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!



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


