Why are the irreducible components $T$-stable?

Question

I'm having trouble with part of a proof (7.1.5) in Springer's Linear Algebraic Groups. Let $r: G \rightarrow \textrm{GL}(V)$ be a rational representation of a linear algebraic group $G$, $B$ a Borel subgroup of $G$, and suppose that there exists a nonzero $v \in V$ such that $V$ is spanned by $g \cdot v : g \in G$, and $$B = \{ g \in G: gv \in kv\}$$ Let $T$ be a maximal torus of $G$ which is one dimensional, $\lambda: k^{\ast} \rightarrow T$ an isomorphism, $e_1, ... , e_n$ a basis for $V$, such that for some integers $m_1 \geq \cdots \geq m_n$, not all equal, we have $$\lambda(a)e_i = a^{m_i}e_i$$ Now $r$ induces an isomorphism of $G/B$ onto a closed subvariety $Z$ of $\mathbb{P}(V)$, the set of one dimensional subspaces of $V$, via $gB \mapsto [g\cdot v]$. Let $U_1$ be the affine open set in $\mathbb{P}(V)$ consisting of points with nonzero first coordinate (with respect to the basis $e_i$). Then $\mathbb{P}(V) \setminus U_1$ is closed, and $$[\mathbb{P}(V) \setminus U_1] \cap Z$$ is closed and clearly $T$-stable. Springer claims that the irreducible components of $[\mathbb{P}(V) \setminus U_1] \cap Z$ are also $T$-stable, but gives no indication of why. Why should this be the case?

Answer 1

Since the basis of $V$ are chosen to be eigenvectors of $T$, the set with first coordinate zero is obviously $T$-stable.

Let $T$ act on $X$, the stabilizer of a component $X_1$ equals to $\cap_{x \in X_1} \{t \in T | t.x \in X_1\}$, is a closed subgroup of $T$, and is of finite index ($t \in T$ permutes components of $X$, easily seen since $T$ is a group, $t^{-1}$ exists). Its identity component is contained in $T$, both are closed irreducible and have the same dimension, they must be equal.

This is true for any connected algebraic group $T$.

Answer 2

$G$ (and hence $T$) acts by projective automorphisms on $\mathbb P(V)$ and they are automorphisms of $\mathbb P(V)$ in algebro-geometric sense thus they preserve irreducible components of any subvariety of $\mathbb P(V)$ they stabilize.

Answer 3

The irreducible components are $T$ stable because $T$ is connected (in the Zariski topology). The action defines a continuous map from $T$ to the discrete set of permutations of the mentioned irreducible components. By the connectedness of$~T$, the image of $T$ is a single permutation of the irreducible components, necessarily the identity permutation (being the image of the identity element of$~T$).