieties. I didn't think it was essential to the question, but I'd be happy to provide more context about that if it would help! $\endgroup$ Apr 24, 2020 at 4:07
  • $\begingroup$ In your example with ${\rm SL}_n$ you seem to abuse the notation for $n$. In this case the dimension of $T'=T$ is $n-1$, so it should act in ${\Bbb A}_k^{n-1}$, not in ${\Bbb A}_k^{n}$. $\endgroup$ Apr 24, 2020 at 6:21

1 Answer 1


The answer to your question, as stated, is No. Take $G={\rm SL}_2$, $P=B$, $T'=T$. Note that you do not specify the isomorphism $T'\to {\Bbb G}_m^n$. Let us take the following isomorphism: $$T'\to {\Bbb G}_m\colon\,{\rm diag}(s,s^{-1})\mapsto s^{-1}\text{ for }s\in k^\times.$$ Then our torus $T=T'$ acts on $\Bbb A^1$ by $${\rm diag}(s,s^{-1})\colon\, x\mapsto s^{-1}x.$$ Now if you take $s=\lambda(t)=t^m$, then your condition (1) that $\lambda$ contracts all of $\mathbb{A}^1_k$ to $0$ means that $m<0$, while your condition (2) that $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ means that $m>0$. Contradiction....

  • $\begingroup$ Oh, of course! I was worried there was some basis compatibility problem, but I got lost in the details and didn't realize it was so simple. Thank you for your help! $\endgroup$ Apr 24, 2020 at 18:23

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