# Element of Weyl chamber contracting $\mathbb{A}^n_k$ to a point

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!

• 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! Apr 24, 2020 at 4:07
• 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 ${\Bbb A}_k^{n}$. Apr 24, 2020 at 6:21

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....